SOLU Algebra Lineal - Stanley Grossman
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Description
Instructor’s Solutions Manual for
EIEKEKTSSY
LinearAlgebra Stanley
I. G r o s s m a n Fifth Edition
Andy D e m e tre F re d G y l y s - C o l w e l l D a v id R a g o z in
sum/EssiTY se msssmsTBH
THOMSON
---------BR O O K S/CO LE
Australia • Cañada • México • Singapore Spain • United Kingdom • United States
C O P Y R I G H T © 1994 Brooks/Cole, a división of Thomson Leaming, Inc. Thomson Learning ™ is a trademark used herein under license. A L L R I G H T S R E S E R V E D . N o part of this work covered by the copyright hereon may be reproduced or used in any form or by any means — graphic, electronic, or mechanical, including photocopying, recording, taping, W e b distribution, or information storage or retrieval systems — without the written permission of the publisher. M
atlab®
is a registered trademark of The MathWorks, Inc.
Brooks/Cole — Thomson Learning 10 Davis Drive Belmont CA 94002-3098 USA For information about our products, contact us:
Thomson Learning Academic Resource Center 1-800-423-0563 http://www.wadsworth.com For permission to use material from this text, contact us by Web: http://www.thomsonrights.com Fax:1-800-730-2215 Phone: 1-800-730-2214 Ragozin: Instructor’s Solutions Manual to accompany E le m e n ta ry L in e a r A lg e b r a , 5e, by Grossman
ISBN 0-03-097355-4 Printed in the United States of America 10 9 8 7 6 5 4 3 2
1
P r e fa c e
This Instructor's Solutions Manual is an ancillary for the fifth edition of Grossman's Elementary Linear Algebra. It contains detailed Solutions to all problems in the text— including the MATLAB and graphing calculator problems— and in the Applications Supplement. Below is an overview of all the ancillaries to accompany the main text. A p p lic a tio n s S u p p le m e n t
• one chapter each on linear programming and on Markov chains and game theory • available packaged with the text or for sepárate purchase • numerous examples and problems • answers to odd-numbered problems are at the back of the Applications Supplement
S tu d en t S o lu tio n s M anual
• complete Solutions to all the odd-numbered problems in the text and the Applications Supplement
MATLAB M anual: C om p u ter L ab oratory E x ercises and M -file d is k
• Computer laboratory exercises and applications using MATLAB. Each section lists objectives, prerequisites, and MATLAB features before the lab exercise is presented. The student is then encouraged to apply concepts interactively and create an edited diary session. An M-file disk containing programs of selected applications in the manual is avaüable free upon request from The MathWorks, Inc., in either Mac or PC versions.
E lem en ta ry L in ea r A lg eb ra T o o lb o x (M -file d isk )
• MATLAB programs that accompany the main text in either PC or Mac versión are available free upon request from The MathWorks, Inc.
H P -4 8 G /G X C alcu lator M anual
• calculator enhancement for science and engineering mathematics using the high-level Hewlett-Packard calculator
A c k n o w le d g m e n ts
This Instructor's Solutions Manual has been prepared with the help of many people. The Solutions to the fourth edition problems were prepared by Rick Miranda of Colorado State University, with the assistance of Howard Thompson and John Symms. These provide the basis for much of the present work. Andy Demetre prepared the Solutions for the new problems in the main body of the text. Fred Gylys-Colwell developed the Solutions for the MATLAB problems in Chapters 1 and 4. David Ragozin provided the Solutions for the MATLAB problems in Chapters 5 and 6, the CALCULATOR box Solutions for the TI-85, and the editorial changes and updated Solutions found throughout the rest of the manual. Michael Ragozin assisted with the TI-85 Solutions.
Mary Sheets produced the TeX files for the manual. The new figures were produced by Fred Gylys-Colwell and David Ragozin using MATLAB, S-Plus, and TI-85 Graph Link. Andy Demetre Fred Gylys-Colwell David L. Ragozin Seattle, Washington March 1994
T able o f C o n te n ts Systems of Linear Equations and Matrices
1
Two Linear Equations in Two Unknowns
1
m Equations in n unknowns: Gauss-Jordan and Gaussian Elimination
6
Homogeneous Systems of Equations
36
Vectors and Matrices
42
Vector and Matrix Products
49
Matrices and Linear Systems of Equations
78
The Inverse of a Square Matrix
88
The Transpose of a Matrix
113
Elementary Matrices and Matrix Inverses
120
LU-Factorizations of a Matrix
134
Graph Theory: An Application of Matrices
154
Review Exercises for Chapter 1
156
Determinants
165
Definitions
165
Properties of Determinants
182
Proofs of Three Important Theorems and Some History
191
Determinants and Inverses
192
Cramer’s Rule
198
Review Exercises for Chapter 2
202
Vectors in ®2 and M3
205
Vectors in the Plañe
205
The Scalar Product and Projections in M2
215
Vectors in Space
223
The Cross Product of Two Vectors
226
Lines and Planes in Space
235
Review Exercises for Chapter 3
240
In stru cto r’s M anual
Vector Spaces
245
Definition and Basic Properties
245
Subspaces
250
Linear Combination and Span
253
Linear Independence
268
Bases and Dimensión
294
The Rank, Nullity, Row Space and Column Space of a Matrix
308
Change of Basis
350
Orthonormal Bases and Projections in IR”
367
Least Squares Approximation
390
Inner Product Spaces and Projections
407
The Foundations of Vector Space Theory: The Existence of a Basis
416
Review Exercises for Chapter 4
417
Linear Transformations
423
Definitions and Examples
423
Properties of a Linear Transformation: Range and Kernel
426
The Matrix Representation of a Linear Transformation
428
Isomorphisms
435
Isometries
437
Review Exercises for Chapter 5
440
Eigenvalues, Eigenvectors and Canonical Forms
443
Eigenvalues and Eigenvectors
443
A Model of Population Growth
477
Similar Matrices and Diagonalization
488
Symmetric Matrices and Orthogonal Diagonalization
500
Quadratic Forms and Conic Sections
507
Jordán Canonical Form
515
An Im portant Application: Matrix Differential Equations
519
A Different Perspective: The Theorems of Cayley-Hamilton and Gershgorin
523
Review Exercises for Chapter 6
535
Instructor's M anual for Elem entary Linear Algebra by S.I.Grossman
Appendices
Contents
539
1.
Mathematical Induction
539
2.
Complex Numbers
543
3.
The Error in Numerical Computations and Computational Complexity
545
4.
Gaussian Elimination with Pivoting
547
Linear Programming
551
1.1
Convex Sets and Linear Inequalities
551
1.2
Linear Programming, The Córner Point Method
559
1.3
Slack Variables
563
1.4
The Simplex Method I: The Standard Maximizing Problem
565
1.5
The Simplex Method II: The Dual Mínimum Problem
572
1.6
The Simplex Method III: Finding a Feasible Solution
576
Review Exercises for Application 1
578
Markov Chains and Game Theory
585
2.1
Two-Person Games: Puré Strategies
585
2.2
Two-Person Games: Mixed Strategies
587
2.3
Matrix Games and Linear Programming
590
2.4
Markov Chains
594
2.5
Absorbing Markov Chains
597
2.6
Queuing Theory and a Model in Psychology
600
Review Exercises for Application 2
602
Application 1.
Application 2.
T w o Linear Equations in T w o Unknowns
C h ap ter 1. S y ste m s o f Linear E qu ation s and M atrices S ectio n 1.2 1.
4x — 12y = 16 —4x -f 2 t/ = 6_____ —10y = 22henee,
y = —11/5 and x = —13/5.
^ 11^22 —012021 = ( 1 )( 2 ) —(—3)(—4) = 2 — 12 = —10 . 2. I4x — 7y — —21 5x + 7y — 4_ 19x = -1 7 henee, x = -1 7 /1 9 and y = 3 + 2x = 3 - 34/19 = 23/19. a n a 22 - ai 2 a2 i = (2)(7) - (-1)(5) - 14 - (-5 ) = 19. 3.
6 x — 24 y = 15 —6 x -f 24y — 16_______ 0 = 31 => no solution.
a n ^22 ~ ai 2 a 2 i — (2)(12) —(—8)(—3) = 24 —24 = 0. 4.
6x — 24 y — 18 —6 x -f 24y = —18____ 0= 0 =£■linescoincide.
an 5.
^22
~ ai 2 a2 i =: (2)(12) —(—8)(—3) = 24 —24 = 0.
6x + y = 3 —4x — y = 8________________ 2x = 11 henee, x — 11/2 and y — 3 —6x = 3 —33 = —30.
a n a22 —a i2a2i = (6)(—1) —(1)(—4) = —6 —(—4) = —2. 6.
9x -f 3y = 0 2x — 3?/ — 0_________ llx = 0 henee, x = 0 and y = —3x = 0 . 011^22 —^ 12^21 = (3)(—3) —(1)(2) = —9 —2 = —1 1 .
7.
4x — 6 y —4x + 6 y
= 0 — 0_______ 0 = 0 => lines coincide; 4x —6y = 0 implies y = (2/3)x for arbitrary x.
a n a 22 ” ai 2 a 2 i — (4)(3) —(—6)(—2) = 12 — 12 = 0. 8.
25x 4- 10y = 15 4x -f 10y = 6_______ 21x = 9 henee, x = 9/21 and y = (3 —5x)/2 = 9/21. an
a 22
•” a i 2fl2i = (5)(5) —(2)(2) = 25 —4 = 21.
Section 1.2
1
2
C h ap ter 1 System s o f Linear Equations and M atrices
In s tru c to r^ M anual
9 . 8x -f 12 y = 16
9x -{- 12y = 15________ = 1
—x
henee, x = —1 and y = (4 — 2 x )/3 = 2.
« n «22 ~ « i 2« 2i = (2)(4) — (3)(3) = 8— 9= —1.
10 .
a x + by = c a x — by = _c_________
2a x
= 2 c henee, x = c /a (assum ing
(assum ing 6
a ^ 0)and y = (c — a x )/6 = 0
0). If a = 0 and 6 ^ 0 , th en there are no Solutions unless c = 0, in which case y = 0
and any x is a solution. If a ^ 0 and 6 = 0, then x = c /a and any y is a solution. Finally, if b o th a an d 6 are zero, th en th ere are no Solutions unless c = 0, too, in which case any x and y gives a solu tion. a i \ d 22 ~ d i 2 d 2 i = a ( —6) — ba = —2ab.
1 1 . a 2x + b2x +
ciby = aby =
ac be____
(a 2 — 62)x = ac — be henee, x = c(a — 6) /( a 2 — 62) = c/(a + 6) (assum ing a2 — b2 ^ 0) an d y = (c — a x )/6 = c /(a + 6) also. If a2 — b2 = 0, th e n a = ± 6; if a = 6 /
0,
th en th e eq u atio n s are th e sam e, and y = (c/a) — x and any x gives a solution. If a = —6, th en there are no Solutions unless c = 0, in which case any x and y give a solution if 6 = 0, and if 6 ^ 0, y = x w ith any x gives a solution. no point of intersection. 18. 12x - 18y = 21 12x - 18v = 24 0 = - 3 =» no point of intersection. 19. 12x - 18y = 30 12x - 18v = 30 0 = 0 => lines coincide. 20.
3x + y = 4 —5x + y = 2 8x = 2 henee, x = 1/4 and y = 4 —3x =
21. 6x + 8 y = 10 6x — 7t/ = 8 15y = 2 henee, y = 2/15 and x = (5 —4t/)/3 = 67/45. 22. Let mi = the slope of L and rri2 = the slope of Lxmi = 1; ni 2 = —1. L : x — y = 6 , and Lx : x + y = 0 Point of intersection: (3, —3). d = xi + a¡j/i \ P o in t ol intersection: =-, z-r=------- 1 \ a 2 + b¿ a¿ + b¿ )
Then d = |axi + by\ — c\¡yja 2 + b2 after a lot of algebra.
30. Let x = number of birds and y = number of beasts. Then x -f y = 60; 2x + Ay = 200. Henee x = 20 and y = 40.
31. If a n a 22 — « i 2« 2i = 0, then a n a 22 = 012021- Assuming a i 2a 22 # 0, and dividing both sides of the equation by a i 2a 22 we get a i i / « i 2 = « 21/ 022- This implies —o n / a i 2 = —021/ 022- Solving each linear equation of system (1) for y we get y = —o n / a i 2x + 61/012 and 2/ = —o 21 / 022^ + 62/ a 22. These lines have slope —o n / a i 2 and —021/022 respectively. Slopes are equal. Therefore the lines are parallel. If «12 =
fhen from o n a 22 = 0 and o n # 0, we get a 22 = 0. So the lines are parallel because they are
both vertical. If a 22 = 0 similar reasoning holds.
32. Suppose otherwise, i.e., suppose that a n a 22 —012021 = 0. Then #31 shows that the lines given in sys tem (1) are parallel. Thus system (1) either has an infinite number of Solutions or no solution. This contradicts the assumption that the system has a unique solution. Result follows.
33. If a n a 22 — a i 2a2i # 0 then onfl 22 # o i 2a 2i. Dividing both sides of the equation by a i 2a 22 we get 011/«12 # « 2i / « 22- Thús —o n / « i 2 # —a 21/ 022- Henee the slopes of the lines in system (1) are not
equal (see solution to #31). Thus the lines are not parallel. Therefore system (1) has a unique solu tion.
T w o Linear Equations ¡n T w o Unknowns
Section 1.2
34. Let x = number of cups and y = number of saucers. Then,
3x -f 2y = 480 25x -f 20y = 4400; henee, x = 80 and y = 120 .
Eq. 2 —10 Eq. 1: —5x = —400
35.
3x + 2 y = 480 Eq. 2 - 5 Eq. 1: 0 = 0 15x + 10?/ = 2400. So, these two lines coincide; henee y = (480 —3x)/2 where 0 < x < 160, to forcé y > 0.
36.
3 x + 2y = 480. Eq. 2 - 5 Eq. 1: 0 = 100, 15x -f 10y = 2500 so this system of equations has no solution.
37. Let x = number of ice-cream sodas and y = number of milk shakes. x -f y = 160; Eq. 2 —4 Eq. 1: —y = —128. 4x -f 3y = 512 Henee, y = 128, x = 160 —y = 32.
Then,
5
In s tru c to r^ M anual
C h ap ter 1 System s o f Linear E quations and M atrices
/?2 —* — 4" R3 —►—2R\ + R3 R\ —►2 R 2 4" R\ / 1 0 1 1 -1 R$ —►—9 R\ 4- R3 0 l o 0 -4
2 1 6 18\ 2. 5 0 8 -16 3 2 -10 -3 / R 2 —►2 R 3 4- R 2
n \ «3 -40 | «2 12/ /I 0 0 1 0 0 lo 0 1
3 -4 -4
H3 —►R\ 4R\ R3
i?2 —* —5/?i 4* R 2 / 1 3 -4 -15 28 R3 —*■2 R\ 4" R3 0 -2 7 \°
15 \ ( 1 3 -4 -16 5 0 8 18/ V-2 1 6 Rs —►—7 R2 + R 3
R 2 ^ Rs
(2 ,-3 ,1 ) is the unique solution.
Use back substitution to
166
find the solution (—4, 7, |) . Ri - ±Ri Í?2 —* —2 Ri + -^2 H3 —►Ri 4- R3 be
4.
Use
a r b itr a r y .
b a ck su b s titu tio n to
R3 —+2 R 2 4* R 3 R2 ^ - l R 2
find the
S o lu tio n s
Ri -* 3^1 R 2 —+ —2R\ 4- R2 R 3 —*■—5Hi 4- R3
3 6 -6 2 -5 4 5 28 -26
-2 1 2 0 1 -8/9 0 0 0
Let
(3 + §£ 3 , §£ 3 , £ 3).
R$ —*■2 R 2 4" R3
0 I . The bot-23
tom row is equivalent to the equation 0 = 23, which is impossible. So the system has no solution.
1 -1 1 5 2 -3
5.
r2
7\
4
R 2 —►—4R\ 4" R 2 R3 —*■—2 R\ 4- R3
0/
- -ifi2
.fíi —►—R 2 4" Ri
1 -1 1 5 1 3
7\ 4 18/
be arbitrary. Then
/1 -51 lo0 0
-1 9 -1
R3 —►—R 3
7
R 2 —►—9H3 4- R 2
-24 -14
R i —*■R$ 4" Ri
1 1 0 21 0 -5 0 -150 0 0 1 14
-9X 30 | . (—9,30,14) is the unique solution. 14,
R 2 —►—4R\ 4* R 2 -1 ( 1 1 0 -5 9 R 3 —+—6 R 1 4- R3 V 0 -5 9 — §£31
7
-24 -24
*3
R3 —R2 4- R3 Ü2 R\ —►—R 2 4" R\
+ § £ 3 ) £ 3) are the Solutions.
10
4/5
0 1 -9/5 0 0
0
. Let
£3
m E q u atio n s ¡n n unknowns: Gauss-Jordan and Gaussian E lim in atio n
Section 1.3
The same row operations as in problem 6 gives the equivalent m atrix 1 0 4/5 0 1 -9/5 0 0
Since 0 ^ 1 , the system has no solution.
0
Rí —►—4Ri + R 2 / 1 -2 3 R3 — —2 R 1 + R3 ( 0 9 - 1 3 _________ „ \ 0 3 -3
R3
/ I -2 3 ( 0 1 -1 0 9 -13
Ri —►2R2 -f- R\
(0,0,0) is the unique solution, by back substitution.
9.
1 1 -1 4 -1 5 6 1 3 tr a r y .
10 .
„
0
0/
Use back
2 5 0 -2 4 0
R 2 —*• —4Ri + R2 .( 1 1 -1 R 3 —*■—6 R 1 + R3 I 0 -5 9 \10 -5 9
°\
s u b s titu tio n to
6\
4 -2 ) R3 -+ -± R 3 Ri —►2R3 + R\ R2 —►—2.5J?3 + R2
f l 2 -1 V3 4 -2
4\ 7J
find the
R 3 —►—R 2 “I" -^3 R 2 - -Í« 2
Let £3 be arbi-
S o lu tio n s ( — | x 3 , § £ 3 , £ 3 ).
R\ í í R2 / 1 0 -2 4\ 2 5 R 3 —►—2R\ + R3 b 6 vo 4 4 r lO /
R2 R 3 —►—4R2 -+- R 3
10
-2
0 1 5/2 0 0-6
34/3 \ -37/6 I . (34/3, —37/6,11/3) is the unique solution. 1 1 /3 /
R2 —►—3 R 1 + fl2 2->
2 2
(1 2 -1 V0 1 -1/2
4\ 5/2 )
Ri —* —2 R 2 + Ri ( 1 0 0 1
0 -
1.2
5/ 5/ 2 ) '
be arb itra ry . T h e n ( —1, | + ^ £ 3 , £ 3) are th e Solutions.
12 .
/ 1 2 -4 4 \ R2 -*2Rt+R2 ( 1 2 - 4 4 \ Let x 2 and a:3 be arbitrary. Then \ - 2 -4 8 - 8 / V0 0 0 O / (4 —2 x 2 + 4 £ 3 , £ 2, £ 3) are the Solutions.
13 { 1 2 -4 U ' \ - 2 -4 8
4 \ R 2 —* 2R\ + R2 ( \ 2 -4 4 \ \ 0 0 0 1 / ’ ^ nce ® ^ -9J -9
R2 —3Ri -f R 2 14 ( l 2 - 1 1 7>) \ 3 6 -3 3 2 1 / (7 —2x2 + X3 —X4 , X2, X3 , X4) are the Solutions.
system
no s°lution.
L e tX 3
C h a p ter 1 System s o f Linear E quations and M atrices
Ri — * R 2 —* — R l
R 3 —*■3ü l + R'2
R3 —►—J3 /Í3 ÍÍ2 —* 5 R3 + ^2 /?! —*• R 3 + Rl
/ 1 3 16. 0 \-l
20/13 \ -28/13 I . Then (y§ - ^ar4, — -45/13 )
-1 1 - 1/3 o -13/3 3
+ ^ X 4 , - f § + 13 * 4 , * 4) are the
where X4 is arbitrary.
2>\
-2 1 1 0 2 -2 -8 1 4-1-1 7/ 6 - 2 0
R3
- 6/42 + R3
Rí
-4R2■+■R*
fl 0 0 \0
R2 R
—S R i a
+ 1ti0 -26
—►R i + R
a
(1 \o -2
1
1
0
1
-1 -5 4 -1 -1
-14 I
4 -1 1
9/
2\ 0.25 -15.5
1 -0.25 -0.25 0 0.5 -3.5 0
1
R2
H3
R2 ~ *\R 2
¡J
«3
2R3
R 0 and X2 = X3 — 5000 > 0. Henee, the populations that can be supported are 5,000 < X3 < 6,000, x\ = 30,000 — 5 x 3 , and #2 =
— 5,000
+
X3.
The solution is not unique.
38. Let d,E d p } and ds denote the number of days spent in the respective countries. The information gives the following system of equations: 30 d s + 20 dp + 20 d s — 340
20c/jr; + 30 dp + 20^5 = 320 10d s 4" 10dp + lOí/5 = 140
m Equations in n unknowns: Gauss-Jordan and Gaussian E lim in atio n
Section 1.3
Upon reducing, we obtain '30 20 20 20 30 20 10 10 10
Henee, de = 6, dp = 4, and ds = 4.
39.
Let sd, s h , and sm denote the respective number of shares. The information gives the following
Sys
tem of equations: —sj) — 1.5s¿/ -f 0.5saí == —350 1.5s£> —0.5s/f + «Af = 600 Writing the system as an augmented m atrix and reducing to echelon form gives -1 -1.5 0.5 -350 \ fl 1.5 -0.5 350 \ ( 1 0 0.4545 390.9 \ c . 1,0 -2.75 1.75 75J V0 1 -0.6364 -27.27) ' Slnce can be 1.5 -0.5 1 600 ) chosen arbitran y, the broker does not have enough information. If sm = 200, then se = 300 and sH = 100.
40. Let / and b denote the number of fighter planes and bombers, respectively. The information gives the following equations: / + b = 60 6 / + 26 = 250 / - 26 = 0
/I 1 6 2 Vi -2
60 250 0 system is inconsistent. Then
1 60 \ 60 /I 1 -4 -110 — 1 20 0 -3 -6 0 / \o -4 -110
60 \ 20 I . Since 0 ^ —30, the -3 0 /
b -a\ JR3 —►Rs + i ?2 -2b-b3a I 5b-5a-hc ) Henee, the system is inconsistent if 36 —2a + c ^ 0.
42.
i?3—► R3~2R2
3 -1 a ( 2 -1 3 b 1 7 -5 c be consistent, we must have —2a + 6 + c = 0.
h\
a-2b I . For the system to -2a+b-fc J
11
C h ap ter 1 System s o f Linear E quations and M atrices
In s tru c to r^ M anual
43. Either o n , 021, or 031 is nonzero, otherwise, the system is either inconsistent or has an infinite number of Solutions. W ithout loss of generality, we may assume a n / 1 a n / a n « 13/«11 0 C*22 OL23 .0 C*32 «33
«11 «12 «13
give
«21 «22 «23 k«31 «32 «33
«23 "" «21«13/«11 j «32 = «32
where an(l «33 = «33
* 0
0. Elementary row operations
*
) = X ] X l a «i *= 1 j —l
5
4 ci3j 4- o>4j) =
84.
2
y^fl3>¿>2 *=2 i = l
i=l
4
3
4
. ^(021^17^7*54022^27*07*5 +023^37*07*5) = ^ y^Q2¿fc»7*07*5
86
7=1
í=l7=l
TV 87. y : (ajb 4- 6jb) = ÜM + k—M
+ OAf+1 4- ¿M+l 4
— om 4 o j v f 4 • • • 4 o>n 4 TV
= ¿
h OjV 4 bjsí
4 &m+i 4 • • • 4 ¿>tv
TV
a* =
Jfc=M
k—M
i. X!(°k~bk)=]X(a*+(-1)M(=}X^a*+X](_1)6fc X^a*~X^ Jfc=M
fc=M
Jb=M
jfe=M
*=A/
k=M
55
56
C h ap ter 1 Systems o f Linear Equations and Matrices N
89. ^ a.k = ifc=M =
clm +
ajií+i H
h a/v
a M + ^ A f + l + ’ * *+ a m - l + a m + Om+1 + * * ' + a N
m
N
=J ak+ X)a* 2
k=M
fc= m + l
Instructor’s Manual
Vector and Matrix Products
Calculator 1.6 57
CALCULATOR SOLUTIONS 1.6 In this section many of the problems have two input matrices, and we append the problem number to the matrix ñame; for instance we use A16nn and B16nn to refer to the left and right factors in the products in problem nn, nn=90,91,92. We assume that the matrices have been entered and we will only show how to obtain the Solutions from these input matrices. [[ -12.0704 -26.8412 ] 90. A1690 (x) B16 90 [ENTER)givesthe product: [ 44.4207 -45.0695 ] [ 91.7485 4.2242 ]] [[
34192
38621 ]
[ 50408
44115 ]
[ 62661
71731 ]
[
59190
55046 ]]
[[
-.6557
-3.3655
]
(x) B 1 6 9 2 (ENTER) gives the product: [
.6907
3.4072
]
5.5459
]]
^
,
91. A 1 6 9 1 (x) B 1 6 9 1 lENTERlgives the product: ^
92.
A1692
1
' 5
F
[ -1.4255
93. (a) To see P 1 6 9 3 and Q1693 are probability matrices compute their row sums by computing their products with the column of all ones: [[1][1][1][1]] [S T O ) ONES [[1 ] r 21 P1693
(x) ONES (ENTER)
[[1 ] [ 2.1
______
:
a n d Q1693
(x) ONES ¡ENTER)
:
[ 1 ]]
.
[ 1 ]]
(b) [[
The product P1693 (x] Q1693 [ENTER]
.31118
.18444.14174
.36264 ]
.32625
.27585 .08454
.31336 ]
.17955
.22651 .19619
.39775 ]
.30047
.15251 .33558
.2 1 1 4 4 ]]
can be saved as PQ1693 via |2nd) (ANS) [S T O ) PQ1693. Then you see the product is a probability matrix [[1 ] [ 2. ] by showing its row sums are all 1 by PQ1693
(x) ONES [ENTER) w h i c h y i e l d s :
^
[ 1 ]] 94. Entering A l 694 Q n , n = 2,5,10,50,100 gives: A16942
A16945
A 1 6 9 4 10
A 1 6 9 4 50
is
is
is
is
[[1 [0
9] 4]]
[[1
93]
[0 3 2 ] ]
[[1
3069]
[0 1 0 2 4 ] ]
[[1 [0
3 . 3 7 8 e15] 1 . 1 2 6 e15]]
A 1 6 9 4 100 is [[1 [0
3 . 8 0 3 e 30] 1 . 2 6 8 e30]]
58 Chapter 1
Systems of Linear Equations and Matrices
Instructor’s Manual
95. From the A2 , A5, A10 results in Problem 94, where the diagonal componente are {12, 2 2}, {15, 2 5} and {l 10, 210}, it appears that the diagonal componente of An are just a n, bn, c 11.
Vecto r and M a trix Products
MATLAB
1.6
M ATLAB 1.6 1. » A = rand(3,4) A = 0.6885 0.7362 0.8682 0.7254 0.6295 0.9995
0.8886 0.2332 0.3063
0.3510 0.5133 0.5911
» B = rand(4,2) B = 0.8460 0.4154 0.4121 0.5373 0.8415 0.4679 0.2693 0.2872 » A*B ans = 1.7281 1.3679 1.3614
1.1982 1.0070 1.1116
» B*A ??? Error using ==> * Inner matrix dimensions must agree.
The product of a 3 x 4 with a 4 x 2 will be a 3 x 2 matrix, so A B is a 3 x 2 matrix. However, the product of a 4 x 2 with a 3 x 4 is not defined since 2, the number of columns of left factor, is not 3, the number of rows of the right factor.
» A = round(10*(2*rand(3)-l)) A = -9 -10 -2 1 -2 4 3 -9 2 » B = round(10*(2j B = 9 -8 4 7 3 8 1 5 -2 » A*B ans = -153 -1 -34
46 -22 -55
-126 8 -50
» B*A ans = -77 -36 4
-110 -148 -51
-42 14 0
The probability that for two random matrices, A B = B A is very small.
60
C h ap ter 1 Systems o f Linear Equations and Matrices
3. (a) » A
=C2 0 7 7
9 -23 0 4 -12 4 5-11 8 -10 4 ];
» b =[34; 24; 15; 33 ]; » z =[-2; 3; 1; 0]; » x =[-5; 10; 2; 2]; » A*x ’ /. This will be b. ans = 34 24 18 33 » A*z ans =
# /, This will be zero.
0 0 0 0
(b) For any scalar s, j4(x + sz) = A x -fThis can be tested by computing the following: » s = rand(l) s = 0.7529 » A * (x+s*z) ans = 34.0000 24.0000 15.0000 33.0000
*/, This will be b
4. »'/, Generate Random matrices: » A = round( 10*( 2*rand(2,4)-l) ) A = -3 3 0 4 -5 4 5 9 » B = round( 10*( 2*rand(4,5)-l) ) B = -1 0 10 1 -5 9 3 -7 0 3 -4 -2 3 7 1 -1 2 -3 -10 8
= A x -f 0 = Ax.
Instructor’s Manual
Vector and M a trix Products » B ( :,2) B = -1 9 -4 -1
MATLAB
1.6
= B ( :,3) 10 -7 3 -3
1 0 7 -10
-5 3 1 8
*/. p a r t (b) : A*B ans = -63 -63 26 -90 -90 12
-43 -60
56 114
10 -7 3 -3
» »
Since columns 2 and 3 were the same in B , they will also be the same in AB. (c) The above can be repeated several times. (d) Proof: Assume that the columns m and n are the same in B. This implies that ■= for any i. If C — A B , what we wish to show is that Cjm = Cjn for any j , i.e., that the m and rath columns of C are the same. To do this, we will write Cjm using notation. Assume that the width of A is p, then: p p Cjm ~ ^ ^ Q-jibim — ^ ^ Qjibjn — Cjn . *= 1 *= 1 Which concludes the proof.
round( 10*( 2*rand(5,6) - 1) ) -5 9 -5 7 4
-1 6 -10 -7 4
4 -1 2 5 10
4 -4 5 -7 -10
0 -8 -7 8 6
5 -1 -10 2 0
round( 10*( 2*rand(6 , D -• 1) x = -9 1 -2 -6 -2 3 »
A*x - ( x(l)*A(:,1) + x(2)*A(:,2) + x(3)*A(:,3) + ... x(4)*A(:,4) + x(5)*A(:,5) + x(6)*A(:,6) ) ans = 0 0 0 0 0
This is essentially the same as expression (10) in the text. The expression inside the parentheses is the definition of m atrix multiplication.
C hapter 1 Systems o f Linear Equations and Matrices
In stru cto r^ Manual
6 . (a) We can write A B == B A as A B — B A = 0. Then, with this choice of A and 5 ,
0 = AB - BA _ í ax i + bx 3 ax 2 + fcx4 \ y cxi 4- dx 3 CX2 + dx 4 J
/ axi 4 - CX2 bx\ + dx 2 A \ 0 x 3 + ex 4 6x 3 + dx 4 y
_ / —cx 2 4- 6^3 \ cxi 4- (d —a)x 3 —cx4
-i- (a - d)x 2 4- bx 4 \ cx 2 —fcx3 J
If we use each of the 4 entries in this matrix as one equation in our system, we will get a 4 x 4 system with coefficient m atrix R and variables x. (b) O) a = 1; b = -i; c = 5; d = 0 -c b 0 -b 0 b a-d d-a -c c 0 -b 0 c
R = C
10 —1
62
R = 0 1 5 0 » rref(R) ans = 1. 0000 0 0 0
-5 5 0 5
-1 0 -5 1
0 -1 -5 0
0- 1. 0000 - 1. 0000 1. 0000 0. 2000 0 0 0 0
0 0 0
= I*.nr = ( X3 + a;4 S So X\ — X3 4" x4, X2 = or BR = V ® S ) = X3( i V o ) + l 4 ( o i ) ' Notice
that thematricesx4 ^ J ^ = x4/ commutewithall matrices, andsotherewill alwaysbe infinitelymanySolutionsfor any A. (ü) » format rat */, Use rat(rref(R), ’s*) in Matlab 3.5 » rref(R) ans = 1 0 -1 1 0 1 1 / 5 0 0 0 0 0 0 0 0 0
If we choose X3 = 5, and x 4 = 1 we will get x\ = 6 , and x 2 = —1. You may choose any other integers, as long as X3 is divisible by 5. (iii) The m atrix B will be » B = [6-1; 5 1] B = 6 -1 5 1
V ecto r and M a trix Products
MATLAB
1.6
» A = [a b; c d] A = 1 -1 5 -4 » A*B - B*A ans = 0 0
0 0
Since A B —B A is zero, we have verified that A B = BA. (iv) This can be repeated for any other choice of £3 and £ 4. (c) We can repeat the above, using the new matrix A: = 1; b = 2; c = 3; d = 4 b 0 -c := [ 0 0 b -b a-d -c 0 d-a c c -b 0] 0 l
R = 0
00 1
-2 3 0
-3 0 3
2 0 3 -2
0 2 -3 0
» format rat % This gives rational numbers in the output, for Matlab 4.0. » rref(R) */, Use rat(rref(R), 1*1) in Matlab 3.5. ans = 1 0 1 1 0 1 -2/3 0 0 0 0 0 0 0 0 0 ' 1 o* r - i 131 + *4 ° 1 . (Again £ 41 are 1 °. possible B ’s.) If we choose £3 = 3, and £4 = 1 we will get x\ — —2, and £2 = 2. The m atrix B will be
As above, we may choose £3 and £4 arbitrarily and B = x 3
» B = [-2 2; 3 1] B =
-2 3 » A = [a b; c d] A = 1 3
2 4
» A*B - B*A ans = 0 0
0 0
Since A B — B A is zero, we have verified that A B = B A . (d) The above may be repeated using any m atrix for A. To avoid round off error, you should use an integer matrix.
C hapter 1 Systems o f Linear Equations and Matrices
In stru cto r^ Manual
» A = round( 10*( 2*rand(2,2) -1) ) A = -6 4 -9 4 » B = B = 9 -2
round( 10*( 2*rand(2,2) -1) ) 0 7
» C = (A+B)~2 C = -35 56 -154 77 » D = A~2 + 2*A*B + B~2 D = -43 48 -192 85 » C-D ans =
8
8
38
-8
In general, it is not true that C — D. However, they will be equal when we use A and B from prob lem 6 : » A = [1 -1; 5 -4]; » B = [6-1; 5 1]; » C = (A+B)*2 C = 29 -8 40 -11 » D = A~2 + 2*A*B + B~2 D = 29 -8 40 -11 » C-D ans = 0 0
0 0
When this is repeated with the matrices from 6 (c), C will again be the same as D. In fact the statement (A + B )2 = ,42 + 2,45 + B 2 if and only if AB = BA. Proof: We may expand (A + B ) 2 as follows: {A + B ) 2 = (A + B)(A + B) = A(A + B) + B(A + B) = AA + AB + BA + BB = A 2 + A B + B A + B 2. If we subtract this from A 2 + 2A B + B 2, we get A B — BA, which is zero whenever AB = BA. Thus we may say th at (A + B ) 2 is ,42 + 2A B + B 2 exactly when A B is BA.
V ecto r and M a trix Products
MATLAB
1.6
8. (a) round( 10*(2*rand(6,5)A = 4 2 9 7 1 -8
-9 5 -3 3 5 10
3 -2 4 8 5 -5
-3 -5 10 4 5 3
-9 3 8 -5 -1 5
0
0
[ 1 0 0 0 0 0] E = 1 »
0
0
3
- 9 - 3
E*A
ans = 4
» E = [0 0 1 0 0 0] E =
0
»
0
1
4
-3
-9
0
0
E*A
ans = 9
10
8
In the first case, E A was the first row of A, in the second case, E A was the third row of A. In general if E is all zeros except a 1 in the ¿th column, E A will be the ¿th row of A. (b) » »
E = [200000]; E*A
ans = 8 » »
6
-18
-6
-18
20
16
E = [ 0 0 2 0 0 0 ] ; E*A
ans = 18
8
-6
As above, E A will be the ¿th row of A , but this time it will be multiplied by 2. (c) (i) » E »
= [1 0 1 0 0 0 ] ; E*A
ans = 13
7
-12
7
-1
(ii) Here E A is the sum of the first and third rows. » E = [ 2 0 1 0 0 0 ] ; » E*A ans = 17 10 -21 4
-10
65
66
C h ap ter 1 Systems o f Linear Equations and Matrices
In stru cto r^ Manual
Here, E A is the twice the first row plus the third row. In general E A will be made up by multiplying the íth row of A by the zth element of E , and then adding these rows together. (d) In general, if E is zero except for a p in the Arth entry, then E A will be the kth row of A multiplied by p. To test this, » A = round( 10*(2*rand(3,5)-l)) A = -5 -9 -4 -1 5 -8 9 5 0 8 9 -9 7 -2 1 » E = » E*A ans = -24
[ 0 3 0]; '/, A 3 in the 2nd entry. */♦ This will be the 2nd row of A, roultiplied by 3. 0
24
27
15
This may be repeated with a different choice of E and A. (e) In general, if E has a p in the kth entry and a q in the jth entry, and zeros elsewhere, EA will be the kth row of A times p plus the jt h row of A times q.
»
E =[0 3 1]; */. 3 in the 2nd entry. 1 in the 3rdentry. » E*A */,This will be 3*(2nd row) + l*(3rd row). ans = -15 -2 25 18 22
This may be repeated with a different choice of E and A. (f) You should find the same results as in (d) and (e), except using columns instead of rows: » F = [0; 0; 2; 0; 0] F = 0 0
X A 2 in the 3rd entry.
2 0 0 » A*F '/• This will be 2 times the 3rd column of A. ans =
-8 16
2 This may be repeated with different A ’s and F ’s. 9. (a) » A = round( 10*(2*rand(3)-l)) A = -7 7 5 -10 3 10 4 5 8 » B = round( 10*(2*rand(3)-l)) B = -5 0 -2 -4 2 7 -3 7 -5
Vecto r and M a trix Products
MATLAB
1.6
» UA = triu(A) */, The upper triangular part of A. UA = -7 7 5 0 3 10
0
0
8
» = triu(B) */, The upper triangular part of B. UB = -5 0 -2 0 2 7 0 0 - 5 » UA*UB ans = 35 0 0
14 6 0
38 -29 -40
This has the property that UA*UB is also upper triangular. When this is repeated for larger matrices, UA+UB will still be upper triangular. (b) If A and B are upper triangular matrices, then C = A B is also an upper triangular matrix. Proof: The m atrix A is upper triangular when atj = 0 for any i > j. Since B is also upper trian gular, bij = 0 when i > j. W hat we wish to show is that c,¿ = 0 for i > j. Assume that i > j. If we use summation notation, n
C{j —^ ^Ctjkbkj. k=l Split this sum into two sums: k < i and k > i: i —1 Cij — ^ ^ Clikbkj k —1
n
“f" ^
^ O-ik^kj •
k=i
In the first sum, we have i > k so a,* = 0. In the second sum, we have k > i > j, so b^j = 0. In either sum, we are just adding up 0, so c,;- = 0. Which implies that C is upper triangular. (c) The product of two lower triangular matrices is also lower triangular. To test this, generate two random lower triangular matrices.: » A = tril( round( A = 0 -2 0 0 0 7 -7 2 -7 » B = tril( round( B = 10 0 0 -2 -5 0 -7 0 -7 » A*B ans = -20 -14 -25
*/• This is ; 0 -35 -10
0 0 49
67
C hapter 1 Systems o f Linear Equations and Matrices
In stru cto r^ Manual
1 0. (a) » A = A = -2 -6 -9 8 -1
round( 10*(2*rand(5)-l)) -7 9 -2 -7 8
-8 -7 -9 -3 -5
-7 6 -1 -3 -1
6 9 3 -6 4
-8 -7 0 0 0
-7 6 -1 0 0
6 9 3 -6 0
triu(A,1) 0 0 0 0 0
-7 0 0 0 0
B is the m atrix made from A with nonzero elements above the main diagonal. If we cali the di agonal above the main diagonal the first diagonal and the one above that the second, and so on, B has zeros below the first diagonal. Type h e lp t r i u for more information about t r i u . » B~2 ans = 0 0 0 0 0
0 0 0 0 0
49 0 0 0 0
-34 7 0 0 0
-45 -57 6 0 0
B 2 has zeros below the second diagonal. Similarly, B 3 will have zeros below the third diagonal. jB5 = 0, so B is nilpotent with index 5. 0 >) » B = triu(A,2) B = 0 0 - 8 - 7 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0
6 9 3 0 0
This time, B has zeros below the second diagonal. » B~2 ans = 0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
-24 0 0 0 0
B 2 has zeros below the fourth diagonal. B 3 will be zero, so B is nilpotent with index 3. (c) If we repeat (a) with a 7 x 7 matrix, we will find that B k will have zeros below the jfcth diagonal. This means th at B will be nilpotent with index 7. If we repeat (b) with a 7 x 7 matrix, we will
Vecto r and M a trix Products
MATLAB
1.6
find that B k will have zeros below the 2&th diagonal. This means that B will be nilpotent with index 4. (d) If we choose B = triu(A, j ) , then B k will have zeros below the j - k th diagonal. For a 6 x 6 matrix, B will be nilpotent with index 3 when we choose j to be the smallest number so that j • 3 > 6 , which is 2 . » A = A = 7 -8 -8 5 3 -6
round( 10*(2*rand(6)-l))
-4 6 6 -2
-4 -6 -10 -6 10 -5
6 -7 -2 2 -6 7
-7 10 -5 -5 -8 -6
» C = triu(A,2) c = 0 0 -5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
-4 -6 0 0 0 0
6 -7 -2 0 0 0
-7 10 -5 -5 0 0
10 0 0 0 0 0
45 30 0 0 0 0
-5
00 i
» C~2 ans = 0 0 0 0 0 0
-6 -2 9 9 -2
*/♦ This is not zero. 0 0 0 0 0 0
0 0 0 0 0 0
» C~3 '/, This is zero . ans = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0
So index oí n: 0 0 0 0 0 0
11. First generate the eight matrices: » A = round( 6*( 2*rand(2) - 1)) A = 3 -2 6 -3 » B = round( 6*( 2*rand(2) - 1)) B = -5 5 2 -3
0 0 0 0 0 0
0 0 0 0 0 0
69
70
C hapter 1 Systems o f Linear Equations and M atrices » C = round( 6*( 2*rand(2) - 1)) C = -I 0 3 -3 » D = round( 6*( 2*rand(2) - 1)) D = -3 -4 0 -2 E = round( 6*( 2*rand(2) - 1)) E = 5 -5 5 5 F = round( 6*( 2*rand(2) - 1)) 0 0
-2 6
G = round( 6*( 2*rand(2) - 1)) 0 -3
-5 5
H = round( 6*( 2*rand(2) - 1)) -5 0
-1 -3
/# The block matrix. AA =: [ A B ; ; C D] # 3 6 -1 3
-2 -3 0 -3
-5 2 -3 -2
5 -3 -4 0
BB == C E F;; G H] */, Another block matrix. 5 5 0 -3
-5 5 -5 5
0 0 -5 0
-2 6 -1 -3
25 -70 0 -20
25 -10 15 10
-28 -23 17 -22
AA*BB -10 24 7 0 K = K = -10 24 7 0
[ A*E+B*G 25 -70 0 -20
25 -10 15 10
A*F+B*H; -28 -23 17 -22
C*E+D*G C*F+D*H]
Instructor's Manual
Vecto r and M a trix Products » AA*BB - K ans = 0 0 0 0 0 0 0 0
0 0 0 0
MATLAB
1.6
0 0 0 0
fact, AA*BB will always be the same as K
12. » A = round( 10*(2*rand(3,4)-l)) A = 5 -10 9 8 -9 7 4 1 5 -7 7 -1 » B = round( 10*(2*rand(4,5)-l)) B = 8 1 0 7 -I 2 -4 6 9 -5 7 -9 5 “6 -2 -7 -7 1 1 1 D = A ( :,1)*B(1,:) + A ( :,,2)*B(2 A ( :,3)*B(3,:) + A(:,,4)*B(4 D = 187 7 96 -4 -9 -35 -112 -42 42 11 -96 99 17 -26 -11 »
» D - A*B ans = 0 0 0 0 0 0
0 0 0
0 0 0
0 0 0
In general, D will be AB. To see this, notice that the kth term in this sum is the product of the kth column of A and the Arth row of B. The ij th entry in this m atrix will be: aikbkj. When we add together all of the matrices, the ijth entry will be Ylkzzi aikbkj, which is the ijth entry of AB. 13. (a) » » » » AB
AB = zeros(3,5); AB(1, [1 2]) = [1 1]; AB(2, [2 3]) = [1 1]; AB(3, [1 4 5]) = [1 1 1] = 1 1 0 0 0 1 1 0 1 0 0 1 1
0 0
71
72
C hapter 1 Systems o f Linear Equations and Matrices » » » » » » BC
BC = zeros(5,8); BC(1, [1 35]) = [1 BC(2, [3 47]) = [1 BC(3,[1 56 8]) = BC(4,8) =1; BC(5, [5 67]) = [1 =
1 0 1 0 0 » » » » » » » »
0 0 0 0 0
1 1 0 0 0
In s tru c to r^ Manual
1 1]; 1 1]; [1 1 1 1]; 1 1]
0 1 0
1 0 1 0
0 0 1
0 1 0
0
0
0 0 1
0
1
0
1
1
1 0
CD = zeros(8,10); CD(1, [1 23]) = [1 1 1]; CD(2,[3 46]) = [1 1 1]; CD(3,[8 9 10]) = [1 1 1]; CD(4, [4 57]) = [1 1 1]; CD(5, [1 46 8]) = [1 1 1 1] ; CD(6, [2 4]) = [1 1]; CD(7,[1 5 9]) = [1 1 1]; >(8,[1
2
6
4
7
9
10])
=
[1
1
1 1 1
1
1]
CD = 1
1
1
0
0
0
0
0
0
0
0
0
1
1
0
1
0
0
0
0
0
0
0
0
0
0
0
1
1
1
0
0
0
1
1
0
1
0
0
0 0
1
0
0
1
0
1
0
1
0
0
1
0
1
0
0
0
0
0
0
1
0
0
0
1
0
0
0
1
0
1
1
0
1
0
1
1
0
1
1
(b) Person i in group 1 will have contact through person j in group 3 through person k in group 2 if ABik = 1 and BCkj — 1 , or in other words ABikBCkj = 1. If we sum over k ) this will tell us how many indirect contacts person i has with person j. So the indirect contact m atrix will be A B • BC. Similarly, to get from group 1 to group 4 via groups 2 and 3, we will multiply all three matrices together: » AD = AB*BC*CD AD = 3
1
1
4
3
1
2 4
2 2
1 2
1 2
3 2
3
3 2
2
5
3
1
4
1
3
1
3
3
2
None of the entries are zero. This signifies that everybody in group 1 has some indirect contact with each person in group 4. Since the (1,5) entry is 2, there are 2 different paths that connecting person 1 in group 1 with person 5 in group 4. Similarly, since the ( 2 , 1 ) entry is 4, there are 4 different paths connecting person 2 in group 1 with person 1 in group 4. (c) In order to find the total number of contacts a person in group 4 has, we add the rows of the in direct contact matrix. From problem 12, we see that a simple way to do this is to multiply by Cl 1 1 ]. » ones(l,3) * AD ans = 12
7
3
10
5
6
4
8
9
6
V ecto r and M a trix Products
MATLAB
1.6
The person with the most indirect contacts is person 1, with 12 contacts. Person 3 has 3 contacts, which is the least. Similarly, we can add the columns together to find out how many indi rect contacts each person in group 1 has. » AD * ones(10,l) ans = 19 25 26
Person 3 has the most indirect contacts with people in group 4, and so is the most dangerous. 14. (a) Column one means that of the households using product 1 , after one month 80/switch to prod uct 3. Column two means that of the households using product 2, after one month 75/switch to prod uct 3. Column three means that of the households using product 3, after one month 90/switch to product 2 . (b) Since x is the distribution of households using the products after 0 months, P x will be the distribution of households using the products after 1 month. Similarly, P fcx will be how many house holds use each product after k months. (c) » »
x = [10000; 10000; 10000]; P = [ .8 .2 .05 .05 .75 .05 .15 .05 .9 ] P = 0.8000 0.2000 0.0500 0.0500 0.7500 0.0500 0.1500 0.0500 0.9000
It may be convenient to round off your answers. One way to do this is to set the output format to bank. This rounds numbers to the nearest hundredth. » format bank » P~5 * x ans = 10275.83 5840.35 13883.83 » P~10* x ans = 9477.30 5141.24 15381.46 » P~15 * x ans = 9142.60 5023.74 15833.66 » P~20 * x ans = 9038.77 5003.99 15957.24
C h ap ter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
» P*25 * x ans = 9010.03 5000.67 15989.30 » P~30 * x ans = 9002.52 5000.11 15997.37 » P~35 * x ans = 9000.62 5000.02 15999.36 » P~40 * x ans = 9000.15 5000.00 15999.85 » P~45 * x ans = 9000.04 5000.00 15999.96 » P~50 * x ans = 9000.01 5000.00 15999.99
As n gets larger, P nx tends to (900, 500,1600). This may be interpreted by saying that eventually every month the same number of households switch to product ¿, as switch from product i, for i = 1 , 2 ,3.
» x = [0; 30000; 0]; » P~5 * x ans = 12056.86 9201.75 8741.39 » P~50 * x ans = 9000.04 5000.00 15999.96
Although for small n, P nx is different from that in (c), for large ra, P nx tends to the same valué: (9000,5000,16000). (e) For any choice of x, P nx will tend to the same vector: P nx —►(9000,5000,16000).
Vecto r and M a trix Products
MATLAB
1.6
(f) » P ‘50 ans = 0.30 0.17 0.53 » 30000 * P~50 ans = 9000.00 5000.00 16000.00
0.30 0.17 0.53
0.30 0.17 0.53
9000.04 5000.00 15999.96
8999.99 5000.00 16000.01
When n is lar ge, the columns of 30000Pn are very cióse to each other, and to the vector (9000, 5000,15000). So that when 300000Pn is multiplied by any vector x, it will be cióse to (xi + X2 + # 3) times this vector.
» P = [.8 .1 .1; .05 » n = 50; */• Try this » 1000 * P~n ans = 333.33 238.10 428.57 » »
format
.75 .1; .15 .15 .8]; for n= 5,10,15,25...
333.33 238.10 428.57
333.33 238.10 428.57
'/, Don't forget to return to normal output format at the end of this problem.
In the limit, the columns of 1000Pn tend to the vector (333.33,238.1,428.57). This will be the long term distribution of cars no m atter what the starting distribution is. A car rental agency could use this information by planning to have a larger parking lot at office 3 than at office 1 , or by planning to hire more mechanics at office 3 than at office 1. 15. (a) Column 1 says that 40% of the fish in group 1 survive to belong to group 2 the next year. Col umn 2 says that 20% of the fish in group 2 stay in group 2, and 50% of the fish in group 2 survive to be in group 3 the next year. This would happen if group 2 covers fish in an age range of more than 1 year. Column 3 says that each fish in group 3 has 2 babies, and that 20% survive and stay in group 3, and 50% survive and enter group 4. Column 4 says that each fish in group 4 has 2 babies, and then 20% survive to stay in group 4 and 40% survive to enter group 5. Column 5 says that each fish in group 5 has a 10% chance of survival to stay in group 5. (b) If x is the distribution of fish at this moment, then y = Sx will be the number of fish after one year. Since y is also a distribution of fish, S y = S • 5 x = S2x will be the number of fish one year later, or two years from now. (c) »
S = [ 0 .4 0 0 0
0 2 2 0 .2 0 0 0 .5 .2 0 0 0 .5 .2 0 0 0 .4 .1 ];
75
76
C h ap ter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
» x =[5000; 10000; 20000; 20000; 5000]; » n = 10; floor(S~n*x) ans = 41016 21666 12949 7754 3709 » */• repeat for n=20,30, » n = 50; floor(S~n*x) ans = 49063 24412 15183 9443 4179
...
After about 10 years, the population grows steadily. The growth rate will be exponential. (d) » S(1,3) = 1; » n = 10; floor(S~n*x) ans = 15774 8573 5572 4100 2105 » '/, repeat for n=20,30, ... » n = 50; floor(S~n*x) ans = 550 305
212 147 71
This time, the population decays exponentially. Not enough new fish are being born to keep the population steady. » S(1,3) = 2; S(3,2) = .3; » n = 10; floor(S~n*x) ans = 13280 8007 3334 2438 1321 » */• repeat for n=20,30, ... » n = 50; floor(S~n*x) ans =
100 58 25 18 9
Vecto r and M a trix Products
MATLAB
1.6
Again, the total population decays when the survival rate from group 2 into group 3 is dropped from 50% down to 30%. Not enough fish survive to be group 3 in order to create new fish. (e) At the end of the first year, the number of fish that have survived will be Sx. If we harvest h of these fish, we will end up with u = Sx — h fish. Similarly, S u will be the number of these remaining fish that survive to the end of the second year. If we harvest h of these, we will have S u — h fish remaining.
(f) »
S(l,3) = 2; S(3,2) = .5;
»
h = [ 0 ; 0 ; 0 ; 0 ; 2000] ;
» u = S*x - h u = 80000 4000 9000 14000 6500 » u = S*u - h u = 46000 32800 3800 7300 4250
'/, This command should be repeated several times,
After two more iterations of the last command, the vector u will have a negative number in the last entry. This means that we may harvest this many fish for 3 years, and at the end of the fourth year, there will be less than 2000 mature fish to be harvested. (g) After repeating the following experiment, for different valúes of n, you will find that u begins to drop for the first few years, and then begins to increase after about 5 years. If n is chosen to be 1530 or smaller, u will never be negative. » n = 1530; » h = [0; 0; 0; 0; n] ; » u = S*x -h u = 80000 4000 9000 14000 6970 » u = S*u -h » u = 46000 32800 3800 7300 4767
'/, This step should be repeated several times, until X u begins to rise again.
(h) If the above experiment is repeated using nonzero valúes in the last two entries of h, the total harvest can be improved. For example, if h = (0,0,0,1500,790), the total harvest will be 2290, and the population of fish will not become negative over 15 years. These valúes can be improved by slightly changing h.
Instructor’s Manual
C h ap ter 1 Systems o f Linear Equations and Matrices
S ectio n 1.7 1.
'2 -
2.
0 (:)■(?)
4.
3.
6.
8.
7. 4xi — X2 4- 5x3 = 4 6 xi 4- X2 + 3x3 = 20
x\ 2xi + 3x2 4- X3 = 2 10. 4x 2 + x 3 = 3
11.
2xi 4- X3 = 2 9. —3xi 4- 4x2 = 3 5x 2 4- 5x 3 = 5
X2 = 2 x\ =3 = 2
x2
= 3
x3
= -5 X4 =
6 xi -f 2x 2 4- %3 = 2 13. —2xi 4- 3x2 4- X3 = 4 0= 2
14.
12.
2xi 4- 3x 2 4- X3 = 0 4xi — X2 4- 5x3 = 0 3xi + 6x 2 — 7 x 3 = 0
15.
7xi 4- 2x 2 = 1 3xi 4- X2 = 2 6x 1 4- 9x 2 = 3
6
3xi 4- x 2 4* 5x 3 = 6 2xi 4* 3x2 4- 2x 3 = 4
5 J , *1 = 3/2, *2 = 5/4, * 3 = - 2 /5
,* , = ( 2 , 0); ( _2 t
| o ) — ( í 'o | o )
= *2(3,1),
x = ( 2, 0) + *a(3 ,l)
18.
1 -1 1
1 -1 1
3 -3 3
0
1 -1
1
3 -3 3
\ )
0 0
0)
( 1 -1 ! 0 \ vo o o 0 )
xp = (6 , 0, 0)
Xfc = (*2 —*3)^2, * 3)
x = (6,0,0) + (*2 - * 3, x 2, * 3)
19.
x = (2,0,0) + *3(—1/3, - 4 /3 ,1 )
1 0 1/3 0 1 4/3 0 0 0
xp = (2 , 0 , 0)
1 0 1/3 0 1 4/3 0 0 0
xfc = *s(—1/3, - 4 /3 ,1 )
Matrices and Linear Systems o f Equations
20.
1 0 1/3 0 1 4/3 0 0 0
Xp — ( 2 , 0 , 0)
1 0 1/3 0 1 4/3 0 0 0
Xh = x 3(—1/3, —4/3,1)
Section 1.7
79
x = (2 , 0 , 0) + * s ( - l /3 , - 4 / 3 , 1 ) 1 1 - 1 2
21.
3 2
1 -1 2
N
fl
1 -1
y
Vo-i
1 1 01
3 2
\ fl 1 - 1 2 / “ V0 - 1 4 - 7
3\ f l O -4 / ~ \ 0 1 -
2
Z\
fl
0
4 '7
"V
\°
1
1 -1
-5 7 -5 7
0 1 )
xp = ( - 1 , 4 , 0 , 0)
XA = (—3x 3 + 5x4, 4x 3 - 7x4, X3, X4) X
= (-1 ,4 ,0 ,0 ) + (—3x 3 -t- 5x4, 4x 3 - 7x4, x3, x4) 10 2-1 0 1 1 0 0 0 - 4 1
1 0 0 - 1/2 0 1 0 1/4 0 0 1 -1/4
xP = ( 5 ,4 ,-3 ,0 ) 10 2-1 0 1 1 0 0 0 - 4 1
1 0 0 - 1/2 0 1 0 1/4 0 0 1 -1/4
x A = x 4( l / 2 , - 1 /4 ,1 /4 ,1 ) x = (5,4, —3,0) + x (l/2 , —1 /4,1/4,1) 23. Plugging y = ciyi + c^yi into the left side of the differential equation gives, since (cy)" — cyn and (yi + 2/2)" = y'{ + v 'í c\y'{ + c2y 2 + a(x)(cu/'1 + C2 J/2 ) + ¿(x )(ciy i + ciyi) = ci(y'i + «K^Oyí + K x )y\) + ^{y'i + a(x)y'2 + b(x)y2) = ci( 0) + c2(0) = o
24. l/p - y" + a(x)(y'p - y'q) + b(x)(yp - yq) = (yp + a(*)yp + K x )yp) - (y'q +
a ( x )i/q
+ b(x)yq)
= /(* ) - /(* ) = 0 Thus, yp(x) —yq(x) solves y"(x) -f a(x)y'(x) + b(x)y(x) = 0.
C hapter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
M A T L A B 1.7 1.
(a) » A = round(10*( 2*rand(3)-l)) A = -6 4 0 -9 9 7 4 -2 -9 » b = round(10*(2*rand(3,1)-1)) b = -9 1 3 » R = rref([A b]) R = 1 . 0000 0 0 1. 0000 0 0
0 0 1.0000
3.8053 3.4579 0.5895
» x = R ( :,4) x = 3.8053 3.4579 0.5895 » A*x */. First find A*x ans = -9.0000 1 . 0000 3.0000 » A*x - b */, Compare Ax with b. ans = 1.0e-14 * 0 -0.2665 0.0888
In theory, A x —b should be zero, but in practice, the Computer will have some round-off error. Here the error is on the order of 10~14. »
y = x(l)*A(:,1) + x(2)*A(:,2) + x(3)*A(:,3)
y = -9.0000 1 . 0000 3.0000 » y-b ans = 1.0e-14 * 0 -0.2665 0.0888
Again, y is the same as A * x, so it will be b up to some round-off error.
M atrices and Linear Systems o f Equations
(b)
MATLAB
1.7
(i) »
A = [4 2 5 9
17 5 5 -1 19 4 23 -4 ]
9 1 9 5
» b = [U; 9; 16; 40] ; » R = rref([A b]) R = 2 -1 0 1 1 1 1 0 0 0 0 0 0 0 0 0
5 -1 0 0
The general solution will have £3 and £4 arbitrary, x\ = —2£3 + £4 + 5, and £2 = —£3 — £4 —1 . We may pick, for example, £3 = 1 and £4 = 0, to get the particular solution: »
x = [3; -2; 1; 0] ;
(ii) » A*x ans = 11 9 16 40 »
% This
should be the same as b.
y = x(l)*A(:,1) + x(2)*A(:,2) + x(3)*A(:,3) + x(4)*A(:,4)
y =
11 9 16 40
As in (a), if x is a solution of the system with [A b] as augmented matrix, Ax = b, then Ax and £1 A(:, 1) + KX4 A(:, 4) = y are both b. (iii) If this is repeated with other choices of £3 and £ 4, the same results will occur: Ax = b and y = b. For example: » x = [ » A*x ans =
4; -3; 1; 1];
11 9 16 40
(iv) »
y = x (1)*A (:,1) + x(2)*A(:,2) + x(3)*A(:,3) + x(4)*A(:,4)
y =
11 9 16 40
(c) The solution of a system of equations represented by [A b] is the same as the solution of the ma trix equation Ax = b. Also, multiplying a matrix by a single column is equivalent to adding múltiples of the columns of this matrix.
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C hapter 1 Systems o f Linear Equations and Matrices
In stru cto r^ Manual
2. (a) The ij entry of A x is the inner product of the ¿th row of A with the j th column of x. Since x has only one column, j must always be 1 , and the first column of x is just x itself. Since A x = 0 , the inner product of the ¿th row of A with x is zero. This means that the ¿th row is orthogonal to x. (b) This can be done by solving A x = 0 where A is the matrix whose rows are the given vectors. » A = [1 2 -3 0 4; » rref(A) ans = 1.0000 0 0 1.0000
5 -5 2 0 1];
-0.7333 -1.1333
0 0
1.4667 1.2667
For a solution, we may choose £ 3 , £ 4, and £5 arbitrarily, and then set x\ = .733Sxs — 1.4667£5 and £2 = 1.1333£3 — 1.2667£5. 3. (a) The m atrix x sol ves the nonhomogeneous system A x = b. The m atrix z was a solution of the homogeneous system Az = 0. If we set y = x + sz, then we found that y was also a solution of the nonhomogeneous system A y = b. The corollary tells us the converse, i.e. that any such solution can be written as y = x H- sh where H is a solution of A h = 0. (b) (i) Refer to the solution of l(b) above. The matrix R is the reduced echelon form of [A b]. Since two variables can be chosen arbitrarily in the solution of A x = b, there are infinitely many Solu tions. (Ü) »
x = A\b
Warning: Matrix is singular to working precisión, x = Inf Inf Inf Inf » » »
*/, Since A is singular, it has no inverse. '/, from the answer to l(b). x = [ 5; -1; 0; 0];
Instead, enter a solution
(iii) » rref(A) ans = 1 0 0 1 0 0 0 0
2 1 0 0
- 1 1 0 0
Solutions of ^4x = 0 are of the form £3 and £4 are arbitrary, £1 = —2£3 + £ 4, and £2 = - £ 3 - £4. Pick one solution, with £3 = 1, £4 = 0 » z = [-2; -1; 1; 0]; » A*(x+z) '/. This should yield b. ans =
11 9 16 40
Matrices and Linear Systems o f Equations
MATLAB
1.7
Another solution, with £3 = 0, £4 = 1. » z = [1; -1; 0; 1]; » A*(x+z) '/• Again, this should be b. ans =
11 9 16 40
This can be repeated two more times by choosing other valúes for £3 and £ 4. 4. (a) »
A = [ 5 5 8 0 45 8 7 39 8 9 9116];
» rref(A) ans =
1 0
0
1
0
0
0
0
0
0
1
0
0 0 0
1
If we were to reduce [A b] for any b, we would still have the same left hand side. Since there is a pivot in every row, the system will be consistent, and there will be a unique solution. (b) If we sol ve A x = b, for x, then we may write b as
b=Ci£i -fC2£2+C3£3+C4£4 where c, is the zth column of A. For example: » b = round( 10*(2*rand(4,1)-1)) b = -6 -9 4 4 » R = rref([A b]) R = 1.0000 0 0 1.0000 0 0 0 o » x = R ( :,5) x = 0.6476 3.5799 -3.3922 -0.3361
0 0 1.0000 0
0 0 0 1.0000
0.6476 3.5799 -3.3922 -0.3361
83
84
C hapter 1 Systems o f Linear Equations and Matrices » »
In structor’s Manual
'/• Now we may write b as a combination of columns of A: y = A ( :, l)*x(l) + A(:,2)*x(2) + A(:,3)*x(3) +A(:,4)*x(4)
y =
- 6.0000 -9.0001 3.9999 3.9999
Up to round off error, y is the same as b. This can be repeated two more times. (c) »
A = [ 5 4 3 9
» rref(A) ans = 1 0 0 0
5 -5 0 5 -6 7 9 -15 9 7 6]; 1
0 1 0 0
1 -2 0 0
0 0 1 0
It is possible that there is a b for which the reduced form of [A b] does not have a zero in the (4,5) entry. Since the left hand side has all zeros in the fourth row, this would mean the system is inconsistent. By experimenting with several possible b you can find that if b = (1, 0, 0, 0)*, there will be no solution: » b = [1; 0; 0; 0]; » rref([A b ]) ans = 1 0 1 0 0 1 2 0
0 0
0
0
0
1
0
0
0
0
0 1
Notice th at the above solution is inconsistent. (d) If you start with a vector b which is a combination of columns of A, then there will always be a solution of A x = b. » k = ronnd( 10*(2*rand(4,1)-1)) */• generate 4 random numbers. k = 9
-2 0 7 » V, Write b as a combination of columns of A: » b = A (:,1)*k(1) + A ( :,2)*k(2) + A(:,3)*k(3) +A(:,4)*k(4) b = 35.0000 23.6475 5.9754 76.9836
M atrices and Linear Systems o f Equations » rref([ A b]) ans = 1.0000 0 0 1.0000 0 0 0 0
1.0000 -2.0000 0 0
0 0 1.0000 0
MATLAB
1.7
9.0000 -2.0000 -0.3361 0
This system is consistent. (e) To see that this system will always have a solution, we only need to show that it has at least one solution. However, writing b as a combination of columns of A using the scalars &,*, is equivalent to the m atrix multiplication A k = b. This means that the vector k will be a solution of A x = b. Since the system has a solution, it is consistent. (a)
(i) For A from 4(c): »
A = [ 5 4 3 9
5 -50 5 -6 7 9 -15 9 1 7 6];
» rref(A) ans = 1 0 0 1
-
1 2
0 0
0
0
0
1
0
0
0
0
(ii) T h e Solutions of th is hom ogeneous system have xs arb itrary , and x \ = —X3 , x 2 = 2 x 3 , and x 4 = 0.
(iii) If we set X3 = 1, then we have x\ = —1 and x 2 = 2. This corresponds to 0 = A x = —le í -f 2c 2 -f IC3 -f 0c4,
where c, is the zth column of A. Which can be rewritten as c 3 = le í - 2 c2. To check this: » 1*A(:,1) - 2*A(:,2) ans = -5
*/, This should be the same as A(:,3).
-6 -15 7 » A(:,3) ans = -5
*/, This is the third column.
-6 -15 7
(iv) This system only allows one arbitrary variable. (v) If we let x be the third column of rref (A) then .Ax will be the third column of A.
86
C hapter 1 Systems o f Linear Equations and Matrices
In structor’s Manual
(b) »
A = [4 2 5 9
9 17 5 1 5-1 9 19 4 5 23 -4 ];
» rref(A) ans = 1 0 0 1 0 0 0 0
1
2 - 1 1 0 0 0 0
The solution of this system will have X 3 and X 4 arbitrary, and x i = — 2 ¿r3 4 - # 4 , and If X3 = 1 and X4 = 0, then we have x\ = —2 and X2 = —1. This corresponds to
X 2 =
— X 3 — X 4 .
c3 = 2ci -f lc2. Similarly, if £3 = 0 and £4 = 1 , then we have x\ = 1 and x 2 = —1. This corresponds to c4 = — leí -f lc2. As above, if x is the third column of rref (A) then A x will be the third column of A. Similarly, if x is fourth column of rref (A) then Ax will be the fourth column of A. To check this: » R = rref(A); » x = R ( :,3) x =
2 1 0 0 » A*x ans = 17 5 19 23
'/, This will be the 3rd column of A.
» x = R(:,4) x = -1 1 0 0 » A*x ans = 5 -1 4 -4
*/• This is the 4th column of A.
Matrices and Linear Systems o f Equations
MATLAB
1.7
(c) First we generate a random m atrix and modify it as directed. » A = round( 10*(2*rand(6,6)-l)); » A (:,3) = 2*A(:,2) - 3*A(:,1); » A ( :,5) = -A(:,1) + 2*A(:,2) - 3*A(:,4); » A ( :,6) = A ( :,2) + 4*A(:,4) A = 33 -28 10 8 -14 -9 27 -38 4 -5 -22 8 -1 0 -7 5 -1 -17 7 5 0 3 5 1 21 -16 -9 0 27 -4 -19 10 -43 35 3 -5 rref(A) R = 1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
-3 2 0 0 0 0
-1 2 -3 0 0 0
0 1 4 0 0 0
This will be the same, for almost every random matrix that we start with. The solution of this system has £ 3 , £ 5 , and £6 arbitrary and £1 = 3£3 + £s, £2 = —2£3 —2£5 —£ 6, and £4 = 3£5 —4£6. As in (b) above, if we set one of the three arbitrary variables to 1 and the others to 0, and write out x we will get: í~ z\
/~ í\
21
x =
0 0
o 0/
1
2 o -3
,x =
\
/° \ or x =
1 0
4
0
0
0/
\ 0/
Writing out A x = 0 as a linear combination of columns of A, we will get the original equations: » » »
A ( :,3) A ( :,5) A ( :,6)
= 2*A(:,2) - 3*A(:,1); = -A(:,1) + 2*A(:,2) - 3*A(:,4); = A ( :,2) + 4*A(:,4)
87
88
C hapter 1 Systems o f Linear Equations and Matrices
In stru ctor’s Manual
S e c tio n 1.8
1. Since det A = 2- 2 —3 1 = 1 ^ 0 , A-1 exists. A-1 = y
„
(2 1
/2 1 1 0 \
0r \3
2 0 1 )
(2 0 (,0 1
4 -2 \ -3 2 )
-3
(,0 1
c
2 -1 \
( 1 0
(,0 1
(2 1
10 \ ^ 1 ) -*
1,0 |
10 \ -3 2 )
_! _ /
2)' S
2
“ ( ,- 3
2- l \ 2) '
2. Since det A = (—1) • (—12) — 1 • 12 = 0, A is not invertible.
3. det j 4 = 0 —1 = —1 ^ 0
= M ) ( _ J “ J) = ( J ¿ )
4. Since det A = 3 —3 = 0, A-1 does not exist. 5. A is not invertible since det A = ab — ba = 0. 1 0
-
1 0
1 /2
0 1 3/2 0 0 1
1 2/3 1/3 0 1 1 0 0 1
1/3
0
5/4
0' 0
0 -1/4
0 0
1/3 -1/3
0 1/2 0
0
0
0
1/3 -1/3 -1 /3 ' 1/2
1/2 1 /2
13/8 - 1 /2 - 1 / 8 ' A " 1 = [ -1 5 /8 1/2 3/8 5/4 0 —1/4
13/8 -1/2 -1 /8 ' -15/8 1/2 3/8 5/4 0 -1/4,
0
-
0 1
1/2
0' 0
0 -1
1/3 - 1 /3 - 1 / 3 ' A-1 =
1
0 -1
0
0
1/2
1
0 - 1
1 -1 0 0 1 -1 0 0 1,
0
9.
6 15
2 9 -18
1 2 1 0 -7
1 0 0' 0.1333 0.0667 0 -3 2 1
A is not invertible. 0 1 1 -3 0 -1 1/8 -1 /4 ' 3/8 3/4 1/4 1/2,
A" i _
1 0
1/4 1/4 0 ' 1/4 -3/4 0
1/2
0 1 -3/2 0 0 2 3/8 1/8 - 1 / 4 ' - 1 /8 - 3 /8 3/4 - 1 /4 1/4 1/2,
-
1/2 1/2 1
T h e Inverse o f a Square M a tr ix
Section 1.8
1 0 -5 2 -1 4 19 -7 3
11.
A is not invertible.
2 3 1 1 1 2
/I i 1 2 13. 1 -1 Vi 3 (1 0 0 1 0 0 \0 0
1 1 -1 2 2 1 3 2 0 2 0 -1/3 1 -2/3 0 3
A-1=
1 0 0 0 -1 1 1 -5
0 1 0 0
0 0 1 0
°\ °l 0 1/ 1 1 -1/3 -2/3 -2/3 -1/3 2 2
1 0 °\ /I 0 1 1 -1 -> 0 1 1 0 0 0 1/ \0 0 1
1 /I i 1 1 0 i -2 1 -1 0 -2 1 0 -1 Vo 2 2 1 -1 /I 0 0 0 °\ 0 0 10 0 0 0 0 10 1/ \0 0 0 1
0 1 0 0
-1 2 0 1 -1 0 -1 1 1
0 /I °v 0 0 10 1 0 0 Vo 0 1/ 7/3 -1/3 -1/3 4/9 -1/9 -4/9 -1/9 -2/9 1/9 -5/3 2/3 2/3
0 1 0 0
1 0 0 0 1 0 0 0 1
3 0 -2 1 -3 2 6 -1
2 -1 0 -1 1 0 -3 2 1 1 -2 0
2/3 \ 1/9 2/9 1 /3 /
7/3 - 1 / 3 - 1 /3 - 2 / 3 ' 4/9 - 1 / 9 - 4 /9 1/9 - 1 / 9 - 2 / 9 1/9 2/9 -5/3 2/3 2/3 1/3>
1 0 2 3 -1 i 0 4 2 i -1 3 V-i 0 5 7
1 0 0 0
0 1 0 0
0 0 1 0
°\ 0 0 1/
/I 0 0 Vo
0 1 1 0
2 3 2 7 -5 -3 7 10
1 1 -2 1
0 1 0 0
0 0 1 0
óv 0 0 1/
/I 0 2 3 0 1 2 7 0 0 -7 -10 Vo 0 0 0
^4 is not invertible. 0 2 1 -2 -6 0 2 5 o 0 1 3 3/ 2 - 5/4 1/2 1/4 -1/2 ;3 /4 Vo 0 - 1/2 1/4 (l 0 0 0
1 -3 3 -12 15. -2 10 \-lL 6 /I 0 0 1 0 0
0 10 0 0 0 10
\0 0 0 1 A' -1 _
0 1 -1 2 -2 -1 -1 1 1
0 0 0\ 1 0 0 |
/ I -3 0 0 -3 -2 0 4 2 o i o r 0 0 1/ Vo 3 1 5/ 2 0 3/ 4 0 \ 1/2 0 1/4 0 -3/2 1 3/4 0 -1/2 0 -3/4 1 /
-2 0 1 1 /I 0 0 Vo
1 0 0 0\ -3 1 0 0 2 0 10 1 0 0 1/ 0 0 1 -2 1 0 1 -1 0 1 -3/2 3 0 0 -1/2 1
3 3 0\ 1 10 -2 -3/2 0 -1 -3/2 1 /
1 1 -3 -2
0 1 -1 -1
0 0 1 1
°\ 0 0 1/
89
90
C h ap ter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
16. A B C C - ' B - ' A - 1 = A B I B " 1A " 1 = A B B ~ l A " 1 = A I A " 1 = A A " 1 = I. Henee, by theorem 8, A B C is invertible and ( A B C ) " 1 = A " 1B ~ l C ~ x. 17. Show A 1 A 2 ’ -A mA ñ 1 ** A ^ A ^ 1 = I. Then by theorem 8, A 1 A 2 • • *^4m is invertible with inverse 4- 1 .. . 4- 1 j - i
i8
f 3 4V 3 V -2 - 3 ) V -2 - 3 /
9- 812-12\ V -6 + 6 - 8 + 9 / ~
19. If A = ± 7 then .A2 = 7. If a n = —022 and 021012 = 1 —o? 1 then ( ü22 012 ^ V °21 «22 / í
=
«22 + a 12a 21 —a22«12 + ai2«22 _ j —022^21 + x =
0.08 - 0 . ? :lo ) ( 1 ) = ( 0 -0.15 0.
Since detv4 = *
Henee, 1 - batches
13
anc* ^ = 0 )
^ nce ^et A = —0-01 ^
Thus’ 4 units of type A and 5 units of type
B are needed. 36. (a) (d)
37. (a)
0.293
(b)
200,000-0.293 = 58,600
0
50,000-0.044 = 2,200 '0.293 0 0' technology m atrix A = j 0.014 0.207 0.017 .0.044 0.010 0.216,
Leontief m atrix = 1 —A =
(b)
(c)
0.707 0 0' -0.014 0.793 -0.017 -0.044 -0.010 0.784,
1.414 0 0' (I - A ) - 1 = | 0.027 1.261 0.027 0.080 0.016 1.276,
13,213 \ / 18,689N x = ( / - A ) - 1 j 17,597 = 22,598 1 ,7 8 6 / 3,615,
V
T h e Inverse o f a Square M a trix
Section 1.8
It would require 18,689 pounds of agricultural producís, 22,598 pounds of manufactured goods, and 3,615 pounds of energy.
38
( oí )
40- ( 3 3 )
1 1/2
invertible
0
1
invertible
(o 0 ) not invertible 1 2 / 3 1 /3 ' 0 1 1 | invertible 0 0 1 1 6 2\ /1 6 2' 0 15 9 1—> I 0 1 3/5 1 not invertible 0-30-18/ \ 0 0 0,
43.
44.
' 1 —1/2 2\ 0 1 —14 j not invertible 0 0 0/ 11 1 1' 0 1 -2 1 0 0 1 - 2 /3 .0 0 0 1.
invertible
/1 0 2 3\ 0 12 7 not invertible 0 0 1 10/7 \000 0/ 46. Since 011022 —012021 ^ 0, then either a n or 012 is nonzero. We may assume without loss of generality that a n ^ 0. 1 /a n 0 í o n 012 1 oN í \ «12/ail \ 021 “22 0 1 / l 0 a22 ~ 012^ 21/^11 - 021/011 1 - ( ¡
022/ det A -a21 / det A
-ü 2 i /d e t A \ 011 / d^t A J
94
C hapter 1 Systems o f Linear Equations and Matrices
In s tru c to r^ Manual
47. (i) Suppose A is invertible. Then A x = 0 implies x = A ~ l 0 = 0. When reducing the augmented m atrix (^4|0) to reduced row echelon form, we must have (A|0) —►( I |0). Using the same elementary row operations will give A I. Conversely, suppose A is row equivalent to I. Write (A\I) and row reduce A to I. Then we will have (A\ I ) —►(I \ B ). Henee, A B — I. We want to show B A = I. It will suffice to show B A x = x for every n-vector x. Note that B is row equivalent to I . Henee, for every x, we can find a y such that B y = x. Thus, B A x = B A ( B y ) = B ( A B )y = B l y = B y = x. Therefore, A is invertible. (ii) Suppose A is invertible. Suppose A x = b = Ay. Multiplying by A ”1 we obtain A ~ l (Ax) = A ~ 1(Ay). It follows that x = y. Conversely, suppose the system A x = b has a unique solution for every n-vector b. This implies A is row equivalent to I. Therefore, by part (i), A is invertible. (iv) Suppose A is invertible. By (i) A is row equivalent to the identity m atrix In is in row ech elon form and has n-pivots. Conversely, suppose that an echelon form of A has n pivots. Then the reduced echelon form of A is the identity matrix In . Thus A is row equivalent to the identity matrix In . By (i) A is invertible.
48. From (^q ^ J A n + A A 21 —
^ = ( ^ O ^ J we ^12 4“ A A 2 2 ==
following equations:
A 21 = 0 and A 22 = I-
Solving for A n and A 12 we get A n = I and A 12 — —A .
49. From ( Al1 P ) f ‘? 11 \ I12 ) = ( í. ® ] we obtain the following equations: A n B n = J, A n B i 2
\^21 ^22 / \ ^ 21 ^*22 J
\U 1 /
O, A 2 1 B 11 4- A 2 2 B 21 = O and A 2 1 B 12 + A 2 2 B 22 == I- Solving for the B¡j we get Bu =
, B 12 = Of B 21 — A 22 ( ^ A 2 i A 1i ) and B 22 = ^ 22*•
T hn, ( Al1 ° V 1 - ^ A" ° \ U 21 A 22) \ A £ ( - A 21A ñ l ) A 221 ) '
The Inverse of a Square Matrix
Calculator 1.8 95
CALCULATOR SOLUTIONS 1.8 The solution for Problem nn assumes the data has been entered into the matrix A18nn. 50. The inverse of A1850, calculated by A 1 8 50 (2nd) (F^ IENTER) is [[
-.076312056738
]
[
.099858156028
-.1 0 1 8 4 3 9 7 1 6 3 1 .272151300236
-.076501182033
-.192056737589
]
[
.143262411348
.075177304965
]]
-.067139479905
.
51. The inverse A1851, calculated by A l 851 (2nd) [F1] [ENTER) is [[
-1 .4 0 7 5 4 0 3 9 4 9 7 .456014362657
.303411131059
]
[
.657091561939
-.166965888689
-.154398563734
] .
[
.439856373429
-.26750448833
-.03231597846
]]
52. The inverse of A1852, calculated by A l 8 52 [2nd] (F1"] [ENTER] is [[
-1 .6 9 7 0 1 0 5 3 6 5 4 1.60958317276
2.30794647641
1.41376091667
]
[
-.793941492883 1.67786019382
.951949895801
.261501231662
]
[
1.95621257025
[
-.6 5 4 4 2 3 0 7 6 0 4 2 .641726876009
-.200637557471
-.462876267722 -.249248482807
.357416829406
]
.536103553562
]]
.010590297574
]
53. The inverse A1853, calculated by A l 853 [2nd] [F1] (ENTER) is [[
.03984485542
.00954069806
.035197499863
[ -.003683920834
.00460111628
4 . 2 6 2 9 8 4 9 5 0 6 e -4 - . 0 0 5 6 0 8 4 5 3 1 2 7
]
[
.018345358489
.009413418129
.008529325192
.00297300599
]
[
.019410170095
.007025671643
.025503332841
.015762697722
]]
*
54. The inverse of A1854, calculated by A l 8 54 [2nd) [x1] [ENTER) is [[
.333333333333
-.208333333333
1.675 -1.42142857143 ]
[ 0
.125
-.325
[0
0
.2
-.114285714286 ]
[0
0
0
- . 1 4 2 8 5 7 1 4 2 8 5 7 ]]
.578571428571
]
which has zeros below the diagonal. 55. The inverse of A l855, calculated b y A 1 8 55 |2nd) [F1] [ENTER) is .126871293496
.203412258426
[ 0
-.068965517241
-.067111605488
.035660968327
.113274788336
]
[ 0
0
-.02688172043
.019100169779
.007849426165
]
[ 0
0
0
.010964912281
-.0 0 3 2 2 6 3 4 9 4 5 6 ]
[ 0
0
0
0
.02132196162
[[
.04329004329
-.125690401552 -.196440007897
]
]]
which has zeros below the diagonal. 56. The results in Problem 54 and 55 suggest that the inverse of an upper triangular matrix is upper triangular.
C h ap ter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
M A T L A B 1.8 1. (a) (i) » A = [1 2 3; 2 5 4; 1 -1 1 0 ] ; » R = [A eye(3)] R = 1 2 3 1 0 2 5 4 0 1
1
-1
10
0
0
54 -16 -7
-23 7 3
» rref(R) ans =
1 0 0
0 1 0
-7
2 1
» S = ans ( :, [4:6] ) S = 54 -23 -7 -16 7 2 -7 3 1
(ü) » S*A ans =
1 0 0
0 1 0
0 0 1
» A*S ans =
1 0 0
0 1 0
Both S A and A S are the identity matrix. Henee, S is the inverse of A. (iii)
»
inv(A) ans = 54.0000 -16.0000 -7.0000
-23.0000 -7.0000 7.0000 2.0000 3.0000 1.0000
This seems to be the same as S although S-inv(A) m a y not be exactly zero due to round off. The c o m m a n d inv(A) computes the inverse of the matrix A. (b) » A = 2*rand(5)-l A = -0.3435 0.8206 0.2653 0.5244 0.5128 -0.4751 -0.9051 0.9821 -0.2693 0.4722
-0.5059 0.9651 0.4453 0.5067 0.3030
-0.8546 0.2633 0.7694 -0.4546 -0.1272
0.5330 -0.0445 -0.5245 -0.4502 -0.2815
T h e Inverse o f a Square M a trix » R = [A eye(5)]; » rref(R) ans = Columns 1 through 7 0 1.0000 0 0 1.0000 0 1.0000 0 0 0 0 0 0 0 0 Columns 8 through 10 0.3383 1.7333 -0.2542 2.2202 0.2963 -0.8329 0.1091 -1.8634 0.2153 0.8678 -0.6154 -0.5434 -1.6150 -0.2000 -1.8254 » S = ans(:,[6:10] ) S = 1.6785 0.1073 2.0062 0.1332 -1.5142 0.9549 0.0574 0.1234 -0.7602 1.0249
0 0 0 1.0000 0
0 0 0 0 1.0000
1.7333 2.2202 -1.8634 0.8678 -1.6150
-0.2542 0.2963 0.2153 -0.6154 -0.2000
0.3383 -0.8329 0.1091 -0.5434 -1.8254
» A*S ans = 1.0000 0.0000 0.0000 0.0000 0.0000
0 1.0000 0.0000 0.0000 0.0000
0.0000 0.0000 1.0000 0.0000 0.0000
0.0000 0.0000 0.0000 1.0000 0.0000
0.0000 0.0000 0.0000 0.0000 1.0000
» S*A ans = 1.0000 0.0000 0.0000 0.0000 0.0000
0.0000 1.0000 0.0000 0.0000 0.0000
0.0000 0.0000 1.0000 0.0000 0.0000
0.0000 0.0000 0.0000 1.0000 0.0000
0.0000 0.0000 0.0000 0.0000 1.0000
1.6785 2.0062 -1.5142 0.0574 -0.7602
» inv(A) - S */• This should be zero up to round off error. ans = 1.0e-15 * 0 -0.4441 0.1943 0.3886 -0.2220 0 -0.0278 0 0.4996 -0.1110 -0.2220 -0.1110 0 -0.1943 0.1943 0.0278 -0.0416 0.1110 0 0 0.1110 -0.2220 0.2220 -0.1388 0.2220
MATLAB
0.1073 0.1332 0.9549 0.1234 1.0249
98
C h ap ter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
2. (i) »
A = (1/13)* [ 2 7 5 ;
098;
740];
» rref (A) */, part (a) ans = 1 0 0 0 1 0
0
0
1
» B = inv(A) This will exist since there are no zero rows above B = -32.0000 20.0000 11.0000 56.0000 -35.0000 -16.0000 -63.0000 41.0000 18.0000 » '/. For part (c): » A*B '/, This should be I . ans =
1.0000
0.0000
0.0000
0
1.0000
0.0000
0
0.0000
1.0000
» B*A ans =
'/, This should also be I.
1.0000 0 0
0.0000 1.0000 0.0000
» b = 2*rand(3,l) - 1 b = 0.0090 0.0326 -0.3619 » rref([A b]) ans =
1.0000 0 0 » »
*/• Choose a random b, with 3 rows.
*/, Solve Ax=b.
0 1.0000 0
x = ans(:,4); y = inv(A)*b
0 0.0000 1.0000
0 0 1.0000
-3.6191 5.1571 -5.7488
'/• Set x equal to the solution. */• Solve Ax=b using inverses.
-3.6191 5.1571 -5.7488 » x-y ans = 1.0e-15 * 0.4441 0.8882 -0.8882
•/, This should be zero up to round off error.
T h e Inverse o f a Square M a trix
MATLAB
1.8
(ii) » A = [2 -4 5; 0 0 8; 7 -14 0] ; » rref (A) */, part (a) ans = -2 0 1 0 1 0 0 0 0 » B = inv(A) •/. ‘ Warning : Matrix is B = Inf Inf Inf Inf Inf Inf Inf Inf Inf »
'/, For part (b) :
(iii) » A = [14-21; ! » rref (A) '/. part ans = 0 2 1 0 1 -1 0 0 0 0 0 0
0 0 1 0
» B = inv(A) */, This will not exist since there are zero rows above. Warning: Matrix is cióse to singular or badly scaled. Results may be inaccurate. RCOND = 7.178166e-18 B = 1.0e+15 * -3.6029 -1.8014 1.8014 1.2867 1.8014 0.9007 -0.9007 -0.6434 1.8014 0.9007 -0.9007 -0.6434
0.0000
0
0.0000
0.0000
For part (b): From the command r r e f , we see that A is actually singular. However, MATLAB gives us an answer since round oíF error during the computation makes A seem to be invertibie. However, MATLAB gives a warning since it could tell “nonsingularity” might be due to round off error. (iv) » A = [1461; 5197; » rref (A) */. part (a) ans =
1 0 0 0
0
1 0 0
0
0
1 0
0
7484;
0757];
0 0
1
» B = inv(A) */, This will exist since there are no zero rows. B = -0.1558 -0.0779 0.2208 -0.0260 0.0115 -0.1609 0.1133 0.0945 0.2121 0.1061 -0.1061 -0.0758 -0.1631 0.0851 -0.0375 0.1025
100
C h ap ter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
» */• For part (c): » A*B */• This should be I. ans = 0
0.0000 1.0000
0.0000 0.0000
0
0.0000 0.0000 1.0000
0.0000
o
1.0000
» B*A ans =
*/, This should also be I.
1.0000 0 0 0.0000
0.0000 1.0000 0 0.0000
» rref ([A b] ) ans =
1.0000 0 0 0
0.0000
0.0000 0.0000 1.0000 0.0000
» b = 2*rand(4,l) - 1 b = 0.9733 - 0.0120 -0.4677 -0.8185
» »
0.0000 0.0000 0.0000 1.0000
o
0.0000 1.0000
% Choose a random b.
*/, Solve Ax=b.
0 1.0000
0 0
0 0
1.0000
0 0 0
0
1.0000
x = ans(:,5); y = inv(A)*b
-0.2327 -0.1172 0.3168 -0.2260
*/• Set x equal to the solution. */. Solve Ax=b using inverses.
y = -0.2327 -0.1172 0.3168 -0.2260 » x-y ans = 1.0e-16 * 0 -0.5551 0 0.2776
'/• This should be zero up to round off error.
(v) »
A = (-1/56) *
[1
2
345
0 -1 2 -1 2 1 0 0 2 - 1 1 1 - 1 1 1 0
0 0
0
4 ];
T h e Inverse o f a Square M a trix » rref (A) ans = 1 0 0 0 0 »
*/. part (a)
B = inv(A)
o o o 0 1
o o 0 1 o
0 0 1 0 o
0 1 0 0 0
'/• This will exist since there are no zero rows above.
B =
8.0000
-12.0000 -
8.0000
-4.0000 0
-40.0000 4 .0 0 0 0 -16.0000
- 8.00 00 40.0000
20 . 0 0 0 0
-24.0000 0
0
8.0000
» */, For part (c): » A*B '/• This should be I. ans = 0 1.0000 0
0.0000
1.0000 0.0000
0.0000 1.0000
0 0 0
0
0.0000 0.0000 o
1.0000
0.0000 1.0000
0
0.0000 0.0000 1.0000
0.0000
0.0000 o
0
» rref([A b]) ans =
1.0000 0 0 0 0
y = 2.0200
14.0423 -26.9510 -10.5699 13.5557
0.0000 0.0000
0.0000
o 1.0000 o
0.0000 0.0000 1.0000
o
*/. Choose a random b.
*/• Solve Ax=b.
0 1.0000 0 0 0
x = ans(:,6); y = inv(A)*b -
0 0 0
0
30.0000
2 0 .0 0 0 0
*/. This should also be I.
» b = 2*rand(5,l) - 1 b = 0.5230 0.5404 0.6556 -0.7493 -0.9683
» »
-18.0000 -14.0000
0 0
0
2 2 .0 0 0 0
-56.0000 -28.0000 0 28.0000 0
0.0000 1.0000
1.0000
» B*A ans
MATLAB
0
0
0 1.0000
0 0 1.0000
0 0
0
0 0
-
0 0 1.0000
Set x equal to the solution. */, Solve Ax=b using inverses.
2.0200
14.0423 -26.9510 -10.5699 13.5557
1.8
101
102
C hapter 1 Systems o f Linear Equations and Matrices » x-y ans = 1.0e-13 * -0.1066 0.0355
In stru cto r^ Manual
*/, This should be zero up to round off error,
0 0 0.0178 »
A = [ 1 2-1 7 5 0-1 2-3 2 1 0 3 1 - 1
1 1 1 4 0
0
» rref (A) ans = 1 0
0
•/. Matrix for (vi)
1 0
4 ];
# /, part (a)
0 0 0
0 1
3 -2
1 3
0 0 0
0 0 0
0 0 0
0 0 1 0 0
» B = inv(A) */• This will not exist since there are zero rows. Warning: Matrix is singular to working precisión. B = Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf »
3.
*/• This is the same warning as in (ii).
(a) » A = round(10*(2*rand(5)-l)) A = 4 10 0 - 5 -6 7 8 2 -2 -7
3
-
5
5 -4 5 - 3 » B = A; » B(3,:) B = 1 0 3 5 1
7
1
-2 -1 7 - 4
1 6 -9
= 3*B(1,:) + 5*B(2,:) 8 2 34 -7 -6
4 -7 -23 -5 -10
-2 -9 -51 9 -5
-6 -4 -38 3 -7
T h e Inverse o f a Square M a trix
MATLAB
1.8
» rref(B) ans =
1.0000 0 0 0 0
0 1.0000 0 0 0
0 0 o 1.0000 o
0 0 1.0000 0 0
-6.8037 3.2398 -4.1014 4.3544
0
Since the bottom row of the reduced matrix is zero, B will not be invertible. (b) » B= A; 2*B(2,:) - B(1,:) » B(4,:) B = 8 4 -2 -6 1 -7 -9 -4 2 0 -8 4 8 7 9 -18 -16 -4 -2 -1 -10 -6 -5 -7 1 » rref(B) ans =
0 1.0000 0 0 0
1.0000 0 0 0 0
0 0 1.0000 0 0
0 0 0 1.0000 o
-6.6780 3.0199 -3.8251 4.0906
0
Again, the bottom row is zero, so B is ñot invertible. (c) Assume that row i is a linear combination of the other rows. Since one of the valid operations in Gaussian Elimination is to add a múltiple of one row to another, we may subtract this linear combination of the other rows from row i. This will leave a m atrix with zeros in row i. By rearranging the rows, we may put this zero row at the bottom. After continuing the Gaussian Elimi nation, this bottom row will still be zero, so the matrix will be singular.
» A = round(10*(2*rand(7)-l)) A = -4 0 1 4 2 -5 3 -2 9 -2 -2 1 0 -7 -6 0 -1 3 -9 -7 9 2 -7 10 8 6 2 -2 -7 -1 7 -7 -6 -5 -7
9 -2 -7 8 -8 -7 -9
-3 -5 -7 6 -1 -3 -1
9 -2 -7 8 -8 -7 -9
-3 -5 -7 6 -1 -3 -1
B = A; B ( :,3) = 2*B(:,1) - B ( :,2) B = 4 -2 -2 0 -7 2 7
2 9 1 -7 10 -2 -7
6 -13 -5 7 -24 6 21
-4 3 -7 3 2 6 -5
0 -2 -6 -9 8 -1 -7
103
C hapter 1 Systems o f Linear Equations and Matrices
»
» » » » D =
Instructor’s Manual
c = A; = C(:,l)+C(:,2)-C(:,3); = 3*C(:,2)
1 -o
2 9 1
0
» C ( :,4) » C ( :,6) C = 4 -2 -2 -7 2 7
10 -2 -7
1 -5 0 -1 9 -7 -6
5 12 -1 -6 -6 7 6
0 -2 -6 -9 8 -1 -7
6 27 3 -21 30 -6 -21
-3 -5 -7 6 -1 -3 -1
D = A; D(:,2) = 3*D(:, i); D ( :,4) = 2*D(:, 1)-D(:, 2)+4*D( :,3); D ( :,5) = D(:,2) - 5*D( :,3)
6 21
1 -5 0 -1 9 -7 -6
0 -18 2 -4 43 -30 -31
7 19 -6 5 -66 41 51
9 -2 -7 8 -8 -7 -9
-3 -5 -7 6 -1 -3 -1
0 1 0 0 0 0 0
2 -1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
» rref(C) ans = 1 0 0 1 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0
1 1 -1 0 0 0 0
0 0 0 1 0 0 0
0 3 0 0 0 0 0
0 0 0 0 1 0 0
» rref(D) ans = 1 3 0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0
-1 4 0 0 0 0 0
3 -5 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
12 -6 -6 0 i to M-
4 -2 -2 0 i
104
2 7 » rref(B) ans = 1 0 0 0 0 0 0
Each r r e f has a row of zeros so the original matrices are not invertible. So we conjecture any m atrix with some columns equal to linear combinations of other columns will not be invertible.
T h e Inverse o f a Square M a tr ix
MATLAB
1.8
(b) Repeat with your own E. (The example E = A; E(: ,3) = 3*E(: ,2) ; E( : ,4) = 2*E( : ,1) + 4*E( : ,3) is interesting.) (c) If column j oí A is a linear combination of columns preceeding it, then there will be no pivot in column j of rref(A), and the non-zero entries in column j of rref (A) may be the coefñcients in the linear combination which represents the j th column. For instance column 4 in rref (D) has —1 in row 1, 4 in row 2 and D(:, 4) = —1D(:, 1) + 4Z}(:, 2). (However, for the E suggested in (b), rref (E) will not recover the (2,4) coefficients for column 4 since column 3 will not be a pivot column.) (d) This is the converse of Problem 5, Section 1.7. There we saw that if column j of rref (A) had no pivot, then column j oí A is a linear combination of the preceding (pivot) columns of A with coefficients given by the entries in column j of rref (A). (a) (i) » A = triu( round(10*(2*rand(5)-l)) ); » A(2,2) = 0 A = 1 -1 -1 -7 5 0 0 6 3 5 0 0 -3 2 -4
0
0
0
-10
-2
0
0
0
0
4
0
0
» rref (A) ans =
1
0
-
*/, part (i)
1
0
0
1
0
0
0
0
0
1
0
0 0
0 0
0 0
0 0
1 0
Since the bottom row is zero, A is not invertible. This can be repeated 4 more times. In general, if A is upper triangular, and there is a zero on the diagonal, A will not be invertible. However, if there is not a zero, it will be invertible: » A = triu( round(10*(2*rand(5)-l)) ) A = -6 -2 10 3 3 0 2 -5 4 3 0 0 -5 6 -9 0 0 0 4 2 0 0 0 0 -5 » rref(A) ans =
1
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
This m atrix is invertible.
106
C h ap ter 1 Systems o f Linear Equations and Matrices
In structor’s M anual
(ü) » B = inv(A) */, part (ii) B = -0.1667 -0.1667 -0.1667 -0.5000 0 0.5000 0 0 -0.2000 0 0 0 0 0 0
0.5417 0.2500 0.3000 0.2500 0
0.3167 1.3000 0.4800 0.1000 -0.2000
The inverse of an upper triangular matrix is also upper triangular. Also, if the diagonal entries of A are an then the diagonal entries of A ~ l will be l/a,¿. Since one of the diagonal entries in (i) was 0, the m atrix will not be invertible since 1/0 is not defined. (iii) Part (iii): To reduce the m atrix [A I] to echelon form, we would divide row one by a, row two by d, and row three by / . After this step, we would have 1/a, 1/d, and 1 / / on the diag onal of the right hand side. We would then add múltiples of row three to rows one and two, which will not change the diagonal on the right. Finally, we would add a múltiple of row two to row one, which again will not change the diagonal on the right. This will leave the I on the left, and A” 1 on the right. As predicted in (ii) the diagonal entries of A " 1 are the inverses of those in A. Also, as predicted in (i), the first step of this process will fail if a, d, or / is zero. (See the solution to Problem 29.) (b) » A = [1 2 3; 4 5 6; 7 8 9 ] ; » rref(A) ans =
1 0 0
0
1
1 2
0
0
A is not invertible. » B = [1 2 3 4; 5 6 7 8;9 10 11 12; 13 14 15 16]; » rref(B) ans =
1
0 - 1 - 2
0
1
0 0
0 0
2
3
0 0
0 0
B is also not invertible. In general, a matrix of this form will not be invertible. If C — ( ) is an n x n m atrix with Cij = j -f* (¿ — l)n , then csj = 2 * C2j —1 * c\j. This means that the third row is 2 times the second row minus the first row. (c) The assertion that there is a unique nth degree polynomial that fits n -f 1 points is the same as saying that the coefficient m atrix is invertible. » x = 2*rand(5,1)-1 x = -0.2330 0.0388 0.6619 -0.9309 -0.8931
T h e Inverse o f a Square M a trix » V= vander(x) V = 0.0029 -0 .0126 0.0000 0 .0001 0.1920 0 .2900 0.7508 -0 .8066 0.6361 -0 .7123 » inv(V) ans = 8.9238 10.0230 -3.7580 -4.7803 0.1907
-6 .5334 -9 .1141 0 .6876 4 .1676 0 .8377
0.0543 0.0015 0.4382 0.8665 0.7976
-0.2330 0.0388 0.6619 -0.9309 -0.8931
1.0000 1.0000 1.0000 1.0000 1.0000
0.7240 1.4612 0.8518 0.1049 -0.0054
24.5579 10.4447 -13.6375 -2.8701 0.1314
-27.6723 -12.8147 15.8561 3.3779 -0.1543
MATLAB
1.8
This may be repeated several times. As long as the points in x ai m atrix V will be invertible. Enter Al, A2,
. A5. Then
» rref(Al) ans = 1 0 0 1 0 0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 0 0 1
We see A l is invertible. The same result will come from rref (A3) and rref (A4), so both A3 and A4 are invertible. » rref(A2) ans =
1
1 0 0 0 0
3
0 0 0
» rref(A5) ans =
1 0 0 0 0
-2
0 0 0 0
0 1 0 0 0
1 -2 1 0 0
Both A2 and A 5 are not invertible. » rref(Al*A2) ans =
1 0 0 0 o
0 1 o o o
3
1
-2
3
0 0 0
0 0 0
107
108
Instructor’s Manual
C hapter 1 Systems o f Linear Equations and Matrices » rref(A1*A3) ans =
1 0 0 0 o
0 1 o o o
o o o 0 1
o 0 1 o o
A l •A2 is not invertible, and A l •AZ is invertible. From the list given A l •A3, A l •A4 , and A3 •A4 are invertible, while the others are not: » rref(A1+A5) ans =
-2
0 0 0 0
0 1 0 0 0
o 0 1 o o
1 -2
1 o o
» rref(A2+A3) ans =
1 0 0 0 0
-2
0 0 0 0
0 1 0 0 0
1 -2 1 0 o
» rref(A2+A4) ans =
1.0000 0 0 0 0
0 1.0000 0 0 0
1.2857 0.2857
-0.0621 0.8509
0 0 0
0 0 0
» rref(A2*A5) ans =
0 0 1 0 0
1 -2
1 0 0
» rref(A3*A5) ans =
0 0 1 0 0
1
-2
1 0 0
» rref(A4*A5) ans =
1 0 0 0 o
0 1 o o o
1
-2
1 0 0
0 0 1.0000 0 0
T h e Inverse o f a Square M a trix
MATLAB
1.8
A product of two matrices will be invertible only if both of the original two matrices are invert ible. In the above, since A 2 and A5 were not invertible, any product involving either of these is not invertible. (b) For A l • AS:
» inv(Al*A3) - inv(Al)*inv(A3) ans = 4.8095 -20.8571 -3.3810 -15.8571 16.5833 -7.8690 16.5238 -4.7976 1.0238 13.0238 7.0476 10.0238 -5.8452 -4.7024 -9.1190 -3.9881 0
0
0
» inv(Al*A3) - inv(A3)*inv(Al) ans = 1.0e-14 0 0.2665 -0.0444 0.1776 0 -0.0333 -0.1776 0.1332 -0.2665 -0.1554 0.1776 0.1332
0
0
0
8.0857 0.7357 -5.7357 2.9500
0
0
0 0.0888 -0.3553 0.1776
0.0888 -0.1332 -0.2665 0.0888
0
0
The second result is 0 to within round off error. Similarly: » inv(Al*A4) - inv(Al)*inv(A4) ans = -16.3571 13.5714 3.8571 3.0000 9.5714 9.5000 9.2857 -9.7143 6.7143 -1.6429 -1.5000 1.5000 -1.4286 -0.2857 0.7857
10.4286 18.3571 8.5714 -0.7143 -2.0000
6.4286 -14.1786 0.7500 “1.3929 0.7857
» inv(Al*A4) - inv(A4)*inv(Al) ans = 1.0e-13 * 0.0089 -0.3730 0.2087 -0.0355 0.3908 -0.1066 0.2309 -0.0488 -0.0355 -0.0644 -0.0266 0.0155 0.0006 0.0600 -0.0377
-0.6040 -0.1243 -0.1599 -0.0822 0.0977
0.3730 0.1599 0.1332 0.0389 -0.0622
» inv(A3*A4) - inv(A3)*inv(A4) ans = -74.6667 223.3333 -265.0000 -531.2333 320.8333 -435.5000 -435.0667 153.6667 -228.0000 158.9667 -20.3333 38.0000 33.6000 -22.6667 30.0000
117.6667 214.5000 119.0000 -22.1667 -14.6667
-63.7333 -48.5667 -1.9333 -16.2667 4.0000
» inv(A3*A4) - inv(A4)*inv(A3) ans = 1.0e-ll * -0.1478 0.0284 -0.2160 0.0284 -0.1364 -0.1307 -0.0114 -0.0625 -0.0966 0.0135 0.0014 0.0199 0.0092 0.0025 0.0181
0.1180 0.0796 0.0554 -0.0117 -0.0107
-0.0142 0.0092 0.0014 -0.0001 0.0006
109
110
C hapter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
From these, we may conjecture that =0
or
(.A S )" 1
In fact, this will be true: see theorem 3 in this section.
» A = [ 1 2 3; 4 5 6; 7 8 9]; » rref(A) ans =
1 0 0
0
-
1
1
2
0
0
Since the bottom row is all zeros, A isn’t invertible. (a) »
format short e '/, Use scientific notation. í = l.e-5; C = A; C(3,3) = A(3,3) + f ; » inv(C) ans = 9.9998e+04 -2.0000e+05 1.0000e+05 -2.0000e+05 4.0000e+05 -2.0000e+05 1.0000e+05 -2.0000e+05 1.0000e+05
»
(b) » f = l.e-7; » inv(C) ans = 1.0000e+07 -2.0000e+07 1.0000e+07
C = A; C(3,3) = A(3,3) + f;
-2.0000e+07 4.0000e+07 -2.0000e+07
1.0000e+07 -2.0000e+07 1.0000e+07
» f = l.e-10; C = A; C(3,3) = A(3,3) + f; » inv(C) ans = 1.0000e+10 -2.0000e+10 1.0000e+10 -2.0000e+10 4.0000e+10 -2.0000e+10 1.0000e+10 -2.0000e+10 1.0000e+10
(c) The entries in C ~ l are roughly the same size as 1/ / . (d) » » »
x= [1; 1; 1] ; f= l.e-5; C = A; C(3,3) = A(3,3) + f; b = [6; 15; 2 4 + f ] ; y= inv(C) * b
y = 1.0000e+00 1.0000e+00 1.0000e+00 » z = x-y z = 0 0
4 . 6566e-10
T h e Inverse o f a Square M a trix » »
f y
MATLAB
1.8
= l.e-7;C = A; 0(3,3) = A(3,3) + f; b= [6; 15; 24+f]; = inv(C) * b
y =
1 .0000e+00 1 .0000e+00 1 .0000e+00 » z = x-y z = 0 -5.9605e-08 -5.9605e-08 » »
f y
= l.e-10; C = A; 0(3,3) = A(3,3) + f; b= [6; 15; 24+f]; = inv(C) * b
y = 9.9997e-01 1.0001e+00 9.9997e-01 » z = x-y z = 3.0518e-05 -6.1035e-05 3.0518e-05 »
format
Return to standard format for the next problem.
As / gets smaller, the error term z becomes larger. This means that the closer C is to a noninvertible matrix, the more error there is in the computation of C ~ 1. In fact the sum of the exponents in / and z is always -15. This is related to the fact that there are about 15 significant digits in MATLAB’s internal computations. 8. (a) Problem 37: » A = [ 0.293 0.014 0.044; 0 0.207 0.010; 0 0.017 0.216]; » L = eye(3) -A '/• This is the Leontief raatrix. L = 7.0700e-01 -1.4000e-02 -4.4000e-02 0 7.9300e-01 -1.0000e-02 0 -1.7000e-02 7.8400e-01 » x = inv(L) * [ 13216; 17597; 1786] x = 1.0e+04 * 1.9305 2.2225 0.2760
Israel needs 19,305 pounds for Agriculture, 22,225 for Manufacturing, and 2,760 for Energy to export the given amounts.
111
112
C h ap ter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
(b) » A = [ .2 .1 .3; .15 .25 .25; .1 .05 0]; » L = eye(3) - A » format long » x = inv(L) * [300000; 200000; 200000] x = 1.0e+05 * 5.37197626654496 4.66453674121406 2.77042446371520 » format
This is the same answer as the one to 9(b) in Section 1.3. 9. If we arrange the message as a sequence of rows, then we need to multiply each row by the encoding matrix. Since these are rows, and not columns, the encoding m atrix must be on the right. Write M for the message and C for the coded message. If encoding the message is done by multiplying by A then C = M • A. Multiplying C by A” 1 will decode the message because C • A ~ l = (M A) - A ~ l = M (A - A " 1) — M I — M. » »
»
A = [ 1 2 - 3 4 5 ; - 2 - 5 8 - 8 - 9 ; 1 2 - 2 7 9 ; 1 1 0 6 12; 2 4 -6 8 11] C = [47 49 -19 257 487 10 -9 63 137 236 79 142 -184 372 536 59 70 -40 332 588]; H = round(C * inv(A)) '/, use round to get rid of any small error term.
M 1 15 22 6
18 21 9 21
»setstr(M +
ans = ARE [Y 0U [HA VING [ FUN[[
5 27 14 14
*k}
27 8 7 27 1)
25 1 27 27
*/. setstr(l:27 + 'A* - 1) is ,ABC...XYZ[ '/, So given command prints the message '/, but with *[* instead of * *.
The message decodes to “ARE YOU HAVING FUN” .
T h e Transpose o f a M a trix
Section 1.9
Section 1.9 ( - ¡ i )
* • & )
T -1 r 2 05 3 4 5,
/
6.
2 2 1 l \
\ - í 4 6 5J
11. Let A =
*■ ( ;- ; i)
’1 2 3 2 4 -5 3 - 5 7,
7.
'a d g' beh c f J>
9.
( o n ^12 • • • aim \ 021 O22 *’ * 02m
10.
O n
- f
621
and B —
611
012 + ¿12
Then (A + B ) %=
00 00 00
¿ lm \
/ ¿11 ¿12 *
V 6ni
\ d n i ü n 2 * * ’ 0 nm J
(
2 1' 4. | - 1 5 06.
021
+
¿21
622
6n2 •
**•
On i
• •
¿ 2 m
• bnm / - f
i > n l \
O22 4" ¿22 • ** On2 4* ¿n2
= A 1+ B X
V 0>lm 4~ ¿lm fl2m 4" ¿2m * * ' a n m 4“ ¿nm /
12. a = 5; /? = 3 13. ciij = ciji and ¿,¿ = ¿¿,- for 1 < i < n and 1 < j < n. Then, symmetric.
-f ¿¿j = Oj¿ + ¿j¿. Thus, A + £ is
= dkj for 1 < j < n, 1 < k < n. And since B is symmetric, ¿*¿ = ¿¿¿ for n n l ^ i ^ n , 1 ^ ¿ ^ fi. Therefore, ^ ^Qjib¿ib* = E ^ oaí- Thus, (A£)* = 5A . Jb= l Jb= l
14. Since A is symmetric,
15. Suppose A is m x n. Then A* is n x m. Then AA* is defined and is an m x m matrix. Note that n n ^^o¿jba; jb = ^^djkdik- That is, the i j th component of AAt is equal to the j i th component of AAt . Jfc= l
Jb= l
Thus, A A t is symemtric. Another proof is that (A A t )t = (A í )t A í = AA*, so AA* is equal to its own transpose, henee is symmetric. 16. If i = j then clearly a,-j = a¿t*. If i ^ j then
symmetric.
= 0 and a;-¿ = 0. Thus, dij = ay¿. Thus, A is
113
114
C hapter 1 Systems o f Linear Equations and Matrices
17. Let U =
/ « l l «12 0 1/22 V 0
« ln \ «2n
be upper triangular.
0 • • • 0 unn J
/un « 1 2
0 « 2 2
• •• 0^ o
-
O
Then U* = V « l n «2n
18. (a) No
Instructor’s Manual
(b) Yes
is lower triangular.
••• 0 /
(c) No
(d) Yes
19. A* = - A and B % = —B. Then (A + B)f = A* + B* = - A - B = - { A + B). Thus, A + B is skew-sy m metric. 20. A* = —A. ciij = —ají. Elements on the main diagonal are of the form a¿¿. o>a = 0. 21. (AB)* = B tA t = (—J5)(—A) = jBA A B is symmetric if and only if (A B )* = Thus A B is symmetric if and only if A and B commute.
= —a¿¿. It follows that
But (A £ )f = BA.
22. The i j th component of (A + A*)/2 = (a¿¿ + a¿¿)/2 and the j i th component of (A + A*)/2 — (aj¿-|-a¿¿)/2. Thus (A -f A x)/2 is symmetric. 23. The i j th component of (^4 —A x)/2 = (a,j —dji)/2 and the j i th component is (dji — dij)/2 = —(dij — a j i ) / 2 . Thus (A — Al ) / 2 is skew-symmetric. 24. Let A, J9, and C be n x n matrices. Suppose A = B + C where B is symmetric and C is skewsymmetric. Then = b¡j + c¡j and dji — bj¡ + Cji• But b¡j — bj¡ and Cji — —cij. Then b{j -f Cij = ciij bij
— Qji
Then = (dij + dji)/2 and c,j = (dij —dji)/2. Note that these Solutions are unique. Thus, any square m atrix can be written in a unique way as the sum of the symmetric m atrix (A + A t)/2 and the skew-symmetric m atrix (A —dx)/2. (This uses the results of Problems 22 and 23.)
25. AA* = ( ° n ° 12) ( a n Q n ) = ( Í 1 + ° ?2 011021 1 T Í ' ) = ( l ? V Thus ^ is invertible \fl21 a22 / \ a 12 a22 / \ailfl21 + ai2fl22 a21 + a22 / *J and A 1 = A *.
26. A1 =: ( 2 4 )
= - é ( _ 3 _í )
M ' ) ' 1 = - i (_ 2 - ? ) = ( - ?
= M -y
T h e Transpose o f a M a tr ix
'3 0 0 \ 28. A* = [ 2 2 0 1 2 —1 /
29. A* =
/1 0 5 \ 125 \ 1 3 1/
A -1=
í ^ " 1)4 =
/ 1 / 3 —1/3 —1 /3 \ 0 1/2 1 \ 0 0 - 1 /
/
(vi"1)4 =
/ 13/8 —15/8 5 /4 ' -1 /2 1/2 0 ) = (vi4) - 1 V-l/8 3/8 - l / 4 y
1/3 0 0 ' - 1 /3 1/2 0 \ —1/3 1 -1
Section 1.9
115
116
C hapter 1 Systems o f Linear Equations and Matrices
In stru cto r^ Manual
M A T L A B 1.9
1. » A = round( 5*(2*rand(4,3)-l)) A = -3 4 -5 -5 -1 -4
2
0
0
2
3
2
» B = round( 5*(2*rand(3,2)-l)) B = -5 -1
-1
2
-4
1
» (A*B)* - A* * B* ??? Error using ==> * Inner matrix dimensions must agree. » (A*B)' ans =
0 0
- B> * k> 0 0
0 0
0 0
It is possible that A tB t is not always defined. However {AB)1 is always the same as B iA t . 2. » A = [1 2 3; 2 5 4; 1 - 1 10]; » inv(A’) ans = 54.0000 -16.0000 -7.0000 -23.0000 7.0000 3.0000 -7.0000 2.0000 1.0000 » inv(A)* ans = 54.0000 -23.0000 -7.0000
*/, This was invertible.
-16.0000 -7.0000 7.0000 3.0000 2.0000 1.0000
This should be repeated for each of the other matrices. In each case A t is invertible if and only if A is invertible. Also, (A f)~1 == (A~1)t . 3.
(a) » A = round( 10*(2*rand(4) -1) ) A = 9 3 5 -3 7 - 2 - 5 3
1 -8
4
8
9
5 5
10
T h e Transpose o f a M a trix » B = A* + A B = 10 18 10 -4 6 -1 -11 11
6 -1 -18 10
MATLAB
1.9
-11 11 10 20
For any m atrix A, B will be symmetric. (b) » C = A' C = 0 4 -4 0 4 -9 5 -5
-4 9 0 0
-5 5 0 0
any A , C will be antisymmetric: C l — —C.
(c) » G = A*A> G = 124 23 23 87 -39 59 -53 -67
'i; For -39 59 123 29
the matrix above, -53 -67 29 253
» A = round( 10*(2*rand(3,4) -1) ) A = -3 4 -9 -5 -5 5 3 -1 10 3 8 5 » G = A*A> G = 131 13 13 60 -115 -16
-115 -16 198
For any m atrix A, G will be symmetric. (d) If the ij entry of A is then the ij entry of A f is a¿¿. The ij entry of B is 6¿¿ = which is the same as bji = aji + atJ-, so B is symmetric. The ij entry of C is —aji, while the ij entry of C %is aji — so C* = —C. The ij entry of G will be gij = J2k aik^jk- The j i entry of G will be Y^kajküik which is the same as gij. Henee G is symmetric. (a) In problem 2 from section 1.7, it was shown that the solution of A x = 0 produces all vectors x that are perpendicular to the rows of A. Since the columns of A are the same as the rows of A*, the solution of A fx = 0 will produces all vectors x that are perpendicular to the colums of A. (b) (i) » A = [2 0 1; 0 2 1; 1 1 1 ; - 1 1 1 ; 1 1 1 ] ; » rref(A’) ans = 1.0000 0 0.5000 0 0.5000 0 1.0000 0.5000 0 0.5000 0 0 0 1.0000 0
118
C h ap ter 1 Systems o f Linear Equations and Matrices
The
Solutions
have
X3 a n d X5 a r b i t r a r y , a n d
Instructor’s M anual
x \ = — .5 x 3 — .5 x 5 , X 2 = — .5 x 3 — . 5 x s , a n d X4 = 0.
(ü) » A = [2 4 5; 0 5 7; 7 8 0; 7 0 4; 9 1 1]; » rref(A*) ans = 1.0000 0 0 5.8462 5.3846 0 1.0000 0 -3.6044 -3.7033 0 0 1.0000 -0.6703 -0.2527
Solutionshave X4 andX5 arbitrary, andx\ andX3 = .6703x4 + .2527x5.
The 3.7033x5,
= —5.8462x4 — 5.3846x5, X2 = 3.6044x4 +
(üí) » A = rand(5,3) A = 0.2661 0.3841 0.0907 0.2771 0.9478 0.9138 0.0737 0.5297 0.5007 0.4644
0.9410 0.0501 0.7615 0.7702 0.8278
» rref(A») ans =
1.0000 0 0
0 1.0000 0
0 0 1.0000
1.0319 1.7249 -0.3771
The Solutions have X4 and X5 arbitrary, and x\ = 0.4801x5, and X3 = 0.3771x4 —0.4142xs. 5. » A = 2*rand(4)-l A = -0.7493 0.2591 -0.9683 0.4724 0.3769 0.4508 0.7365 0.9989 »
0.7771 -0.5336 -0.3874 -0.2980
0.0265 0.1822 0.6920 -0.1758
0.7859 -0.5678 -0.1922 0.1515
-0.2077 0.1184 -0.8858 0.3977
Q = orth(A)
Q= 0.5071 0.6553 -0.2551 -0.4984
0.2864 0.4839 0.3366 0.7553
(a) » x = 2*rand(4,1)-1 x = 0.6830 -0.4614 -0.1692 0.0746
0.5701 -0.4801 0.4142
-1.0319x4 - 0.5701x5, x 2 = -1.7249x4 +
T h e Transpose o f a M a tr ix »
MATLAB
1.9
y = 2*rand(4,1)-1
y = -0.0642 -0.4256 -0.6433 -0.6926 » s = x* * y s = 0.2097
'/• The scalar product of x and y.
» r = ( Q*x )* s = 0.2097
* (Q*y)
*/. The scalar product of Qx and Qy.
» format short e » s-r ans = 5.5511e-17
Since Q is orthogonal, x • y is the same as (Qx) • (Qy). (b) If the above steps are repeated, even if A is complex, the inner product of x and y will always be the same as that of Qx and Qy. (For complex A this depends on the fact that A! — (A)*. Also you should reverse the x, y variables as x • y = y7* x for complex vectors.)
» x = Q(:,l); y = Q(:,2); */, Let x be the lst column, and y the 2nd. » sqrt(x* * x) */, the length of the lst column. ans =
1.0000
» x* * y
*/, The inner product of the lst and 2nd column. ans = 5.5511e-17
This is zero up to round off and the same results follow for all the other columns. (d) » inv(Q) ans = 0.5071 0.2864 0.7859 -0.2078
0.6553 0.4839 -0.5678 0.1184
-0.2551 0.3366 -0.1923 -0.8858
-0.4984 0.7553 0.1515 0.3976
Q 1 and Q' are the same for any orthogonal matrix. (e) To show that x • y = Qx • Qy, we will use Q” 1 = Q%. (If Q is complex replace í with '.) Q x - Q y = (Q xY Qy = yiiQiQy = x* —>3i?2, so not an elementary m atrix 6. R\~¿LR2 is an elementary matrix 7. Two operations, R ¡ 7 Í R 2 , R 2 TÍ.R3 , so not an elementary matrix 8. Two operations needed, R 2 —* R 2 + 2/?i, R 3 —►R$ + 3 /íi, so not an elementary matrix 9. R 2 + 2R\ is an elementary m atrix
10. Ra + R 2 is an elementary m atrix
11. Two operations needed. R 2 —>R 2 + Ri, R 4
R 4 + R 3 , so not an elementary m atrix
12. i?i — R '2 is an elementary matrix
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
38.
39.
40.
100 00 1 0 10.
1 -2 0 0 10 .0 0 1
Elem entary Matrices and M a trix Inverses
/
x
(
32)
A l —* ¿>Rl
/
(Jl/a)
^ -« 2 -3 ^ 20 o 1) ( ; : )
A =
/l2\ ( 34J
R 2 —* 2 Ü 2
x
/
\
(¿J)
(;■ ';)
«2 - «2 - 3iíi / 1
2\ ►l ° - 2 /
n
R2 - - i f i 2 n \ D /10\ 2\o iy
, , 1 O\ ( 1 O w 1 2 A — 3 1 / VO—2 / V01 /
\
R 3 —> R í -
5H i
R3
fil _ R2 \ O O
\ 5 5 1/
/100\ A = [ O1 O I V5 O 1 /
-4 /
/ 1 0 0 \ / 1 1 0\ [ 0 2 0 ]j oi OJ \ 001/ \0° V
ñl "* ñl + 0-5H3 \ O O 1
/ 1 0 0 \ / 1 O —1 / 2 \ ( O I 00 1 0 \ 0 0 - 4/ \0°
/3 0 0 \ 010 \0 O1 /
/ 1 0 0 \ / 1 2/3 O\ / 1 O 0 \ / 1 O 0 \ 020 O 10 0 1 O 011 \0 0 1 / \0 0 1 / \ 0 0 —1 / \ 0 0 1 /
1 0
( 1
0
oo\ 10
V 1 o 0/ V o - l 1/
1 \
0 0 01 -- 11 / )
/10
01
/2 0 0\ 010 \001/
/100\ 010 V3 0 1 /
o
, o
fl-fl
o\ /1 o 1\ 0 010
\ 0 0 - l / Vo
R 3 —> ■R 3 —3Ri o 1 l ) r 3 ^ r 3 + r 2 y Qo _ 4 j
,4=
0\
0 1 3/ 2
1/ \ 0 0 1/
/ 1 O —1 /3 \ 01 O \00 1/
R 3 —)- - / ? 3
(
/001\ ,4= 010
/ 1O
H3 —►—R3 R 2 ►-^2 ~ -^3 i?l —►iíl + f#3
«i í?2 —^ 0.5/?2 Hi —►/?1 —f i?2
A=
- jÜ 3
0 1/
/ 1 0 0 \
fí I\ Q0 01 l0J) Ri 0\ 01 - 1
/1 o
\00
1/
R> - R> - R> (11 °o) R i^R i-2R 3 y 0 0 í j
/ I 0 0 \ / 1 0 0\ / 1 0 0\ / 1 0 2\ 0 1 0 0 1 0 0 1 1 0 1 0 Vo - 1 1 / \ o o —4 / \ o o 1 / \ o o 1 /
Section 1.10
122
C h ap ter 1 Systems o f Linear Equations and Matrices
/2 0 47. A = 0 \0 (2 0 48. 0 Vo
0 0 0\ 10 0 0 10 0 0 1/
/ I 0 0 0^ 0 3 0 0
0 0 10 Vo o o i /
/I 0 0 Vo
R x - *o. s Rx / 1 0 —1 /4 0 \ 1 0 0\ R 2 — 0 . 5 r 2 2 10 1 0 1 1/ 2 0 0 2 1 I Ri —►R\ —O.5 R 2 0 0 2 1 ] 0 0 2/ voo 02/
100 0 10
1/8 \ R - 1 /4 ]
001
1/2 I
'2 0 0 ,0 (l 0 o Vo
0 0 0' 10 0 0 10 0 0 1, 0 -1 /4 1 0 o 1 o o
0 0 0\ 10 0 0 10 0 0 5/ R 3 - » O.5 H 3 R 2 ^ R 2 - 0 .5 R3
Ri —►Ri + \R 3
H4 —►O.5H4 _ R _ 05fí / 1 0 0 0^ „ „ [ 0100 0010,
2 / «X - «1 -
,000
A =
/1 0 0 0\ 01 00 00-40 Vo o 0 1 /
Instructor’s Manual
10 02 00 .0 0
Voooi/
0 0^ / I 1/2 0 0\ / I 0 0 0 0 10 0 00 0 10 0 10 0 01 0 1 0020 0 1 / \0 0 0 1 / Vo 0 0 1
/1 O
00'
0 11/2 0
00 V oo
10 01,
0\ 0 o 1/ 0\ / I OO0\ / 1 0 0 o 0 10 0 1 0 1 0 A = 0 0 1 1 /2 0 0 1 0 I Vo o o 2 / V o o o 1/
/100 O' 0 1 O - 1 /4 001 o VOOO 1>
10 0 1 /8 ' o o ,0 0 0 1, 010 001
49. ac / O implies a / O and c / 0. Row reducing A to I 2 by back elimination via c-'Ri í a 0\ a 1 Ri ( 1 0 R\ —►R\ —6i?2 o1 ( - ) \0 1 / — shows I 2 = (a ~ 1 Ri )(Ri —bR2)(c l R 2 )A. So solving for A gives A = (cR 2 )(Rx + bR 2 )(aRx), or
M S cM ¡°c)
0 0
( ;?
1b 0 c0 ( S ! )
50. adf / 0 implies a, d and / are nonzero. Row reducing A to I 3 by back elimination gives / a b c\ ( Ode )
\ 00/ /
^ R _ r
2
2
( a b 0\
6 3 J 0d0 I
Ri - * Ri ~ cR3 \ 0 0 1 /
d~ 1 R 2 / a 0 0\ R x - R x - bR2 ( 0 1 0 1 , \ 0 0 1/
_j °
^3
"
E lem entary M atrices and M a trix Inverses
Section 1.10
So taking all the inverse operations applied to I 3 (in the opposite order) shows
1 0 c \ / I b 0 \ ( a 0 0' = | 0 1 e 1 OdO 010 00//
V o 0 1/
\ 0 0 1,
51. Let U = ( « u ^ i ) ( ^ 22* ^ 2) • *'(unnRn)A. Then U is an n x n matrix with l ’s down the diagonal and 0’s below it. Note that U is row equivalent to I and henee, by theorem 4, can be written as a product of elementary matrices. If U = E\ E 2 • • •Ek , where each Ei is an elementary matrix, then A = (unnR n) • • • (u 2 2 R 2 ) ( u n R i ) E i E 2 • • Ek- By theorem 4, A is invertible. 52. By problem 51, A is invertible. As in problem 51, let U = (u Since u has l ’s down the diagonal and 0’s below it, when row reducing U to I we need only add múltiples of a row to those rows above it. Henee, we can write U as a product of upper triangular elementary matrices: U — E 1 E 2 - - Ek- Note that E f 1 is upper triangular for each i. Show that the product of two upper triangular matrices is upper triangular. Then deduce A” 1 = E k 1 • • • E ^ E ^ ^ u n * i ) (u221i? 2) • • - (UnnRn) is upper triangular. 53. A * is upper triangular, so (A*)”*1 is upper triangular by the result of problem 52. But (A*)-1 = (A ~ 1)t so (A-1 )* is upper triangular, which means that A” 1 = [(A”"1)*]* is lower triangular.
54. PijA — C where crs = y ^Prk^ks- If r =
then
= ^^Pikdks = üjs since pik is 1 if k = j and 0
k-l
kzzl
otherwise. If r = j, then C j s = since pjk is 1 if k = i and 0 otherwise. If r / i and r / j, then crs = ars. Henee, PijA is the matrix obtained by permuting the ith and j th rows.
55. AijA = B where brs = ^^ci^aks- If r ^ j , then brs = ars since afrk is 1 if k = r and 0 otherwise. If 1
r — j, then bjs = y ^ a '-fcajba = + djs since af-k is c if k — i, 1 if k = j , and 0 otherwise. Henee, k=l AijA is the m atrix obtained from A by multiplying the ith row by c and adding it to the j th row. m 56. Mi A = B where brs =
If r = ¿, then &¿5 = ca¿5 since rriik is c if Ar = ¿ and 0 otherwise. k - l
If r / i, then brs = ars since m rk is 1 if Ar = r and 0 otherwise. It follows that M,A is the matrix obtained from A by multiplying the ith row by c. /12\ ( 2 4J
R 2 -+R2 - 2 R! ( \ 2 \ __________
( 00) ’
.
/10\/12\
A = {2l){t)l> )
123
124
C h ap ter 1 Systems o f Linear Equations and Matrices
(?s)- (¿s) 60.
/1-12\ /l-1 2 \ 2 1 4 I -f I 0 3 0 \ 4 —1 8 / \0 0 0/ / 1 0 0\ -A = I 2 1 0 I \0 01/
61.
/ 1 0 0 \ / 1 0 0 \ /I I 01 0 J Í O I O J j O \ 4 0 1 / \ 0 1 1/ \ 0
\ 30 0 0/
—1 2
/ 1 —3 3 \ / 1 —3 3\ ( 0 —3 1 J —>- ( 0 1 - 1 /3 \1 02/ \0 0 0/ .4=
/100\ 010 \1 01 /
/1 0 0 \ / 1 0 0 \ 0 -3 0 010 \ 0 0 1 / \ 0 3 1/
/ 1 —3 3\ 0 1 - 1 /3 \0 0 0/
/
62.
1 0 0 \ / 1 0 0\ 2 3 0 j —► | 0 10 | [-140/ \0 0 0 j /100\ .4=1210) \ 0 0 1/
/ 1 0 0 \ /1 0 0\ / 1 0 0 \ / 1 0 0 \ I O io) I0 3 0 ) IOIOJ ( o i o ) \ —1 0 1 / \ 0 0 1 / \ 0 4 1 / \ 0 0 0 /
In structor’s M anual
Elem entary Matrices and M a trix Inverses
MATLAB
1.10
M ATLAB 1.10 1. (a) » A = round(10 * (2*rand(4)-l)) A = -6 9 -9 -10 -9 -2 -9 -2 -9 4 0 1 3 -2 4 7 (i)
» F=eye(4); F(3,3) = 4; » F*A ans = -6 9 -9 -10 -9 -2 -9 -2 16 0 4 -36 4 7 3 -2 (ii)
» F=eye(4); F(1,2) = -3 » F*A ans = 21 15 18 -4 -9 -2 -9 -2 -9 4 0 1 3 4 7 -2 (iii)
» F=eye(4); F([l 4] ,:) = » F*A ans = 3 4 7 -2 -9 -9 -2 -2 -9 4 0 1 -6 9 -9 -10 (b)
» F=eye(4); F(3,3) = 4; » inv(F) ans =
1.0000 0
0 1.0000
0
0
0
0
•/. */• This is R3 — > 1/4 R3, i.e. divide R3 by 4.
0
0
0 0
0
0.2500
0
1.0000
125
126
C h ap ter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
00 » F=eye(4); F(l,2) = -3; » inv(F) ans =
*/. R1 — > R1 - 3 R 2 , # /# This is R1 — > R1 + 3 R2, i.e. add 3R2 to R 1 .
0 0 1 0 » F=eye(4); F([l 4],:) = F([4,l],:); */. Part (iii): interchange 1 and 4. */. This also interchanges 1 and 4. » inv(F) ans =
0 0 0 1
1 0 0 0
In each case, the inverse of F represents a row operation that is the reverse of the original operation. 2. (a) » A = [ 7 2 3 ; -1 0 4; 2 1 1] A = 3 7 2 4 -1 0 2 1 1 » B = A; » c = -B(2,1)/B(1,1); » F1 = eye(3) ; Fl(2,1) : » »
B = F1*B; F = Fl;
'/, Store this matrix in B. '/, Compute the multiplier to eliminate B(2,l). */, Generate the elementary matrix which '/, subtracts c*row 1 from row 2. '/* Apply the matrix F1 to B. '/, F will be the product of the elemetary matrices.
» c = -B(3,1)/B(1,1); » F2 = eye(3) ; F2(3,1) : » B = F2*B B = 2.0000 3.0000 7.0000 4.4286 0.2857 0 0.1429 0.4286 0 » F = F2*F;
'/, Compute the multiplier to eliminate B(3,l). */, Form elementary matrix for R3-cRl '/, Finish column 1 elimination
» c = -B(3,,2)/B(2,2) » F3 = eye(3); F3(3,2) = c; » » B = F3*B B = 3.0000 2.0000 7.0000 4.4286 0.2857 0 -6.5000 0 0 » F = F3*F¡\
'/, Next, eliminate B(3,2) in column 2. '/, This will finish forward elimination '/, Apply the matrix to B.
» » » »
'/, Start backelimination, divide row 3 by B(3,3).
c = 1/B(3,3); F4 = eye(3); F4(3,3) = c; B = F4*B; F = F4*F;
'/, F accumulates the product of F i ’s.
'/, Keep track of F.
'/, Normalize R3 to start with 1. */, Keep track of F.
Elem entary Matrices and M a trix Inverses » » » »
c = -B(2,3)/B(3,3); F5 = eye(3); F5(2,3) = c; B = F5*B; F = F5*F;
» c = -B(l,3)/B(3,3); » F6 = eye(3); F6(l,3) = c; » B = F6*B B = 7.0000 2.0000 0 0 0.2857 0
0 »
0
0 0
'/, Next, backeliminate B(l,3) to finish colnmn 3. */, Apply the matrix to B.
Keep track of F. */• Next, divide row 2 by B(2,2). '/, Normalize row 2 to start with 1.
1 0
% !%Keep
» c = -B(l,2)/B(2,2); » F8 = eye(3); F8(l,2) = c; » B = F8*B B = 7 0
0 0 »
backeliminate B(2,3).
1.0000
F = F7*F;
track of F.
V%Next,
backeliminate B(l,2).
% Apply
the matrix to B, finish column 2.
1 0 '/• Keep track of F.
F = F8*F;
» c = 1/B(1,1); » F9 = eye(3); F9(l,l) = c; » B = F9*B B =
1 0 0
0 1 0
Finish by dividing row 1 by B(l,l). '/, Normalizing RI gives rref (A)
0 0 1
» F = F9*F F = 0.3077 -0.0769 -0.6923 -0.0769 0.0769 0.2308
'/, Form the final product of F i ’s. -0.6154 2.3846 -0.1538
(b) » F*A ans =
1.0000 0.0000 0
0.0000 1.0000 0
0.0000 0.0000 1.0000
0 1.0000 0.0000
0.0000 0.0000 1.0000
» A*F ans =
1.0000 0 0.0000
1.10
*/. Apply the matrix to B. */• Keep track of F.
F = F6*F;
» c = 1/B(2,2); » F7 = eye(3); F7(2,2) = c; » B = F7*B B = 7 2
»
% Now
MATLAB
127
128
C hapter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
Both F A and A F are the identity, so F = A " 1. F A = J3 since F = (F9)(F8)(F7)...(F2){F1) and all the steps showing the row echelon form of A is / 3 can be achieved by one left multiplication by F instead of the individual multiplications by F l , F 2 , .... (c) »
D = inv(Fl)*inv(F2)*inv(F3)*inv(F4)*inv(F5)*inv(F6)* ... inv(F7)* inv(F8)* inv(F9)
D = 7.0000 -1.0000
2.0000 0
3.0000 4.0000
2.0000
1.0000
1.0000
D is the inverse of F = A ~l , so it is the same as A. (D = F ~ l since the inverse of a product is the product of the inverses in the opposite order.) (d) » »
A = [0 2 3; 1 1 4; 2 4 1] ; B = A;
» Fl = eye(3); Fl([1,2],:) » B = F1*B B =
1
1
0 2 » » »
3 1 */♦ F will be the product of the elemetary matrices. '/, Compute the multiplier. # /, R3-cRl will eliminate B(3,l). '/. Apply the matrix F2 to B.
» B = F2*B B =
»
Fl([2,l],:); */. interchange 1 and 2. '/, Apply the matrix Fl to B.
4 2 4
F = Fl; c = -B(3,1)/B(1,1); F2 = eye(3); F2(3,l) = c
1 0 0
'/, Store this matrix in B. '/• Since B(l,l) is zero, we must first '/, interchange it with another row, so that we '/, do not get a divide by zero error. This is '/, the only difference between the steps here '/, and those in part (a) .
1 2 2
4 3 -7
F = F2*F;
*/• F is now the product of F2*F1.
» c = 1/B(1,1); » F3 = eye(3); F3(l,l) = c; » B = F3*B B = 4 3 -7
'/, Next, divide row 1 by B(l,l).
» » » »
'/, Keep track of F. */, Next, eliminate B(l,2).
F = F3*F; c = -B(l,2)/B(2,2); F4 = eye(3); F4(l,2) = c; B = F4*B
'/. Apply the matrix to B.
*/, Apply the matrix to B.
Elem entary Matrices and M a trix Inverses
MATLAB
1.10
B =
1.0000 0 0 » » » »
0 2.0000 2.0000
2.5000 3.0000 -7.0000 '/, Keep track of F. '/, Next, elimínate B(3,2).
F = F4*F; c = -B(3,2)/B(2,2); F5 = eye(3); F5(3,2) = c; B = F5*B
Apply the matrix to B.
B =
1.0000 0 0
0 2.0000 0
2.5000 3.0000 -
10 .0 0 0 0
» F = F5*F; » c = 1/B(2,2); » F6 = eye(3); F6(2,2) = c; » B = F6*B B = 1.0000 0 2.5000 0 1.0000 1.5000 0 0 - 10.0000 » » » »
Keep track of F. */, Next, divide row 2 by B(2,2). '/, Apply the matrix to B.
*/, Keep track of F. '/, Next, elimínate B(l,3).
F = F6*F; c = -B(l,3)/B(3,3); F7 = eye(3); F7(l,3) = c; B = F7*B
'/, Apply the matrix to B.
B =
» » » »
1.0000
0
0
0 0
1.0000 0
1.5000 10.0000
-
F = F7*F; c = -B(2,3)/B(3,3); F8 = eye(3); F8(2,3) = c; B = F8*B
'/. Keep track of F. */. Next, elimínate B(2,3). */, Apply the matrix to B.
B =
1 0 o » » » »
0 1 o
o o -10
F = F8*F; c = 1/B(3,3); F9 = eye(3); F9(3,3) = c; B = F9*B
'/, Keep track of F. */, finally, divide row 3 by B(3,3). */, Apply the matrix to B.
B =
1 O O » F = F9*F F = -0.7500 0.3500
0.1000
O 1 O '/, Look at final form of F. 0.5000 -0.3000
0.2500 0.1500
0.2000
-0.1000
As in (b) and (c), we find th a t F*A=A*F=I, so F is the inverse of A, and th a t D = in v (F l)* in v (F 2 ) * . . . * in v ( F 9 ) is th e inverse of F, so it is the sam e as A.
129
130
C hapter 1 Systems o f Linear Equations and Matrices
Instructor's Manual
3. (a) » A=[ 1 2 3 ; 1 1 7; 2 4 5]; » U=A; '/• Store A in U; after » c=-U(2,l)/U(l,l) */. Eliminate U(2,l). c =
reduction U will be "echelon" form
-1 » F1 = eye(3); Fl(2,l)=c F1 =
1 -
0 1
0 »
0 1 0 0 1
U = F1*U
*/, Apply F1 to U, to eliminate.
» c = -U(3,l)/U(l,l) c =
'/, Eliminate U(3,l).
-2 » F2 = eye(3); F2(3,l) = c F2 =
0 0 1
1 0 -2 »
*/. Apply F2 to U.
U = F2*U
U= 1 0 0
2 -1 0
3 4
-1
Note: U is now upper triangular.
» F = F2*F1 F =
1 -1 -2
'/• F is the product of the elementary matrices.
0
0 0 1
1 0
(b) » L = inv(Fl)*inv(F2) L =
1 1 2
0 1
0 0 0
1
The matrix L is lower triangular. For each of the entries below the diagonal in L , we see that it is the same valué as — c, where c was the multiplier used to eliminate the same entry in A. In fact the entries are just the entries in the inverses inv(Fl), inv(F2). Since we can recover Fl, F2 from L (move down column 1 for each successive Fi).
Elem entary Matrices and M a trix Inverses
» L*U ans = 1
1
MATLAB
1.10
'/, This is the same as A. 2
3
1
2
7 4
5
Since F is the product of the elementary matrices that reduce A to 7/, we know that F A = U. Since L = F ~ \ we have LU = LFA = F ~ l F A = A.
»
» »
A = [6 2 7 3; 8 10 1 4; 1 0 7 6 8; 4 8 9 5]; */, Store A in U, and work with U. A; */, Eliminate U(2,l). -U(2,1)/U(1,1) c
U
C=
“ 1..3333 » F1 = eye(4); Fl(2,l) = c F1 = 0 0 1 .0000 1.0000 0 -1..3333 0 0 1.0000 0 0 0 U = F1*U; » c = -U(3,1)/U(1,1) c = -1 .6667 » F2 = eye(4); F2(3,l) = c F2 = 0 1 .0000
0 0 0 1.0000
»
»
0
1.0000
-1 .6667 0
0 0
'/. Eliminate U(3,l).
0 0
1.0000
0 0 0
0
1.0000
U = F2*U;
» c = -U(4,1)/U(1,1) c = -0 ,6667
'/, Eliminate U(4,l). This finishes column 1.
» F3 = eye(4); F3(4,l) = c F3 = 0 0 1 0000 0 1.0000 0 1.0000 0 0 0 .6667 0
» U = F3*U U= .0000 0 0 0
2.0000 7.3333 3.6667 6.6667
7.0000 -8.3333 -5.6667 4.3333
0 0 0
1.0000
3.0000 0 3.0000 3.0000
132
C hapter 1 Systems o f Linear Equations and Matrices » c = -U(3,2)/U(2,2) c = -0 5000 » F4 = eye(4); F4(3,2) = c F4 = o 1 .0000
»
In stru cto r^ Manual
*/, Eliminate U(3,2), moving on to column 2.
0
1.0000
o o
0
-0.5000
1.0000
0 0 0
0
0
0
1.0000
U = F4*U;
» c -U(4,2)/U(2,2) c = -0 ,9091
'/♦ Eliminate U(4,2), finish column 2.
» F5 = eye(4); F5(4,2) = c F5 = 0 1 0000
0 0
0
1.0000
0 0
0 -0.9091
1.0000
0 0 0
0
1.0000
2.0000
7.0000 -8.3333 -1.5000 11.9091
3.0000 0 3.0000 3.0000
» U = F5*U U= .0000 0 0 0
7.3333
0 0
» c = -U(4,3)/U(3,3) c = 7r.9394
*/, Eliminate U(4,3), moving on to column 3.
» F6 = eye(4); F6(4,3) = c F6 =
1 0000 0 0 0 » U =: F6*U u = 6 0000 0 0 0
0 1.0000
0 0
0 0
1.0000
0 0 0
7.9394
1.0000
7.0000 -8.3333 -1.5000 0
3.0000 0 3.0000 26.8182
2.0000 7.3333 0 0
» F = F6*F5*F4*F3*F2*F1 F = 0 0 1 .0000 0 1.0000 -1 .3333 1.0000 -0.5000 -1 .0000 -4.8788 7.9394 -7 .3939 »
*/, The product of the F*s 0 0 0
1.0000
L = inv(Fl)*inv(F2)*inv(F3)*inv(F4)*inv(F5)*inv(F6) */, This is the inverse of F.
Elem entary Matrices and M a trix Inverses
MATLAB
1.10
L = o
1.0000 1.3333 1.6667 0.6667
1.0000 0.5000 0.9091
o 0 1.0000 -7.9394
o O O 1.0000
If you print in v (F i) you see its non-zero entry below diagonal is just the negative of corresponding entry in F i and is equal to corresponding entry in L. » L*U ans = 6 8 10 4
'/• As in (c), this is the same as A 2 10 7 8
7 1 6 9
3 4 8 5
As in (b), we see that both L and F are lower triangular, and that LU = A. As in (c), we each entry in L is the negative of the multiplier used to eliminate the corresponding entry in {/, or the non-zero entries in i n v ( F l ) ,.. .,inv(F 6).
133
134
C hapter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
S e c tio n 1.11
In problems 1-8, the m atrix L is constructed from the identity m atrix in the following way: For each row operation on A of the form R, —* Ri —kRj, put a k in the i , j position of the identity matrix. , , (l 2\ 1 Á = {S4) 12 03
2. A -
R2 -*R2 - 3 R i ( l 2 \ TJ _ r flO \ > U - 2 j =t/ ThUSL=l 3 l J '
= U. Thus L
—1 5 ^
3. A
H2 - R 2 + 6 Ri f - 1
5 ►(\ “ ¿0 33 33) = 0
637
4. A
Thusi = ( - ¿ ? )
R3 —►R 3 —3Ri
R 2 —►i ?2 —2 R\
=
0
5. ^ = ( 4 3 5 )
_9U „.T hus¿= [2 0-3.
!2 1.
R 2 —*■R2 —2R\
= u.
\101J
3. A
=
7. A
-
(l 2-1 (1 2 - 1 4\ R 4 — ► R\ — ’ i R 2 0-1 5 0-1 5 0 0 - 2 -12 0 0 - 2 -12 \ 0 0 —8 —24 > \ 0 —3 7 0/ /I 2-1 4 / I 0 0 0\ 0 - 1 5 8 1 _ .. ™ r _ I 0100 00 - 2 - 1 2 I 12 1 1 0 \0 0 0 24 / V1 3 4 1 / —►R$ —R 2
9 -2' 0 - 2 1 12 0 -6 f
R¿
JR4 —4H3
LU -Factor¡zat¡ons o f a M a tr ix
A
Section 1.11
2 3-1 0 1 4-11 -2 5 - 2 0 0 - 4 5 2/ ’2 3 - 1 -1 6^ /?4 —►Ri -f0 1 4-11 0 1 4-11 0 0 -35 94 0 0 -35 94 ,0 0 21 -42 \0 -4 5 2/ 1 0 00 R2 —►/?2 —2R\
=
R 3 —*■R 3
2 1 0 0 1 8 10 0 —4 —| 1-
U. Thus L =
_2^* The system Ly = ie > yields e9ua^ ons yi = —2 and 3yi + 2/2 = 4. Solving we get y2 = 10. Now,
9. From problem 1 we have A = LU where L = ( 3 ^ ) and U
(31) (y1) ” ( from C/x = y, i.e.,
d)
^
(a*1) = ( l o ) WG
Xl
=
^x2 =
and ~^X 2 = 10. Backsolving
we get xi = 8, x2 — —5. The solution is ^
10. From problem 2 we have A = LC/ where L = 10\ 01 (o 3 )
( —1 2/2
= (
and U = ^0 3 ^ ’
yields the equations yi = —1, y2 = 4. Now, from í/x = y, i.e.,
(1)
11. From problem 3 we have A = LU where L = ^ 6 l ) and U = [ 6 l ) ( y 1) = ^ 5 ) yields
5 33
. . .
£i = — ,X 2 = — . The solution is
q 33 ) • The system Ly — b, i.e.,
e9ua^ ons Jh = 0, —6j/i + j /2 = 5. Solving we get y2 = 5. Now,
from Ux = y, i.e., ^ 0 33^ ( a ;1 ) = ( ü ) W6
25 33
~
4 ) W6 °k*'£dn Xl 4- 2x2 = —1 and 3x2 = 4. Backsolving we get xi = ——
and x 2 = - . The solution is o
(
sys^em ^
~ Xl
®x 2 =
33^2 = 5. Backsolving we get
/ ' I,
’1 0 0 \ / I 4 6' 12. From problem 4 we have A = LU where L = | 2 1 0 ) , £ / = [ 0 —9 —9 | . The system L y = b, i.e. 0 0 -3 3 f l -1 10 2/2 I —I 7 yields the equations yi = - 1 , 2yi + y2 = 7, 3yi + — y2 + y3 = 2. Solving
yz
\ 2/
9
135
136
Instructor’s Manual
C h ap ter 1 Systems o f Linear Equations and Matrices
we get
(1 yi = —1, 3/2 = 9, 3/3 = —5. Now, from Ux = y, i.e., I 0 \ 0
x\ + 4*2 + 6 x 3 = - 1 , - 9 x 2 - 9x 3 = 9, —3 x 3 =
13. From problem 5we have A
- 5 . Backsolving
4 6\ /x A /-l\ —9 —9 I I x 2 I = I 9 j we obtain
\x3J
0 - 3 /
we get
/ 10 0 \ / 2 LU where L — I 2 10 1 and Í7 = I 0 \ 1 0 1 / \0
1
\ - 5 /
—8
—1
x x = — , x 2 = — , x3 =
5 -.
7\ 1 —9 I . The system Ly = b, 0- 1 /
( l0 0 \fy A / 6\ i.e., 1 2 1 0 ] I 3/2 ] = ( 1 ) yields the equations 3/1 = 6,2t/i + 3/2 = 1, 2/1 + 2/3 = 1- Solving we
\ i o 1 1 \ y 3) get yi
— 6, 2/2 = —1 1 , 2/3 =
2xi + x 2 +
= 6, x2 -63 '
7x3
solution is I
\i) —5.
9x3
Now, from Ux
=
(21 7\ ( x A y, i.e., 10 1 —9 1 I x 2 1 =
/
\0 o - 1
\
J \ x 3J
= —11, —x 3 = - 5 . Backsolving we get xj =
I
6\ —11 1 we obtain -5 /
x 2 = 34, x 3 =
5.
The
34
14. From problem 6 we have A = LU where L =
1 0 0\ /3 9 —2 \ I 2 1 0 1and U = I 0 —21 12 j . The system Ly = .! £ ! / \0 0 f j
1° ° \ ( yi\ ( 3\ 4 2 b, i.e., | 2 10 II y 2 I = I 16 I yields the equations yi = 3, 2yx + y2 = 10, —3/1 + - y 2 + yz = 4. 3 21 \y^J \ 4/ '3 9 - 2 \ ( x 1 ' Backsolving we get 3/1 = 3, 3/2 = 4, y3 = — . Now, from Ux = y, i.e., | 0 —21 12 | j x 2 7 1,0 0 |&r / Vx 3 9 —8 275 4 I we obtain 3xi + 9 x 2 - 2x 3 = 3, -2 1 x 2 + 12x3 -- 4, — x 3 = — . Solving we get xx = ——, x 2 = _8 I 21 7 uo -12
-8
,x3 = 7
3
/ 275/63 \ The solution is I —12/7 I . V -8 /3 /
( \ 2 -1 4^ (\ 0 0 0\ 0 10 0 0 - 1 5 8 and U = The system 15. From problem 7 we have A = LU where L = 2 110 0 0 - 2 - 12 \ 1 3 4 1/ \ 0 0 0 24/ / 1 0 0 0 \ (V A 0 10 0 Vi yields the equations 3/1 = 3, y2 = - 1 1 , 23/1 + 3/2 + 2/3 = Ly = b, i.e. 2 110 2/3 \ 1 3 4 1 / \ 2/4 / 4 , 3 / 1 + 3y2 + 4j/3 + 1/4 = -5. Solving we get 3/1 = 3 , 3 / 2 = —1 1 , 3/3 = 9 , 3/4 = —11. Now, from t / x = y,
LU -Factor¡zat¡ons o f a M a trix
Section 1.11
/ I 2 -1 4 \ / Xl\ 0 —1 5 8 X2 í.e., 0 0 - 2 -1 2 *3 VO 0 0 2 4 / V * 4 /
3\ -1 1 we obtain *i + 2*2 —x3 + 4*4 = 3, —*2 + 5*3 + 8*4 = —11, 9 V-ll/ . . . 71 -1 7 -7 -1 1 ^ - 2*3 — 12*44 = 9, 9 , 24*. 24*4 = —11- Backsolving we get *1 = ^ 2>x 2 = = “ 24“ ' /
solution is
71/12 \ -1 7 /1 2 | - 7 /4 • —11/24/ 1 0 0 0' 2 1 0 0 -1 8 10 0-4^1,
16. From problem 8 we have A = LU where L =
(2 3 -1 6\ 0 1 4 -1 1 and U = . The 0 0 -3 5 94 \0 0 0 ^ /
1 0 00 2 1 0 0 -1 8 10 0-4=2 1
/1\ 0 yields the equations 2/1 = 1, 2y\ + 2/2 = 0, system Ly = b, that is, 0 \4/ 3 31 - 2/1 + 82/2 + 2/3 = 0, - 4 j /2 - - 2/3 + 2/4 = 4. Solving we get jq = 1, y2 = - 2 , 2/3 = 17, y4 = — • Now from 5 5 /2 3 -1 6^ / Xí\ 0 1 4 -1 1 *2 we obtain 2*i + 3*2 —*3 + 6 * 4 = 1, *2 + 4 * 3 —1 1 * 4 = Ux = y, i.e., 0 0 - 3 5 94 *3 V oo 0 ^ / \ *4 / 72 31 „ , . -565 5 169 31 ^ - 2 , - 35*3 + 94*4 = 17, — x 4 = y . Solving we get *x = ^Q Q g .^ = ^ , * 3 = 2 5 2 »*4 = The
solution is
’—565/1008\ 5/72 169/252 31 /7 2 /
17- m ( ¡ ¡ )
= £/ T l' “ í > =
(?¿)
p 4 =
( i ¡ ) ( l 24 )
=
( J 2)
= "
or
P A — LU where L - a ? ) -
b) LUx = P + x = P b = ^ J ^ ^ _ g ^ L y = ^ Q ^ ^ 2/1^ = ^ that £/x = y = ^
3 ^ we
= ^
^ = —
We seek a y such that Ly = ^ = 3 ans ^ q 2 ) ( *2 ) = (
get *1 = —11, * 2 = ^. The solution is ^
*3 ^ .
From
3 ^ • Now, we seek an x such
3 ) - ^ e Set xi + 4 x 2 =
and 2 * 2 = 3. Solving
137
138
In structor’s M anual
C hapter 1 Systems o f Linear Equations and Matrices 3
3 0
0
32
3 2,
¡ 1-1
2 2' 0 0
2\
/O 1 0 \
2 4 . Then P = 0-4,
1 0 0 , and P A = \0 0 l j
1 - 1 2' 1 —1 2 \ 0 2 4 1. Now we row reduce P A to an upper triangular matrix. I 0 2 4 0 32/ V0 3 2 1 -1 2 \ /I 0 0 0 2 4 I = U. Thus PA = LU where L = | 0 1 0 0 0-4. / O 1 0 \ / —1 \ / 2\ / 2\ b) LU x — P A x = P h = 1 1 0 0 1 I 2 1 = 1 — 1 1. We seek a y such that Ly = I — 1 I. From \0 0 1 / \ 4 / \ 4/ V 4/ /1 0 0\ / ¡/i \ / 2\ 2 22 ( 0 1 0 I i/2 1 I - 1 we get 2/1 = 2, 2/2 = - 1 , - 2/2 + íte = 4. Solvingwe get j/3 = — , \ 0 | 1 / \ 2/3 / \ 4/ / 2\ / 2\ /l- l 2\ / « A and thus y = I —1 I . Weseek an x such that £7x = I —1 I . That is I 0 2 4 1 I X2 1 = V 1 1 /2 / \ 1 1 /2 / \0 0 - 4 / V*3/ Ly =
2\ - 1 . We get xi —1 2 + 2 a¡3 = 2, 1 1 /2 /
-1 1
19. (a)
/
9/4 1. 11/8 /
/0 2 4 \ 037 \4 1 5/
b)LUx / I
R3
R3^ R3
¡ o3 7 | _ u 0 0 ^
Rz —rR2
4 1 03
5 /0 0 1 \ 7 1. Then P = ( 0 1 0 , and P A = \1 0 0 / o o ir
Thus P Á _ L[/ where ¿ _ | o l o
/ 0 0 1 \ / —1 \ í 2\ PA x = P b = I 0 1 0 I I 0 1 = 1 0 1 . We seek \ 1 0 0/ \ 2 / \ - l / 0
0 \ /
2/1
\
9
Solving we get xi =7, x 2 =- , x 3
7\
. The solution is I \
/ 4 ! 5\ I 00 33 77 I \0 2 4/
11
2x 2 + 4x 3 = - 1 , - 4 x 3 = — . 4
/
Ly =
2 \
ay such that Ly =
From
2
I 0 1 0 )I 2/2 I= 0 I we get 2/1 = 2, 2/2 = 0, - 2/2+ 2/3 = -1 - Solving we get 2/3 = - 1 , and V o i l / W V- 1 / 3 / 2\ / 2\ /4 1 5\ / n \ / 2\ thus y = ( 0 I . Now, we find anx such that í/x = ( 0 ) . That is I 0 3 7 ) [ x 2 ) = [ 0 1. V-l/ \-l/ \ o o ^ ) \ x 3J V-1/ 2 —1 —7 3 We get 4xi + x 2 + 5x 3 = 2,3x 2 + 7x 3 = 0,— x 3= - 1 . Backsolving weget xj = — x 2 = — , x 3 = - . The solution is I —7/2 I. V
3 /2 /
LU -Factor¡zat¡ons o f a M a tr ix
'0 5 - 1 20. (a) | 2 3 5 4 6 -7
/ I 0 0\ 0 5 -1 = U. Thus P A = LU where 1 = 0 1 0 0 0-17/ \ 2 0 1/
7*3 — * 7*3 - 2Ri
Ly=
23 5\ / 010\ 0 5 - 1 . Then P = 1 0 0 and P A = .00-17/ \001/
7*3 —►7*3 — 2 R i
b) LU x = P A x = P b -
Section 1.11
/O 1 0 I 10 0 \ 0 0 1,
/I 0 0 0 10 \ 201
-3 \ / —3 10 I . We seek a y such that L y = I 10 5/
\
From
5
-3 ' 10 | we get 2/1 = —3 , 2/2 = 10,2j/i + y3 = 5. Solving we get 2/3 = 11 and 5.
—3 \ /2 3 5 \ /x i \ / —3 \ 10 I . Now, we find an x such that U x — I 10 I . That is í 0 5 —1 I I xi I= ( 10 1 .We 11/ V 11/ \0 0 - n J \ x a J \ll) —457 159 —11 get 2x i + 3 x2 + 5x3 = - 3 , 5x2- x 3 = 10, -1 7 x 3 = 11. Solving we get xi = x2 = ——, x3 = ■. 170 85 17 '-4 5 7 /1 7 0 \ The solution is | 159/85 1. - 1 1 /1 7 /
R 2 —2Ri
21. (a)
,0
0
and P A
0 =¿22/ =
4 5 6\ 2 3 1! 4-15
/° o 0 -7 3 2 0 31 V i -4 5 6/
1 -4 5 6\ 0 0 -7 3 0 23 1 0 00 =£S2/ / 1 —4 5 —►i ?3 — 2 f í 2 0 2 3 0 0-7 2 0 3
6\ 1 3 0 -1 9 - 1 5 /
R i -
\0 6\ 1 3 1/
0 7*4 —►7*4 — 27 *i
y -R z
= U . Thus P A = LU where L =
10 01 02 ,2 4
0 0\ 0 0 10 ^ 1 /
/ 0 0 0 1\ 10 0 0 0 10 0 \0 0 1 0 / /I -4 5 0 2 3 0 0-7 \0 8-7-11,
139
140
Instructor’s M anual
C hapter 1 Systems o f Linear Equations and Matrices
( 0001^
3\
J?üo
0 0 10/
(1 0
00
. We seek a y such that Ly =
\2 4 ^ 1
v i
4'
. Now, we find an x such that C/x =
Solving we get 2/3 = —7, 2/4 = 1, and y
v i. is
1 0 0
\0
-4 5 2 3 0 -7
6
0 0
’-73/162 \ 4/81 53/54 - 7 /1 6 2 /
solution is
0 0 -2 5 0 -6
2 0
4\ 3 . That -7 1/
4\ 3 . We get xi —4 x 2 + 5x 3 + 6x 4 = 4, 2x 2 + 3x 3 + x 4 = 3, -7 1/ . . . -7 3 4 53 -7 x 4 = 1. Backsolving we get xx = ,x 2 = — , x 3 = ^ > x4 = The
1 3
-7x3 + 3 x 4 = —7
22 . (a)
. From
4\ 19 3 we get j/i = 4, 2/2 = 3, 2y2 + 2/3 = - 1 , 2j/i + 4t/2 + y J /3 + 2/4=2. -1 2/
0 1 0 0 0 2 10
Ly =
4'
-1 2 4/
3' 4 1
'0 4 - 2 5\ 4 it 3 - Ha - | h 2 | 5 0 - 6 ir - i s 2 0 1-2 0 0 T -5 ,0 0 —2 3/ \ 0 0 —2 3 / / 5 0 —6 4 /0 1 0 0 \ /5 0 -6 0 4-2 5 0 4 -2 0 0 0 1 . Then P = and P A — 0 0 ±7
/0 4 -2 Ri^Rt 5 0 - 6
1-2
0 4 -2
5,
5 0 -6 4\ «1^ 2 | 0 4 - 2 5 00 £ + P ,0 0 - 2 3/
5\ 4
0 0 10
\0 0
/ 5 0 —6 04-2 «3 - f Ri
4\ 5
\ 0 0 —2
3/
ooff
Ri
0
0 00 1
b) LUx = P A x = P b =
0 0 10
Vi 0 0 0/ 0 0\ / 10 00 0 1 Ly = 10 I 0 V0 0 - i f 1/
/ 2/i \ 2/2
2/3 Vy4/
1-2
/ ~ 2\ 4 5 7/
0 0\ 00 10 ^ i /
We seek a y such that Ly =
From
( we get 2/1 = 4, y2
_ 2 -1 0 7, jr 2/i + 2/3 = 5, =[7- 2/3 + 2/4 = - 2 .
4\
17
2 0
\0 0 -2
■Ri + ¥fR3
/ 10 / 5 0 —6 4\ 0 1 0 4 -2 5 = U. Thus P A = LU where L = 0 0 17 - 18 1° V o o \ 0 0 0 is 17 /0 1 0 0 \
\1 0 0 0 /
if/
Solving we get y3 = — , y4 = 0, and y = D
7 17/5 0/
Now, find an x such that Ux =
4\ 7 17/5 0/
4\ 5 3/
LU -Factor¡zat¡ons o f a M a trix
/5 0 -6 0 4 -2 T hat is oo \0 0 0
4\ 5
Section 1.11
We get 5xi — 6x3 + 4x4 = 4, 4x2 — 2 x 3 + 5x4 = 7,
if/
17 15 17 18 ~ -x 3 ---- “~x4 = — , — x 4 = 0. Solving we get x\ = 2, x 2 = 7 , x 3 = 1, x 4 = 0. The solution is 5 5 5 17 4
2\
9/ 4
1
0/
23. (a)
0 U . Thus P A = LU where L =
1
0
0
1 0
V2
( °\ 6
2/3
5
\ y
4
/
/6 ' I 1 lo V5/
—x 3 —6x4 = 5, —1 4 x 4 í
12\ 13/2 7 \ -2 /
0
/0 \ 6 . We seek a y such that L y = 5 \ 1/
/° \ 6 . From 5 \ 1/
. We get yi = 0, y2 = 6 , - 2 y i + y3 = 5, - 2 í /2 - 3t/3 + y4 = 1.
\ 1/
Solving we get 2/3 = 5, r/4 = 28, and y = /I 2 -3 2\ 0 -2 3 1 is 0 0 -1 -6 \0 0 0 -1 4 /
0 0\
- 2 0 1 0 0 - 2 - 3 1/
/0 0 1 0 \ 10 00 1 b) L [/x = P A x = P b = 0001 I \0 1 0 0 / 1 0 0 0\ /yi\ 0 1 00 1 -2 0 10 V 0 - 2 -3 1/
2\ 1 -6 0-14/
6 . Now, find an x such that Í7x = 5 \ 28 /
0\ 6 . That 5 \2 8 /
°\ 6 . We get xi + 2x2 — 3x3 + 2x4 = 0, —2x2 + 3x3 + x4 = 6 , 5 \ 28 / 13 = 28. Backsolving we get x\ = 12, x 2 = — x 3 = 7, x 4 = —2. The solution is
141
142
Ch ap ter 1 Systems o f Linear Equations and M atrices
Instructoras Manual
24. Let B = LM. bu = Y2^ikTnki‘ Since L and M are lower triangular with ones on the diagonal, we
k=i
have the following conditions: Uk = 0 if k > i, m*,- = 0 if i > k ,
= 1, and mu = 1. So if k < i n
or k > i we have that Uk^ki = 0. If k = ¿ we have lama = 1. Thus
= Y j ikTriki = 1. Now Jb= l
mjbj = 0 if j > k, Uk = 0 if k > i. Suppose j > i. If k < ¿, then k < j and henee k=i rrikj = 0. If k > i then Uk = 0. Thus if j > i then Uk^kj — 0 and henee = 0. Therefore L M is lower triangular with ones on the diagonal. 25. Suppose that L and M are upper triangular. Then if j <
Uj — 0 and
= 0. Let B = LM,
n
bij — Y j lkm kj . Suppose j < i. If k < j, then k < i and henee Uk — 0. If k > j , then mkj — 0. Thus k-i
if j < *, then liktrikj = 0 and henee bij = 0. Therefore L M is upper triangular.
R 2 —¥R2
26. A =
R3
2/?i l
10 0 000 | = | Also - 2 10 \ - 4 x 1) \ 0 0 0 LC/-factorization of the m atrix is not unique.
R3 + 4Ri
-1
2
= U. Thus A = LU 1
2 —4 —2 | for any real number x. So the 4-8-4i
/ 3 —3 2 5 \ / I 0 0 0 \ /3 -3 2 5 \ 2 1 —6 0 _ a l OO 0 u v w . This yields the equations 3a = 2, —3a + u = 1, 27. A = 5-2-4 5 “ 6c 10 0 0xy \ 1 —4 8 5 / \ d e f 1/ \ 0 0 0 z J 2a + v = —6, 5a + w = 0, 36 = 5, —35 + cu = —2, 2b+ cv + x - —4, 55 + cw + y = 5, 3d = 1, —3d + eu = —4, 2d + ev -f f x = 8, hd + ew + f y + z = 5. Solving we get a = 5= c = 1, d = e = —1, / = any real number, u = 3, v = a: = 0, y = 0, z = 0. Thus A = LU where / 3 —3 2 5' ( 1 0 0 0' | 10 0 0 z=p=$* where / can be any real number, and U = Since there is more L = 0 0 0 0 I 1 10 \o o o 0/ \ f —1 / 1 , than one possibility for / , the LU factorization of A is not unique.
28. A
R2
■ & )
29. A R 3 —►R3 —R 2
R2 ~~2Rl
l 2 ) = U . Thus A = LU where L 0 0) - ( !!) ■
i?2 —*R 2 "t 2Rl
- 1 2 3' 0 5 13 1 3 10,
R3 —^► R3 + Ri
1 0 01 0 5 13 I = U. Thus A = LU where L = I —2 10 00 0/ \ —1 1 1
- 1 2 3' 0 5 13 0 5 13
LU -F acto rization s o f a M a tr ix
-1
14
6\
-1 14 0 18 30. A = 0 31 5 I -------------- ► 0 3 1 135 1 35 13/ / - I 1 4 6\ r 3 ^ r 3 - 3R2 i 0 1 8 14 R 4 —►R 4 —4R2 _________ ► 0 0- 2 3 - 3 7 V 04 9 19/ 2 - 1 0
¡-1 0 0 0
31. A
2
j
R 2 ^ R 2 4-2Rl
Section 1.11
6\ - 1 1 4 6' 14 | R* - Ri + Ri 0 1 8 14 5 I -------------- ► 0 3 1 5 13/ 0 4 9 19/ 4 6\ R 4 —►R 4 —r 3 8 14 0 0 -2 3 -3 7 0 0 -2 3 - 3 7 /
1 4 6\ 1 0 0 0' 1 8 14 -2100 = U. Thus A = LU where L = 0 -23 -37 0 3 10 \ - l 4 1 1, 0 0 0/ ¿ 2 - 1 1 7\ 3 2 1 6 1 3 0-1 \ 4 51 5 /
=
7\
2 -1 R 4 —►R\ —2fíi
0
0 \0
/2 -1 0 I 1 3 \4 5
R 2 —►R 2 —
9
'2 f 9 I '2 7 -1 - 9 /
/?4 —►/Í4 —2/Í2
7\
9
2
2 -1 0
R3 -
«3
-1 5/
7\ 9 '2 9
o \4
/ 2 —1 1 7 \ ou 12 - ¿2 - a2 o o o \o o 0/
5/
= U . Thus A = LU where L —
1 0 o 0\ 1 100 ± 11o 2 2 0 1/ 2 - 1 0 2^ 0 00 0 -2 1 0 - 2 6 -3 0 6/
(
32. A
Ü4
=
2-10 2 4-2 0 4 -2 1 0 - 2 \ 6-3 0 6 2 -1 0 2\ 00 o j _ v
I o
«4
,0
33. / = ^
Thus A = LU where L _
000/
i ?3 —►.R3 +
2 -1 0 2 \ 0 000 0 000 6 -3 0 6 /
10 0 0\ 2 10 0 - 1 0 10 3 0 0 1/
1 2 4 ) = ( a l ) ( o w ! ) ' This yields *he equations o = —1, 2a + u = 2, 3a + i; = 4. Solving
we get u = 4, v = 7. Thus A = L í/ where L = ^
2 1\ 34. A = | —1 4 1 60/
/ I 0 0 \ /2 1 \ = lalOJ \ 6 c 1/ \0 0 /
1 1 ^ anc^ u = ( o 4 7 ) '
Í O u j . This yields the equations 2a = —1, a + u = 4, 26 = 6, b + cu = 0.
-1 -2 9 í Solving we get a = — ,6 = 3,c = — u — - Thus A — LU where L — I 2
3
2
V
1
3
°°\ 1 0 I and
-i
1/
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Instructor’s Manual
C h ap ter 1 Systems o f Linear Equations and Matrices
35. A =
7 13 4\ -2 5 6 8 /
^ ) ( I ^ ^ ^ I • This yields the equations 7a = —2, a + u = 5, 3a + v = 6, a l y yO u v w J
—2 37 48 64 4a -f w = 8. Solving we get a = -y-, a = — , v = — , w — — . Thus A = 1
where L
= ( - f l ) and
4\
í / _ ( o 37/7 48/7 6 4 / 7 / '4-1 2 l \ / I 0 0 \ / 4 - 1 2 1' 36. A = | 2 1 6 5 1= I a 10 1 í 0 3 2-17 / V 6 c 1 7 V0 0 * r
u u tu ]. This yields the equations 4a = 2, —a + u = l,2 a + u =
(4 -1 2 1\ £ 7 = 1 0 3/2 5 9/2 . \0 0 35/3 - 2 / /10000\ 5 1 3\ 0u v a 1000 0 0 w . This yields the equations 5a = —2, a + u = 4, 3a+u = 2, 56 = 6 c 100 37. 00 0 -2 2 0 de/10 \ g h i j 1/ Vo 0 0 / 5-31/ 1 ,6+cu — 6, 36+cu+ta = 1,5d = —2, d+eu = 2 ,3d+ eu+ /tu = 0 ,5y = 5, g+hu — —3 ,3y+/iu+tu;+y 0 = , o . • -2 , 1 29 , - 2 6 1 . 10 . 5 1. Solvmg we get a = — ,6 = - , c = — , d = — ,e = — , / = - , g = 1 ,/i = - j - p * = ; = any real 5 -2 1
13\ 42 6 1
number, u = — , v = — ,w — ———. Thus A — LU where L = 5 5 11
1 0 000\ - 2 /5 1 0 00 1/5 29/22 1 0 0 , j any real - 2 /5 6/11 1/7 10 1—10/11 —5/21 j l f
(5 1 3\ 0 22/5 16/5 0 -4 2/11 number, and U - 0 0 0 0 \0 0 0/ / I 0 0 0 0 \ / - I 2 1\ / " 121\ 0u u al 0 00 165 0 0 tu . This yields the equations —a = l,2 a + u = 6,a + u = 6 c 10 0 38. A = -2 3 7 00 0 d e / 10 102 \ y /t i j 1 / 0 0 0/ 4 15/ 5, —b = —2,26 + cu = 3,6 + cu + tu = 7, —d — 1 ,2d+eu = 0, d+ eu + /tu = 2, —g = 4 ,2g + hu — l,g + hv + —1 1 6 9 9 iw + j- = 5. Solving we get a = - 1 ,6 = 2,c = — , d = - l , e = - , / = — ,g = - 4 , h = - , i = — , j =
LU -F acto rization s o f a M a trix
( = 8 , 1; = 6, w
(-12 1\ 08 6 number, and U = 0 0 23/4 00 0 V 00 0/
23 4 ‘
1 -1
Section 1.11
0
0 0 0\
1
000
2 -1 /8 100 - 1 1/4 6/23 1 0 \-4 9/8 9/23 i 1 /
j any real
145
146 Chapter 1
Systems of Linear Equations and Matrices
Instructor’s Manual
CALCULATOR SOLUTIONS 1.11 Problems 39-44 ask for a PA = LU factorization. This is computed on the TI-85 by the LU function, which is invoked with four arguments in the form LU (A, L, U, P ) , where A is the input argument - the ñame of the variable containing the matrix to be factored, and L, U, P are output arguments, telling the function ñames for the variables in which to store the respective parts of the factorization. You can invoke this function by entering the alphabetic characters shown, including the three commas or you can use the MATRX MATH (MORE) LU menú entry, and then entering the "A, L, u , P" entries. As usual we will assume the input matrices are entered in variables A l l l n n , nn = 3 9 , * - * , 4 4 . The LU factorization computed on the TI-85 is actually the "Crout" LU-factorization which gives an upper triangular U with ones on the diagonal, as if we had found an echelon form of PA by forward elimination without row interchanges. 39. After entering LU (Al 1139, L11139, U11139, P11139) [ENTER] , we find that
L11139 (ENTER) yields
U l l 1 3 9 [ENTER] yields
[ [ 30 0 ] [ 0 2 0 ] , [ 2 -1 .6 6 6 6 6 6 6 6 6 6 7 8 .5 ]] [[ 1 .333333333333 2 .3 3 3 3 3 3 3 3 3 3 3 ] [ 0 1 2 .5 ] , [ 0 0
1
]]
[ [ 0 1 0 ] and for the permutation matrix P 1 1 1 39 [ENTER] is [ 1 0 0 ] . [ 0 0 1 ] ] 40.
After LU (Al 1140, L11140, U11140, P11140) [ENTER] , we find that [[ [ L11140 [ENTER] yields 1 [ [
16 4 13 2
[[ ,— ——> [ U11140 ENTER] yields 1---------------------- [ [
1 0 0 0
0 0 0 ] 1 0 .7 5 0 0 ] - 2 .0 6 2 5 1 4 .9 5 3 4 8 8 3 7 2 1 0 ] -1 .6 2 5 8 .7 2 0 9 3 0 2 3 2 5 6 8 .2 9 2 3 7 9 4 7 1 2 3 ]
.3125 1 0 0
-.5 .25 1 .6 7 4 4 1 8 6 0 4 6 5 - .8 3 7 2 0 9 3 02326 1 -.1 3 2 1 9 2 8 4 6 0 3 4 0 1
t— 77'r-y and for the permutation matnx P1114 0 [ENTER | gives
] ] ] ]]
[ [ 0 0 0 1 ] [ 0 1 0 0 ] j- 0 o 1 0 ] * [ 1 0 0 0 ] ]
41. After LU (Al 1141, L11141, U11141, P11141) [ENTER] , we find that [[ [ L11141 (ENTER) yields 1---------------------- [ [
16 0 0 0 - 7 0 5 4 .5 6 2 5 8 .1 6 9 6 4 2 8 5 7 1 4 2 -.3 7 5 - 1 .5 8 9 2 8 5 7 1 4 2 9
0 ] 0 ] 0 ] 2 .9 7 5 9 5 6 2 8 4 1 5 ]
LU-Factorizations of a Matrix
[[ ,--------------. [ U11141 ENTER yields 1 [ [
1 0 0 0
- . 3 1 2 5 .6875 1 -.5 7 1 4 2 8 5 7 1 4 2 9 0 1 0 0
.5 -.1 4 2 8 5 7 1 4 2 8 5 7 .018579234973 1
Calculator 1.11 147
] ] ] ]]
[ [ 0 0 0 1 ] ,
and for the permutation matnx P11141 [ENTER ] gives
j
[ 1
0
00
l o
0]
0]
[ 0 0 1 0 ] ] 42. After LU (A l1 1 4 2 , L i l i 4 2 , U11 1 4 2 , P11142) (ENTER) , we find that L i l i 42 (ENTER) yields [[710 0 0 0 ] [ 35 6 9 .6 7 6 0 5 6 3 3 8 0 0 0 ] [ 14 3 8 .0 7 0 4 2 2 5 3 5 2 7 9 .5 1 5 4 6 3 9 1 7 5 0 0 ] , [ 14 1 4 .0 7 0 4 2 2 5 3 5 2 1 9 .5 9 6 7 2 5 2 8 8 1 -3 3 .4 5 7 5 2 3 9 6 6 4 0 ] [ 23 2 4 .9 0 1 4 0 8 4 5 0 7 2 4 .2 3 0 2 4 0 5 4 9 8 - 3 1 .3 4 7 4 6 5 3 1 8 3 1 4 .7 7 7 0 3 8 2 4 1 6 ]] U11142 (ENTER) yields [[ 1 - .6 4 7 8 8 7 3 2 3 9 4 4 [ 0 1 [0 0 [0 0 [ 0 0
.830985915493 -1 .5 7 9 9 4 7 4 4 2 8 9 1 0 0
.915492957746 -.1 2 9 7 7 5 6 2 1 5 8 9 .227926876702 1 0
-.3 0 9 8 5 9 1 5 4 9 3 -.5 6 1 9 5 6 7 4 1 4 5 9 1 .4 8 0 6 1 7 1 3 9 8 9 .112687 6 6 8 7 9 5 1
] ] ] , ] ]]
[ [ 0 0 1 0 0 ] [ 0 0 0
and for the permutation matrix P11142 (ENTER) gives
10 ]
[0 0 0 0 1 ] . [ 0 1 0 00 ] [ 1 0 0 0 0 ] ]
43. After LU (Al 1143 , L11143 , U11143 #P i l i 43 ) (ENTER) , we find that [[ [ L11143 (ENTER) yields [ [
.91 .83 .46 .21
0 0 .50021978022 0 -.0 3 6 2 6 3 7 3 6 2 6 4 .189244288225 .266923076923 .063804920914
[[ 1 .252747252747 U1U 4 3 (EÑTBR) yields 1 0 1 ---------[ 00 [ 0 0
.175824175824
- .2 1 9 7 8 0 2 1 9 7 8
-1.65114235501 1.90399824253 - 2 .2 1 8 5 8 7 4 8 1 4 3 0 1
1
[ [ 0 1 0 0 ] and for the permutation matrix P i l i 43 [ENTER] gives
[
q
0 1 0 ] ‘
[ 1 0 0 0 ] ] 44.
0 ] 0 ] 0 ] .0 4 9 489575595 ]]
After LU(A 1 1 1 4 2 ,L 1 1 1 4 2 , U11142, P11142) [ENTER] , we find that
] ] ] ]]
148 Chapter!
Instructor’s Manual
Systems of Linear Equations and Matrices
[ [ 1 0 0 0 0 [ 0 5 0 0 0 L11144 [ENTER) yields [ 1 0 3 0 0 [ 0 0 0 5 0 [ 0 0 2 3 - 1 .6 [ [ 1 2 0
[
U11144 (ENTER) yields
0 1 1 .2 [ 00 1 [ 0 0 0 [ 00 0
0 0
.2
0
0
0
1 0
1. 2 1
] ] ] ] , ] ]] [ [ 1 0 0 0 0 ] [ 0
and for the permutation matrix P11144 (ENTER) gives
0
1 0
0]
[0 1 0 0 0 ] . [ 0 0 0 0 1]
[ 0 0 0 1 0 ]]
LU -Factor¡zat¡ons o f a M a trix
MATLAB
1.11
M A T L A B 1.11
1. We eliminate down each column, and accumulate the product of the inverses of the elementary m atri ces. » A » U » c » F » U » L L =
= = = = = =
[8 2 -4 6; 10 1 -8 9; 4 7 10 3]; A; */, Eliminate U(2,l). -U(2,1)/U(1,1) i eye(3); F(2,l) = c; F*U; inv(F)
1.0000 1.2500 0
0 0 1.0000
0 1.0000 0
» c = -U(3,l)/U(l,l) » » F = eye(3); F(3,l) = c; » U = F*U U = -4.0000 8.0000 2.0000 0 -1.5000 -3.0000 12.0000 0 6.0000 » L = L*inv(F) L = 1.0000 0 1.2500 1.0000 0.5000 0
'/, Eliminate U(3,l).
6.0000 1.5000 0
0 0 1.0000
» c = -U(3,2)/U(2,2) i » F = eye(3); F(3,2) = c; » U = F*U U = 8.0000 2.0000 -4.0000 0 -1.5000 -3.0000 0 0 0
*/, Eliminate U(3,2).
» L = L*inv(F) L = 1.0000 0 1.2500 1.0000 0.5000 -4.0000
*/, Not ice that L is lower triangular
» L*U ans = 8 10 4
'/. Not ice that U is in echelon form. 6.0000 1.5000 6.0000
0 0 1.0000 '/, This should be the same as A.
2 - 4 1 -8 7 10
6 9 3
149
150
C h ap ter 1 Systems o f Linear Equations and Matrices
In stru cto r^ M anual
2. » A = rand(5); */, A random 5x5 matrix. » b = rand(5,l); */, A random 5 vector. » flops(O); rref([A b]); frref=flops frref = */, This number will beslightly higher in MATLAB 3.5 1986 */, It may also vary slightly for different random A » flops(O), x = A\b, flu=flops x = -0.9377 0.7523 0.8705 -0.0748 0.1154 flu = 271
(b) The previous code can be repeated several more times for part (b). (c) The number of operations was fewer for the second versión. However, very similar operation counts should apply for the A\b and rref methods. The great discrepency here results from the fact that rref incurs an enormous overhead when it attem pts to produce nice (rational) results if appropriate. You can enter the MATLAB command type rref to see the code, which always includes at least one cali to rat (A). If you compute flops(O); [num,den] =rat (A); all(all(A==num./den)) ;flops to de termine the number of flops involved in deciding rationality, you will find that, say for a 5x6, there are about only 210 flops of the 1900 reported above which actually arise from rref. Thus rref ( [A b] ) and A\b do, infact take almost the same number of operations. Even without the calis to ‘ra t’, the computed counts indicate a bit of overhead. If you did rref by hand, you’d do less than 200 multiplications, divisions and additions. (In fact less than 125 if you avoid unnecessary operations, see Appendix 3, page A-27 and the solution to A3.3). Similar counts would apply to LU-factorization followed by forward and back elimination on the right hand side, even when partial pivoting is done. 3. (a) » A = 2*rand(3)-l A = 0.4834 0.0500 -0.9618 -0.0734 -0.8696 0.7721 » [L, U, P] L = 1.0000 -0.8027 -0.5026 U = -0.9618 0 0 P =
0 0 1
1 0 0
0.4268 -0.0221 0.3354
= lu(A) 0 1.0000 -0.0141
0 0 1.0000
-0.0734 -0.9285 0
-0.0221 0.3176 0.4202
0 1 0
L U -Facto rizatio ns o f a M a trix
*/, This should be the same as P*A below.
» L*U ans = -0.9618 0.7721 0.4834 » P*A ans = -0.9618 0.7721 0.4834
MATLAB
-0.0734 -0.8696 0.0500
-
0.0221 0.3354 0.4268 X This should be the same as L*U above.
-0.0734 -0.8696 0.0500
-
0.0221 0.3354 0.4268
(b) » A = round(10*(2*rand(4)-l)) A = 4 8 10 0 -6 1 -6 8 8 -7 -1 -1 7 -1 -4 -1 X, U, P] = lu(A) L = 1.0000 0.5000 0.8750 ■0.7500
0 1.0000 0.4457 -0.3696
0 0 1.0000 0.3677
0 0 0 1.0000
8.0000 0 0 0
-7.0000 11.5000 0 0
-1.0000 10.5000 -7.8043 0
-1.0000 0.5000 -0.3478 7.5627
U=
P = 0 1 0 0 :=
0 0 0 1
1 0 0 0
0 0 1 0
-7 8 -1 1
-1 10 -4 -6
-1 0 -1 8
p *a
C = 8 4 7 -6 » » » L2
» »
c = -C(2,1)/C(1,1); F = eye(4); F(2,l) = c; */. Eliminate C(2,l). C = F*C; L2 = inv(F) '/, L2 should end up the same as L. =
c C
1.0000
0
0.5000
1.0000
0
0
0 0
0
1.0000
0 1.0000
0
0
0
0
= -C(3,1)/C(1,1); F = eye(4); F(3,l) = c; */. Eliminate C(3,l). = F*C;
1.11
151
152
In stru cto r^ Manual
C hapter 1 Systems o f Linear Equations and Matrices » L2 = L2*inv(F); » c = -C(4,1)/C(1,1); » C = F*C C = -7.0000 8.0000 11.5000 0 5.1250 0 -4.2500 0
'/• Accumulate product of inv(F)'s. F = eye(4); F(4,l) = c; */, Elimínate C(4,l), end col. 1.
-1.0000
-1.0000
10.5000 -3.1250 -6.7500
0.5000 -0.1250 7.2500 # /, Note colman 1 of L and L2 agree.
» L2 = L2*inv(F) L2 =
1.0000 0.5000 0.8750
0
0
1.0000 0 -0.7500
0 0
0 1.0000 0
1.0000
» » »
c = -C(3,2)/C(2,2); F = eye(4); F(3,2) = c; C = F*C;
*/, Elimínate C(3,2), start col. 2 '/• Note C(2,2) is largest in column 2.
»
L2 = L2*inv(F);
» »
c = -C(4,2)/C(2,2); F = eye(4); F(4,2) = c; */. Elimínate C(4,2). '/, This finishes column 2. F*C C
c = 8.0000
0 0 0
-7.0000 11.5000
0 0
-1.0000
-1.0000
10.5000 -7.8043 -2.8696
0.5000 -0.3478 7.4348
» L2 = L2*inv(F) '/, Now column 2 of L2 and L agree. L2 =
1.0000 0.5000 0.8750 -0.7500
0 1.0000 0.4457 -0.3696
» c = -C(4,3)/C(3,3); » C = F*C C = -7.0000 8.0000 11.5000 0 0 0
0
0
0 1.0000 0
F = eye(4); F(4,3) = c;
-1.0000
-1.0000
10.5000 -7.8043
0.5000 -0.3478 7.5627
0
0
0 0 1.0000
% Elimínate C(4,3). */, Note c(3,3) is largest.
» L2 = L2*inv(F) L2 =
1.0000
0
0.5000 0.8750 -0.7500
1.0000 0.4457 -0.3696
0 0 1.0000 0.3677
0 0 0 1.0000
Notice that C reduces to U and that L2 is the same as L, as predicted. At each step we can check that the pivot was the largest number (in absolute valué) when compared to those below it in the same column.
L U -F acto rization s o f a M a trix
» A = rand(3) A = 0.7734 0.4177 0.7273 0.6825 0.3192 0.6806 » [ L, U, P] = lu(A) L = 1.0000 0 0.4127 1.0000 0.9405 0.5700 U = 0.7734 0.4177 0 0.5082 0 0 P = 1 0 0 0 0 1 0 1 0 » B = P*A B = 0.7734 0.3192 0.7273
•/. A randora matrix. 0.2053 0.8364 0.7089
0 0 1.0000 0.2053 0.6242 0.2876
'/, Store P*A in B, and work wi1 0.4177 0.6806 0.6825
0.2053 0.7089 0.8364
» B(2,:) = B(2,:) - L(2,1)*B(1 ,:) '/• Eliminate B(2,l) using B = '/, L(2,l) as the multiplier 0.2053 0.7734 0.4177 0 0.5082 0.6242 0.7273 0.6825 0.8364 » B(3,:) = B(3,:) - L(3,1)*B(1 , :) */, Eliminate B(3,l). B = 0.7734 0.4177 0.2053 0 0.5082 0.6242 0.0000 0.2896 0.6434 » B(3,:) = B(3,:) - L(3,2)*B(2 ,:) */, Eliminate B(3,2). B = 0.7734 0.4177 0.2053 0 0.5082 0.6242 0.2876 0.0000 0.0000
Notice that B was reduced to U, as expected.
MATLAB
1.11
153
154
C hapter 1 Systems o f Linear Equations and Matrices
S e c tio n 1.12
/ 0 1 10 0\
/O 1 1 0 ^ 0000 1. 110 0 \l 0 0 0/
0 0 0 1 0
2.
\1 0 1 0 0 /
/0 0 1 2 0 0 \ 100100 000101 110010 000100 V ioioio/
I X
7.
3
1 1 0 10 0 0 10 1
1 R3 —3i?i
2
fía + I 6H2 ^ 2 5 2 ^ 0 1 7 = í/. Thus A = LU where L = \ 0 0 120
/
The system Ly = b, i.e.,
yields the equations y\ = 3, 2yi -+-1/2 = 0, 3yi — 16t/2 + 2/3 = 7. Solving we
164
C h ap ter 1 Systems o f Linear Equations and Matrices
Instructor’s Manual
3\ ( 2 5 —2 \ —6 I . Now from Ux = y, i.e., [ 0 1 7 I -9 8 / Vo 0 120/ 167 obtain 2 xi+ 5x2-2a;3 = 3,X2+7*3 = -6,120*3 = -9 8 . Solving we get xj = — lzU /1 6 7 /1 2 0 \ The solution is | —17/60 I . \ -4 9 /6 0 / get j/2 = —6, j/3 = —98 and y = |
1 61.
\
/
— 5 —
( Xl\ / 3\ í X2 I = I —6 I we \* 3 / \-9 8 / —17 —49 x2 = - tt- , x3 = . oU oU
Definitions
Section 2.1
C h ap ter 2. D eterm in a n ts S ectio n 2.1 1.
103 0 1 4 = (1)(1)(0) + (0)(4)(2) + (3)(0)(1) - (3)(1)(2) - (1)(4)(1) - (0)(0)(0) 2 10 = - 6 - 4 = -10 -1 1 0 2 1 4 .= (—1)(1)(6) + (1)(4)(1) + (0)(2)(5) - (1)(1)(0) - (5 )(4 )(-l) - (6)(2)(1) 156 = - 6 + 4 + 2 0 -1 2 =
3.
3-14 6 3 5 = (3)(3)(6) + (—1)(5)(2) + (4)(6)(—1) - (2)(3)(4) - (-1)(5)(3) - (6 )(6 )(-l) 2-16 = 54 - 10 - 24 - 24 + 15 + 36 = 47
-10
4.
6
02 4 1 2 -3
(—1)(2)(—3) + (0)(4)(1) + (6)(0)(2) - (1)(2)(6) - (2 )(4 )(-l) - (-3)(0)(0)
= 6-12+ 8 = 2
-2 3 1 4 6 5 = (—2)(6)(1) + (3)(5)(0) + (1)(4)(2) - (0)(6)(1) - (2)(5)(-2) - (1)(4)(3) 0 2 1
= -1 2 + 8 + 2 0 - 1 2 = 4
6.
5 -2 1 6 0 3 = (5)(0)(4) + (—2)(3)(-2) + (1)(6)(1) - (—2)(0)(1) - (1)(3)(5) - (4)(6)(-2) -2
14
1 2 + 6 - 1 5 + 48 = 51 2031 0 14 2 0 0 15
12 3 0
14 2 0 14 012 = 2 0 15 + 3 00 5 - 1 00 1 230 120 123
= (2)(21) + (3)(5)- ( ! ) ( ! ) = 56.
165
166
C h ap ter 2 D eterm in an ts
-3 -4 5 2
9.
0 00 7 00 8-10 3 06
Instructor’s Manual
(—3)(7)(—1)(6) = 126
-2 0 0 7 2-14 12-14 = (-2 ) 0 - 1 5 - 7 3 0-15 2 30 42 30
1 2 -1
3 0-1 42 3
2 24 + (—2)(—5) 2 20
-1
3
- 7 (-3 )
2 -1
2
3
= - 1 6 + 8 0 + 168 + 42 = 274
10 .
2 3 ■- 1 4 1 7 8 0 4 •- 1 0 0 -- 2
0 0 0 0
0
2 2 = (2)(1)(4)(—2)(6) = -9 6 8 0 6
0
11. Let A =
5
(an 0 0 022 \
0
0\ 0
...
(bu and B =
ann J
(< * 1 1 6 1 1
0
bnn '
0\ o
0
0 O22^22
Then A B =
0\ o
0
0 622
6 •* o,nnbnn J det A = o n 022 • ■•o-nni det B = 611622 • • • bnn] det A B = (o h 6ii)(o 22622) • • • (o,nnbnn) = (
c = A(2,1)/A(1,1); 1)/A(1,1); A = 1.0000 6.0000 4.1667 0 1.0000 0 0.8333 0 o II >
»
V V
2.0000 1.3333 -3.0000 1.6667
3.0000 1.8000 1.0000 4.5000
II
» c = A(3,2)/A(2,2); A(3,:) = A(3,:) A(4,:) » c = A(4,2)/A(2,2); A = 3.0000 1.0000 2.0000 6.0000 1.3333 1.5000 4.1667 0 0 -3.3200 0.6400 0 0 1.4000 4.2000 0 >
188
» c = A(4,3)/A(3,3); A(4,:) = A(4,:) - c*A(3,:) A = 2.0000 3.0000 1.0000 6.0000 1.5000 4.1667 1.3333 0 -3.3200 0.6400 0 0 4.4699 0 0 0 » det(A) ans = -371
'/, Eliminate A(4,3).
Properties of Determinants
M A T L A B 2.2
There were no row interchanges, and det (.4) was unchanged.
G>) » A = [012; » det(A) ans = 0 »
345;
123];
A ([1 3],:) = A( [3 1], :)
*/• The first row interchange.
A = 1 3 0
2 4 1
3 5 2
» c = A(2,1)/A(1,1); A = 1 2 3 0 -2 -4 0 1 2
A(2,:) = A(2,:) - c*A(l,:) •/. Eliminate A(2,l).
» c = A(3,2)/A(2,2); A = 1 2 3 0 -2 -4
A(3,:) = A(3,:) - c*A(2,:) */. Eliminate A(3,2).
0
0
0
» det(A) ans = 0 There was one interchange, so (-l)*det(t/).
k
= 1 and (— 1)* = — 1. Since 0 = — 1 • 0, we have det (A) =
(c) » A = [ 1 2 3; 4 5 6; -2 14]; » det(A) ans = 0 » A ( [1 2],:) = A([2 1],:) A = 4 5 6 1 2 3 -2 1 4
'/, First interchange.
» c = A(2,1)/A(1,1) ; A(2,:) = A(2,:) - c*A(l,:); » c = A(3,1)/A(1,1) ; A(3,:) = A(3,:) - c*A(l,:) A = 4.0000 5.0000 6.0000 0 0.7500 1.5000 0 3.5000 7.0000 » A([2 3],:) A = 4.0000 0 0
= A([3 2],:) 5.0000 3.5000 0.7500
6.0000 7.0000 1.5000
7.
Eliminate A(2,l). */. Eliminate A(3,l).
'/, Second interchange.
189
Chapter 2 Determinants
Instructor’s Manual
» c = A(3,2)/A(2,2); A = 4.0000 5.0000 0 3.5000 0 0
A(3,:) = A(3,:) - c*A(2,:)
'/,
Eliminate A(3,2).
6.0000 7.0000 0
» det(A) ans = 0
There were two interchages, so (—1)* = +1. Again, 0 = 1 - 0 . (d)
i O
» A = round( 10*(2*(rand(4) -1)) ) A = -14 -18 -12 -11 0 -1 -1 -14 -19 -10 -19 -2 -5 -9 i Men
190
[L,U,P] = lu(A) L = 1.0000 0.6667 0 0.9333
0 1.0000 0.0811 0.7027
0 0 1.0000 0.4475
0 0 0 1.0000
15.0000 0 0 0
--10.0000 --12.3333 0 0
-9.0000 4.0000 -14.3243 0
-5.0000 -15.6667 0.2703 4.5547
U =
P = 0 0 0 1 »
0 0 1 0
0 1 0 0
1 0 0 0
det(A)
ans = -12070 » det(U) ans = -1.2070e+04 » det(P) ans = 1
The m atrix P represents the swapping the first row with the fourth, and the second with the third. Since this is two interchanges, k — 2, and (—1)* = 1. Thus, det (A) = det ((7). The determinant of P is exactly the same as (—1)*.
Proofs of Three Important Theorems and Some History
Section 2.3
S ec tio n 2.3
1. Note that E B is m atrix B with the ¿th and .;th rows switched. Then det E B = —det B. But, det E = —1. Thus, det E B = det E • det B. 2. Note that E B is the m atrix B with row j replaced with row j plus ctimes row det B. But, det E = 1. Thus, det E B = det E • det B.
i.Then det E B =
3. Notethat E B is the m atrix B with row j replaced with c times row j. Then det E B But, det E = c. Thus, det E B = det E* det i?.
= c*det B.
192
Instructor’s Manual
Chapter 2 Determinants S e c tio n 2.4 1 . det A — 4
=^
adj A
1/2 ■1/4
-1/2 3/4
2. det A — 0; A is not invertible 3. det A = - 1
adj A
4. det A = —8
adj A —
5. det .4 = —12
A-
1-1 0
adj A —
det A = —8
adj A =
0'
0
1
A~ 1 =
adj A
/ —21 -4
12. det A = —1
adj 4 =
3 1
3 6\ 4 -1
2 - 1-2 15 - 6 - 6 - 3 / 1
/
3/8 1/8 —1/4 \ —1/8 - 3 /8 3/4 ,- 1 / 4 1/4 1 / 2 /
10. det 4 = 0; A is not invertible
9. det A = 0; A is not invertible
1 1 . det A = - 9
1/3 - 1 /4 - 1 /6 0 1/4 1/2 0 1/4 - 1 /2
1 —1 | = A~
adj .4 = | 0
7. det A = —1
13/8 - 1 /2 —1/8 \ = I -1 5 /8 1/2 3/8 5/4 0 -1 /4 /
A -1= I
adj 4 =
6 . det A — \
1
0 —1 -1 1
0 - 2\ 2 -2
0 -1 -3 2 -2 -3
3 2/
7/3 - 1 /3 - 1 /3 - 2 /3 \ 4/9 - 1 /9 - 4 / 9 1/9 -1/9 - 2 /9 1/9 2/9 —5/3 2/3 2/3 1 /3 /
A~l =
0
1
0
1 - 1-2
A~x =
0 —2
2
\
2
1 3 - 3 2 3 -2/
13. By theorem 2.2.3, det .4 = det 4*. Henee, det A is nonzero if and only if detj4‘ is nonzero. By theorem 4, A is invertible if and only if A* is invertible. A -i-l A ~ 3
14. det v4 = 3
a -3 4 1 —a — a
17.
=
1
det A 1 = 1/3
det A '
15. det A = -2 8
16.
5 -1 -2
oí
1
=- — 28
— a 2 -f 12 = 0. If a = 4 or —3 then the matrix is not invertible.
a — 1 a -f 1
1 2 3 2 —a a -|-3 a - f 7 invertible.
0 3a —1 4a + 1 1 2 3 = 0 (R i = R 3 ). Henee, for all valúes of a, the m atrix is not 0 3a —1 4a + 1
Determinants and Inverses
Section 2.4
18. By theorem 2, (>l)(adj A) = (det A)I. By theorem 4, det A = 0. It follows that (A )(adjA ) is the zero matrix. 19. Let A = ( C°SÜ i . det A = eos2 9 + sin 2 0 = 1. By theorem 3, this m atrix is invertible. y —sin0 cos0 J . i _
sin ¿A _ ( eos 9 —s in 0 \ 4 ^ ( cos 0 —siní - adj A = 1 . . n :os 9 J ~~ \ sin 9 eos 9 J det A V sin 0 cos 1
193
194
Instructoras Manual
Chapter 2 Determinants
M A T L A B 2.4 1. » n = 5; m = 4; » A = 2*rand(n,m)-l; » det(A * * A) ans = 4.8868 » n = 4; m = 5; » A = 2*rand(n,m)-l; » det(A* * A) ans = 4.8441e-17
When n > m, det (A* A) > 0. However, when n < m, det (A1A) = 0 up to roundoff error.
» A = round(10*(2*rand(4)-l)) A = -4 0 7 -3 -9 -1 4 -9 0 - 7 2 10
-
1
-
4
4
1
» »
flops(O); C = zeros(4); c ( l . l ) = det(A( [2 3 4 ] , C (l,3 ) = det(A( [2 3 4 ] , C (2 ,1) = -det(A([1 3 4 ] , C (2,3) = -det(A( [1 3 4 ] , C (3 ,1) = det(A( [1 2 4 ] , C (3,3) = det(A( [1 2 4 ] , C (4 ,1) = -det(A([1 2 3 ] , C (4 ,3) = -det(A( [1 2 3 ] , C = C>
(a)
» » » » » » » » »
[2 3 4 ] ) ) ; C ( l , 2 ) = ■-det(A([2 [ 1 2 4 ] ) ) ; C(1,4) = ■-det(A([2 [ 2 3 4 ] ) ) ; C(2 , 2 ) = det(A([1 [ 1 2 4 ] ) ) ; C (2,4) = det(A( [1 [ 2 3 4 ] ) ) ; C (3,2) = ■-det(A([1 [ 1 2 4 ] ) ) ; C (3,4) = ■-det(A([1 [ 2 3 4 ] ) ) ; c (4 ,2 ) = det(A([1 [ 1 2 4 ] ) ) ; C (4,4) = det(A( [1
171 76 111 31
223 -62 43 -197
4 ], 4 ], 4 ], 4 ], 4 ], 4 ], 3 ], 3 ],
2 2 2 2 '/, Don't forget the transpose.
C = 86 -284 -224 -154
3 3 3 3
-433 452 -103 337
» s = flops s = 200
(b) » í lops(0); » D = det(A)*inv(A) D = 86.0000 171.0000 223.0000 -433.0000 -284.0000 76.0000 -62.0000 452.0000 43.0000 -103.0000 -224.0000 111.0000 -154.0000 31.0000 -197.0000 337.0000
[1 [1 [1 [1 [1 [1 [1 [1
3 4 ]) 2 3])
3 4 ]) 2 3])
3 4 ]) 2 3])
3 4 ]) 2 3])
Determinants and Inverses
M A T L A B 2.4
» ss = ílops ss = 232
(c) » C-D ans = 1.0e-12 -0.0568 0 -0.0284 0.0568
0.0284 0 0.0284 -0.0071
0.0568 -0.0568 0.0142 -0.1137
0.0284 0.0284 -0.0142 0.0568
The m atrix D is the same as C except for some small round off error. This is expected by equation (8). Method (a) uses fewer flops than method (b).
»
C = 20 * [7 7 -7 2 5 6; 5 7 - 9 5 2 0; c = 40 140 140 -140 80 0 100 -200 60 180 140 -100 100 100 140 -180 180 100 40 20 20 180 -340 80 » rref(C) ans = 1 0 0 0 0 0
0 5 -10 4 8 5 2 1 9 10 : 100 160 80 40 200 40
0;
7]
120 120 0 0 160 140 •/, . This has a zero row, so C is not invertible. *
0 1 0 0 0 0
1 -2 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
» A=C; A(3 ,3) = C(3,3) + 1. e- 10; » rref(A) */. This ans = 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 » det(A) ans = 6.5509
The determinant of A is not very cióse to 0. This indicates that the “natural assumption” is false.
196
Instructor’s Manual
Chapter 2 Determinants
4. (a) One way to generate an upper triangular m atrix with determinant 1 is to create a random matrix with zeros below the first diagonal, and to add the identity matrix to this. Type help triu for more information. » A = triu(round( 6*(2*rand(5)-l)),1) + eye(5); » det(A) ans = 1
» » » » » »
A(4, A(5, A(2, A(3, A(4, A(5, A = 1 3 -4 1
-3
) ) ) ) ) )
= = = = = =
A(4, A(5, A(2, A(3, A(4, A(5,
-2 -5 8 -4 6
) ) ) ) ) )
2*A(2,:) 2*A(3,:) 3*A(1,:) 4*A(1,:) A(1,:) 3*A(1,:)
-3
”3 13 -15 11
'/, Perform several row operations to A.
3 9 -15 0 -14
-5 -13 25 -8 25
(b) » det(A) ans = 1
We expect det (A) = 1 since we started off with det (A) = 1, and we only used elementary row operations whose determinants were 1. Since det(A) = l and all the det’s used to form adj(A) will be integers, inv(A) = (l/l)*adj (A) should have integer entries. (c) MATLAB is able to convert between characters and numbers. Type help strfun for more infor mation. » s = [ 'SMALL >». > PLAS'; 'TIC T'; 'HINGS'; ' M0VE'; 'WHEN ' . 'YOU D >1
}ONT
L ' ; '00K
»
M = S - 'A' + 1 M = 19 13 1 16 12 -32 20 3 9 14 8 9 15 -32 13 5 23 8 15 21 25 20 15 14 15 15 11
12 1 -32 7 22 14 -32 -32 -32
»
12 19 20 19 5 -32 4 12 -32 */, Encode the message.
C = M*A C = 30 -88 -57 -71 -46 137 -58 -91 80
O; */, Convert to numbers.
-71 190 187 137 61 -294 195 260 -81
-131 398 652 235 -23 -590 677 785 181
-35 719 614 612 315 1006 561 799 -539
-9 -398 -184 -371 -274 514 -161 -297 463
Determinants and Inverses » M2 = C* inv(A); » setstr(round(M2) + ans = SMALL PLAS TIC T HINGS MOVE WHEN YOU D ONT L OOK
fk*
1)
M A T L A B 2.4
'/, The instructer should decode the message. '/, setstr converts numbers to characters.
197
Instructor’s Manual
Chapter 2 Determinants
S e c tio n 2.5 - 1
2 -1
3
47 4 X \
—
8
2 3 - 7
-14 5 =
+
21
“
29
0 + 5
_5_
1
6
10
2’
x2 =
-5;
2 . xj
3 - 1
1 2
5
2
1
1
8
x\
1 5 - 0
x2 = 1 0 "
- 3
11
3 - 2
4.
+ 4
4 5
1
5 - 2 xi =
87 29
10
15
10
“
2
6
3.
7 _ 29
3 4
2
4
94
47 29
4
0 -1 5
- 7
=
~ 50
=
o
2;
x
2
=
2
6
3
5 - 3
8
11 5 K
J
2
1
1
6
3 - 2 -1 25 =
-2 5
-2 5
=
r 5;
8 X3 =
5
2 11
J—
75 =
or
- 3
-2 5
- 3
2
5
8 1 1 -2 4 -1 2 0 -1
62
31 x
2
1 8 1 0 -2 -1 0 2 3 =
1 0 -22
J
/->
11 A
3 5 X3 =
J
1
8 1
1 .
- 4 - 1 4 1
to
198
-1
0 =
_5
13
4 -1 -1
2
7 2 1 0 2 -1 1 1 3 5.
1 1 0 -1 -1 1 3
-11
12 0 -1 1 1
23
~T5
13
T3~
13
45
xi = 13’
- 1
6 . xi =
5 - 1
3 1
3
0 2
0
2
5 - 1
7.
x\ =
“
= 0; x 2 =
2 5 -1
4
4
-2
-5 2
3
0
1 2 0 1 - 1
1 1
2 4 1 2
5
0 1
2 1 0
-
1 0 1
1 1
0
0
-
0
- o } x3 —
1 1 —
1 5
- 3
3
3 0
3
0 1 1 1 1 2 0- 1-1 0 0 3 6 10 0 -1
9
10 “
3 ’
X2
1 1
4 - 1 - 1
0 3 30
-2 2
0
3
6
15 0 - 1 — 9—
=
9 = 0 ;
-5 2
—52
—5 2
-1
-2
-2
16
3
2 1 4 1 0 2 0 -1 1
- 1
2
1
-5 2
5
4 0
5 0
3
^-2
2 ’ X2
-2
3
—52
4 1
6 1
5. X i =
0
2 -1 -1
-2 2 4
2 2 7
2 7
1
= 1
Cramer’s Rule 1 2 0 1
1
1 1 6
2 0 4 - 1
6
0 0 3
x3 =
10
9
2 - 3
xi =
0
x3 =
X4
0 5
21
-29
- 5
9
3
7 0 - 1
2 1
4 - 3
-21
-1 5
_
0
29’
0
0 0 -171
2 3 - 5
^=29
X2
171
-2 9
29 ’
0 3 - 5 0
7
2
2
4 - 1
0
3 ;
4
3 3
1 0
2 0 3 -5 1 0 0 -1 0 2 1 0 4 -1 0 0 1 0
13 "
1 6 - 1
0 0-1 2 1 0 -10 0
7
9.
39
5 - 1
1 0 0 0
Section 2.5
- 3
0 -2 9
2
-1 0 0
4
0 0 2 1 -1 0
0
0 3
1
0 284
- 5 -
-2 9
-2 84 ~
29
’
X4~
-2 9
7
2 - 3
2
182 ”
-2 9
-182 -
29
10. (a) From Figure 2.2, it is easy to see that b • eos A + a • eos B = c. Similarly, if we insert the perpen dicular from the vertex at angle A to the opposite side and insert the perpendicular from the vertex of angle B to the opposite side, then we obtain c • eos A + a • eos C — b and c • eos B + 6 • eos C — a. c0a (b) 6 a 0 = abe + abe = 2abc ^ 0 0 c b c0 6 ba c a 2 -f 62 —c2 a2c + b2c Oca (c) eos C ■ 2abe 2ab 2 bac (d) 2ab • eos C = a 2 + 62 —c2, or c2 _ a2 -f b2 — 2ab • eos C
199
Instructor's Manual
Chapter 2 Determinants
M A T L A B 2.5
» A = 2*rand(5)-l A = 0.2591 -0.5336 0.4724 -0.3874 0.4508 -0.2980 0.0265 0.9989 0.7771 0.1822 » b b =
0.6920 -0.1758 0.6830 -0.4614 -0.1692
0.0746 -0.0642 -0.4256 -0.6433 -0.6926
0.1433 0.6048 -0.9339 0.0689 -0.0030
2*rand(5,1)-1 0.9107 0.4966 0.1092 0.7815 0.2497
(a) » flops(0); » d = det(A) d = 0.0387
x = 4.4245 2.4047 1.0314 4.9862 -0.2662 » s = flops s = 425
(b) » »
flops(0); z = A\b 4.4245 2.4047 1.0314 4.9862 -0.2662
II
K)
X = zeros(5,1); C = A; C ( :,1) = b; b; C = c = A; C ( :,3) = b; c = A; C ( :,4) = b; c = A; C ( :,5) = b; X = x / d o
» » » » » » »
>
200
x(l) x(2) x(3) x(4) x(5)
= = = = =
det(C) det(C) det(C) det(C) det(C) */. Di
by d.
Cramer's Rule
MATLAB
2.5
» ss = flops ss = 266
(c) » format short e » x-z ans = -2.6645e-15 0 -2.2204e-16 -1.7764e-15 0
T h e tw o Solutions are very cióse to each other. T he num ber of flops was alm ost twice as m uch for C ra m e r’s rule as for th e b u ilt-in m ethod.
(d) For a 7 x 7 matrix, Cramer’s rule takes about 1680 flops versus about 600 for the built-in method. It appears A \ b becomes more efficient as the size of the system increases. (In fact the built-in method takes about (2/3)n 3 flops for large N, while Cramer’s rule, using the (efficient) built-in method for det takes about (2/3)n 4 flops for large n.
201
202
Chapter 2 Determinants
Instructor’s Manual
R e v ie w E xercises for C h a p ter 2 -1 2
04
-3 5 = -1 2 + 35 = 23 -7 4
= _4 _ o = - 4
3.
1 -2 3 0 4 5 = (1)(4)(6) - 24 0 06
5.
1 - •1 2 3 4 2 = 16 + 4 + 18 + 16 + 12 - 6 = 60 -2 3 4
4.
5 00 6 2 0 = (5)(2)(6) = 60 10 100 6
3 1 -2 4 0 5 = - 3 0 - 8 - 1 2 - 15 = -6 5 -6 1 3
7.
1 - 12 3 1 -1 2 1 -1 3 -1 2 3 4 025 = -4 2 3 7 - 2 -1 2 7 + 5 -1 2 3 -1 237 5 14 5 10 104 5 104 = -4(23) - 2(—17) + 5(—40) = 92 + 142 - 200 = 34 3 15 17 19 0 2 21 60 = (3 )( 2 )( 1 )(—1 ) = - 6 0 0 1 50 0 0 0 -1
9.
10 .
-3 4 2 1
= —2 —8 = —11; A n = 1; A 12 = —2; A 21 - —4; A 2 2
+.33
, / - 6 -1 5 —34 \ 0 - 9 -1 2 = 6; A 1 = —— 0
0
1 -1 2
12 .
3 1 4 = 8 —20 —6 —10 + 24 + 4 = 0. No inverse. 5 -1 8 1 1 1 1 0 1 0 1 1
1 - 1 = ■ - i ; A n = --1 ; A 12 = —1; A 13
= - i; A' 2
13.
1 -4 -1 1 V -2 - 3
3 -5 7 0 2 4 = (3)(2)(—3) = -1 8 ; A n = -6 ; A 12 = 0; A 13 = 0 ; A 2Í = -1 5 ; A 22 = - 9 ; A 23 = 0 ; 0 0 -3 ^31 = —34; A 3 2 = —12;
11.
-3 ; A - 1 =
1 0
0
0 - 1 3 0 1 0
3
0 -2
0 - 1 0
= 2
1
-1
0 1\ ( 1 0 / - 1 1 0 -1 - 1 = 1 1 V 1 —1 - 1 ) V-1
0 3 0 -1 3 0 0 0 - 2 - 1 1 0 - -2 = 2(2) -- 1 ( 3 -1 0 0 ■ -1 0
6
Review Exercises for Chapter 2
-Aii = 2 ; A 12 — 18; Ais — 6 ; A 1 4 — 1 ; A 21 — 2 ; A 22 — —4; A 2 3 — 6 ; A 24 = 1 ; A 31 = 1 ; ^32 = 0; ( 2 2 0 " g
3 -1 2 4 1 10 3 = -1 -2 4 15 6 -4 1 2
14.
1
324 -1 2 4 3 -1 2 4 1 5 + -2 1 5 + 3 -2 4 1 -4 1 2 6 1 2 6 -4 1
= 21 4- 27 —48 = 0; No inverse.
15. x\
■
16. xi =
3 -1 5 2
17. X! =
7 ’
2 -1 3 2 1 7 -1 4 0 -5 2 3 -1
2 -1
18. x\ =
1
3 -1 2 3
4 -1
*3
123 19 ,X 2 ”
24 #3 19 ’
0 2 -1
19
3
14 56
1 X2 =
4’
—
2 5 -1 -1 0 3 4 -1 1 55-----
3 -1 3
0 2 -1 -1 2 -1
1
7
17 1 2 4 -5
1 -1 1 2 0 -5 0 3 -1
5
23 35 X2 — 7
11
1 -1 7 2 04 0 3 2
34
19
19
70 56
5 4’
1
5
2 0 4 -1 -1
56 1 7 0 -1 -1 2 2 -3 0 -1 -1 0 2 1 4 0 1 0 -1 1 0 2 2 -3 4 -1 -1 0 -2 1 4 0 0 7 1 2 - 1 -3 4 -1 0 0 -2 1 2 0
42 56
-3 T
66
22
39
13’
1 7 - 1 1 0-1 2-3 4 0 - 1 0 -2 2 4 0 39
1 0
282 ~39~
94 13’
1 0 - 1 7 0
-1 8 -6 39 “ 13
X4
—
2
2 -1
4 -1 -1 0 - 2 1 4 2 3S
189 30
63 13
q
1 -11
“
203
204
Notes
Instructors
Vectors in the Plañe
Section 3.1
C h a p te r 3. V e c to r s in M2 a n d M3 Section 3.1 N ote t a n - 1 y /x is alw ays betw een —7r / 2 and 7t / 2 . To get 9 bew teen 0 and 2ir will require adding w or 2ir for som e (x, y) pairs.
1. |v| = %/42 + 42 = 4\/2
0 — tan 1 1 = — 4
2. |v| = >/(—4)2 + 42 = 4^ 2
37t 9 — 7r -f tan _1(—1) = — ( 7r added as in 2nd quadrant.)
3. |v| ■ yjA2 + ( - 4 ) 2 = 4a/2
77r # = 27r + tan _1(—1) = — (27r added for 4th quadrant.)
57r 4. |v| = \ / ( —4)2 + (—4)2 = 4V2 9 = 7r 4* tan-1 1 = — (7r added for 3rd quadrant.) 5. |v| = ^/(V 3)2 + l 2 = 2
# = tan -1 1/ a/3 = 77
6 . |v| = x/ l 2 + (V3 )2 = 2
9 = tan - 1 \/3 = ^ o
6
7. M = v 'c - i )2 + (>/3>2 = 2
27T 9 = 7r -f tan - 1 (—\/3) = — (2nd quadrant)
8 . lv l = •\/l 2 + (—\/3 )2 = 2
# = 27r -f tan _ 1 (-^\/3) =
9. lv l = \ / (
# = 7r + tan - 1 V3 = — (3rd quadrant) ó
l ) 2 + (—\/3 )2 = 2
0
57T (4th quadrant) o
10 . |v| = \ / l 2 + 22 = VE
6 = tan - 1 2 » 1.1071
1 1 . |v| = y / ( - 5 ) 2 + 82 = y/89
0 = 7r + tan - 1 (—8/5) » 2.1294 (2nd quadrant)
12 . |v| = ^ / l l 2 + (—14)2 = \/317
0 = 27r + tan x(—14/11) w 5.3784 (4th quadrant)
13. (a) (6,9)
(b) ( -3 ,7 )
(c) ( -7 ,1 )
14. (a) —2i + 3j (b) 6i - 9j
(c) 6i - 9j
(d) 28i - 42j
15. |i| = |(1,0)| = \ / l 2 + O2 = 1
|j| = Vo 2 + l 2 = 1
(d) (39, - 22 ) (e) 28i - 42j
(f) -2 8 i + 42j
16. \(1/V2)i + ( 1 /V 2 )j| = \ / ( l / V 2)2 + ( 1 / V 2)2 = 1 17. |u| = J a 2/( a 2 + b2) + b2/(a2 -f b2) = 1. Since quadrant of u, v are the same, their directions will b th e sam e since ta n 1
18. |v| = \/Í3 19. |v| = \/2 20 . M — ^
u \J a2 -f b2
= ta n 1 —.
a
u = v / l v l = (2/V Í3)i + (3/VT3)j u = v /|v | = ( l / \ / 2)i - ( 1 / v ^ j u = v /|v | = (—3/5)i + (4/5)j
21.
o II ~v_
22.
Use th e definition of sin # and eos#, as the y or x coordínate of a p o int on u n it circle, # radians counter clockwise from th e x-axis.
u = v /|v | = ( l / \ / 2)i + ( l / \ / 5 ) j u = —( 1 / a / 2 )í —(1 / v^2 )j
if a > 0 if a < 0
205
206
Chapter 3 Vectors in ffi2 and l 3
23. |v| = y/22 + (-3)2 = \/l3 24. |v| = a/ 73 25. |u| = V 2 26. |u| = \/Í3 27. |u| = 5 28. |u| = y / ñ
sin0 = 8/V73
sin 9 = - 3 /V l3
Instructor’s Manual
eos 9 = 2 / V ñ
cos0 = - 3 /\/7 3
v = ~ ( l / V 2 ) i - {l/y/2)S v = —(2/-v/Í3)i + (3 /\/Í3 )j (Notice direction v =direction u —7r!). v = (3/5)i - (4/5)j v = (2/\/r3)i - (3/V l3)j
29. (a) u + v = i - j (u + v ) /|u + v| = (l/y/2)i - ( l / \ / 2)j (b) 2u —3v = 7i —12j (2u - 3 v )/|2 u - 3v| = (7/VT93)i - (12/x/l93)j (c) 3u + 8v = - 2 i + 7j (3u + 8 v )/|3 u + 8v| = -(2 /\/5 3 )i + (7/\/53)j 30. | PQ | = y/(c + a — c)2 + (d + b — d)2 = V a 2 + b2 31. An equation of the line passing through the points O and R is bx —ay = 0. An equation of the line passing through the points P and Q is bx — ay + ad —be = 0. Since the lines are parallel, the direction of PQ is the same as the direction of the vector (a, 6). 32. v = ( 3 cos 7r / 6 , 3 sin 7r / 6 ) = (3 \/3 /2 ,3 /2 ) 33. v = (8 eos 7t / 3 , 8 sin 7r / 3 ) = (4,4\/3) 34. v = (eos 7t / 4 , sin 7r / 4 ) = (1 /\/2, l/\/2 ) 35. v = (6 eos 27r/3, 6 sin27r/3) = (—3,3>/3) 36. Let u = (i¿i, 1/ 2) and v = (tq ,^ ) - Show that u\V\ + « 2^2 < V ( ui + u2)(ví + ^2) sides. Then |u + v |2 = |(« 1 , u2) + (vi,v2)l2 = («1 + v { f + (u2 + v 2)2
squaring both
= u\ + u\ + 2(uiVi + u2v2) + v\ + v\ < M 2 + 2 \J(u2 + ul)(v2 + v 2) + |v |2 = (lu l+ |v |)2 Taking square roots, we obtain |u + v| < |u| + |v|. 37. Let u = (^i,i¿ 2) and v = (^ 1 ,^ 2)- Suppose |u + v| = |u| + |v|. The proof in problem 36 shows that |u + v |2= |u | 2 4 - 1v | 2 + 2 (uiiq + ^ 2^2); henee we must have (uiVi -\-u2 V2 ) = |u ||v | > 0 , asneither u ñor v is the zero vector. Squaring both sides of (uiiq -f ^ 2^2) = lu ||v l and simplifying gives U1 V2 = U2 Viu1 We may assume u, ^ 0. Then u = ( u i, u 2) — — (^ 1 ,^ 2) = a v - But plugging into uiVi + u2v2 > 0 ui gives a > 0. Conversely, suppose u = a v for some a > 0. Then |u + v| = |(a + l)v | = (a + l)|v | = a |v | + |v| = |a v | + |v| = |u| + |v|.
Vectors in the Plañe
Calculator 3.1 207
CALCULATOR SOLUTIONS 3.1 The problems in this section ask you to compute the magnitude and direction (in radians and degrees) for four groups of vectors; within each group the vectors differ only by the signs of their coordinates. One way to solve these is to solve all four problems (in radian form) by: [x-value,y-value] [STO») varname [ENTER) to enter the first in the group (and to check the accuracy of your entry). varname (2nd) [VECTR) (F4) ¡F3] [ENTER] to convert to polar form (magnitude and angle - say in radians) (2nd) [ENTRY] to recall the previous entry (which will include the function » P o l) Then use the arrow keys and [DEL] or ¡2nd) (INS) to delete or insert the - signs in the appropriate places, and then (ENTER) will produce the (same) magnitude and new direction. Repeat the [2nd] [ENTRY) and editing steps for the remaining sign changes. Now change to degree mode by (2nd) (MODE) 0 0 0 0 [ENTER) [EXIT) (or just omit the 0 if TI-85 is already in degree mode and you wish to retum to radian m ode). Then repeat the [2nd] [ENTRY) and editing steps to get the magnitudes and directions in degrees for the four problems in the group. In our Solutions below we indicate input and answers in radians and degrees for each sepárate problem, without describing the mechanism for changing between these two modes, or recalling the previous entry (or answer) for editing. 38. Enter [1.735,2.437] |STO»] A 313 8 [ENTER] and then in radian mode A 3 1 3 8 yields: [2 .9 9 1 5 2 0 3 4 9 2 5 Z . 95210120347] or in degrees: [2 .9 9 1 5 2 0 3 4925Z 54 . 5513 80 6263o] 39. Enter [1.735,-2.437] [STO») A 3139 [ENTER) and then A 3139 (the menú item) yields: [ 2 . 9 9 1 5 2 0 3 4 9 2 5 Z -. 95210120347] orin degrees: [2 . 9 9 1 5 2 0 3 4 9 2 5 Z -5 4 . 5513806263o] . N o t i c e negating the y coordinate in problem 38 just negates the angle. 40. Enter [-1.735,2.437] (STO») A 3140 (ENTER) and then A 31 40 (the menú item) yields: [ 2 . 9 9 1 5 2 0 3 4 9 2 5 Z 2 .18949145015] orin degrees: [ 2 . 99152034925Z 125 . 44861937 4o] . Notice that negating both coordinates (from A313 9) just adds k radians (or 180°) to the (negative) angle. 41. Enter [-1.735,-2.437] (STO») A 3141 ¡ENTER) and then A 3 1 4 1 (the menú item) yields: [2 .9915203 4 9 2 5 Z -2 .189 49145015] orin degrees: [2 . 9915203 4 9 2 5 Z - 125 .448619374°] . Notice that negating the x coordinate (from A313 9) just complements the (negative) angle with respect to - k radians (or -180°). 42. Enter [-58,99] [STO») A3142 [ENTER) and then A3142 yields:
[114 .73 883 3 879Z2 .10 07 5283 072] orin degrees: [ 114 .73 8833 879Z120 . 3 64271o] 43. Enter [-58,-99] (STO») A3143 (ENTER) and then A3143 yields:
208 Chapter 3
Vectors ¡n R2 and R3
Instructor’s Manual
[1 1 4 .7 3 8 8 3 38 7 9 Z -2 .1 0 0 7 5 2 8 3 0 7 2 ] orindegrees: [1 1 4 .7 3 8 8 3 3 8 7 9 Z - 1 2 0 . 364271o] 44. Enter[58, 99] [STO» 1 A 3 1 4 4 [ENTER] and then A 3144 < » P o l> yields:
[ . 8 4 8 0 6 6 6 2 4 7 4 1 Z 1 .0 4 0 8 3 9 8 2 2 8 7 ] orindegrees: [ 114 .7 3 8 8 3 3 879Z59 . 6357289998o] 45. Enter [5 8 ,-9 9 ] [STO» | A 3145 [ENTER] and then A 3145 < » P o l> yields:
[114 .738833 8 7 9 Z - 1 .0 4 0 8 3 9 8 2 2 8 7 ] orindegrees: [114 .7 3 8 8 3 3 8 7 9 Z -5 9 . 63 57289998o] 46. Enter [0.01468,-0.08517] [STO» ] A 3146 [ENTER] and then A 3146 < » P o l> yields: [ . 0 8 6 4 2 5 8 7 1 7 0 5 Z -1 .4 0 0 1 1 2 2 2 9 4 1 ] orindegrees: [ . 0 8 6 4 2 5 8 7 1 7 0 5 Z -8 0 . 2205215899o] 47. Enter [0.014168,0.08517] [STO» ] A 3147 [ENTER] and then A 3147 < » P o l> yields:
[ .0 8 6 4 2 5 8 7 1 7 0 5 Z 1 .40011222941] orindegrees: [. 086425871705Z 80 . 220521 5 8 9 9 o] 48. Enter [-0.014168,-0.08517] [STO» | A 3148 [ENTER] and then A 3148 < » P o l> yields: [ .0 8 6 4 2 5 8 7 1 7 0 5 Z - 1 .74148042418] orindegrees: [ . 0 8 6 4 2 5 8 7 1 7 0 5 Z -9 9 . 7794784101o] 49. Enter [-0 .0 1 4 1 6 8 ,0 .0 8 5 1 7 ] [STO» | A 3145 [ENTER] and then A 3145 < » P o l> yields:
[ .0 8 6 4 2 5 8 7 1 7 0 5 Z - 1 .74148042418] orindegrees: [ . 086425871705Z 99 .7794784101°]
Vectors in the Plañe
M A T L A B 3.1
M ATLAB 3.1 1. (a) Notice that for vectors in the second or third quadrant, we must add 7r to the arctangent in order to get the direction. » v = [4; 4]; » norm(v) ans = 5.6569
'/• Problem 1.
» direction = atan( v(2)/v(l) ) direction = 0.7854 » deg = direction*180/pi deg = 45 » v = [4; -4]; » norm(v) ans = 5.6569
'/. Problem 3.
» direction = atan( v(2)/v(l) ) direction = -0.7854 » deg = direction*180/pi deg = -45 » v = [sqrt(3); 1]; » norm(v) ans = 2
'/. Problem 5.
» direction = atan( v(2)/v(l) ) direction = 0.5236 » deg = direction*180/pi deg = 30.0000 » v = [-1; sqrt(3)]; » norm(v) ans = 2
'/. Problem 7.
» direction = atan( v(2)/v(l) ) + pi direction = 2.0944 » deg = direction*180/pi deg = 120.0000 » v = [-1; -sqrt(3)]; » norm(v) ans =
2
*/• Problem 9.
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Chapter 3 Vectors in M2 and M3
Instructor's Manual
» direction = atan( v(2)/v(l) ) + pi direction = 4.1888 » deg = direction*180/pi deg = 240.0000 » v = [-5; -8]; » norm(v) ans = 9.4340
*/, Problem 11.
» direction = atan( v(2)/v(l) ) + pi direction = 4.1538 » deg = direction*180/pi deg = 237.9946
(b) » v = [ 1.735; 2.437]; » norm(v) ans = 2.9915
*/. Problem 38.
» direction = atan( v(2)/v(l) ) direction = 0.9521 » deg = direction*180/pi deg = 54.5514 » v = [ -1.735; 2.437]; » norm(v) ans = 2.9915
*/. Problem 40.
» direction = atan( v(2)/v(l) ) + pi direction = 2.1895 » deg = direction*180/pi deg = 125.4486 » v = [ -58; 99] ; » norm(v) ans = 114.7388
*/. Problem 42.
» direction = atan( v(2)/v(l) ) + pi direction = 2.1008 » deg = direction*180/pi deg = 120.3643
Vectors in the Plañe » v = [58; 99]; » norm(v) ans = 114.7388
*/, Problem 44.
» direction = atan( v(2)/v(l) ) direction = 1.0408 » deg = direction*180/pi deg = 59.6357 » v = [ 0.01468; -0.08517]; » norm(v) ans = 0.0864
% Problem
46.
» direction = atan( v(2)/v(l) ) direction = -1.4001 » deg = direction*180/pi deg = -80.2205 » v = [ -0.01468; -0.08517]; » norm(v) ans = 0.0864
*/. Problem 48.
» direction = atan( v(2)/v(l) ) + pi direction = 4.5417 » deg = direction*180/pi deg = 260.2205
2 . (a)
»
u = [2;.5]; v = [-1;2]; ww = u-v; aa = Cu’ v ’ w ’ w w ’]; M = max(abs(aa)); axis('square’); axis([-M M -M M]) hold on plot( [0 v(l)], [0 v(2)], [0 u(l)], [0 u(2)] ) grid
» w = u+v; » » » »
M A T L A B 3.1
211
212
Chapter 3 Vectors in M2 and K 3
» » » » »
a = 1; b = 1; z = a*u+b*v; plot( [0 z(l)], [0 z(2)], ’:c5’) a = .5; b = 1; z = a*u+b*v;
» » » » »
plot([0 z(l)], [0 z(2)], ’:c5’) a = .5; b = .5; z = a*u+b*v; plot([0 z(l)], [0 z(2)], ’:c5>) a = .2; b = .8;
» » » » »
z = a*u+b*v; plot([0 z(l)], [0 z(2)] , >:c5>) a = .7; b = .2; z = a*u+b*v; plot([0 z(l)], [0 z(2)], >:c5*>
Instructor’s Manual
If many more a and b were chosen, with 0 < a,b < 1, then the parallelagram between u and v would start to be filled in. » » » »
a = 1; b = -1; z = a*u+b*v; plot([0 z(l)], [0 z(2)] , ’:c6’) a = .1; b = -1;
Vectors in the Plañe » » » » »
z = a*u+b*v; plot([0 z(l)], [0 z(2)], '— a = .5; b = -.5; z = a*u+b*v; plot([0 z(l)], [0 z(2)], '—
» » » » » »
a = .2; b = -.8; z = a*u+b*v; plot([0 z(l)], [0 z(2)], c6>) a = .7; b = -.2; z = a*u+b*v; plot([0 z(l)], [0 z(2)], *— c6})
M A T L A B 3.1
c6*) c6*)
If many more a and b were chosen, with 0 < a < 1 and —1 < b < 0, then the parallelagram between u and —v would start to be filled in. » » »
a = -1; b = 1; z = a*u+b*v; plot([0 z(l)], [0 z(2)], '-.cS1)
» a = -.5; b = 1; >> z = a*u+b*v; » plot( [0 z(l)], [0 z(2)], '-.ce1) » » »
a = -1; b = .5; z = a*u+b*v; plot([0 z(l)], [0 z(2)],
» » »
a = - . 6 ; b = .2; z = a*u+b*v; plot([0 z(l)], [0 z(2)], ,~.c6))
» » »
a = -.4; b = .7; z = a*u+b*v; plot([0 z(l)], [0 z(2)],
*-. 0 6 *)
»
a = -1; b = 0 ; z = a*u+b*v; plot([0 z(l)], [0 z(2)],
*-.c 6 *)
» »
*-.c6})
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Chapter 3 Vectors in M2 and M3
Instructor's Manual
As above, these fill in the parallelogram between —u and v. If we allow both a and b to be negative, then we will cover the fourth quarter of this parallelo gram. If we allow a and b to grow, then we would get larger and larger región, which eventually covers everything in the plañe. (b) If the above is repeated when u and v are parallel, the linear combinations of u and v will all lie on the same line through the origin.
» » »
u = [2;.5]; v = [-1;2]; w = [ .5; —3] ; lincomb(u,v,w)
*/, Pick two vectors. */, Pick a third vector, or use rand(2,l).
U S [2,0.5] v = [-1 ;2] w = [0.5;-3]
T h e Scalar Product and Projections in M2 Section 3.2
S ec tio n 3.2
1. u • v = (1)(1) + ( 1)(—1) = 0; eosp = 0 3. u v = (—5)(0) + (0)(18) = 0; eos p = 0
2. u . v = (3)(0) 4- (0)(—7) = 0; 4. u • v = (a)(0) + (0)(/?) = 0;
20
5. u v = (2)(5) + (5)(2) = 20;
eos v = ~ ^ =
6 . u • v = (2)(5) 4- (5)(—2) = 0;
eos p = 0
7. u v = (-3 )(-2 ) + (4)(-7) = -22 8 . u v = (4)(5) 4- (5 )(—4) = 0
eos* =
cosp = 0 eos p = 0
20
= -
—22
-22
^
eos p = 0
9. u • v = (a)(/?) 4- (/?)(—a ) = 0 = > c o s ^ = 0=>v? = 7r / 2 => u and v are orthogonal. 10. T h e scalar p ro d u c t is an o p eratio n in which th e in p u t is two vectors and th e o u tp u t is a num ber. T h en u • v w is n o t defined: once we take the first scalar p ro duct, we would th en need to take the scalar p ro d u c t of a n u m b er and a vector, which does not m ake sense.
1 1 . v = —2 u => u and v are parallel. 12. u
v = (2)(6) 4- (3 )(—4) = 0 => eos p = 0 => p = 7t / 2 => u and v are orthogonal.
13. u • v = (2)(6) 4- (3)(4) = 24 14. u
24
eos u and v are not parallel and not orthogonal. 26
v = (2 )(—6) 4- (3)(4) = 0 =£• eos p = 7t / 2 => u and v are orthogonal.
15. u • v = (7)(0) + (0 )(—23) = 0 => eos u and v are orthogonal. 16. v = —u / 2 => u and v are parallel. 17. (a) We need u - v = 3 4 - 4 a = 0 = > 4 a = - 3 = > a = —3 /4 (b) u / 3 = i + (4 /3 )j =» a = 4 /3 (c) We need e o s 36 + 9 6 a + 6 4 a 2 = 2 5 (2 a 2 + 2) =*■
1 4 a 2 4- 9 6 a — 14 = 0 => l a 2 4- 4 8 a — 7 = 0 => (7 a — l ) ( a 4- 7) = 0 => a = 1/7 or a = —7. a = 1 /7 => p — 37r / 4 , so only a = 1/7. (d) We need eos p = ■ = - =>• 6 4- 8a = 5 \ / a 2 4-1 => 36 4- 9 6 a 4- 6 4 a 2 = 2 5 a 2 4- 25 => 5v a 2 4 - 1 2
i 25\/3 => = 7r / 3 . (The other gives a with 39a 2 4- 96a 4- 11 = 0 => a = ------ — áy
= ~ ).
18. (a) We need u • v = —2a — 10 = 0 => a = —5 (b) —2u/5 = (4/5)i - 2j => a = 4/5 (c) We need eos p =
/—°/ = -7 “ => Iba 4- 160a 4- 400 = 29a 4- 116 ==> 13a2 — 160a —284 = V 2 9 v a 2 4-4 2 80± 58x/3 . /Q 0 => a = ----- —------ => p = 27r/3 lo
(d) We need eos p = — v 2 9 \/a 2 4- 4
2
from c), we can see that there is no solution.
19. Since the i components of u and v are both positive and fixed, it is impossible for u and v to have opposite directions. 20. Since the j component of u is positive and the j component of v is negative, it is impossible for u and v to have the same direction.
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Instructor's Manual
Chapter 3 Vectors ¡n M2 and M3
In 21-31 we use projv 3
2 1 . projv
23. projv
=
u
u=
3 3 - ( i + j) = - i + - j
u =
0,
asu • v
—1 “ 3J) =
27. projv
« =
24. projv u
= 0
—2
25. projv u =
_5
5
5 -j
44.
11.
22 . projv u = — ( i + j ) = — i -
3 + jljJ
1 1 /,. -x (4i + j )
= —
=
— i + — J
5 10 15 26‘ Pr°ív u = Í 3 (2i + 3-í) = 1 3 *+ 133
+ J)
28- Proív u = a“ ^ ^2 (QÍ + # 0
oí —Q 29. projv u = —2 —
ex. —¡3 30' ProJ v u = ~ 2 — (i + í)
31. projv u = a i ° 2 ~1~ — (nní -)- ¿jj). In order for v and projv u to have the same direction, we need V «2 + b2 a\ü2 4* > 0. 32. In order
for v and projv u to
^
h
have opposite directions, we need ai a2c2 + 2abed 4- b2d2 = a2c2 + a2d2 4- b2c2 + V a 2 -f b2y/c2 -f d 2
35. Let
u
b2d2 => a2d2—2abcd+b2c2 = 0 => (ad—bc)2 = 0 => ad—bc = 0 => c = -ra * Then - u = — i-fdj = ci+dj. b b b d . . Then a = —. Suppose u = ai -f ój and v = aai -f abj for some constant a. Then 6
^
aa2 -f Oíb2 yja2 4 - b2y/a2a 2 + a 262
a(a2 -f ó2) |a |(a 2 + &2)
Thus, the nonzero vectors u and v are parallel if and only if v = a u for some constant a. xi • v 36. Suppose u and v are orthogonal. Then (p = 7t / 2 . Then cos(7t / 2) = j—— - = 0 => u • v = 0. Suppose u •v =
0.
Then eos p = 7r/2 => u and v are orthogonal.
37. Note that (0, —c/b) and (—c/a,0 ) are points on the line. Then the direction vector for the line is u = c e - i — - j. Then u v = c —c = 0 => v is orthogonal to the line ax 4- by -f c = 0 . a b c. c. ab 38. The direction vector of the line is v = - 1 — t J . Then — v = 61 —aj = u => u is parallel to the line a b c ax -f by + c = 0 .
T h e Scalar Product and Projections ¡n M2 Section 3.2
217
39. Let A, B } and C represent the points (1,3), (4 ,-2 ), and (—3,6), respectively. Also, let A , B , and C represent the angles at the corresponding vértices. AB = 3i — 5j; AC = —4i -f 3j; -2 7 -2 7 co^ = V 5 W Ü = ¡ v ! ; BA = - 3 l+ 5 * eos B =
= - 6*.~ . 73842’ 52 _ 52 eos C — 7257113 ~~ 57113'
CA = 4i - 3j; CB = 7i - 8j;
40. Let A, B, C represent the points ( a ,, ¿ i), ( 0 2 , 62 ) , and resent the angles at the corresponding vértices.
( 0 3 , 63)
respectively. Also let A, B, and C rep-
AB = (a 2 - a i)i + ( 62 - 6, )j; AC = (a 3 - oi)i + (63 - 6i)j; (q 2 - ai)(a3 - Qi) + ( h ~ h ) ( h - 61 ) eos A = y/(a2 — a i ) 2 - f ( 6 2 — b\)2yj{as — a i ) 2 + ( 6 3 — b\)2 (ai - a 2)(Q3 ~ ^ 2) A (bi - b2)(b3 - b2) Similarly, eos B = \/( a i - a 2)2 + (61 - b2)2y/(a3 - a 2)2 + (63 - ó2)2 (ai —a 3)(a 2 —a3) -f (b\ —b3)(b2 —63)_____ and eos C = \ / ( a i —a 3)2 -f (61 —63) 2 ^/(a 2 —a 3)2 + (62 —b2)2
u |u v| |u||v|
v
u v |u||v| |u v| |u||v|. u v
u
v.
41. Let = a ii + a2j and = 6ii -f 6^ . • = eos * v) *v
'/, Problem 25.
P = 0 0
» »
u = p =
3];
( u* *
v)/(v*
P = 2.5882 0.6471
» »
u = p =
P = -0.1538 0.2308
» »
[ 1 ; 1] ; v = [ 2 ; 3 ] ; ( u* * vJ/Cv* * v) *v
u = p =
*/, Problem 2 6 .
P = 0.7692 1.1538
2.
(i) » »
u = [2; 1] ; v= [3; 0] ; p = ( u* * v)/(v* * v)
* v
•/, Part (a), the projection computed.
P = 2 0 »
prjtn(u,v)
*/, Part (b)
T h e Scalar Product and Projections ¡n M2
P is Ihe projection of U onto V
P ends at purple o; V ends at yellow *
(ü) » »
u = [2;3] ; v=[-3;0]; p = ( u* * v)/(v> * v)
* v 7, Part (a).
p = 2 0 »
prjtn(u.v)
*/, Part (b). P is Ihe projection of U onto V
P ends at purple o; V ends at yellow *
(iii) » »
u = [2; 1] ; v= [-1; 2] ; p = ( u» * v)/(v * * v)
* v
'/. Part (a).
P = 0 0 »
prjtn(u,v)
'/. Part (b).
MATLAB
3.2
221
222
Instructor’s Manual
Chapter 3 Vectors in R 2 and M3 P is th« projectíon of U onto V
» »
u = [2; 3]; v=[-l;-2]; p = ( u* * v)/(v' * v)
* v
•/• Part (a).
P = 1.6000 3.2000 »
prjtn(u,v)
'/. Part (b) . P is the projectíon of U onto V
(c) The vector
u—pis the component of uwhich is orthogonal to v.
Vectors ¡n Space
Section 3.3
Section 3.3 1. PQ = 7 ( 3 - 3 )2 + ( - 4 - 2)2 + (3 - 5)2 = 2 7 Í0 2. PQ = v"(3 - 3 )2 + ( - 4 + 4)2 + (7 - 9)2 = 2 3. PQ = V ( - 2 - 4) 2 + (1 - l )2 + (3 - 3)2 = 6 4. |v | — 3
v/|v|= j
cosa = 0
5 . |v| = 3
v / | v | = —i
eos a = 4 / 7 Í 7
eos¡3 = 0
cosa = —1
6. |v | = \ / 4 2 + (—l )2 = V Ü
co s7 = 0
cos/? = l
eos 7 = 0
v / |v | = ( 4 /V Í7 )i - ( l / V Í 7 ) j eos 7 = 0
eos/? = —1 / 7 Í 7
7. |v | = \ / l 2 + 22 = y/b v /|v | = ( l / \ / 5 ) i + (2 /\/5 )k c o sa = l/\/5 cos/? = 0 co s7 = 2 / \ / 5 8 . |v| - ^ 1 2 + ( - 1)2 + 12 = V3
v /|v | = (1/V 5)i - (1/V3)j + (1/V 5)k
cos(3 = —1/\/3
c o s o = l/\/3
9. |v| = > /l 2 + l 2 + ( - l )2 = s/5 eos a — 1/ y/3
c o s 7 = 1 /\/3 v /|v | = (1/V 5)i + (1/V3)j - (1/V 5)k
eos ¡ 3 = 1 / y/Z
eos 7 = —1 /\/3
10. |v| = VS v /|v | = —( l/\/3 ) i + (l/\/3 )j + (l/\/3 )k eos a = - l / \ / 3 c o s /? = l/ \/ 3 cos 7 = l /\/3 1 1 . |v| — \/3
cosa = l / \ / 3
cos/?= —1/\/3
cos 7 = —1/\/3
12. |v| — \/3
cosa = —1 /\/3
cos(3=l/y/%
cos^ = —\/y/Z
13. |v| = a/ 3
cosa = —1 /\/3
eos/? = —1/\/3
cos7
14. |v| = v^3
cosa = —1/\/3
eos/? = —1 /\/3
cos 7 = —1/\/3
15. |v| = >/2 2 + 52 + (—7 )2 = y/78 eos a = 2 /7 7 8
v /|v | = (2/>/78)i + (5/V?8)j - (7/>/78)k
eos /? = 5/V78
eos 7 = - 7 /7 7 8
16. |v| = 7 ( - 3 ) 2 + (—3)2 + 82 = ^82 eos a = eos /? = —3 /7 8 2
v / | v | = —(3/V§2)i - (3/\/82)j + (8 /7 8 2 )k
eos 7 = 8 /78 2
17. |v| = 7 ( - 2 ) 2 + (-3)2 + ( - 4 ) 2 = 729 eos a = - 2 / 7 2 9
= 1 /\ /3
eos ¡3 = - 3 /7 2 9
v /|v | = - (2 /7 2 9 )i - (3/V^9)j - (4 /729)k eos 7 = - 4 /7 2 9
18. Let u = a i + a j + ak . As u is a unit vector, we must have 3 a 2 = 1. Since the direction angles are between 0 and 7r / 2 , then a = 1 /73. 19. 12 [(l/7 3 )i + ( l/ 7 3 ) j + (l/7 3 )k ] = 473i + 473j + 4 73k __ 9 7T 9 7T 9 ^" 3 1 1 3 , . 20. eos —+ eos —+ eos - = - + - + - = - ¿ 1 . 21. PQ = (1, - 3 ,4 ) 22 . PQ = (11,0,0)
| PQ | = 726 u
PQ / | PQ | = (1/756, - 3 /7 5 6 ,4 /7 2 6 )
= (-1 ,0 ,0 )
23. Let R = (a, b, c). Then PR = (a + 3, b — 1, c — 7). We want PR • PQ = (a -f 3) = 0. It follows that b and c are arbitrary and a = —3. Henee, all points of the form (—3, 6, c) satisfy the condition. 24. | PR | = 1 implies (6 — l )2 -f (c —7)2 = 1, the equation of a circle.
223
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Instructor’s Manual
Chapter 3 Vectors in M2 and M3
2, u
v
u• v < |u||v|. If u= u v |u||v|. Then
25. By theorem if and are two nonzero vectors then Henee, for all vectors and we have • <
|u||v|.
u
v,
0 or
v=
0, then
u• v =
|u+ v|2 = (u+ v) •(u+v) = |u|2 + 2 u•v+ |v|2 < lu |2 +2 |u||v| + |v|2 = (|u| + |v|)2 Taking square roots, we obtain
|u+ v| 0. We conclude only if = or = for some > 0. Thus nonnegative scalar múltiple of the other.
u v |u||v| v au u av v au
27. 29. 31. 33.
a
- 6j + 9k 8i - 14j + 9k 16i + 29j + 42k y / l 2 + (-7)2 + 3 2 = Vb9
35. 3 a 2 - 1 9 2 a + 368 = 0 =>a = 96 ± 52^ 3
._ ^
14 39. projv u = — (i + j ) = 7i + 7j 5
15
14 40. projv u = y ( i ~ J) = 7 i - 7j 10
—3
—3
Q
41. projv u = - ( 3 ¡ + 2j) = - i + I j j
42. p,ojv u = — (i - 3j) = — i + - j
oq _q 7 43. projv u = - ( - 3 i - 7 j ) = T i - - j
_7 oí 7 44. proj v u = — (-31 - j) = - 1 + - j
—► 45. PQ
=
i + 9j; RS
=
—► — ► —33________________33 3i - 4j; p ro j-q RS = — (i + 9j) = — i -
297 — j;
■ ^ - 3 3 /n. , . x - 9 9 . 132. proJñ PQ = “ T * 3’ " ^ = T ' + T J 46. V ( 4 T 5 p T H ^ ? l l ^ ^ = VIÓT 47. y / ( - 2 - O) 2 + (4 - O) 2 + ( - 8 - 6)2 = y/216 = 6y/G 48. y/(2 - O) 2 + ( - 7 - 5)2 + (0 + 8)2 = y/2Í2 = 2v/53
n
3
49. |v| = V^130; cosa = 0; eos/? = —j = \ eos 7 = Vm ' 1 V m 50. |v| = x/Í4; cosa = ■
1
; eos/? =
—2
—3
y / Ü ' ^ H ~ y / Ü ’ C° S 1 ~ y / Ü
51.
|v| = \/53; cosa = — eos/? -- ■-^ = ; eos 7 = y/ 5 3 ’ ^ ~ y/ 5 3 ’ ' y/53
52. PQ = - 7 i + 2j + 5k; | PQ | = x/78 -7 . 2 . 5 , u = v ?8 , - 7 ? S j ' W 53.
PQ
= - 8i + 4j- 4k; | PQ |= v/96 = 4\/6 2
.
U~ ^ 1
1 .
1 ,
V 6J + V 6
54. u —v = 4i —4j —2k 55. 3v + 5w = (-91 + 6j + 15k) + (lOi - 20j + 5k) = i - 14j + 20k —9 27 9 45 56. projv » = j a M ¡ + 2j + 5k) = - jjj - jjk 13. „. , , 26. 52. 13, 57. proJw u = - ( 2 . - 4j + k) = - 1 - - j + - k 58. 2u —4v + 7w = (2i - 4j + 6k) - (—121 + 8j + 20k) + (141 - 28j + 7k) = 28i - 20j - 7k 59. u v - w v = 13 - ( -9 ) = 22
241
242
Chapter 3 Vectors ¡n IR2 and R 3
60. eos ¡p = 61. eos ■
3 I 05 0
5
3 j 05
t — — -— ; —-— = —2 —4s => s = —10/7, and —2 —4s = — - — => s = —17/22; Thus there is no
point of intersection.
Review Exerdses for Chapter 3
73. The parametric equation of the line: x = 3 + 2í, y = 1 —t, z = 5 + í. Then 2(3 + 2t) —(1 —í) + (5 + t) 0 = * 6 í + 1 0 - 0 = M = 5/3, * = 19/3, y = - 2 /3 , z = 20/3; d =
361 4 400 _ ^ T + 9 + ~9~-yM
74. vx = 4i 4- 3j - 2k; v 2 = 5i + j + 4k. ij k 4 3 - 2 = 14i —26j —l l k 5 1 4 The line is: x = —1+ 14í, y = 2 — 26t, z = 4 —l l í .
V = Vi X V2 =
75. l(x - 1) + 0(y - 3)+ l(z + 2) = 0 =¡> x + z = - 1 76. 0(x - 1) + 2(y + 4)
- 3(z - 6 ) = 0
2y - Zz = -2 6
77. 2(x + 4) - 3(y - 1) + 5(z - 6) = 0 => 2x - 3y + 5z = 19 78. P = ( - 2 ,4,1), Q = (3, -7 ,5 ), R = ( - 1 , - 2 , -1 ); PQ = 5i — l l j + 4k; QR = - 4 i + 5j - 6k; i j k 5 -1 1 4 = 46i + 14j + 69k -4 5 -6 46(® + 2) + 14(«/ ~ 4) + 69(z - 1) = 0 => 46x + 14y + 69z = 33 n = PQ x QR =
79.
1
3\
1 -1
\ - 4 2 -7
5J
0 -2 -11
(-1 1
9. x = 1- — 2
80. n i x n 2 =
81.
1/ 2 \ 7/2 )
1 1 *y =- — —i,z — t
- t7,
2
/ I 0 9/2 V° 1 11/2
-1
2
2
’
1 j k - 4 6 8 = Oi + Oj + Ok =>■ no points of intersection 2 -3 -4
3 -1 4 8\ ( 1 -1/3 4/3 -3 -1 -11 0 / (0 -2 -7 4 15 7 x = - — —t, y = —4 — - t , z — t 3 6 2
8 /3 l =*
( 01
0 15/6 1 7/2
4 /3 ' ■2)
82. (3,0,0) is a point in the plañe. Let p = (1 - 3)i + ( - 2 - 0)j + (3 - 0)k = - 2 i - 2j + 3k, n = 2i - j - k —5. . -1 0 . 5. 5, proJnP = -g - (2 i-J -k ) = — :i + - j + g k Í50 _ _5_
d = I Pr°jn p | = 83.
n i
= - i + j + k;
COS
(p —
6 n 2
i
j
1 -2
3
= —4i + 2j —7k
- 1 , ; ^ = co8- ( ^ L ) , which is approximately 94®. 3 \/2 3 ’
\/3\/69 84. n = u x v =
\/6
k 1 = 4i + 6j + 8k
2 -3
Plañe: 4(x — 1) + 6(y + 2) + 8(z —1) = 0 ^ 4x + 6y + 8z = 0 Then 4(9) + 6 (—2) + 8(—3) = 36 —12 —24 = 0. Thus u, v, and w are coplanar.
244
Notes
Instructors
Definition and Basic Properties
Section 4.2
C h ap ter 4. Vector Spaces S ectio n 4.2 1 . yes; (i) the sum of two diagonal matrices is diagonal; the rest of the axioms follow from theorem 1.5.1. 2 . no; (iv) not every diagonal m atrix has a multiplicative inverse.
3. no: (iv) if (x, y) £ V, y < 0 then (—a?, —y) g ' V since —y > 0; (vi) does not hold if a < 0 and y < 0. 4. no; (iv) if (x,y) is strictly in the first quadrant, then (—x ,—y) is in the third quadrant; (vi) does not hold if a < 0 . 5. yes; (i) (x, x, x) + (y, y, y) = (x + y, x + í/, x + y) G V; (iii) (0, 0, 0) £ V; (iv) if (x, x, x) £ V , then (—x, —x, —x) £ K; (vi) a(x, x, x) = (ax, ax, ax) £ V*; the rest of the axioms follow from theorem 1.5.1. 6 . no; (i) x 4 —x 4 = 0 ; (iii) 0 £ V
7. yes; the axioms follow from theorems 1.9.1 and 1.5.1.
8- yes; W ( f e o )
( / ? o ) = (& + / ? a + o ) ;(üÍ) ( o o ) e V '’ a (°bo) = { a b ao ) e V ’ the restof
the axioms follow from theorem 1.5.1. 9' no; «
Q
l) + ( l í )
= (/? + 6 a +
2)
^ F;(ü i) (
00)
^ F;(Ív)- ( ¿ “ ) = ( - ^ - l )
* V’
( Y Í ) a ( l ai ) = ( a b Z ) < ¿ V Í Í a ¿ 1' 10 . yes; it is a trivial vector space. 1 1 . yes; (i) the sum of two polynomials with a zero constant term will have a zero constant term; (iii) 0 £
V] (iv) if p{x) £ Vy then —p{x) £ V; (vi) ap(x) has a zero constant term for every scalar a; the rest of the axioms follow from the usual rules of addition and scalar multiplication of polynomials. 1 2 . no; (iii) 0 ^ V\ (iv) if p(x) £ V, then —p(x) £ V since it does not have a positive constant term; (vi) does not hold if a < 0 .
13. yes; (i) if / £ V and g £ Vy then / + g is continuous and /(0 ) + y(0) = /(1 ) + g( 1) = 0; (iii) 0 £ V; (iv) if f E Vy then —/ is continuous and ( ~ /) ( 0) = (—/ ) ( 1 ) = 0 ; the rest of the axioms follow from the usual rules of adding functions and multiplying them by real numbers. 14. yes; (i) ¿i(a, 6 , c) + t 2(a, 6, c) = (*1 + *2)(a, 6, c) £ V; (iii) (0, 0,0) £ K; (iv) if (a, /?, 7 ) = t(a, b, c) £ V, then ( - a , —/?, - 7 ) = (—t)(a, 6, c) £ V; (vi) a[t(a, 6, c)] = (at)(a, b,c) £ V for every a £ M; the rest of the axioms follow from section 1.5. 15. no; (i) forexample, ( 1 , 0 , —1 ) + (2 , 2 , 0) = (3 , 2 , - 1 ) is not on the line; (iii) (0, 0, 0) 0? V\ (iv) (—1 ,0,1) is not on the line; (vi)( 2 , 0 , —2 ) is not on the line. 16. no; (vii) if a / 1 , then a (x + y) = a((x i, x2) + (yi, y2)) = (axi + ayi + a, a x 2+ ay 2 + oc) ± a x + a y = (axi + ayi, a x 2 + a y 2 + 1); (viii)
(a + P)x = ((a + P)xx, (a + P)x2) / a x + /? x = ((a + p)x 1 + 1, (a + P)x2 4- 1 )
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Instructor^ Manual
17. yes; (iii) the zero vector is ( - 1 ,- 1 ) ; (iv) if x = (#>y) € V, then the additive inverse of x is (—x —2 , —y —2 ); the rest of the axioms follow, after some careful algebra. 18. yes; it is a trivial vector space. 19. yes; (i) the sum of two differentiable functions defined on [0,1] is differentiable on [0,1]; (vi) if / is differentiable on [0, 1 ], then a f is differentiable on [0, 1 ] for every scalar a; the rest of the axioms follow from the usual rules of adding functions and multiplying them by real numbers. 20 . yes, providing we understand that scalar now means rational number; (i) (a 4- by/ 2 ) 4- (c+ dy/2) = (a -f
c) + (b + d)y/2 £ V since the sum of two rational numbers is rational; (vi) a(a + by/2) = a a -f aby/2 £ V since the product of two rational numbers is rational; the rest of the axioms follow as special cases of the rules for addition and multiplication for rational numbers. 21. Suppose 0 and 0 ' are both additive identities. Then 0 = 0 4 - 0' and 0' = 0 4 - 0'. Thus, 0 = 0 '. 22 . Suppose y
4- ( x +
x y )
4- y = 0 and
=
y
4-
( x
x - f z = 0 . So x 4 y = x + z . Adding y to both sides of the equation gives 4- z ) . Using properties (ii) and (v), we obtain y 4- 0 = 0 4- z . Thus, y = z .
23. Define z = ( — x ) 4- y. By properties (i) and (iv), z exists. Adding x to both sides, we obtain x 4- z = x 4- ( ( — x ) 4- y) = ( x —x ) 4- y = 0 4- y = y . Suppose z and z ' are such that x 4- z = y and x + z ' = y . Then adding ( — x ) to both yields z = ( — x ) 4- y = z ' . 24. (i) if x > 0 and y > 0, then x 4- y = xy > 0; (ii) (x 4- y) 4- z = xy 4- z = xyz = x 4- yz = x 4- (y 4- z)\ (iii) x 4*l = x T = x = l 4-® = l , x; (iv) x + x _1 = x • x - 1 = 1 ; (v) x 4- y = xy = yx = y 4- x; (vi) if x > 0 , then ax = x a > 0 for any a; (vii) a(x 4- y) = otxy = (xy)a = x aya = xa 4- ya = a x 4-ay; (viii) (a 4- 0)x = x^a+^ = x axf3 = xQ 4 ^ = ax 4 Px\ (ix) a(/?x) = (/3x)a —(x aY — = (a/3)x; (x) lx = x 1 = x 25. (i) Suppose yi and y2 are Solutions. Then
(j/i + yi)" + a(x)(yi + yz)' + Kx)(yi + yi) -- y" + a(x)y[ + b{x)yx + y'¿ + a(x)y'2 + b(x)y2 = 0 + 0 =
0.
Thus yi 4- y2 is a solution. Similarly, (aj/i)" + a(x)(ayi)' + b(x)(ayx) = a(y" + a(x)y[ + b(x)yx) = 0 . Henee, we have closure under addition and scalar multiplication. The additive inverse of yi is (—1 ) 2/1 = —yi. The rest of the axioms fpllow from the usual rules for addition and scalar multiplication of func tions.
Definition and Basic Properties
M A T L A B 4.2
M ATLAB 4.2 1. This problem is a demo from vetrsp.m. 2. (a) The zero vector will be the m atrix with all zero elements. */, Choose valúes for m and » n = 3; m = 4; » X = round(10*(2*rand(n,m)-l)) '/. Some random matrices. X = 7 -5 -1 -7 -2 -2 -4 1 7 1 -6 6 » Y = round(10*(2*rand(n,m)-l)) Y = -9 9 8 -7 -6 1 5 2 0 1 7 4 » Z = round(10*(2*rand(n,m)-l)) Z = -7 -10 4 -6 -8 -2 9 -4 -S -9 -5 8 n d A A
*/• A random scalar.
2*rand(l)-l
a = 0.3041 » b = 2*rand(l)-2 b = -1.6993 » X+Y ans = -2 -1 7
'/, (i) This should be an nxm matrix. 4 3 2
7 -2 1
-14 -5 10
» (X+Y)+Z > X+(Y+Z) ans = -9 -6 -20 11 -9 1 7 -9 2 -7 -4 18 ans = -9 -6 -20 11 -9 1 7 -9 2 -7 -4 18 » X+zeros(3,,4), zeros(3. ans = 7 -5 -1 -7 -2 -2 -4 1 7 1 -6 6 ans = 7 -5 -1 -7 -2 -2 -4 1 7 1 -6 6
'!% (ii) These should be the same.
'/, Note zeros(n, m) is the additive identity.
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Instructor’s Manual
Chapter 4 Vector Spaces » M = -X M = -7 5 2 2 -7 -1 » X + M ans = 0 0 0
1 4 6
7 -1 -6 */, This should be zero.
0 0 0
» X+Y, Y+X ans = -2 4 -1 3 7 2 ans = -2 4 -1 3 7 2 » a*X ans = 2.1288 -0.6082 2.1288
*/, (iv) This should be an nxra matrix
0 0 0
0 0 0 */# (v) These should be the same.
7 -2 1
-14 -5 10
7 -2 1
-14 -5 10 •/* (vi) This should be an nxm matrix
-1.5206 -0.6082 0.3041
-0.3041 -1.2165 -1.8247
-2.1288 0.3041 1.8247 '/, (vii) These should be the same.
» a*(X+Y), a*X + a*Y ans = 1.2165 -0.6082 -0.3041 0.9124 2.1288 0.6082
2.1288 -0.6082 0.3041
-4.2576 -1.5206 3.0412
ans = -0.6082 -0.3041 2.1288
2.1288 -0.6082 0.3041
-4.2576 -1.5206 3.0412
1.2165 0.9124 0.6082
» (a+b)*X, a*X + b*X ans = -9.7665 6.9761 2.7904 2.7904 -9.7665 -1.3952 ans = -9.7665 6.9761 2.7904 2.7904 -1.3952 -9.7665 » a*(b*X), (a*b)*X ans = 2.5840 -3.6176 1.0336 1.0336 -0.5168 -3.6176 ans = 2.5840 -3.6176 1.0336 1.0336 -3.6176 -0.5168
'/, (viii) These should be the same. 1.3952 5.5809 8.3713
9.7665 -1.3952 -8.3713
1.3952 5.5809 8.3713
9.7665 -1.3952 -8.3713 */, (ix) These should be the same.
0.5168 2.0672 3.1008
3.6176 -0.5168 -3.1008
0.5168 2.0672 3.1008
3.6176 -0.5168 -3.1008
Definition and Basic Properties » 1*X ans = 7 -2 7
M A T L A B 4.2
% (x) This should be X. -5 -2 1
-1 -4 - 6
-7 1 6
(b) To prove that Mnm is a vector space, we must check all ten of the vector space axioms. Let X = (xij)} Y = (yij), and Z = (z¿j) be any m x n matrices and let a and ¡3 be any scalars. From the defmition of m atrix addition in Section 1.5, we know (i) X 4- Y is in M nm, and that (ii) (X -f Y) + Z = X + (Y -f Z). (iii) The zero vector will be the m x n m atrix with zero entries, so that 1 + 0 = 0 + X = X . (iv) The negative of X is the matrix whose entries are the negatives of those in X . (v) The entries of X + Y are X{j -f Vij = Vij 4* %ij which are those of Y + X . (vi) The scalar multiplication aX , was defined as the m x n matrix whose entries are axij. (vii) The entries of a ( X + Y ) are a(x{j + yij) = a x ^ + ayij, which are the entries of a X -f aY . (viii) The entries of (a + P)X are (a + P)x%j = otXij 4* /?#»/> which are the entries of a X + PX. (ix) The entries of a((3X) are a(/3xij) = (a/3)xij which are the entries of (a/3)X. (x) The entries of I X are 1Xij = which are those of X . (Section 1.5, Problems 41-43 prove some of these.) (c) Part (a) gives evidence that Mnm is probably a vector space, but it does not prove that it is a vector space, since most assertions involve all X , Y, Z , a, ¡3 and (a) only gave some examples.
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Chapter 4 Vector Spaces
S e c tio n 4.3 For each problem in which H is a subspace you should explain why Theorem 1 holds. 1. H is not a subspace. For a < 0, a(x, y) = (ax, ay) H , since ay < 0 for y > 0. 2. H is a subspace 3. H is a subspace. 4. H is not a subspace. (1,0) ^ H , but 2(1, 0) = (2,0) ^ H. 6 . H is a subspace. 5. H is a subspace. 8 . H is a subspace. 7. H is a subspace. 9. H is a subspace. 10. H is not a subspace. 11. 13. 15. 17. 19. 21
H H H H H
fa l + a\ \0 0J
fb l + 6 0 \0
o. -f- b 2 + a + 6 12. 14. 16. 18. 20.
is a subspace. is a subspace. is not a subspace. H does not contain 0. is not a subspace. H does not contain 0. is a subspace.
( “ “ ) + ( “ / ) = ( t + °c ° + / ) e 'Ib ) + (~c of V.
0 aa 0
aa
d)
=
H is H is H is H is H is
“ (° °) =
¿H.
0
0
not a subspace. H does not contain 0. a subspace. a subspace. a subspace. not a subspace. H does not contain 0. («6
(~(bT+)c ¡ t d ) eHl’ a i ~ l Í )
Te) e H' S° H' is a subsp,ce °! V
= ( ~ a a T b ) e f f 2 S o J Í 2 ‘s a »“ bsPa“
G Hi D H 2. So H\ fl H 2 is a subspace of V.
22. Since every polynomial has a continuous first derivative, H\ fl H 2 = H\. As shown in example 10, H\ is a subspace. 23. Suppose x £ H and y G H . Then A( x + y) = A x -f A y = 0 + 0 = 0 . Thus, x + y a • Ax = a • 0 = 0 . Thus a x G H for every scalar a. Then H is a subspace of Mm.
EH. A(ax)
=
24. H is not a subspace since H does not contain 0 . 25. Note that ( a , 6,c, d) £ H since a2 + b2 + c2 + d2 > 0. So H is a proper subset of M4. (si,yi> *i, wi) £ H an(1 (* 2 , 2/2, z2) w2) G H. Then (xx + x 2, 2/i + 2/2 , ^1 + z2, w x + ^ 2) a (xx -f x2) + 6( 2/1 -f- 2/2) 4- c(z\ + z2) + d(wi -f ^ 2) = 0 + 0 = 0. Also, a (x x, yx, 21 , wx) a (a x i) + 6(ayx) -f c(azx) -f d(au;x) = a - 0 = 0. Thus H is a proper subspace of M4. (Or 23 with A — (a b c d).)
Suppose G # since G H since use Problem
26. Note that (ax, a2, . . . , an) ^ H since af -f+ ••• + > 0 . So H is a proper subset of Mn. Given ( x i,x 2, .. .,x n) and (yx, 2/2, •. •, Vn) G ff then (xx, x2, . . . , xn) + (yx, y2, • • •, 2/n) G -ff since i € K}, which is a line passing through the origin. 20. Since vi x V2 is perpendicular to v i, V2, henee to the plañe spanned by v i, V2, vi x V2 • x = 0 is the equation of a plañe, which contains aiVi 4 0,2 V2- Expanding the cross product shows {yi z 2 — z\y 2 )x 4- (^ 1 X2 —x \ z 2)y 4 (x i ?/2 —%2 yi)z = 0, is an equation of a plañe passing through the origin, which contains H = span {vi, V2}. Since v i, V2 are not parallel vi x V2 / 0 so this equation is the equation of a plañe (and not just 0 = 0). 21. Let v E V. Then there are scalars a i, a 2, .. . , a n such that v = aiV i 4 ( q —l ) ’ ( l o ) ’ anc^ ( —1 o ) ' ^ ote ^ at eac^ m atr*x *s invertible.
(:i) - i [{i í) * (¡ -D] -1[(¡ -í) - (::)] +
23. Since each v¿ G span {ui, U2, .. u„}, then span {vi, v2, . . . , vn} C s p a n { u i,u 2, .. .,u „} . Let A = / Ul \ / vi \ U2 v2 . As j4w = z and det A ^ 0, we have w = A 1z. Let , and z = (üij), w = \U „ /
\v j
A 1 = B = (b¡j). For each u*, u* = ^ 6¿*v¿ G span {vi, v 2). . . , v„}. So span { u i,u 2, .. ., u n} C •=i span { v i, v 2>.. .,v „} . Henee, sp a n { u i,u 2>.. .,u „} = sp a n { v !,v 2, .. .,v „} .
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Instructor's Manual
Chapter 4 Vector Spaces
M A T L A B 4.4 1. (a) See answers in Section 3.1. (b) This problem should be worked interactively. The following commands will produce the graph below. »
u l = [1; 2] ; u2 = [ -2 ;3 ] ; u3 = [5 ;4 ]; a= -2 ; b=2; c= -1 ;
»
combo(ul,u2,u3,a,b,c)
2. (a) (i) We wish to solve ci ^ 2 ) ^ C2 (
3)
~ ( l) ’
can
wr^tten 35 ^ 2cj
+ 3c2 ) ~
3 This system of equations has the augmented matrix 1 -1 3\
2
3 1/
which is the same as [u v | w]. (ii) We wish to solve ci ^ 4 ^ + c2 ^ 2 ) ~ (
6) ’
can
2ci + ( - l ) c 2 wr^tten as ^ 4 c
g ^ . This system of equations has the augmented matrix
2 -1 -1
4
2
6
which is the same as [u v | w]. (iii) We wish to solve c\
=
w^ich can be written as
). This system of equations has the augmented matrix ( 12 8/3 \ V -l 1 -5 /3 /
Linear Combination and Span
M A T L A B 4 .4
which is the same as [u v | w]. (b) This generates iteractive graphics. »
»
u = [1 ;2] ; v = [-1 ; 3 ]; w = [3; 1] ; lincom b(u,v,w )
Two other pictures should be generated with sets (ii) and (iii). 3. (a) (i) We wish to solve C i(j)+C 2(
} )+ c 3 (o) = ( _ 4)
which can be written as
which has the augmented m atrix [viV2V3 |w]. This system may be solved using MATLAB: » A = [ 1 - 1 3 1; 1 1 0 -4]; » rref(A) ans = 1.0000 0 1.5000 0 1.0000 -1.5000
*/, Enter the augmented matrix.
-1.5000 -2.5000
The solution has C3 arbitrary, and c\ — —1.5 —1 .5 c3 and c2 = —2.5 + I . 5 C3. (ii) Similarly » A = [ 1 -2 5 -4; 2 3 4 -1]; » rref(A) ans = 1.0000 0 3.2857 0 1.0000 -0.8571
-2.0000 1.0000
The solution has C3 arbitrary, and c\ = —2 —3.2857C3 and c2 = 1 + .8571c3. (b) (i) If c3 = 0 in system (a.i), then c\ = —1.5 and c2 = —2.5, so that w = —1.5vi —2.5v2. If c3 = 0 in system (a.ii), then ci = —2 and c2 = —1 , so that w = —2v i + v 2. (ii) If c2 = 0 in system (a.i), then C3 = 2.5/1.5 = 1.6667, and c\ = —4, so that w = —4vj + 1 .6667 V3 . In system (a.ii), C3 = —1/.8571 = —1.1667, and c\ = 1.8334, so that w = 1.8334vi - 1.1667v3.
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Chapter 4 Vector Spaces
(iii) If c\ — 0, in system (a.i), then c3 = 1.5/(—1.5) = —1, and C2 = —4, so that w = —4 v 2 —V3 . In system (a.ii), c3 = 2 /(—3.2857) = —0.6987, and c2 = .4783, so that w = .4783v2 —0.6987v3. (c) Here are the four plots produced by com bine2(vl, v2,v3,w) for the data in (a)(ii) above: (note that v i is labelled u i in these plots). »
» »
vl=[l 2]*; v2= [-2 3] 9;v3=[5 4 ] , ;w=[-4 -1] ';
com bine2(vl,v2,v3,w ) p r i n t -deps fig 4 4 3 c .e p s
w=
w=
4.
11/23
(a) The equation w =
ciVi
* u2 +
-f
C2V2
-14/23
+
c 3v 3
*u3
w=
-2
* u1 +
1
11/6
* u1 +
-7/6
*u2
*u3
is a system with augmented matrix. /4 2
7 3 3\ 1-2-3
\9 -8
4 25/
and variables C2, c3. The definition of linear combination says the vector w is a linear combination of the v ’s exactly when the equation (henee the system) has a solution.
Linear Combination and Span
M A T L A B 4.4
(b) (i) » A = [4733; » rref(A) ans = 1 0 0
0 1 0
2 1 -2 -3; 9 -8 4 25];
0 1
0
1 1 2
» c = a n s ( :,4 ) c = 1 -1 2
(ü) » A = [4 7 3 3; 2 0 - 2 - 3 ; 9 13 4 25] ; » rref(A) ans = 1 0 0
»
0
-
1
1 0
0 0
1 0
1
'/• No solution exists
(iii) » A = [8 5 10 10.5; » rref(A) ans = 1 0 0 0
»
0
5 -3 -3 2;
0 1 0 0
-9 5 10 3.5];
0 0
0 1 0
-5 3 -5 -14;
0 1
'/, No solution exists.
(iv) » A = [8 5 10 1; » rref(A) ans = 1 0 0 0
»
0
5 -3 -3 1 ;
0 1 0 0
-9 5 10 1 ] ;
0 0
0 1 0
-5 3 -5 1 ;
0 1
*/. No solution.
(v) » A = [4 3 5 -3 -19; 5 8 2 -7 -9; 3 -5 11 0 -46; -9 -1 -17 8 74]; » rref(A) ans = 1 0 2 0 -7 0 1 - 1 0 5 0 0 0 1 2 0
0
0
0
0
259
260
Instructoras Manual
Chapter 4 Vector Spaces » c = [-7; 5; 0; 2] c = -7 5
*/• Pick c3 = 0.
0 2
(vi) 21 -2 1 ; 9 - 8 4 1 ] ;
» A = [4731; » rref(A) ans = 1. 0000 0 0
0 1. 0000 0
0 0 1 .0 0 00
0.2656 0.0754 -0.1967
» A = [1 -1 1 3; 2 0 -1 2]; » rref(A) ans = 1.0000 0 -0.5000 0 1.0000 -1.5000
1.0000 -2.0000
» c = ans(:,4) c = 0.2656 0.0754 -0.1967
(vii)
»
c = [1; -2; 0]
*/• Pick c3 = 0.
c = 1 -2 0
(c) Since there was a solution to the system, w was in the span of the v ’s in parts (i), (v), (vi) and (vii). Since there was no solution to the system, w was not in the span of the v ’s in parts (ii), (iii) and (iv). In each case where a solution existesd, w - c ( l) * v l- c ( 2 )*v 2- . . . will give 0 up to roundoff error. 5.
(a) (i)
» A = [ 4 7 3; 2 1 -2; 9 - 8 4] ; » rref(A) ans = 1 0 0
0
0 1 0
0 1
(ü) » A = [ 9 5 - 1 0 3; - 9 7 4 5; 5 - 7 7 5 ] ; » rref(A) ans = 0 0 1. 03 38 1. 0000 1 .0 0 00 0 1 .3092 0 0 1. 00 0 0 1. 28 5 0 0
Linear Combination and Span
M A T L A B 4.4
Since in both cases, the row echelon form of A has no zero rows, any system of the form [A w] will have a solution. Since Solutions of this system tell us how to write w as a linear combination of the vectors {vi, v 2, . . . , v*}, if the system has a solution, then w will be in the span of this set. Since this system has a solution for any w in l n, the set will span all of M3. (b) (0 » »
A = [10 9 -4 ; 0 -9 rre f(A )
; -5 0 1;
- 8
-2 -1 ];
o 0
0 1 0 0
1 0 0 0
8
1
o
(Ü) » »
A = [ 4 3 5 3; 5 rre f(A ) 2 -1 0 0
1 0 0 0
8
2 -7 ; 3 -5 11 0; >9 -1 -17
8
];
0 0 1 0
(íü ) » A = [9 5 14 -4 ; -9 7 -2 16; 5 7 12 2]; » rre f(A ) ans = 1 0 0
0 1 0
1
1 - 1 1 0 0
In each of these systems, the row echelon form of A has a zero row, so it is possible to pick a w so that there will be no solution. Since the system cannot be solved, w cannot be written as a linear combination of this set. This means that w is not in the span, so the span is not By experimentation, one can find such a w, for example: for sets (i) and (ii), w = is not in the span and for set (iii) w ==
6 . See the solution for problem 2 in Section 1.8. The matrices in (i), (iv), and (v) were invertible, and
when they were reduced to row echelon form, they had no zero rows. The matrices in (ii), (iii), and (vi) were not invertible, and had zero rows in their row echelon form. If the row echelon form of A has no zero rows, then the row echelon form of [A w] will always be consistent, and so will always have a solution. This means that the columns of A will span all of Mn. A square m atrix is invertible if and only if its columns span Mn.
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Instructor's Manual
Chapter 4 Vector Spaces
7. (a) To solve this problem, we enter the matrix of v¿’s and reduce it to row echelon form.
» A = [ 3 -2 7 14 1; -7 0 2 -5 -5; 4 -7 9 27 0; -2 2 1 -5 -1]; » rref(A) ans = 1 0 0 1 0 0 1 0 2 0 0 0 1 1 0 0 0 0 0 1
Since there are no zero rows, the system [A w] will always havea solution for any w. This means the set will span all of M4. SinceC4 may bechosen arbitrarily, there will be alwaysan infinite number of Solutions. (b) For the first w: (i) » w = [23; -15; » rref([ A w]) ans = 1 0 0 1 0 0 0 0 0
33; -5];
0 2 1
0 0 0 0 1
2
1
1 0
1 2
1
The solution has C4 arbitrary as no pivot in column 4 and c\ = 2 —C4 , C2 = —1 + 2c4, C3 = 2 —C4, and C5 = 1 . (ii) W ith c4 = 0, w = 2vi —lv 2 + 2v 3 + l v 5. (iii) To verify this, recall that v¿ is the same as A (: , i ) . » 2*A(:,1) - 1*A(:,2) + 2*A(:,3) + 1*A(:,5) */, This should be w. ans = 23 -15 33 -5
For the second w: (0
» »
w = [-13; 18; -45; 18]; rref([ A w])
ans = 1 0
0
0 1
0
1 2
-
0
0
1
0
0
0
1
0
0
0
0
3 6 1
1
1
The solution is C4 is arbitrary, and ci = —3 —c4, c2 = 6 + 2 c4, C3 = 1 —C4 , and c5 = 1. (ii) W ith C4 = 0, w = —3vj + 6V2 + IV3 + IV5 .
Linear Combination and Span
M A T L A B 4.4
(iii) » -3*A(:,1) + 6*A(:,2) + 1*A(:,3) + 1*A(:,5) */. This should be w. ans = -13 18 -45 18
(c) The fourth vector was not needed, because we could always choose C4 to be zero. This can be recognized by the fact that the fourth column had no pivot. (d) The m atrix formed from the new set will be: » »
B = [ 3 -2 7 1; -7 0 2 -5; 4 -7 9 0; -2 2 1 -1] ; rref(B)
ans = 1 0 0 0
0
0 1 0 0 1 0 0
0 0 0 1
Sincethis has no rows of zeros, any system of the form [B w] will havea solution. Since there are no columns without a pivot,this solution will be unique. This corresponds to the statement that any vector w will be in the span of columns of the new m atrix and that the coefficients for the linear combination will be unique. (e) First enter the m atrix of vectors: » »
A = [ 10 0 -10 -6 32; 8 2 - 4 - 7 32; -5 7 19 1 -5]; rref(A)
ans = 1 0
0
1
0
1 2
0
0 0
0
2 1
1
-2
As above, for any w in M3, the system [A w] will have a solution, and the coefficients C3 and C5 may be chosen arbitrarily. For the first w in (b) (i)
» »
w = [26; 31; rref( [A w] )
17];
ans = 1 0
0
0 1
0
2
1
0
0 0
1
2
2
1 -2
4 -1
The solution has C3 and C5 arbitrary and c\ = 2 +IC 3 —2cs c2 = 4—2c$—les and C4 = —l-f2cs. (ii) W ith C3 and C5 chosen to be zero, w = 2vi + 4 v 2 — IV4. To verify this: (íü ) » 2*A(:,1) +4*A(:,2) -1*A(:,4) '/. This should be w. ans = 26 31 17
263
264
Instructor^ Manual
Chapter 4 Vector Spaces
For the second w in (b)
» w = [2; 20; 52]; » rref( [A w] ) ans = 1 0
0 1
0
0
-
1
0
2
0
2 - 1 0
1
1
-2
7 -2
The solution has c3 and c5 arbitrary and c\ = —1 + IC3 —2cs C2 = 7 — 2c3 — les and C4 = — 2 + 2 c5 .
(ii) W ith C3 and C5 chosen to be zero, w = —liq + 7 i >2 —2 ^4 . To verify this: (ni) » -1*A(:, 1) +7*A(:,2) -2*A(:,4) ans =
7. This should be w.
2 20
52
For (c): the third and fifth vectors are not needed. We may span all of M3 with the first, second and fourth vectors. To check this, enter the matrix formed by the new set: » B = [10 0 -6; 8 2 -7; -5 7 1]; » rref(B) ans = 1 0 0
0
0 1 0
0 1
As in (d) above, there are no rows of zeros, and every column has a pivot, so the system [B w] will always have a unique solution.
(a) »
A = [ 20; 10; 20; 10; 0];
7. Mix A has 20 parts cement, 10 parts water,
•/. 20 parts sand, 10 parts gravel, and •/. 0 parts fly ash. » B = [ 18; 10; 25; 5; 2]; 7. Mix B. » C = [ 12; 10; 15; 15; 8]; 7. Mix C. » v = [ 1000; 200; 1000; 500; 300]; 7. The custom mix that we want, in grams. » rref([ A B C v]) % Solve c_l A + c_2 B + c_3 C = v. ans = 1 0 0 0 0
0
0 1 0 0 0
0 0
0 1 0 0
0 1 0
Since the fourth row has zeros on the left and a one on the right, there is no solution. Henee, this mix cannot be made. (b) Let v be the vector of components in our custom mix. That the total weight is 5000 means
Linear Combination and Span
M A T L A B 4 .4
265
The amount of cement is v\ = 1250. The ratio of water to cement is 2 to 3, or V2 ¡v\ = 2/3, which can be written as V2 = |^ i = 833.33. The amount of sand and gravel tells us v3 = 1500, and V4 = 1500. Plugging these valúes into the equation for the total weight, and solving for 1*5 , the fly ash, we get: 1250 + 833.333 + 1500 + 1500 + t;5 = 5000, whose solution is v$ = —83.3333. Sinceit this cannot be made as a custom blend.
is
impossible to have a negative amount of something,
9. (a) The vector u represents the polynomial whose constant term is —5, linear term is 3, x 2 term is 0 and whose x 3 term is 1. This is the polynomial q. (b) We add polynomials by adding their terms together: r(x) = ( 2 * 5 - 3 * l ) x 3 + ( 2 * 4 - 3 * 0 ) x 2 + ( 2 * 3 - 3 * 3 ) x + ( 2 * l - 3 * ( - 5 ) ) = 7x 3 + Sx 2 - 3x + 17 » »
v = [ 1; 3; 4; 5]; w = 2*v - 3*u
u = [-5; 3;
0; 1];
w= 17
-3 8
7
0l l—
to
M-
II 1 —1 1
V V
The vector w represents the polynomial r(x) because the constant term of r(x) is 17, its x term is —3, its x 2 term is 8 and its x 3 term is 7. (c) Polynomials of degree two may be represented by vectors in M3.
» vi = [ -2; 0; » v2 = [ 8; -9; » v3 = [ 9; -7; » rref( [ vi v2 ans = 1 0 0
0 1 0
0
-5]; -6]; -1] ; v3 p]) 0 1
'/, This represents p(x) . */, This represents the first poly. in the set '/, The second poly. '/, The third poly. '/• Reduce the augmented matrix.
1 1 1
Yes, p(x) is in the span of this set. p(x) = 1(—5x 2 —2) —1(—6x 2 —9x + 8) + 1(—x 2 —7x + 9) This set does span all of P2 since the system with augmented m atrix [vi v2 v3 | b] would have a solution for any right hand side. (d) As above, we will represent P 3 by vectors in M4. » p = [ -17; 29; 3; 1]; » vi = [-8; 8; -7; -2]; » rreí([ vi v2 v3 p]) ans = 1 0 0 3 0 0 0
1 0 0
0 1 0
v2 = [5; 3; 9; 7];
v3 = [ -3; -1; 6; -7] ;
2 1 0
Yes, p(x) is in the span of this set: p(x) = 3pi(x)-h 2p 2(a?) + lp 3 (a?) where p*(x) has the coefficients in Vj. This set does not span all of P 3 . Since there is a row of zeros in the row echelon form of
266
Instructor's Manual
Chapter 4 Vector Spaces
A = [ v i V2 V3] , it will be possible to pick a vector (which represents a polynomial) so that the system [A p] does not have a solution. (e) » »
A = [2 1 1 1 ; rref(A)
-1 3 2 0; 0 1 1 -1; 1 1 2 0];
ans = 1 0 0 0
0
0 1 0 0
0 0
0 1 0
0 1
Since the m atrix of vectors representing these polynomials reduces to a row echelon form with no zero rows, this set will span all of P 3 . 10. (a) The m atrix C — A — 2B is ( a\ — 2 a 2 C\ —2 c 2 —2 e 2 \ \ bi — 2b2 di —2d2 fi — 2/ 2 ) C is represented by the vector ( a i - 2a2\ 61 — 2 6 2
c1 -
2c2
di — 2d2 ’ ti — 2e2 V f i —2/2 / which is the sum v + 2 w. (b) » w = [1; 29; 3; -17] w=
*/• This v e c to r r e p re s e n ts th e f i r s t m a trix .
1
29 3 -17 » v i = [ -2 ; 8 ; -7 ; 8 ] ; » v2 = [ 7; 3; 9; 5]; » v3 = [ -7 ; -1 ; 6 ; - 3 ]; » r r e f ( [ v i v2 v3 w]) ans = 1 0 0 0
0 1 0 0
0
*/. Solve e l
v i + c2
of v e c to rs .
v2 + c3 v3 = w.
0 0
0 1
'/. Next, e n te r th e s e t
0 0
1
This system does not have a solution, so the first matrix is not in the span of the others. From this we can conclude that this set does not span all of M22, since we have found an element of M 22 that is not in their span.
Linear Combination and Span
M A T L A B 4.4
» w = [4; -2 ; 7; - 6 ; -10; 1]; */, The f i r s t m atrix . » vi = [ 6 ; 9; 5; 3; -1 ; - 1 ] ; */. The s e t of m a tric e s. » v2 = [ 6 ; 10; 4; 9; 4; 7 ]; v3 = [-4 ; - 8 ; 1; -2 ; 0; 2 ]; » v4 = [ 8 ; 7; -1 ; 4; 5; 6 ] ; v5 = [ 4;8 ; 5; 0; -10; - 1 ] ; » v6 = [ -9 ; 3; 4; 4; 0; - 6 ] ; » r r e f ( [v i v2 v3 v4 v5 v 6 w]) ans = 1
0
0
0
0
0
1
0
1
0
0
0
0
-1
0 0 0 0
0 0 0 0
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
2 1 1 0
Yes, w is in the span, w
= lvi -
lv
2 + 2v 3 +
l v 4 -f l v 5.
This set does span all of M 23 because the system above would have a solution for any right hand side.
» v i = [1; -1 ; » v3 = [ 2; 2; » r r e f ( [ v i v2 ans = 1 0 0 0
0
0; 2 ]; v2 = [ 1; 3; 1; 1]; 1; 1]; v4 = [ 0; 0; -1 ; 1]; v3v 4 ]) 0
1 0 0
0 1 0
0 0 0 1
Since there are no zero rows, this system would have a solution for any right hand side. This means that we can write anything in M 22 as a linear combination of these matrices, so they do span M22.
268
Instructor’s Manual
Chapter 4 Vector Spaces
S ection 4.5 so linear ly independent.
1
Linear ly independent, as only the trivial solution.
Linearly dependent, as —2 [ —1 ) -f 1
3.
4.
ro-(í)
so linearly independent.
5. Linearly dependent, as 3 vectors in M2 always dependent (Theorem 2). 6.
1 0 1 0 1 1 = —2; Linearly independent, (Theorem 5.) 1 1 0 1 00 0 1 0 = 1; Linearly independent. 00 1
-3 7 1 4 - 1 2 = —140; Linearly independent. 2 38 9.
1 0.
-3 7 1 4 - 1 1 = —163; Linearly independent. 2 38 1 3 0 5 -2 0 4 0 = 0; Linearly dependent. 1 2 - 1 3 1 -2 -1 -1
11.
1 3 0 5 -2 0 4 0 = 4; Linearly independent. 1 2 - 1 3 1-2
1-1
12. Linearly dependent, as 4 vectors in M3 always dependent (Theorem 2). 13. c i(l —x) + C2 X = 0; ci + (c2 —ci)x = 0 =>■ ci = 0; C2 —ci = 0; So c\ = C2- Linearly independent. 14. -cía; + c2(a;2 - 2a;) + c3(3x + 5a;2) = 0; ( - c i - 2c2 + 3c3)x + (c2 + 5c3 )x 2 = 0; ^ ^ 1 0 -13 0 1 5
^ => Linearly dependent. 1 1 0
15. c i(l —x) + c2 (l-f-x) + c3 x 2 = 0; (ci +C 2) + (—c i + c 2)x-|-c 3 x 2 = 0; | -1 1 0 0 0 1
Linearly independent.
Linear Independence
Section 4.5
/O -1 -1 16. cix + c2(x2—x) + C3 (x3 —x) = 0; ( c i —C2 —c3)x + C2 X2+c3x 3 = 0; I 0 1 0 \0
0
1
Linearly independent. 17. 2c\x + c2(x 3 —3)c3(1 + x — 4x3) + Ci{x3 + 18x - 9) = 0; /O -3 1 -9 (—3c2+c 3 —9c4)+ ( 2c i+ c 3+ 18c4)x+(c 2 —4 c3+c 4)x^ = 0 ; I 2 0 1 18 \0 1 -4 1 '0 0 1
1 -1 0 6
19. ci
°\
0 o o 0 1 0 o o o 1 o \o o o 1 20. ci (
_
0 0
0/ + c2
1 -1 1 0 -1 0 1 1 0 3 -1 1 6 1 2 0
/I
9
Linearly dependent.
°\ o o o/
°\ 0
2ci + 4c3 —ci —3c2 + c3 \ _ 5c2 —5c3 ) ~
5 ) + C 3 ( ? —5 )
¿ ) +C2(?
0 4 2 -1 -3 1 4 1 7 0 5 -5
101/2 0 1 - 4 1
6/1 1
1 0 0 96/11 0 1 0 35/11
18: ci ( 4
0 0 -11 -6
/I 0 2 0 -3 3 0 1 -1 \ 0 5 -5
°\ ° 1 0
0/
( o o ) ° r ( 4ci + c2 + 7c3 0 2 1 -1 0 0 0 0
/I 0 0 Vo
°\ 0 0 0/
=> Linearly dependent.
-1 0
3 1 -1 1 0 0 -1 2 1 0 3 -1 1 Vo 7 -4 0 /I
_
°
0/
/I 0 1 0
°V 0 0 0/
-
00 00
0
0
Vo 0
oo As in Solution 18. oo -1 -1 ° \ / I 0 0 -1/5 -2 -1 0 0 1 0 3/5 5 4 0 0 0 1 4/5 10 7 0 / 1 Vo 0 0
Linearly independent.
j ^ ) + c2 ( 7 J
) +C3(7
3) +C5( - l O
Linearly dependent. 21. c\ sinz + C2 cos x = 0; Set x = 0 => C2 = 0; Then c\ must also be 0. => Linearly independent. 22 . c\x -f c^y/x -f c3-\fx — 0 ; x = 1
ci + c2 + c3 = 0; x = 1/64 => ^- c i + \ c 2 + 7 c3 = 0; 04
1
x = 1/729 ^ 7 k Cl + ¿ C2 + r 3 = 0;
23.
a c = ad — be = 0, by Theorem 5. bd
1
O
4
1
1/64 1/8 1/4 ^ 0 => Linearly independent. 1/729 1/27 1/9 24.
a n 012 a 13 a 21 a22 a23 = 0, see Solution 1.4.16. 0>32 a33
^31
.
2a 5a
a —6 1 =>2a-|-13 = 0 = > a = —13/2. _ 13 5
. Thus the set of vectors is linearly dependent for all real a.
Instructoras Manual
Chapter 4 Vector Spaces
27. If A = (v iv 2 • • •v n), then Ac = 0 says c\Vi -f c2v 2 - f h cnv n = 0. Henee Ac = 0 has a non-trivial solution if and only if v i , . . . , vn dependent by the definition of dependence. 28. If v i , . . . , v n are linearly dependent, then there exists a nontrivial solution (ci , . . . , cn) of ciVx + h cnvn = 0. Then ( c i , . . . , cn, 0) is a nontrivial solution of ciVi + • • • -f cnv n -f cn+ iv n+i = 0. Thus v i , • • • >vn , are linearly dependent. 29. Suppose v i , . . . , v* are linearly dependent. Then there exists a nontrivial solution ( ci , . . . , c*) of ciVi + 1- Ck Vjb = 0. Then (ci, . . . , c*, 0 , . . . , 0) is a nontrivial solution of ciVi -|------ 1- c^v*. + • • • -f cn vn = 0. But this implies v i , . . . , v n are linearly dependent and this is a contradiction. Thus v i , . . . , v* are lin early independent. 30. Suppose vi and v 2 are linearly dependent. Then v 2 = a v x for some a ^ 0. Then vi • v 2 = a |v i |2 ^ 0, since vi is a nonzero vector. This contradicts vi and v 2 being orthogonal. Thus v x and v 2 are lin early independent. 31. If civi + c2v 2 -j- c3v 3 = 0 then 0 = Ovi = c i|v i |2 + c2v 2 • v x -f C 3 V 3 • vi = c i|v i|2. So ci = 0 as v x / 0. Then 0 = c2 |v 2|2 + c3v 3 - v 2 = c2 |v2|2, so c2 = 0. And finally, 0 = c3v 3 • v 3, or c3 = 0. Thus v i, v 2 and V3 are linearly independent, as only ci = c2 = C3 = 0 solves. 32. Suppose v i, v 2, . . . , vn are linearly independent. Then ai vi + a 2v 2 + - • •■fanv n = 0 has only the trivial solution. So no arbitrary variables in solving. Thus every column in row echelon form has a pivot. Since n rows and n columns, every row has a pivot, i.e. no zero rows in echelon form. Conversely, if the row echelon form of A does not contain a row of zeros, this implies that the only solution to Ax = 0 is x = 0, since n non-zero rows implies n pivots. Thus v i, v 2, . . . , vn are linearly independent. since x\ = —X2 — X3 .
33.
(l
34.
0
13/5 -2/5 3/5
^0 1 - 22/5
So solving in terms of £ 3 , X4
yields
36.
/£i\ £2 I
,£3 \ X4 /
- 2/5 \
x 2
7/5 \ - 2 /5
-3/5
1
x3
+ £4
o \
0/
1 j +£3 0
0/
1 0 -13 0 1 6
f l 1 1 -1 -1 0 \ \ 0 5 3 2 -8 0 / “
í ~ 2\
=
J
2 -1 1 6
\ )
-1 -1 4 -6
1 -2
/ xi \ x2 and X3 X4 \ x 5/
37.
fl
O
J
1 2 -1
2 5 4
O
35.
0
270
0 1 \ 0/
+ X4
/ 1 0 2/5 -7/5 3/5 V0 1 3/5 2/5 -8/5 /-3/5\ 8/5
0 + £5 1 0/
0 0 1/
as x\ — —2 x 2 + 3 x 3 —5 x 4.
. So X3 , X4 , £5 arbitrary
Linear Independence
Section 4.5
38. (a) See 39 (a) below, it does not depend on specific u. (b) From u • x = x\ + 2x 2 4- 3 x 3 = 0, get x = (—2,1,0) and y = (—3,0,1). i j k —2 1 0 = i 4- 2j 4- 3k. -3 0 1 (d) Note that w = u. (Other choices for x, y would yield other w.) (e) See 39(e) below. (c) w =
39. (a) Suppose x ,y E H. Then u ' ( x + y) = u * x - f u - y = 0 - f 0 = 0. Then x 4- y E H . Suppose a E M. Then u - (ax) = a ( u - x ) = a(0) = 0. Then a x E H. Therefore, H is a subspace of ffi3. (b) Suppose that u = (a, 6, c); Then since u # 0 , at least one of a, 6, or c is non-zero. Suppose that a is not zero. Then x = (—6, a, 0) and y = (—c, 0, a) are linearly independent vectors in H . A similar process works if 6 / 0 or c # 0 . i j k (c) w =
—6 a 0 = a 2i 4- abj 4- ack. (Key point: w = x x y is orthogonal to x and y.) —c 0 a
(d) Note that w = au. (e) H consists of all the vectors which are perpendicular to u. Then H will be the plañe for which u is a normal vector, w = x x y is also a vector which is perpendicular to the plañe. Since u and w are both perpendicular to the same plañe in M3, they must be linearly dependent. 40. Consider fi ( x ) = a ix 2 4- b\x 4- c1} f 2 (x) = a 2x 2 4- b2x 4- c2, h ( x ) = a 3x 2 4- b3x 4- c3 and / 4(x) = a4x 2 4 64X4-C4 in P2. If we wish to solve for (¿ 1 , k2) Ar3, k4) in the equation ki(fi(x)) + k2( f 2(x)) + k3(f 3(x)) + Ár4( / 4(x)) = 0 , we must equate coefRcients of l , x , x 2 to 0 and we get three homogeneous equations with four unknowns. In this case, there will always be a nontrivial solution for (&i, ¿2, &3, k4). Thus any four polynomials in P 2 are linearly dependent. 41. Suppose fi( x) and f 2 (x) span P2. Then for any /(x ) = ax 2 4- bx 4- c in P2, we would need & i(/i(x)) 4k 2 ( f 2 (x)) = f ( x ), for some A?i, Ar2 G M. Therefore, if fi(x) = a ix 2+ bix + ci and f 2 (x) = a 2x 24- 62x-hc2, we would have, equating coefficients of 1 , x, x2: kiai 4- k2a2 = a M i + M 2= b M i + k2c2 = c. W ith three equations and two unknowns it is always possible to choose a, b and c so that no solu tion exists. Thus f \ and f 2 cannot span P2. (Specifically there will be a row of zeros in echelon form whose third column will say 0 is a non-trivial linear combination of a, ó, c. Not all a, 6, c will satisfy this condition.) 42. Suppose / 1 , / 2. . . . , / n , /n+ i, /n +2 are in Pn . (Note: we are using the same notation as in 40.) If we consider k \ f \ 4- k2f 2 4- ... 4- kn+2f n+2 = 0, then, as in #40, if we equate coefRcients we get n 4-1 equations in n 4- 2 unknowns. Then there will always be a nontrivial solution for k2, . . . , kn+2). Thus any n 4-2 polynomials in Pn are linearly dependent. 43. Note that if we are given any set of linearly dependent vectors, then if any vectors are added to the set we still have a set of linearly dependent vectors. Thus if any set has a subset which is linearly de pendent, then the original set is linearly dependent. Then any linearly independent set cannot have a subset which is linearly dependent. Thus, any subset of a linearly independent set is linearly indepen dent. 44. Suppose A i }A 2y A3, A 4, A 5 , A q and A 7 are in M32. Consider solving ai Ai 4- a2A 2 4- a3A 3 4- a4A 4 4«5^5 4- cl^A q 4- ay A y = O for (ai, a2, a3la4la^}ae, ay). This generates six homogeneous equations with seven unknowns. Then, regar dless of the given matrices, there will always be a nontrivial solution. Thus any seven matrices of M32 are linearly dependent.
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45. Suppose that Ax, . . . , A mn+i are in Mmn. Consider solving ]Ca¿A¿ = O for numbers {a¿}; this is mn homogeneous equations in mn + 1 unknowns. Therefore there will always be a nontrivial solution. Thus any mn + 1 matrices of Mmn are linearly dependent. 46. Note that 5 Xfl 52 is a subset of both 5 X and 5 2 , each of which is a linearly independent set. Then by problem 43, Si O S 2 is linearly independent. (Note that the empty set of vectors is linearly indepen dent, so you need not require S\ fl S 2 to be non-empty.) 47. Clearly this is true for n — 1. Assume that 1, x, x 2, . . . , x ” ” 1 are linearly independent. Then consider ao + aix-\ \-an- \ x n~l + anx n = 0. Note that an must be zero by properties of polynomial addition (i.e., only add like terms) or take n derivatives to get n\an = 0. Then by our assumption we have aQ= 0, ai = 0 , . . . , an_i = 0. Thus 1, a?, x2, . . . , xn~l yx n are linearly independent. 48. Consider a xv x + a 2(vx + v 2) H b an(v x + v 2 H b v n) = 0. Then we have (ax + a2 H b an)vi + (a 2 -f • • • -f an)v 2 -b • • • + flnVn = 0. Since {vi, v2, . . . , vn} is a linearly independent set, we have a i + a 2 + • • * -b fln — 0 02 + • • • + an = 0
an = 0 By backward substitution, we have an = 0, an_i = 0, . . . , ü 2 = 0, a x = 0. Thus v x, v x + v 2, . . . , v x + V2 + • • • + vn are linearly independent. 49. Since v x,V2, .. ., v n are linearly dependent, there exist 6X, 62, • • •, bn, where 6iv x -f 62v 2 -f • • • + bn\ n = 0 with at least two of the 6, ’s nonzero since v x, V2 , . . . , vn are nonzero. Choose k to be the largest i such that b{ ^ 0. Note then that 1 < k < n and, if we let a¿ = —b¡/bk, then v* = a xv x + a 2 ^ 2 + • • • + aA;_ xv A;_ x. 50. If all the vectors are the zero vector then we are done. If not, then without loss of generality, assume v x is a nonzero vector. Since {vx,V2} is linearly dependent, V2 = a xv x for some a x. Since {vx, V3} is linearly dependent, V3 = 0 ,2 v x for some 02- Continuing on with this process, we find that each vector is a múltiple of v x. 51. f ( x) = cg(x) for some c € M. Then f ( x ) = cg'(x). Then W( f , g)(x) = f i ( x)
Í 2 (x) --
f [(x) 52.
cg(x) g(x) cg'(x) g'{x)
f n(x) /'(ar)
■■- , f n) —
f {r
x\ x ) f c - l\ x ) - - á n- i){x)
53. Consider a x(u -f v) + a2(u + v) + aa(v + w) = 0 (ax -b a2)u -b (ai + a3)v + (a 2 + fl3)w = 0 Since u, v and w are linearly independent, we have a i “b a 2
— 0
-b a 3 = 0 } => ai = 0, 02 = 0 , a3 = 0 , by elimination.
ax a 2
"b
0 ,3
—0
Thus u -b v, u -b w and v -b w are linearly independent.
0.
Linear Independence
54. We need
Section 4.5
1 —c 1 + c = (1 —c)2 —(1 + c)2 = —4c ^ 0. Thus the vectors are linearly independent if 1 4- c 1 —c
c ^ 0. 55. 1 1 1 1 1 1 6 —a c —a a b c = 0 b — a c —a = 6(6 —a) c(c —a) 0 b2 — ab c2 — ca a2 b2 c2 = (6 —a)(c —a)
1 1
6c
= (6 —a)(c —a)(c —6) / O , if a 6, a ^ c and 6 ^ c. Thus the vectors are linearly independent. 56. Suppose {vi, v 2, . . . , v n , v} is a linearly dependent set. Since {vi, v 2, • . v n} is a linearly indepen dent set, then v = aiVi + • • • -f anv n for some a i , . . . , an £ M by the solution to Problem 49. Then v G span { v i, v 2, . . . , vn }, which is a contradiction. Thus {vi, v2>. . . , vn , v} is a linearly independent set. 57.
2 -1 a y (Eliminate in I 1 3 6 1, then choose any a, 6, c making bottom row not all 2 4 cÁ
1
zeros.) 58. 1 —x2, 1 + x2, x. (Any quadratic with non-zero x term will work.) 59. (a) Note that since the vectors u, v and w are coplanar, the points (1*1 , 112*^3), (^1 *^2, ^3)*( ^ 1 , ^ 2, ^ 3) and (0, 0, 0) are all contained in some plañe. Let ax -f by + cz = 0 be the equation of such a plañe with n = (a, 6, c) a normal to the plañe. Note that a, 6 and c are not all zero. Then we have aui + bv.2 + cus = 0 av 1 -|- bv 2 + cvs = 0 aw \ + bu)2 + cws — 0.
ui u2 u3 \ i a\ v \ V 2 v 3 ) ;det A / 0 0 is the only solution to A x = 0. But, x — I 6 1 ^ 0 is a wi w2w3 J \c J solution to Ax = 0. Thus det A= 0. (c) Note that det A* = det A = 0. Thus, A*x = 0 has a nontrivial solution. Thus, by Theorem 3, u, v and w are linearly dependent.
(b) Let A
= I
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M A T L A B 4.5 1. In each case, A the augmented matrix is entered, if r r e f (A) has a column on the left without a pivot in it, then the vectors are dependent, otherwise they are independent. » A = [ 1 -1 0; 2 -3 0]; » rre f (A ) ans = 1 0
0
*/. Problem 1.
0 1
0
» A = [ 2 4 0; - 1 - 2 0; 4 7 0]; */. Problem 2. » rref(A)
ans = 1 0
0
0
» »
0 1
0
0
0
A = [ 2 4 0; - 1 - 2 0 ; 4 8 0]; 7. Problem 3. rref(A)
ans = 1
0 0 0
2
0 0
0 0
» A = [ - 2 4 0; 3 7 0]; » rref(A)
*/. Problem 4.
ans = 1 0 » »
0
0 1
0
A = [ -3 1 4 0; 2 10 - 5 0] ; 7. Problem 5. rref(A)
ans = 1.0000 0 » »
0 1.0000
-1.4062 -0.2188
Ó 0
A = [ 1 0 1 0; 0 1 1 0; 1 1 0 0];*/, Problem 6 . rref(A)
ans = 1 0 0
» »
0
0 1 0
0 0
0 1
0
A = [ 1 0 0 0; 0 1 0 0; 0 0 1 0]; ’ /, Problem 7. rref(A)
ans = 1 0 0
» » ans
0
0 1 0
0 0
0 1
0
A = [-3 7 1 0; 4 - 1 2 0; 2 3 8 0]; X Problem 8. rref(A) =
1
0
0
0
1
0
0
0
0
0 1
0
Linear Independence » »
M A T L A B 4.5
A = [ -3 7 1 0; 4 -1 1 0; 2 3 8 0]; '/, Problem 9. rref(A)
ans = 1 0 0
» »
0
0 1 0
0 0
0 1
0
A = [ 1 3 0 5 0; -2 0 4 0 0; 1 2 - 1 3 0 ; rref(A)
1 -2 -1 -1 0]; '/, Problem 10.
ans =
» »
1
0
0 0 0
1 0 0
0
2
0
1 1 0
0 0 0
1 0
0
A = [ 1 3 0 5 0; -2 0 4 0 0; 1 2 - 1 3 0 ; rref(A)
1-21-10];
*/. Problem 11.
ans = 1 0
» »
0 1
0 0
0
0
1
0
0
0
0 0
0 0
1
0
0
0
A = [ 1 4 -2 7 0; - 1 0 3 10; 2 0 5 2 0];*/. Problem 12. rref(A)
ans = 1.0000 0 0
0 1.0000 0
0 0 1.0000
0.0909 1.9091 0.3636
0 0 0
The sets which are linearly independent are 1, 2, 4, 6 , 7, 8 , 9, and 11. The sets which are linearly dependent are 3, 5, 10, and 12. 2. In the text, it is stated “Three vectors in M3 are linearly dependent if and only if they are coplanar.” (a) Since they are linearly independent, they must not be coplanar. (b) We need only show they are linearly dependent. To do this, we reduce the augmented matrix formed from the homogeneous equation. (i)
» »
A = [1230; rreí(A)
2130;
1340];
ans = 1 0 0
0
1 1 0
1
0 0
0
0
( ii)
» A = [1-120 » rref(A)
;20 6 0
; 1 1 4 0];
ans = 1 0 0
0
3 1 0
1
0 0
0
0
In both cases, there was a column on the left without a pivot, so the sets were dependent.
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3. » m = 4; n = 3; » A = 2*rand(n,ra) -1 A= -0.5621 -0.9059 0.3577
*/. Choose valúes for m and n.
0.3586 0.8694 -0.2330
0.0388 0.6619 -0.9309
-0.8931 0.0594 0.3423
0 1 . 0000 0
0 0 1.0000
4.5903 4.6803 0.2248
» rref(A) ans = 1 . 0000 0 0
The fourth column does not have a pivot, so the columns are linearly dependent.Conjecture: If a m atrix has more columns than rows, the columns will always be linearly dependent. Proof:This follows directly from Theorem 2. 4. Refer to the answer to problem 2 in Section 1.8. The matrices in part (i), (iv), and (v) were invert ible, and since they reduced to the identity matrix, their columns are linearly independent. The ma trices in part (ii), (iii), and (vi) were not invertible, and their columns were dependent. We now check for linear independence of the rows, by reducing A i . » A = (1/13)* [2 7 5; 0 9 8; 7 4 0] ; */. Matrix for (i) » rref(A') ans = 1 0 0 0 1 0 0
0
1
» A = [2 -4 5; 0 0 8; 7 -14 0]; » rref(A*) ans = 1.0000 0 3.5000 0 1.0000 -2.1875 0
0
# /, Matrix for (ii)
0
» A = [1 4 -2 1; 5 1 9 7; 7 4 10 4; 0 7 -7 7]; » rref(A*) ans = 1.0000 0 0 2.8000 0 1.0000 0 1.4000 0 0 1.0000 -1.4000 0 0 0 0 » A = [1 4 6 1; 5 1 9 7; 7 4 8 4; 0 7 5 7]; » rref(A1) ans = 1 0 0 0
0
0 1 0 0
0 0
0 1 0
0
1
*/• Matrix for (iii)
% Matrix for (iv)
Linear Independence »
A = (-1/56) *
[1
2
0- 1
2-1
1
0
0
M A T L A B 4.5
345 */. Matrix for (v) 2
2 -1
1 1 - 1 1 1
0 »
0
0
0
4 ];
rref(A')
ans =
»
1 0
0 1
0 0 0
0 0 0
0
0 0
0 0
1 0 0
0 1 0
0 0 1
0
A = [ 1 2 -1 7 5 0-1 2-3 2 1 0 3 1 - 1
1 1 1 4 » ans
0 0 rref(A1)
0
'/• Matrix for (vi)
1
0
4 ];
=
1.0000 0 0
0 1.0000 0
0 0 1.0000
0.4000 -0.2000 0.6000
0.4000 0.8000 -0.4000
0 0
0 0
0 0
0 0
0 0
In each case, the invertible matrices had linearly independent rows, and the singular matrices had lin early dependent rows. This is proved in the summing up theorem. 5. (a) If the entries in z are { z \ }Z2 , . . . , ¿m}, and the columns of A are {vi, V2 , . . . , vm}, then w = Az = 2 iv'i + Z2 V2 + ... + 2 mVm. Since the z,-’s are scalars, this shows that w is a linear combination of the v¿’s. 0 >) (i):
» A = [ 8 1 10; 7 - 7 - 6 ; -8-1 -1] ; '/• The matrix of vectors. » z = round(10*(2*rand(3,1)-1)) '/• Generate a random vector z. z = -6
-9 4 » w = A*z w = -17 -3 53
*/• w is a linear combination of columns ofA
» rref([ A w] ) ans =
'/♦ Test if this set is linearly dependent.
1
0
0 - 6
0
1
0
0
0
-
9
1
4
Since there is a column without a pivot in the row echelon form, the set {vi, v 2, V3 , w} is linearly dependent. This can be repeated for several other random vectors z.
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* (ü) » A = [ 1 - 1 2 ; 0 2 - 1 ; 1 3 0 ; 1 1 4 ] ; * / . The m atrix of v e c to rs . » z = ro u n d (1 0 * (2 * ra n d (3 ,1 )-1 )) '/• G enerate a random v e c to r z . z = 4 9 -2
» w = A*z w= -9
*/• w i s a l in e a r com bination of
20
31 5 » r r e f ( [ A w] ) ans = 1
*/. T est i f t h i s s e t i s l in e a r l y dependent. 0
0
0
1
0
0
0
4 9
0 1 - 2
0
0
0
Since there is a column without a pivot in the row echelon form, the set {vi, V2 , v 3 ,w} is linearly dependent. (iii) » A = [ 4 10 6 3; 3 2 2 2; 2 8 8 1; 0 1 2 2; 2 4 10 6 ] ; '/, The m atrix . » z = ro u n d (1 0 * (2 * ra n d (4 ,1 )-1 )) '/• G enerate a random v e c to r z. z = 0
7 -9 -9 » y = A*z y =
*/• w is a l in e a r com bination of
-11 - 22
-25 -29 -116 » r r e f ( [ A w] ) ans =
'/• T est i f t h i s s e t i s l in e a r l y dependent.
1
0
0
1
0
0
0
0
0
1 0
0
0
0
0
0
0
7
0
0
1 0
9 -9 0
Since there is a column without a pivot in the row echelon form, the set { v i, V2 , V3 , V4 , w} is linearly dependent. (c) If w is in the span of { v i , V 2 , . . . , v m } , then { v i , V 2 , . . . , v m , w} is a linearly dependent set.
Linear Independence
M A T L A B 4.5
6 . (a) In each case, the m atrix A is made with the vectors as its columns. Since the system with aug mented m atrix S = [A 0] has a nonzero solution, the vectors are linearly dependent.
» » ans
S = [ 1 -1 3 0; 1 1 0 0] ; */, For problem 3i. rref(S) =
1.0000 0
0 1.0000
1.5000 -1.5000
0 0
» S = [ 1 -2 5 0; 2 3 4 0] ; */, For problem 3ii. » rref(S) ans = 1.0000 0 3.2857 0 0 1.0000 -0.8571 0 » » » ans
'/• For problem 7a. S = [ 3 -2 7 14 1 0 ; -7 0 2 -5 -5 0 ; 4 -7 9 27 0 0 ; - 2 2 1 - 5 - 1 0 ] ; rref(S) =
1 0 0 0 » » ans
0
0 1 0 0
0
1 0
1 2 1 0
0
0 0
0
0
0 0
1
S = [ 10 0 -10 -6 32 0; 8 2 -4 -7 32 0; -5 7 19 1 -5 0]; V. For problem 7e. rref(S) =
1
0
0
-
1 0
0
1
0
2
0
0
1
2
0 1 -2
0
0
(b) The equation w = ciVi -I-C2V2 H hcjbV* is equivalent to the system [A w] where A is the matrix whose columns are the vectors v¿. If this system reduces to [R u] in row echelon form, then [A 0] reduces to [R 0]. If there are an infinite number of Solutions to [A w], then R will have a column without a pivot in it. In this case, the system [A 0] will also have an infinite number of Solutions. Since [A 0] has a nontrivial solution, the columns of A are not linearly independent. 7. (a) » m = 3; n =4; » A = 2 * ran d (n ,m )-l A= 0.0594 -0.8663 0.3423 -0.1650 -0.9846 0.3735 -0.2332 0.1780 » r r e f (A) ans = 1 0 0 0
'/. Choose v alú es f o r m and n.
0.8609 0.6923 0.0539 -0.8161 '/, Check f o r l in e a r dependence.
0
0 1 0 0
0 1 0
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Since every column has a pivot, the columns of A are linearly independent. This will be true for almost all randomly chosen matrices. » B = A; » B (: ,3 ) = 3*B(: ,1 ) - 2*B(:,2) B = 0.0594 0.3423 -0.9846 -0.2332
-0.8663 -0.1650 0.3735 0.1780
» r r e f ( B) ans = 1 0 0 0 0
(b) (c) (d) (e) (f)
1.9108 1.3570 -3.7009 -1.0554
3 1
-
0 0
2
0 0
The third column does not have a pivot, so the columns of B are dependent. This corresponds to choosing the third column of B to be a linear combination of the first and second. The above can be repeated several times. If a column of A is a linear combination of other columns of A, then the columns are linearly de pendent. See solution to problem 5 in MATLAB 1.7 If the columns of A are linearly dependent, then one column can be written as a linearcombina tion of the others. This is the converse of the statement in (c). Proof: Let the columns of A be the vectors v i, V2 , . . . , vn . The columns of A are linearly depen dent if and only if there is a nontrivial solution of ciVi -f C2 V2 + ... + cnv n = 0. Pick one of the coefficients that is nonzero, for example, c*. This equation can be rewritten as
-CkVk =
C1V 1
+
. . . +
Ck-lVk-l + Ck+lVk +l +
• • • +
cnv n.
If we divide this by —c*, then we get V * = - ( c i / c jt) v 1 -
... -
( c k -l/ C k )v k- l -
(c¡fc+ l/c¡fc)v¡fc+ i -
... -
(c„/cj fe)vn ,
which means that v* is a linear combination of the other columns. Conversely, If v* is a linear combination of the other columns, then Vjfe = C1V1 + . . . + C k - l V k - 1 + Cjb+iVjb+ 1 + . . . + c n v n .
Bring Vjb to the other side, and letting Ck = —1, we get a nontrivial solution of 0 = CiVi + ... + cnv n , which means that the columns are linearly dependent. 8. (a) (0
» A = [1 0 2; 2 3 1; -1 1 - 3 ]; » r r e f ( A) ans = 1 0 0
0
2 1 - 1 0 0
Linear Independence
M A T L A B 4.5
The third column is 2 times the first plus —1 times the second. To verify this: » 2*A(:,1) -1* A(:,2) ans =
*/, This should be the third column.
2 1
-3 (ii) » A = [10 O -10 -6 32; 8 2 - 4 - 7 32; -5 7 19 1 -5]; » rref(A) ans = 1
0
0 0
-
1 0
1
0
2
2
0 1
0
1 -2
» -1*A(:,1) + 2*A(:,2) ans = -1 0 -4 19
'/, This should be the third column.
» 2*A(:,1) + 1*A(:,2) -2*A(:,4) */, This should be the fifth column. ans = 32 32 -5
(iii) » »
A = [7 r r e f ( A)
6
11 3 5;
1 - 5 -20 9; 7
8
6
11 3
8
;
8
2 -2 -16
6
; 7 3 2
ans =
»
1
0
0
1
3
-3 4
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
-1
-1*A (:, 1 ) + 3*A(: , 2 )
0 0
•/• This should be th e t h i r d column.
ans = 11 -5 11
-2 2 » -3*A(:,1) + 4*A(:,2) ans = 3 -2 0 3 -16 -9
This should be the fourth column.
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(iv) » »
A = [ 1 3 1 13; - 2 4 0 1 - 1 ; R = r r e f ( A)
0-2-319;
11215];
R = 1.0 0 0 0
0
0
0
0
1.0 00 0
0
0
0
0
1.0 00 0
0
0
0
0
1.0 00 0
1.0526 -1.1579 -0.3158 5.7368
» c = R(:,5); '/• The fiíth column of R gives the coef. » */, This should be the fifth column of A: » c(l)*A(:,1) + c(2)*A(:,2) + c(3)*A(:,3) + c(4)*A(:,4) ans = 3.0000
- 1 . 0000 9.0000 5.0000
(b) See answer to problem 49. 9. (a) In each case, the m atrix of vectors is reduced to row echelon form. Since every column has a pivot, the set is independent. Since the bottom row is all zeros, it is possible to pick a w so that the system [A w] does not have a solution, which means that w is not in the span of the columns of A. (i)
» A = [ -1; 2]; » rref(A) ans = 1 0 (ii)
» A = [1 -1 2; 0 2 -1; 1 3 0 ; » rref(A) ans = 1 0
0 0
0
114];
0 1
0
0 0
1 0
(íü ) »
A = [ 4 10 6 3; 3 2 2 2 ; » rref(A) ans = 1 0 0 0
(b)
0 0
1 0
0
0
0
0
1
0
0
0
0
2881;
0122;
2 4 10 6] ;
0 1
0
As above, the m atrix of vectors is reduced to row echelon form. In each case, there is a column without a pivot, so the vectors are linearly dependent. However each row has a pivot; the system [A w] will always have a solution, so the set spans all of Mn.
Linear Independence
MATLAB
4.5
» A = C 1 3 -1; 2 -1 » rref(A) ans = 1.0000
0
0
1.0000
o_j i
(i)
-0.1429 -0.2857
(ü) » A = [1 » rref(A) ans =
-1
2
1
;
0
2 - 1 1; 1 3 0 4] ;
1.0000
0 1.0000
1 . 50 00 -0.5000
0
0 0
0
0
1.0000
0
(iii) » A = [4 3 0 7 1 1 ; » rref(A) ans =
- 1 2 1 2 1
1; 3 2 2 7 - 1 1 ;
1 2 2 5 0
- 0.1111 0.5556
1.0000
0
0
1.0000
0
0
1.0000
0
1 .0 0 00
0
0
0
1.0000
1.0000
0
0
0
0
0
0
1.0000
- 0.2222
(c) No it is not possible. In part (a), there was a pivot in every column, so the number of pivots was the same as the number of columns. However, there was a row without a pivot, so the number of rows was more than the number of pivots. This means that the number of rows is more than the number of columns. Similarly in part (b), the number of rows is less than the number of columns. This cannot happen for n vectors in Mn, where the m atrix A will have the same num ber of rows and columns. (d) If m < n, a set of m vectors will never span Mn. If m = tí, the set is linearly independent if and only if it spans all of Mn. If m > ti, the set will never be linearly independent. The proof of all three of these come from reducing the matrix of these vectors to row echelon form. The set spans IRn when every row has a pivot, and the set is linearly independent when every column has a pivot. If m < n, there can at most be m pivots, so at least one of the n rows will not have a pivot. If m = 7i, then every row has a pivot if and only if there are n pivots. This happens if and only if every column has a pivot. If m > tí, there can be at most n pivots, so at least one of the m columns will not have a pivot. 10. (a) First, the m atrix V whose columns are the vectors v¿ is entered. In each case, every column of the reduced row echelon form of V has a pivot. (i) »
v i = C i -1 2; 0 2 -1; 1 3 0; 1 1 4];
» rref(Vi) ans = 1 0 0 0
0 1 0 0
0 0 1 0
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(ü) » Vii = [ 4 10 6 3; 3 2 2 2; 2 8 8 1; 0 1 2 2]; » rreí(Vii) ans = 1
0
0 0 0
0
1 0 0
0
0
0 1 0
0 1
(iii) » Viii = [-1 -1; 0 0; 2 1; 3 5]; » rref(Viii) ans = 1 0 0 0
0 1 0 0
(iv) » Viv = round( 10*(2*rand(4)-l)) Viv = -9 -10 9 -6 -9 -2 -9 -2 1 -9 0 4 -2 3 4 7 » rref(Viv) ans = 1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
(b) First, a random m atrix is entered. To verify that it is invertible, we check that det (A ) is nonzero. Then the m atrix B is created whose columns are Av,-. This is done using B = AV. Next the ma trix B is reduced to see if the vectors are independent. In each case, the vectors { A v i , A v 2 , • • •, Avk} are linearly independent. » A = round( 10*(2*rand(4)-l)) A = -9 4 1 4 5 8 2 -8 -3 9 3 5 -5 3 7 -2 » det(A) ans = -7290
*/. Check that A is invertible.
Linear Independence
(i) » B = A*Vi B = -1 1 15 11 11 9 5 -23 » rref(B) ans = 1 0 0 0
-29 32 3 28
0 1 0 0
0 0 1 0
» B = A*Vii B = 65 27 0 73 55 133 12 29
40 70 94 4
0 8 32 18
0 0 1 0
0 0 0 1
-68 77 -112 -41
-60 -86 -135 -27
(ii)
» rref(B) ans = 1 0 0 0
0 1 0 0
(iii) » B = A*Viii B = -23 -45 29 31 -8 -19 -8 3 » rref(B) ans = 1 0 0 0
0 1 0 0
(iv) » B = A*Viv B = -53 -29 112 69 -73 54 -32 88
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C hapter 4 V e c to r Spaces » rref(B) ans =
o 0
1 0 0 0
o o o
1 0
1
(c) This is done in the same way that (b) was. »
A(: , 3 ) = 2*A(: ,1 ) - A(:,2)
A = 4 2 9 7
1 -8 3 -2
-9 5 -3 3
7 12 15 16
*/, This should be zero, for A to be singular.
» det(A) ans = 0
(i) » B = A*Vi B == 2 10 19 23 39 21 26 40
-29 32 3 28
ref(B) 1 0 0 0
0 1 0 0
i =
A*Vii
33 8 75 54
89 105 213 197
0 0 1 0
(ü) B = 64 102 174 172
3 12 42 39
» rref(B) ans = 1. 0000 0 0 0
(iii) » B = A*Viii B = -17 -42 37 35 12
-9
34
24
0 1. 0000 0 0
0 0 1. 0000 0
1.3846 -1.8718 1.9359 0
Linear Independence
MATLAB
4.5
» rref(B) ans = 1
O
O O O
O O
1
(iv) » B = A*Viv B = -29 -41 128 69 -33 54 52 88
-87 -122 -225 -216
-65 81 -102 -20
» rref(B) ans = 1. 0000
O O O
O
O O
1. 0000
O O
1. 0000
O
-3.0229 0.0629 3.2171
O
The vectors from parts (i) and (iii) were still linearly independent. However, the vectors in part (ii) and (iv) were no longer linearly independent. (d) If the m atrix A is invertible, and a set {vi, V2, . . . , v*} is linearly independent, then the set { A v i , ^4v 2, . . . , A v k } is also linearly independent. 11. To solve this problem, each polynomial in Pn is represented by a vector in Mn+1, as in problem 9 in MATLAB 4.4. Then the matrix of these vectors is reduced to row echelon form. Finally, if any col umn doesn’t have a pivot, the corresponding vector is written in terms of the other vectors. » » » » »
pl = [1; -1; 0]; p2 = [0; 1; 0]; A = [pl p2] ; rref(A)
•/. Problem 13. The first polynomial 1 */, The second poly. 0 + l x
V,
lx
+ 0 x ~2
+ 0x ~2
ans = 1 0 0
0 1 0
Every column has a pivot, so these polynomials are linearly independent. » »
A = [0 -1 0 » rref(A) ans = 1 0 0
'/,Problem 14. */,The constant terms of the polynomials. */,The x terms. The x~2 terms.
00 -2 3 15];
1
0
-13 5
0
0
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These are linearly dependent, and the third column can be written in terms of the first two. » -13*A(:,1) + 5*A(:,2) ans = 0
3
5 In terms of the polynomials (3x + 5x2) = —13(—x) + 5(x 2 —2x). » »
*/, Problem 15. 5C The constant terms. •/, The x terms. •/, The x~2 terms.
A = [ 1 1 0 - 1 1 0 0 0 1 ];
» rref(A) ans = 1 0 0
0 0 1
0 1 0
These are linearly independent. » »
A = [0
0
'/, Problem 16. */, The Constant terms. */, The x terms. •/, The x~2 terms. */, The x~3 terms.
o
1 -1 -1 0 1 o 0 o i];
» rref(A) ans =
o 0
1 0 0 0
1
o
These are linearly independent. » »
A = [ 0 -3 2 0 0
0 » rref(A) ans = 1 . 0000 0 00 0
0
•/, Problem 17. '/• The Constant terms. '/, The x terms. */, The x~2 terms. */, The x~3 terms.
1 -9 1 18 0
1-4
0
1];
0 1. 0000 1. 0000 0
0 0
8.7273 3.1818 0.5455
0
0
These are linearly dependent. The forth polynomial is 8.7273 times the first plus 3.1818 times the second plus 0.5455 times the third. To verify this: » 8 . 7 2 7 3 * A ( : , 1 ) + 3 . 1818*A( : , 2 ) + 0 . 5 45 5 *A (: , 3 ) ans = -8.9999 18.0001 0
0.9998
Notice we only used 4 decimal places for the solution, and ans only agrees with A (: ,4) to about 4 digits.
Linear Independence
1 2.
In order to work this problem, the matrices will be entered as vectors. » MI = [ 2 - 1 ; 40]; */• Question 18. » M2 = [ 0 -3; 1 5]; M3 = [ 4 1; 7 -5]; » vi = [ Ml(:,1) ; Ml(:,2)] vi = 2
4 -1 0 » v2 = [ M2(:,1) ; M2(:,2)] v2 = 0 1 -3 5 » v3 = [ M3(:,1) ; M3(:,2)] v3 = 4 7 1 -5 » rref( [ vi v2 v3]) ans = 1 0 2 0 1 - 1 0 0
0 0
0 0
These are dependent: M 3 = 2M\ —IM 2. » 2*M1 - 1*M2 ans = 4 1 7 -5
*/• This should be M3.
» MI = [1 -1; 0 6]; M2 = [ -1 0; » M3 = [1 1; -1 2]; M4 = [ 0 1; 1 » vi = [ Ml(:,1) ; Ml(:,2)]; v2 = » v3 = [ M3(:,1) ; M3(:,2)]; v4 = » rref( [ vi v2 v3 v4]) ans = 1 0 0 0
0
0 1 0 0
0 1 0 0
0 0 1
3 1]; */. Question 19. 0]; [ M2(:,l) ; M2(:,2)]; [ M4(:,l) ; M4(:,2)];
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These are linearly independent. » MI = [ -1 0; 1 2]; M2 = [2 3; 7 -4]; */. Question 20. » M3 = [ 8 - 5 ; 7 6 ] ; M4 = [ 4 - 1 ; 2 3 ] ; M5 = [ 2 3; - 1 4 ] ; » vi = [ Ml(:,1) ; Ml(: ,2 ) ] ; v2 = [ M 2 ( : , l ) ; M 2 ( : , 2 ) ] ; » v3 = [ M3(: , 1 ) ; M 3 ( : , 2 ) ] ; v4 = [ M 4 ( : , l ) ; M 4 ( : , 2 ) ] ; » v5 = [ M5(: , 1 ) ; M 5 ( : , 2 ) ] ; » R = rref( [ vi v2 v3 v4 v5]) R = 0 0.9697 0 0 1.0000 0 0.0303 0 1.0000 0 1.0000 0 -1.2121 0 0 1.0000 3.1515 0 0 0
These are dependent. The fifth m atrix is a linear combination of the first four. » c = R(:,5); '/. The coefficients. » c(l)*Ml + c(2)*M2 + c(3)*M3 + c(4)*M4 '/, This should be M5. ans = 2.0000 3.0000 -1.0000 4.0000
13. (a) In order to work this problem, we will need to convert the 2 x 2 matrices into a vectors in M4.
» A = round(10*(2*rand(2)-l)) '/• Generate a random matrix. A = 4 5 8 -5 » a = A(:) a = 4
'/• This converts a matrix to a single column.
8
5 -5
The two commands above are repeated for B , C, D , and E. Next we check the linear dependence of a, 6, c, d, and e. » M M =
= [abcde] 4 8 5 -5
9
-2
0
9 1
7
-10 -2 9 9 -2
-9 3
4 2 7
» R = rref(M) R = 1. 0000 0 00
o
0 1. 0000 1. 0000
0 0
o
o
0 0 0 1.0000
7.2042 5.3266 4.2263 3.4719
Linear Independence
MATLAB
4.5
From R , we see that e is a linear combination of the first four vectors. This corresponds to E being a linear combination of the first four matrices. To verify this: » E E =
'/. The matrix E. 4 2
9 7
»
R(1,5)*A + R(2,5)*B + R(3,5)*C + R(4,5)*D */,This isE as •/•ofA, B, C and ans = 4.0000 9.0000 1.9999 7.0001
a combination D.
This can be repeated for two other sets of random matrices. (b) As in (a), we will convert the 2 x 3 matrices into a vectors in M6. » A = round(10*(2*rand(2,3)-l)) */• Generate a random matrix. A = -9 -3 5 5 3 10 » a = [ A(:,l); A(:,2); A(:,3) ] */, Convert a matrix into one column vector, a = -9 5 -3 3 5 10
This is repeated for B through G. [= [ a b e -9 5 -3 3 5 10
-3 -5 10 4 5 3
d e * g] -9 3 8 -5 -1 5
» R = rref(M) R = 0000 0 0 1.0000 0 0 0 0 0 0 0 0
0 -5 -5 -3 -7 0
8 8 -9 8 0 0
0 0 1.0000 0 0 0
-4 10 0 -5 -8 9
0 0 0 1.0000 0 0
-9 0 -2 -4 8 1
0 0 0 0 1.0000 0
0 0 0 0 0 1. 0000
0.8926 -0.9856 -0.3569 -0.7539 -1.0689 -0.3539
From R , we see that g is a linear combination of the first six vectors. This corresponds to G be ing a linear combination of the first six matrices. To verify this, compare G with the linear com bination: » G G = -9 -2 0 - 4
•/, The matrix G. 8 1
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R (1 , 7) * A + R ( 2, 7 )* B + R( 3, 7) *C + R( 4, 7) *D + R( 5, 7) *E+ R ( 6 , 7 ) * F
ans = -9.0000 0.0000
-2.0000 -4.0000
8.0000 1. 00 0 0
(c) Any set of 9 or more matrices in M 42 is linearly dependent. This can be tested in the same manner as in (a) and (b). (d) See solution for 45. 14. (a)
» A = zeros(6,8); » A(1,1) = -1 A(2,1) = » A(2,2) = -1 A(3,2) = » A(4,3) = -1 A(5,3) = » A(5,4) = -1 A(6,4) = » A(l,5) = -1 A(6,5) = » A(5,6) = -1 A(1,6) = » A(5,7) = -1 A(2,7) = » A(3,8) = -1 A(4,8) = A = 0 0 -1 0 1 -1 0 0 0 1 0 0 0 0 -1 0 0 0 1 -1 0 0 0 1
1 1 1 1 1 1 1 1
•/, There are 6 nodes and 8 edges. % Edge 1 leaves node 1 and enters •/. Edge 2. */. Edge 3. •/. Edge 4. •/. Edge 5. •/. Edge 6. •/. Edge 7. •/. Edge 8. - 1 1 0 0 0 1 0 0 0 0 0 0 0 - 1 - 1 1 0 0
0 0 -1 1 0 0
0 >)
» rref([A(:,1) A(:,7) A(:,4) A(:,5)]) ans = 0 1 1 0 0 1 0 -1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0
» r r e f ( [ A ( : , l ) A(:,7) A ( : , 6 ) ] ) ans = 1 0 0 0 0 0
0
1 0 0 0 0
1 1 0 0 0 0
In each case, and for any closed loop, the columns are linearly dependent.
Linear Independence
MATLAB
4.5
(c) One such set has edges 1, 2, 7, and 8 . » rref([ A(:,l) A(:,2) A(:,7) A(:,8) ]) ans = 1 0 0 0 0 0
0
0 1 0 0 0 0
0 0
0 1 0 0 0
0 1 0 0
As long as there are no closed loops, the set columns will be linearly independent. (d) »
A = zeros(5,8);
» » » » »
= = = = =
A(l,l) A ( 2 , 1) A(3,4) A(4,2) A(5,8)
-1; A (l,3 ) 1; A ( 2 , 2 ) 1; A ( 3 , 7 ) 1; A ( 4 , 3 ) -1; A(5,6)
*/. There are 5 nodes and 8 edges. -i; */, Node 1. = 1; A ( 1 , 4 ) = ■ •/.Node 2. •/•Node 3. = -1 1 1; */• Node 4. = 1; A ( 4 , 8 ) = 1 •/.Node 5.
= 1; A ( 1 ,6) = -1 » = -1 ; A(3,5) = -1 ; A(4,7) = - 1; A ( 5 , S )
A = -1 i 0 0 0 »
0 - 1 0 1 0
1
- 1
0 -
0 1 0
0 1 0 0
0 0 -1 0 1
1 0 0 0 -1
0 0 -1 1 0
0 0 0 1 -1
r r e f ( [ A( : , 4 ) A( : , 5 ) A ( : , 6 ) ] ) */. Edge:s 4, 5 and
ans = 1 0 0 0 0
0
- 1 1 - 1 0 0 0 0 0 0
» rref([ A(:,l) A(:,3) A(:,6) A(:,4)] ) '/, Edges 1, 3, 6 and 4 have no loops, ans = 1 0 0 0 0
0
0 1 0 0 0
0 0
0 1
0 0 0
1 0
The columns corresponding to a loop were dependent, and those corresponding to a set with no loops were independent. (e) If A is the incidence m atrix for a digraph, then its columns are linearly independent if and only if the digraph has no cycles.
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S e c tio n 4.6
1. no; two polynomials cannot span P-¿, see Solution 4.5.41.
/
0
\ 01 1/
and (
(
0 1-5\
2 . yes; they are independent since det I —3 0
= 3(1 + 5) = 18
0\
/ 1'
0; they span since I —3 1 , 1 0
V 0/ \1
-5 \ 0 I span IR3.
3. no; as x 2 — 3 = (x 2 — 1 ) -f (x 2 —2), they are dependent. /I 1 1 1' 4. yes; as det
0010 I
they are independent and span the 4 dimensional space P 3 , by Theo-
\0 0 0 1
rem 5. 5. no; three polynomials cannot span P 3 . (Any linear combination of these three will have ax 3 —4ax -f bx2 + c. So x3 or x not in span.) —
( ; s) < -
a4 ^ 0 - 0
{(2
- C i )
; ) ■ & ) •
( 1
¿) •(s . })}■
■■
( i ( i
¡ h ’ t i i6
,+
can not always be solved.)
7. As abcd ^ 0, then a, 6, c, and d are all nonzero. Thus, c\
0 implies c, = 0 for each
i.
+ c2
+ c3 ^
+ c4
^
=
( “ ^) = ^ (0 0) + f (0 0) + í ( ^0) + 5 (0 2)' HenCe they are
independent and span, so form a basis for M 228 . As M 22 has a basis consisting of 4 matrices, then theorem 2 implies every basis for M 22 contains 4 matrices. Thus the given set of matrices is not a basis for M 22. (Any 5 matrices in M 2 are depen
dent.) 9 . yes; given (x, y) E P , then (x, y) = (x, - x ) = x (l, - 1 )
10. no; they are dependent since (—3, 3) = —3(1, —1), so not a basis.
11
(!)=(>,_i)■*(s)+'(J);{(0■ (J)}-*b — /« \
/(2 /3 )„ -2 « \
12
(;)■(
13
(i)=( /s = 3; i/ = 3 —3 = 0
p=
2 ; í /
=
3
—
2
=
1.
= > p = l ; i / = 3 —1 = 2
1 -1 2 3\ 143 =^p = 2;j/ = 4 - 2 = 2 0 0 0 0/ 0
1 -1 2 3\ 0 143 =* p = 3 ; í/ = 4 - 3 = 1 0 0 0 1/ 2 3\ 0 5/2 =>p = 2;t/ = 2 —2 = 0 0 0/
10.
= - 1 ^ 0 = > p = 4;i/ = 4 - 4 = 0
11.
/l- l 2 3\ 0 - 1 3 3 0 1-3-3 \ 0 —1 3 3 /
12.
13.
1-1 2 3\ -2 2 - 4 - 6 2 - 2 4 6 3 -3 6 9/
0 0\
0\
/ -1 -1
0
3
0
1
0 -4 -2 1 0 -4 0 4 /
4/ •/o = 3; v = 4 - 3 = l
0
Vo
000
=>p = 2;í/ = 4 —2 = 2
0 0 0>
/ 1 —1 2 3' 0 000 = > / ? = 1 ; f = 4 —1 = 3 0 000 \0 0 0 0,
/ - i -1
0 0 2 4 0 -2 3 -1 0
/I -1 2 1' 0-133
0
23
/ -1 -1 0 0 -4 -2 0 0
1
0 2 3 0 -2 - 3 /
-1 0 0\ 0 -4 -2 1 0 0 2 3 0 0 0 0/
(~l —*
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
Section 4 .7
309
14. p = 2;i/ = 3 —2 = 1
/ 1 2 3\ / 123\ 15.I 0 0 4 I —►( 0 0 4 I =>/j = 2 ;í/ = 3 —2 = 1 \0 0 6 / V0 0 0 / For 16-21 choose basis for Range Aas columns ofA with pivots in echelon form or transposes of non-zero rows in echelon form of A*. 16. Basis for Range A = { ( j ) , (
})};
(3
1
0 |o)
(o 1
-6 | o )
(o
1 -3/2 | o ) ; Basis
for N a = 1 3 5\ 0 4 4 1; Basis for Range A = 0 0 0/
17. A* =
1 0 0 1
0 0
-
3/2
0'
1/2
0 ]; Basis for N a 0
0
18. Note that c-¡, = —2ci, and C3 = —ci, then Basis for Range A = °\ 0 1 => x\ = 2x 2 + £ 3 ; Basis for N a 0/
19. Note th at the first, second and fourth columns of A are linearly independent. Basis for Range A =
10 6 6 0 14 3 0 0 0 1
10
6 0
0 14 0 0 0 0 1
20. Note that by problem 11, dimCU = 2 and the first two columns of A are linearly independent. Basis for Range A <
( \ 0 - 1 -2 0 1 -3 -3 0 0 \0 0
f
/ 2\ 3
í x\
< Basis for N a = <
3 1 \ 0/
’
0
\lJ
j
0 0
0 0
°\
0 0
o/
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{( Basis for Range A =
21.
1\
—2 2 V 3/
f X2 — 2 x 3 —3x4 ^ Basis for N a — <
*2
>, as
gives iV^.
^3 V
X4
/
22. By problem 13, dim(7a = 3. Note that the first three columns of A are linearly independent. Basis °\ 2 -2 0/
for Range A = < /-i 0 4 3
-1 0 0 0 2 3 0 -2 1 -1 0 4
/I i °\ 0 1 _> 0 -4 0 0 ° 0/ Vo 0
0 0 -2 1 2 3
0 0
0 0 0/
Continuing the reduction in solution 13
/I 0 * 0 Vo
0 1 0 0
1 0 0 -1 1 3/2 0 0
°\ 0 0 0/
Basis for N a =
Basis
23. ( l —2 3^ 2-14 24. 3-3 3 2 1 0/ /I -1 2 0 25. 4-2 \7 -3
Basis:
1
0
1
2
1
3 -1/
/I 0 01\ 0 1 10 1 26. 1 -2 -2 1 1 VO 2 2 1 /
/l- l 1 - 1\ 0 2 - 2 3 0 2 - 2 5 \0 4 -4 6/
/I 0 0 1\ 0 110 0220 V0 2 2 1 /
/1 -1 1 - 1\ 0 2 - 2 3
0 Vo
0
0
2
0
0
0/
/ I 0 0 1\ 01 101 oooo Vo o o i /
n
Basis:
. Bas,s:
/A
A °\ -1 2 I > 1 -2 V—i / 3/
í°\
’
0 0
>
\2/ j
f°\
(°M 1 0 . ) ’ 1 0 1 A i J V o/ V I / -
10 0
p(A) = 2 ^ 3 = p((A, b)) =>■ No
p(A) = 2 = p{{A, b)) =>• Solution exists. /1 6 -1 -5
29.
4 -1 -1 Solution exists.
0 0 \0
-2 1
1
6-1-5 01-7 0 1 -7
p{A) = 3 = K ( A b ) )
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
(1 -2 1 1 3 0 2 -2 30. 0 4 -1 -1 U 0 3 -1 p((A ,b)) => No
2\
-8 | 1 0/
-2 1 1 0 6 -1 -5 0 4 -1 -1 \ 0 10 -2 -6
/I
2\ ’ 14 1 1 I -12 /
1 -2 1 1 0 6 -1 -5 0 0 1 -7 \ 0 0 1 -7
Section 4 .7
2
-14 -31 -34
P(A) = 3 /
4 =
solution.
31. Since A is a diagonal matrix, the nonzero columns are linearly independent. Then the number of nonzero components on the diagonal is equal to the number of linearly independent columns of A, which is the rank of A. 32. Since A is an upper triangular square m atrix with zeros on the diagonal, bottom row is all zeros, so less than n-pivots in echelon form. Thus p(A) < n. 33. p(A) — dim CU = dim R a = dim CU< = p(At ). 34. (a) p{Á) = dimi^yi < m — number of rows (b) v(Á) = n — p(A) > n — m 35. Let {vi, v 2, . . . , Vfc} be a basis for Range A. Since B is invertible, {Bv i, B v 2, . . . , Bvk} is a linearly independent set in R m and thus is a basis for BA. Then p(A) = p(BA). Since C is invertible, Range C = R n. Then if v £ Range A , there is an x £ R n such that Ax = v. But there also exists y £ R n such that C y = x. Then A C y = v. Then Range A C Range AC. If v £ Range AC, there is an x £ R n such that ACx = v. But then v £ Range A. So Range AC C Range A. Thus Range AC = Range A. Thus p(A) = p(AC). 36. Suppose b £ C a b • Then A B x = b for some x. Then b £ Ca because A y = b for y = Bx. Then Cab Q C a • Thus p(AB) < p(Á). Next, note that the íth row of A B is a combination of the rows of B. Then Ra b Q R b • Thus p(AB) < p(B). Thus p(AB) < min(/9(A ),p(B)). 37. Since p{A) — 5, p{A, b) = 5 for any 5-vector b. Then, by Theorem 7, Ax = b has at least one solution for every 5-vector b. 38, Let M \ , .. ., Mr be the matrices which represent the elementary row operations which would convert A to the reduced echelon form E\ That is, Mr • • •M \ A — E\. Let N i , . . . , N s be the matrices which represent the elementary row operations which would convert B to the reduced echelon form E 2 . Then N s • •N \ B — E 2 . Note that M i , . . . , Mr , N i , . . . , N s are all invertible matrices. Since p(A) = p(B), the number of nonzero pivot elements of E\ equals the number of nonzero pivot elements of E 2 , and thus the first p(Á) rows of each Ei have pivot columns with leading 1. Now elemen tary column operations on E \ , E 2 will bring both into the same form with &-ones on the diagonals of pivot rows and zeros elsewhere. Column operations are right multiplication by elementary matrices. Thus M r , . . . , M\AC\ ■C, = N , ■• •N i B D i •..£>* or ( Wf 1 • • - N ^ M r • ■■M1)A(CÍ ■CiD~k l Di ) = B. 39. This follows from problem 35. 40. Since any k + 1 rows of A are linearly dependent, p(A) < k. Since any k rows of A are linearly inde pendent, p{A) > k. Thus p(A) = k. 41. Suppose p(A) < n. If Ax = 0 has ony x = 0 as a solution, then by Theorem 8 , p(A) = n. This is a contradiction, so there must exist x ^ 0 such that Ax = 0 . Suppose there exists x / 0 such that Ax = 0. If p(A) = n, then, by Theorem 8 , x = 0 is the only solution to Ax = 0 . This is a contradiction, so p(A) < n. 42. Hypotheses mean Range A = R m. Then dim Range A = m = p{A).
311
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Ch ap ter 4 V ecto r Spaces
In stru cto r^ Manual
43. Suppose that B , the row echelon form of A, has k pivots in its first k rows. Since there are no other pivots, all the entries below the first k rows are zero. Let &2,m2>• • • > denote the pivots; let r i, r 2 , . . . , rjb denote the first k rows of B and suppose that c iri -f c2r 2 -f • • • + CkVk = 0. By the definition of a pivot, the m\ component in the vector 0 = c\Ti + h is ci a i >mi. Since &i,mi ^ 0, we conclude that c\ ='0. The m 2 component of the vector is ci&ip -\-c2b2)Tn2. Since c\ — 0 and b2>m2 ^ 0, we conclude that C2 = 0. Continuing in this manner, we see that Cj = 0 for j = 1 , 2, . . . , ib so the first k rows of B are linearly independent. Since all other rows in the row echelon form of A are zero, we conclude that p(A) = dim /í^ = Jb} as r i , . . . , r¿ are a basis for R a Now, suppose that p(A) = k. Let B equal the row echelon form of A. As above, the first k rows of B are linearly independent and all entries below the first k rows are zero. The first nonzero entry in each of the first k rows of B is a pivot, for if not, it would have been made zero by the row reduction of A to its row echelon form. Thus B has k pivots.
Rank and Nullity
Solutions
Calculator 4.7 313
CALCULATOR SOLUTIONS 4.7
The problems in this section ask you to compute the rank, range, row space and nullity of the given matrices. As usual, our Solutions for the TI-85 assume the matrix is in A4Inn. Then we compute r r e f Ann and read off the Solutions from the reduced row echelon form and the original matrix as follows: The nullity o f A (v(A)) is the number o f non-pivot columns in r r e f A. This follows from the fact that the description o f the nullspace obtained from r r e f A shows that each non-pivot column contributes one vector to a basis o f the nullspace by setting that non-pivot column variable to
1
and all other non-pivot variables to 0. So
dim (Na) = number o f non-pivot columns. A basis for the Range o f A (= CA - the column space of A) is given by the columns o f A corresponding to the
pivots in r r e f A. (This follows from the fact that the each vector in the basis for the nullspace described above shows how to write each non-pivot column vector from the original A as a linear combination of the pivot columns o f the original A. This shows the non-pivot columns are redundant and the pivot columns span CA. The pivot columns are independent since any linear combination of columns equal to 0 with zero coeffi cients in the non-pivot columns also must have zeros in the pivot columns from the description of N A in terms o f r r e f A). The rank o f A (p(A) or dim(CA)) is computed as the number of pivots in r r e f A, since each pivot con tributes one column o f the original matrix to the basis for CA described above. (Altematively you could use p(A ) + v ( A ) f # o f columns o f A.) A basis for the row space o f A ( R A) is given by the non-zero rows in r r e f A, since those rows are indepen dent (look at the pivot columns in those rows) and span the space R A ( = ^ r re f (A) by Theorem 5).
[[ . 3 7 [ .46 44. For A 4744: [ .,52 [ .,67 [[ 1 0 2 [ 0 1 -3 yields: [ 0 o 0 [ o o 0
.48
-.7
-.3 9
2.09
.83
]
.87
-1.57
1.04
]
.35
.29
-.3 3
]]
-1.16] r r e f A 4 7 4 4 [ENTER]
0 ] 0 ]
] ]]
So />(a4744) = 3, since there are 3 pivot columns and the ( ^4744 has a basis:
OO
- 1 . 16\ .46 -.39 .83 chosen as the columns of A4744 corresponding to pivots in the reduced echelon » ’ .52 .87 1.04 33 yj L-67J l *35J ^ - . jj form. r a4 744 has the first three (non-zero) rows of r r e f A 4 7 4 4 as a basis: { [ 1 0 2 0 ] , [ 0 1 -3 0 ] , [ 0 0 0 1] }. Finally, v(a 4744) = 4 - p(A4744) = 1 ( which also is the number of non-pivot columns in A4 74 4 ). [[ 45. F o r A 47 4 5 =
187 [- 3 5
[
-46
512
653
512
51 - 2 3 3 - 2 0 7 - 3 2 5
[[
1
0
2
3
[
0
1
-3
-2
257 - 1 4 8 958 1 0 6 7 1162 1.40809890249 ] -5.40620663555 ]
[
0
0
0
0
0
] ] , r r e f A 4 7 4 5 [ÍNTER] yields: ]]
]]
columns and the ran ge o f A4745 (Ca474 5) has a basis: -
r187i f _46"l 35 * 51 ► chosen as the columns of A4745 cor£57> 148>
314 Chapter 4
Vector Spaces
Instructors
responding to pivots in the reduced echelon form. Ra474s has the non-zero rows o f r r e f A4 7 4 5 as a basis: {[
0 2 3 1.40809890249
1
],
[ 0 1 -3 -2
-5.40620663555
]}.Finally,
v (a 4 7 4 5 )= 5 - p(a4745) = 3 ( which also is the number of non-pivot columns in A4 7 4 5). 37
46. ForA 4 7 4 6 =
[[ [ [ [ [ [
81 - 2 9
-48
91 3 0 6
53
215 - 4 7
-85
58
33 - 1 9
102 ]
38 2 0 5
0
-58 ]
-11
-38
423
99 ]
10 3 3 5 - 2 0
172
-80
316 594
-71
46
19 - 1 6 0 ]
, r r e f A474 6 [ENTER 1 yields:
7 339 442 - 1 1 9 ]
- 4 1 6 - 8 3 201
-88
144 ]]
1 0 0 0
-3 .10 3 8 4 73 4 7 3 2.93920187539
-.151801945062 ]
0 1 0 0
.625586755864 1.18104420224
.446191742079 ]
-.1 9 8 695556927 .553790907848
-.469903022938 ]
0 0 1 0 0 0 0 1
1.57599402542
-3.57508816282
.99737850334 ]
0 0 0 0
0
0
0]
0 0 0 0
0
0
0]]
f 37> r 81> í - 2 9 1 ( 5 8 1 306 38 -48 91 53 215 -47 -11 So />(a4746) = 4, since there are 4 pivot columns and the Ca4746 has a basis: > > > 335 -20 -85 10 594 -80 316 7 W7lJ v 46> 1-4 lóJ 1-83J chosen as the columns of A4746 corresponding to pivots in the reduced echelon form. Ra4746 has the four nonzero rows of r r e f A4 7 4 6 as a basis. Finally, v(A4746) - 1 - p(A4746)= 3 ( which also is the number of nonpivot columns in A4 7 4 6). [ 47. For A 4 7 4 7 =
.0284
-.0 3 1 1
- .0 2 0 7 .0431 .0615
]
-.0511
-.1216
- .1 8 1 1 .0904 .031
]
-.0965
-.4 2 7
.0795
-.011
-.5847
.0905
.1604
-.3365
-.4243
.3574 .216
] , r r e f A 4 7 4 7 [ENTER]
- . 4 7 3 .0305
]
.3101
.521
]]
yields: [ [ 1 0 0 0
45.7500000001
[ 0 1 0 0
87.7022036283
[ 0 0 1 0
-71.9680450647
[ 0 0 0 1
[ 0 0 0 0
- 1 . 6 8 3 5 0 1 68351E-12
0
r*H
t-H
O f
So p(A4747) = 4, since there are 4 pivot columns and the Ca4747 has a basis .0284> f-.0311> f-.0207> r.043iV .0904 -.0511 - . 1216 -.1811 - -.0965 -.427 » -.5847 * .3574 - chosen as the columns o f A4747 corresponding to pivots in the .0795 .1604 -.473 .0905 ,-.3365, 1-.4243J ,.310lJ reduced echelon form. RA4747 has the 4 non-zero rows of r r e f A4 7 4 7 as a basis. Finally, v (a 4 7 4 7 )= 5 - p( a 4 7 4 7 )= 1 ( which also is the number of non-pivot columns in A4 74 7) .
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
MATLAB
4 .7
M A T L A B 4.7 1. (i) Problem 7. » » ans
A = [1-123; rref(A)
1066];
=
1 0 0 (a)
0143;
0
6 1 0
4 0
6 3 0
The solution of A x = 0 is
/-ex
/-6x
X = £3 [ — 41 | 0/
-3
0
+X4
1/
So a basis of the nuil space of A is ' /-ex
/-ex
-4 1
-3
(b)
Let B be the m atrix whose columns are these vectors. We use the reduced echelon form of B to check that they are linearly independent. » »
B = [ -6 -6; -4 -3; 1 0; 0 1]; r r e f (B )
ans =
1 0 0 0
0 1 0 0
(c) See below. (d) The dimensión is 2. (ii) Problem 8. » A = [ 1 -1 2 3; 0 1 4 3; 1 0 6 5]; » rref(A) ans =
1 0 0 (a)
>
0 1/ >
1 ’ 0/
0
6 1 0
4 0
0 0 1
The solution of A x = 0 is x = x3
/-ex -4 \
0/
315
316
Instructor's Manual
Ch ap ter 4 V ecto r Spaces
So a basis of the nuil space of A is J
/-6\ ~4 1 . V 0/ j
(b) Let B be the matrix whose columns are these vectors. We use the reduced echelon form of B to check that they are linearly independent. » »
B = [ -6; -4; 1; 0]; r re f (B )
ans =
1 0 0 0 (c) See below. (d) The dimensión is 1. (iii) Problem 10. » »
A = [ 1 -1 2 3; 0 1 0 1; 1 0 1 0; 0 0 0 1]; rref(A)
ans = 1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
(a) The solution of Ax = 0 is x =
The basis is the empty set. (b) The dimensión is 0. (iv) Problem 11. » »
A = [1 -1 2 1; -1 0 1 2; 1 -2 5 4; 2 -1 1 -1] ; rref(A)
ans = 0 1 0 0
-2 -3
0 0
(a) The solution of Ax = 0 is (1\ ( 2\ 3 1 , 3 1 x = x3 + *4 0 1 U / \l)
T h e Rank, Nullity, Row Space and Column Space o f a M a tr ix
MATLAB
4 .7
So a basis of the nuil space of A is
(b)
Let B be the m atrix whose columns are these vectors. We use the reduced echelon form of B to check that they are linearly independent. » B = [ 1 2; 3 3; 1 0; 0 1]; » rref(B) ans =
1 0 0 0
0 1 0 0
(c) See below. (d) The dimensión is 2. (v) Problem 12. » A = [1 -1 2 3; ”2 2 -4 -6; 2 - 2 4 6; 3 - 3 6 9]; » rref(A) ans = 1 - 1 2 3
0 0 0 (a)
0 0 0
0 0 0
0 0 0
The solution of A x = 0 is 1 \ 1
/ - 3\ |
0 + *3
0
0/
1 0/
0 0
| + X4 V
i/
So a basis of the nuil space of A is
(b)
Let B be the m atrix whose columns are these vectors. We use the reduced echelon form of B to check that they are linearly independent. » B = [1-2-3; » rref(B) ans =
1 0 0 0
0
1 0 0 ; 0 10;
0 1 0 0
0 1 0
0 0 1] ;
317
318
C hapter 4 V ecto r Spaces
Instructor's Manual
(c) See below. (d) The dimensión is 3. (vi) Problem 13. » A = [-1 -1 0 0; 0 0 2 3; 4 0 - 2 1; 3 - 1 0 4 ] ; » rref(A) ans = 1.0000 0 0 1 .0000 1.0000 0 0 -1 .0000 0 1.0000 0 1 .5000 0 0 0 0
(a) The solution of A x = 0 is = x4
--1 1 -1 .5 1
So a basis of the nuil space of A is <
(b)
1 > 1.5 V 1/ J
Let B be the m atrix whose columns are these vectors. We use the reduced echelon form of B to check that they are linearly independent. » B = [ -1 ; 1 ; - 1 . 5 ; 1 ]; » rref(B) ans = 1
0 0 0 (c) See below. (d) The dimensión is 1. (vii) » A = [-6 -2 -18 -2 -10; -9 0 -18 4 -5; 4 7 29 2 13]; » rref(A) ans =
0 1 0
2 3
0
0 0 1
(a) The solution of A x = 0 is
x = x3
/-í) ~ 2\ -3 -1 1 + *5 0 0 -1 0/ 1J
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
MATLAB
4 .7
So a basis of the nuil space of A is - 2\ ( -- 11) -3 i 1 ) 0 -1 0 \) \ 0/
> >
(b) Let B be the m atrix whose columns are these vectors. We use the reduced echelon form of B to check that they are linearly independent. » B = [ -2 -1; -3 -1; 1 0; 0 -1; 0 1]; » rref(B) ans = 1 0 0 1 0 0 0 0 0 0
(c) See below. (d) The dimensión is 2. (c) Since we wrote the general solution of A x = 0 as a linear combination of these vectors, where the coefficients in the combination were the arbitrary variables, the general solution of A x = 0 is in the span of these vectors. Since any vector in the nuil space is a solution of A x = 0, these vectors span the nuil space. (d) The dimensión of a vector space is the number of vectors in its basis. In this case, that is the number of arbitrary variables in the solution to system A x — 0. (a)
(i) See answer to Problem l(vi) above. (ii) » R = rref(A); » B = [ -R(l,4); -R(2,4); -R(3,4); 1] B = -1.0000 1.0000 -1.5000 1.0000 »
'/. Notice that B is the same as the answer in Problem l(vi).
» A*B ans = 0 0 0 0 A B = 0 since B is in the nuil space. (b) (i) See the answer to Problem l(vii).
320
C h ap ter 4 V e c to r Spaces
Instructor’s M anual
(ü) » R = rref(A); » B = [[ -R(l,3); -R(2,3); 1;0;0] B =
»
-2
-1
-3
-1
1 0 0
0 -1 1
[ -R(l,5); -R(2,5); 0; -R(3,5); 1]]
'/• These were the same as the vectors in Problem l(vii).
» A*B ans =
0 0 0
0 0 0
We expect A B = 0 since the columns of B are in the nuil space of A. (] )
(ii) (a)
» A = [ 1 -1 0 3 -1 4; 2 0 1 7 2 1/2; 3 5 1 4 15]; » R = rref(A) R = 1.0000 0 0 2.0000 -1.0000 4.0833 0 1.0000 0 -1.0000 0 0.0833 0 0 1.0000 3.0000 4.0000 -7.6667 » C = R( [1:3] ,:) C = 1.0000 0 0 1.0000 0 0
0 0 1.0000
2.0000 -1.0000 3.0000
-1.0000 0 4.0000
4.0833 0.0833 -7.6667
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
MATLAB
4 .7
(b) » B = C'; » '/, Use rref(B) to check that the columns of B are linearly independent. » rref(B) ans =
1 0 0 0 0 0
0
0 1 0 0 0 0
0 1 0 0 0
(c) » »
*/• To check that each original vector is a linear combination */• of the vectors in B, we use rref to solve [B A(j,:)* ] for each j. » rref( [ B A(l , : ) ' ] ) ans = 1
0
0
0
1
0
0 0 0
0
1
-
1
0
1
0
0 0 0
0 0 0
0 0 0
» rref ( [ B A(2, :)>]) ans = 1
0
0
0 0 0
1 0 0
0 0
0 0
2
0
0 1
1
0
0
0 0
0 0
» rref( [ B A(3,:)’]) ans = 1 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0
0 (i)
0
0
3 5 0 0
0
(a)
»
A = [ 1 2 -1 3 1;
54-502;
-10120;...
123-20;
» R = rref(A) R = 1.0000 0 0 1.0000 0 0
0 0
0 0
68-233];
0 0 1.0000
-3.2500 2.5000 -1.2500
-0.2500 0.5000 -0.2500
0 0
0 0
0 0
327
328
C h ap ter 4 V ecto r Spaces
»
Instructor’s Manual
C = R( [ 1 : 3 ] , : )
C =
1.0000 0
0 1.0000
0 0 1.0000
-3.2500 2.5000 -1.2500
-0.2500 0.5000 -0.2500
0>) B = C*; '/. Use rref(B) to check that the columns of B are linearly independent. » rref(B) ans = 0 »
»
0 1 0 0
(c) » '/• To check that each original vector is a linear combination » '/• of the vectors in B, we use rref to solve [B A(j,:)* ] for each » rref( [ B A(l,:)']) ans = 1 0 0 1 2 0 0 1 0 1 -1 0 0 0 0 0 0 0 0 0 »
rref( [ B A(2,:)’])
0 0 1 0 0
-1 0 1 0 0
» rref( [ B A(3,:)']) ans = 5 0 o 1 4 0 0 1 -5 o 0 1 0 o 0 o 0 o o o » ans
rref( [ B A(4,:)’3) =
1 0 0 0 o
0 1 o o o
o 0 1 o o
1
2 3 0 0
T h e Rank, Nullity, Row Space and Column Space o f a M a trix » a n s
MATLAB
4.7
rref( [ B A(5,:)*]) =
1 0 0 0 0
0
0 1 0
0
6 8
1 - 2 0 0 0 0
0 0
For part (a), we expect the columns of B , which are the rows of C to be a basis for the span of the vectors by the argument in example 6. For part (b), since each column has a pivot, the vectors are linearly independent. For part (c), each of the systems [B v] had a unique solution. The coefRcients of the linear combination were the first three entries of the original vector. This is because the first three rows of B are e i, e 2 , and e 3 . 7. (a) Since the range of a m atrix is the same as its column space, by theorem 3, we may follow exam ple 6. The nonzero rows of rref (A1) form a basis for the row space of A f . Taking transposes gives a basis for the range of A. (b) From problem 7. (i) » A = [1-123; 0 1 4 3 ; » R = rref(A*) R = 1 0 1 0 1 1
0 0
0 0
1066];
0 0
» B = ( R( [1:2] , :) ) * B = 1 0 0 1 1 1
*/. Turn the nonzero rows into columns.
» */, Check that the each column of A is a combination of the columns of B: » rref([ B A(:,l)]) '/• The first column of A. ans = 1 0 1 0 1 0
0
0
0
» rref([ B A(:,2)]) ans = 1 0 - 1 0 1 1
0
0
0
» rref([ B A(:,3)]) ans = 1 0 2
0 0
1 0
0
*/, The third column of A.
4 0
» rref([ B A(:,4)]) ans = 1 0 3 0 1 3
0
'/. The second column of A.
0
*/• The fourth column of A.
329
330
C h ap ter 4 V ecto r Spaces
In stru cto r^ Manual
(ii) From 11. » '/. In each case above, there was a unique solution to [B A(:,j)] » A [ 1 - 1 2 1; -1 0 1 2; 1 -2 5 4; 2 -1 1 -1]; » R rref(A*) R =
1 -1 0 0
0 1 0 0
» B = ( R([1:2] , :) ) 9 B =
1 0 2 1
'/• Turn the nonzero rows into columns.
0 1 1 -1
» '/• Check that the each column of A is a combination of the columns of B: '/, The first column of A. » rref([ B A(:,1)]) ans =
1 0 0 0
0 1 0 0
1 -1 0 0
» rref([ B A(:,2)] ) ans = 0 1 -1 o 0 1 o o 0 o o 0 » rref([ B A(:,3)]) ans = 0 1 0 1 0 0 0 0 » rref([ B A(:,4)]) ans = 0 1 0 1 0 0 0 0
*/, The second column of A.
'/, The third column of A.
'/, The fourth column of A.
(iii) From 12. » '/, In each case above, there was a unique solution to [B A(:,j)]. » A = [ 1 - 1 2 3; -2 2 -4 -6; 2 - 2 4 6 ; 3 - 3 6 9 ] ; » R = rref(A*) R = 3 -2 1 0 0 0 0 0 0 0 0 0
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
»
B = ( R([1], :) ) ’
MATLAB
4.7
'/, Turn the nonzero rows into columns.
B =
1 -2 2 3 » */, Check that the each column of A is a combination of the columns of B: » rref([ B A(:,1)]) */. The first column of A. ans =
1 0 0 0
1 0 0 0
» rref([ B A(:,2)]) ans =
*/, The second column of A.
» rreí([ B A(:,3)]) ans =
'/. The third column of A.
» rref([ B A(:,4)]) ans = 3
'/, The fourth column of A.
0 0 0 (iv) From 13. »
*/, In each case above, there was a unique solution to [B A(:,j)]. [ - 1 - 1 0 0 ; 0 0 2 3; 4 0 - 2 1; 3 - 1 0 4 ] ; » A » R rref(A1) R =
0 1 0 0 » B = ( R( [ 1 :3 ] , B = 0 1 0 1
:) ) ’
*/, Turn the nonzero rows into columns.
331
332
Instructor’s Manual
C h ap ter 4 Vecto r Spaces
» V, Check that the each column of A is a combination of the columns of B: •/. The first column of A. » rref([ B A(:,l)]) ans = 0 -1 1 0 0 1 0 4 o 0 0 o */, The second column of A.
» rref([ B A(: ,2)]) ans -1
0 0
0 */• The third column of A. ans =
1 0 0 0 »
0 1 0 0
0 2 -2
o 0 1 o
0 */, The fourth column of A.
rref([ B A(:,4)])
ans
1 0 0 0
0 1 0 0
0 3 1 0
0 0 1 0
(v) » */, In each case above, there was a unique solution to [B A(:,j)] » A = round(10*(2*rand(5)-l)); A (:,2) = .5*A(:,1); » A(:,4) = A(:,1) - 1/3 *A(:,3) A = -6.0000 -3.0000 1.0000 -6.3333 1.0000 -9.0000 -4.5000 3.0000 -10.0000 -8.0000 4.0000 2.0000 -10.0000 7.3333 3.0000 4.0000 2.0000 -2.0000 4.6667 -2.0000 9.0000 4.5000 -9.0000 12.0000 4.0000 » R = rref(A1) R = 1.0000 0 0 1.0000 0 0
0 0
0 0
0 0 1.0000
-0.8617 0.2084 0.1764
-0.5651 -0.2866 0.7575
0 0
0 0
0 0
T h e Rank, Nullity, Row Space and Column Space of a Matrix » B = ( R([1:3] , :) ) * B =
1.0000 0 0
0 1.0000 0
M A T L A B 4.7
'/• Tura the nonzero rows into columns.
0 0 1.0000
-0.8617 0.2084 0.1764 -0.5651 -0.2866 0.7575 » */, Check that the each columnof A is a combination of the columns of B » rref([ BA(:,l)]) '/• The first column of A. ans = 1 0 0 - 6 0 1 0 - 9 0 0 1 4 0 0 0 0 0 0 0 0
» r r e f ( [ BA ( : , 2 ) ] ) ans =
'/• The second column of A.
1.0000 0
0 1.0000
0 0
-3.0000 -4.5000
0
0
1.0000
2.0000
0 0
0 0
0 0
0 0
» rref([ BA(:,3)]) ans = 1
0
0
'/. The third column of A.
0
1
1
0
3
0
0
1 - 1 0
0 0
0 0
0 0
0 0
» r r e f ( [ BA( :,4)3 ) ans =
'/. The f o u r t h column of A.
1.0000
0
0
1.0000
0 0 0
0 0 0
0 1.0000 0 0
» rref([ BA(:,5)]) ans = 1 0 0 1 0 1 0 - 8 0 0 1 3
0 0 »
0 0
0 0
-6.3333
0 -10.0000 7.3333 0 0 '/. The fifth column of A.
0 0
% In each case above, there was a unique solution to [B A(:,j)].
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8. (i) From problem 7. »
A = [1 -1 2 3; 0 1 4 3; 1 0 6 6] ;
(a) » R = rre f(A ) R = 1
0 0 »
0
1 0
6
6
0
3 0
4
'/• See problem 7 for rref(A*) and answer to part (b).
(b) » B = R( [1:2] ,:) B = 1 0 6 0 1 4
'/.a basis for the row space.
6 3
(d) The dimensions of the row space and the column space were both 2. (e) The number of pivots in rref (A) and rref (A*) are the same. (ii) From 11. »
A = [ 1 - 1 2 1; - 1 0 1 2; 1 - 2 5 4; 2 - 1 1 - 1] ;
(a) » R = rref(A) R =
»
1
0
0 0 0
1 0 0
-1 -
3
-2 -
3
0 0
0 0
'/. See problem 7 for rref (A1) and answer to part (b).
(c) A basis for the row space is given by the nonzero rows in R: » B = R( [1:2] ,:) B =
1
0 - 1 - 2
0
1
-
3
-
3
(d) The dimensions of the row space and the column space were both 2. (iii) From 12. »
A = [1-123; -22-4-6; 2 - 2 4 6 ; 3-369];
(a) » R = rref(A) R = 1 - 1 2 0 0 0 0 0 0
0 0 0
3 0 0 0
T h e Rank, Nullity, Row Space and Column Space o f a M a tr ix
»
'/, See problem 7 for rref (A*) and answer to part (b) .
(c) A basis for the row space. » B = R( [1] ,: ) B = 1 - 1 2
3
(d) The dimensions of the row space and the column space were both 1. (iv) From 13. »
A = [-1 -1 0 0; 0 0 2 3; 4 0 -2 1; 3 -1 0 4];
(a) » R = rref(A) R =
1.0000 0
»
0 1.0000
0 0
1.0000 -1.0000
0
0
1.0000
1.5000
0
0
0
0
'/, See problem 7 for rref (A1) and answer to part (b) .
(c) A basis for the row space. »
B = R ( [1:3] , : ) B =
(d)
1.0000 0
0 1.0000
0 0
1.0000 -1.0000
0
0
1.0000
1.5000
The dimensions of the row space and the column space were both 3.
(v) » A = round(10*(2*rand(5)-l)); A(:,2) = .5*A(:,1); » A(:,4) = A(:,1) - 1/3 *A(:,3) A = 4.0000 -5.0000 9.6667 5.0000 8 .00 00 2.5000 10.0000 1.6667 5.0000 0 4.0000 -6.3333 -5.0000 -5.0000 -2.5000 -5.0000 5.0000 -10.6667 -9.0000 -4.5000 3.0000 4.0000 -3.0000 5.0000 2.5000
(a) » R = rref(A) R = 1.0000 0.5000 0 0
0 0 0 »
0
0 1 . 0 0 00
0 0 0
1. 00 0 0 -0.3333
0 0
0 0
1.0000 0 0
0 0 0
*/• See problem 7 for rref (A1) and answer to part (b) .
MATLAB
4 .7
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(c) A basis for the row space. » B = R( [1:3] ,: ) B = 1.0000 0.5000 0 0 0 0
0 1.0000 0
1.0000 -0.3333 0
0 0 1.0000
(d) The dimensions of the row space and the column space were both 3. (d) The dimensions of the row space and the column space of A will always be the same. (e) The number of nonzero rows in rref (A) and rref (A*) is the same. The number of nonzero rows is the dimensión of the row space of a matrix. So this verifies (d) as row space of A! “is” column space of A after taking transposes. 9. (a) See MATLAB 4.4, Problems 3 and 7. (b) In each case, C will be the matrix formed as in problem 6. (i) » » A = [1-23; » rref(A) ans = 1 - 2
0 0
0 0
» B = A(:,[1 3] ) B = 1 1
-2
0
3
1
•/, The first matrix. - 2 4 - 6 ; 1 0 1] *; '/, Note the 9 to make rows into columns
0 1 0 % B is the lst and 3rd columns of A.
(ü) » rref(B) ans = 1 0 0 1 0 0
'/• These should be linearly independent.
( üí) »
R = rref(A*); C = R([l:2],:),j */, From problem 6, R has 2 non-zero rows.
(iv) » rref([B C] ), rref([C B] ) */• Both of these systems should be solvable. ans = 1.0000 0 0 -0.5000 0 1.0000 1.0000 0.5000 0 0 0 0 ans = 1 0 1 1 0 1 - 2 0 0 0 0 0
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
MATLAB
4 .7
(i) The 2nd matrix.
ii
-1 0 3 - 1 4 ;
0 1 0 0 0 0
2 0 1
0 0 1 0 0 0 '/• B is the first 3 columns of A.
,[1 2 3])
>
tu
V V
» A = [ 1 » rref(A) ans = 1 0 0 0 0 0
B = 1.0000 -1.0000 0 3.0000 -1.0000 4.0000
3.0000 5.0000 1.0000 4.0000 1.0000 5.0000
2.0000 0 1.0000 7.0000 2.0000 0.5000
(ii) '/• These should be linearly independent.
» rref(B) ans =
0 0 1
1 0 0 0 0 0
0 0 0
(iii) »
R = rref(A*); C = R( [1:3] ,:) *; */, From problem 6, R has 3 non-zero rows.
(iv) » rref([B C]), rref([C B] ) ans =
1.0000 0 0
0 0 1 0 0 o o o
0 0 1.0000 0 0 0
0 1.0000 0 0 0 0
0
0 1 o o o o
o 0 1 o o o
1 -1 o o o o
'/. Both of these systems should be solvable. 0.8333 -0.1667 0.1667
-0.1667 -0.1667 0.1667
0 0 0
0
3 5 1 0 0 o
0 0
-1.6667 1.3333 -0.3333 0 0 0
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(i) For 3rd Matrix. »
A = [ 1 2 -1 3 1; - 1 0 1 2 0 ; 5 4 - 5 0 2; 1 2 3 -2 0; 6 8 -2 3 3] » rref(A) ans = 3 1 0 -2 0 1
0 0 0
1 0 0
0 0 0
'/♦ B is the lst, 2nd and 4th columns of A. */, Since those have pivots.
» B = A(:,[1 2 4]) B = 1 1
2
2
-1 3 1
3 -2
0
(ii) */, These should be linearly independent.
» rref(B) ans =
0 0 1 0 0
1 0 0
0 0 (üí) »
R = rref(A*); C = R([1:3] ,:) 1; '/• From problem 6, R has 3 non-zero rows.
(iv) » rref([B C] ), rref([C B] ) ans = 0 1.0000
0
1.0000
0 0 0
0 0 0
Both of these systems should be solvable. 0 0
-0.2500
1.0000
0.2500 0
0
-1.0000
ans = 0 1 0 0 0
0 0 1 0 0
-1 0 1 0 0
1
2 3 0 0
0.5000 0.5000 0 0 0
-0.2500 0 0.2500 0 0
T h e Rank, Nullity, Row Space and Column Space of a Matrix
M A T L A B 4.7
(c) (Solutions only given for part (i) of (b). For (b.ii) note independence of columns of B follows from the fact that rref (B) consists of pivot columns of rref (A). For (c) count columns and compare with Solutions to MATLAB Problem 7. » A = [1 -1 2 3; 0 1 4 : » rref(A) ans = 6 6 1 0 0 1 3 4 0
0
» B = A(: ,[ B = 1
0
0
2] )
1
-1
0
1
1
0
» A = [ 1 » rref(A) ans = 1 0 0 0
-1
;
2! 1
0 1 0 0
-1
0
-2 -3 0 0
-1 -3 0 0
» B = A(:,[1 2] ) B = 1 -1 -1 0 1 -2 2 -1 » A = [1 -1 2 3; -2 2 -4 -6; 2 -2 4 6; 3 -3 6 9]; */, Matrix (iii) from #12. » rref(A) ans = 1 - 1 2 3
0 0
0 0
0 0
0 0
0
0
0
0
» B = A(:, [1]) B = 1
-2 2 3 » A = [-1 - 1 0 0; 0 0 2 3; 4 0 -2 1; 3 - 1 0 4 ] ; '/, Matrix (iv) from #13 » rref(A) ans =
1.0000 0
0 1.0000
0 0
1.0000 -1.0000
0 0
0 0
1.0000 0
1.5000 0
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C h ap ter 4 V ecto r Spaces » B B =
A( :, [1 2 3])
-1 0
0
4 3
-2
2
0
» A = round(10*(2*rand(5)-l)); A(:,2) = .5*A(:,1); '/• Matrix (v). » A(:,4) = A(:,1) - 1/3 *A(:,3) A = 1.0000 -3.0000 1.0000 -6.3333 -6.0000 - 8.00 00 -4.5000 3.0000 -10.0000 -9.0000 3.0000 2.0000 -10.0000 7.3333 4.0000 2 .0 0 0 0 2.0000 -2.0000 4.6667 4.0000 4.0000 4.5000 -9.0000 12.0000 9.0000 » rref(A) ans =
1.0000 0 0 0 0
0.5000
0 0 0 0
0 1.0000 o o o
1.0000 -0.3333
0 0 0
0 0 1.0000 0
» B = A(:, Cl 3 5]) B = -6 1 1 -9 3 -8 3 -10 4 -2 -2 4 -9 4 9
10. (i) (a) » A = round(10*( 2 * r a n d ( 5 , 3 ) - l ) ) ; » B = [A eye(5)] B = 1 0 3 -1 7 0 1 5 9 -7 0 0 -9 -10 5 0 0 10 5 4 0 0 8 5 7 » rref(B) ans = Columns 1 through 7 0 1.0000 1.0000 0 0 0 0 0 0 0 Column 8 -0.3785 0.3628 0.0442 -3.0505 5.7256
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
•/, Every row has a pivot so columns spai
0 0 1.0000 0 0
0 0 0 1.0000 0
0 0 0 0 1.0000
-0.1199 0.0315 0.0473 -0.4826 1.0631
0.3628 -0.3060 0.0410 2.3817 -5.6120
T h e Rank, Nullity, Row Space and Column Space o f a M a tr ix
MATLAB
4.7
(b) » c = B ( : , E l: : 5] ) c = -1 7 3 9 -7 5 - 10 -9 5 5 4 10 5 7 8
1 0 0 0 0
» */, The f i r s t t h r e e colum ns » rre f(C ) ans = 0 1 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0
0 1 0 0 0 o f C a re th e same as A . */, T h is s h o u ld be th e i d e n t i t y : */♦ I n f a c t i t s r r e f ( B ) ( : , 1 : 5 ) T h e r e f o r e colum ns o f C a xe a b a s i s . 0 0 0 0 1
(a) » A = [ 1 2 3 1; 2 8 9 3 ; » B = [ A e y e (4 )] B = 1 1 2 -1 0 2 8 1 0 3 9 -3 0 3 -1 1
1 1 -3 - 1 ] » ;
0 1 0 0
0 0 1 0
0 0 0 1
*/. E v e r y ro w has a p i v o t
» rre f(B ) ans =
0 0 1.0000 0
0 1.0000 0 0
1.0000 0 0 0
3.6667
0.3333
-1.0000
0
0.6667
0.3333
0 0 0
0
0
.0000
-3 .3 3 3 3
1.0000 -1 .3 3 3 3 -3 .0 0 0 0
(b ) » C = B (:, C =
[1 2 3 6 ] )
2 8 9
-1 1 -3
1
3
-1
0
0
0
» */, The f i r s t t h r e e colum ns o f C a re th e same as A . » rre f(C ) •/, T h is s h o u ld be th e i d e n t i t y . ans = 0 0 1 0 0 0 0 0 1 0 1 0 (c ) Since the colum ns m a k in g up the id e n tity span M n, the to ta l set o f vectors w ill s till span M n . Henee the set can be reduced to a basis b y ta k in g colum ns in B w ith p ivo ts in r r e f (B ) . Since the o rig in a l set is lin e a rly independent, the colum ns correspond ing to these vectors w ill have p ivo ts in them .
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11. (i) From problem 7. » » » C
A =[ 1 - 1 2 3; 0 1 4 3; 1 0 6 6 ] ; R =r r e f (A *); B = ( R ( [ l: 2 ] , :) )*; */• S o lu tio n from problem 7. C =orth(A ) % This i s an o rthogonal b a s is . = 0.2673 0.7715 0.5345 -0.6172 0.8018 0.1543
» r r e f ( [ C B ]) , r r e f ( [ B C]) '/• Both of th e s e show unique S o lu tio n s , sin c e ans = '/• P iv o ts a re in columns of th e f i r s t m atrix . 1.0000 0 1.0690 1.3363 0 1.0000 0.9258 -0.4629 0 0 0 0 ans = 1.0000 0 0.2673 0.7715 0 1.0000 0.5345 -0.6172 0 0 0 0 (ii) From 11. » A = [ 1 -1 2 1; -1 0 1 2; 1 -2 5 4; 2 -1 1 - 1 ]; » R = r r e f ( A 1); B = ( R ( [ l: 2 ] , :) ) ’ ; '/, S o lu tio n from problem 7. '/, This i s an o rthogonal b a s is . » orth(A ) C 0.2178 0.3592 0.1796 -0.5663 0.8980 -0.1307 0.1796 0.7841 »
r r e f ( [ C B] ) , r r e f ( [ B C] )
'/• Both of th e s e show unique S o lu tio n s .
0 0 0
0 1.0000 0 0
2.3349 0.7405 0 0
0.8980 -1.4811 0 0
ans = 1.0000 0 0 0
0 1.0000 0 0
0.3592 0.1796 0 0
0.2178 -0.5663 0 0
.0 0 0 0
(iii) From 12. » » » C
A = [1 -1 2 3; -2 2 -4 - 6 ; 2 -2 4 6 ; 3 -3 6 9]; R = r r e f ( A ') ; B = ( R ( [ l] , :) ) ’ ; '/. S o lu tio n from problem 7. C = orth(A ) */• This i s an o rthogonal b a s is . = -0.2357 0.4714 -0.4714 -0.7071
T h e Rank, Nullity, Row Space and Column Space o f a M a tr ix
» r r e f ( [ C B] ) , r r e f ( [ B C] ) ans = 1.0000 -4.2426 0 0 0 0 0 0 ans = 1.0000 -0.2357 0 0 0 0 0 0
MATLAB
4 .7
*/• Both o í th e s e show unique S o lu tio n s .
(iv) From 13. » » » C
»
A = [-1 -1 0 0; 0 0 2 3; 4 0 -2 1; 3 - 1 0 4 ] ; R = r re f (A O ; B= ( R ( [ l: 3 ] , :) )*; */♦ S o lu tio n from problem 7. '/, This i s an o rthogonal b a s is . C = orth(A ) = 0.1961 0.1531 -0.8295 0 0.7464 0.4392 -0.7845 -0.3636 0.0488 -0.5883 0.5359 -0.3416 r r e f ( [ C B] ) , r r e f ( [ B C]) .0000
0 0 0 ans = 1.0000 0 0 0
'/. Both of th e s e show unique S o lu tio n s.
0 1.0000 0 0
0 0 1.0000 o
-0.3922 0.6890 -1.1711
-0.5883 1.2823 0.0976
0
0
-1.3728 0.1723 -0.2928 0
0
o o 1.0000 o
0.1961 0 -0.7845
0.1531 0.7464 -0.3636 0
-0.8295 0.4392 0.0488 0
1.0000 0
0
0
12. We will use the method from problem 9, once we convert to vector terms as in earlier sections. (a) » A = [3 0 4 -1 ; - 1 0 0 -1 ; 0 -2 1 0; 4 1 3 0] ' ¡ » rre f(A ) ans =
1 0 0
0
0
»
1 0
0
0
0
1
0
0
'/,
1
allow s e n try as rows
0 0
1
'/, Since t h i s s e t i s l i n e a r l y independent.
I t i s a b a s is f o r i t s span.
(b) » A = [ - 6 4 -9 4; -2 7 0 -9 ; » rre f(A ) ans =
1 0
0
1
3
2
0 0
0
0
0
1
0
0
0
0
-18 29 -18 -19;
-2 2 4 0 ] ’ ;
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» B = A ( : , [ l 2 4 ]) B = -6
-2
4 -9 4
'/• The b a s is i s th e l s t , 2nd and 4 th elem ents.
-2
7
2
0
4
-9
0
For part (a), the set of all four polynomials is a basis. For part (b), the set of the first, second, and fourth m atrix form the basis. 13. (a)
» n = 4; A = round(1 0 * (2 * ran d (n )-T )) A= -5 2 -5 -4 -4 7 -2 -6 -3 -2 1 -7 0 7 - 1 1 » rre f(A ) ans =
1 0 0
0
0
1 0
0 0
0 1
0 0
0 0 1
» rank(A) ans = 4 (b) » B = A; B (:,2 ) = B (:,1 ) - 3 * B (:,4 ); » r re f(B ) ans = 1.0000 0 0 0.3333 0 1.0000 0 -0.3333 0 0 1.0000 0 0 0 0 0 » rank(B) ans = 3 B is not invertible since the columns are dependent, due to row of zeros.
(c) » B ( : ,3 ) = 2 * B (:,2 ); » r re f(B ) ans = 1.0000 0 0 1.0000 0 0 0 0
0 2.0000 0 0
0.3333 -0.3333 0 0
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
MATLAB
4 .7
» rank(B) ans = 2 B is still not invertible for the same reason. (d) Repeat for other n ’s. (e) The rank of A is the number of pivots in r r e f (A). (f) The n x n m atrix A is invertible if and only if the rank of A is n. (g) » » » »
n = 5; B (:,2 ) B ( : ,3) B (:,5 )
B = = =
= ro u n d (1 0 * (2 * ra n d (n )-l)); '/, C reate a random m atrix of s iz e 5. B ( : , l ) - 3 * B (:,4 ); */• Set 5-2 columns to be m ú ltip le s of o th e rs . 2 * B (:,2 ); 2*B(:,1 ) + 2 * B (:,2 );
» ra n k (B ) ans =
2 » » »
n = 6 ; B = round(10*(2* ran d (n )- 1 ) ) ; C reate a random m atrix of s iz e 6 . B (:,2 ) = B ( : , l ) - 3 * B (:,4 ); '/♦ Set 6-4 columns to be m ú ltip le s of o th e rs . B ( : ,3) = 2 * B (:,2 );
» ra n k (B ) ans =
4 14. (a) » A = ro u n d ( 1 0 * (2 * r a n d ( 2 ,3 )-l)) ; » r a n k ( A ) , ra n k (A * ) ans =
2 ans =
2 » A = r o u n d ( 1 0 * ( 2 * r a n d ( 4 ,5 ) -l) ) ; » r a n k ( A ) , ra n k (A * ) ans = 4 ans = 4 » A = ro u n d ( 1 0 * (2 * r a n d ( 7 ,3 )-l)) ; » rc m k (A ), ra n k (A * ) ans = 3 ans = 3
(b) »
n = 6;
» A = r o u n d ( 1 0 * ( 2 * r a n d ( n ) -l) ) ; » r a n k ( A ) , r^ L n k (A , ) ans =
6 ans = 6
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» A = round( 1 0 * (2 * ra n d (n )-l)); » A (:,2 ) = A ( : ,l ) - 3 * A (:,4 ); %reduce th e rank of A. » A ( :,3) = 2*A (:,2 ) ; » rank(A ), ran k íA 1) ans = 4 ans = 4 » A = round( 1 0 * (2 * ra n d (n )-l)); » A (:,3 ) = 2 * A (:,2 ); */. reduce th e rank of A. » A (:,4 ) = -1*A (: ,3 ) ; » A (:,5 ) = A (:,3 )-2 * A (:,2 ); » rank(A ), rank(A*) ans = 3 ans = 3 (c) rankCA^rankCA*). (d) Since rank(^4) is the dimensión of the column space, by problem 8, this should be the same as rank(j4;), which is the dimensión of the row space. 15. Using A forthe augmented matrix, and C for the coefficient matrix: » A= [ 1 - 2 3 11; 4 1 - 1 4 ; » C = A (: , [ 1 : 3 ] ) ; » rank(A ),rank(C ) ans = 3 ans = 3 » A= [-2 1 6 18; 5 0 » C= A( : , [1:3] ) ; » rank(A ),rank(C ) ans = 3 ans = 3
8
» » A = [3 6 - 6 9; 2 -5 4 » C = A (: , [ 1 : 3 ] ) ; » rank(A ),rank(C ) ans = 3 ans = 2 » A = [ 1 1 - 1 7; 4 » C = A (: , [ 1 : 3 ] ) ; » rank(A ),rank(C ) ans = 3 ans = 2
-1
2 - 1 3 10]; */. For problem 1, s e c tio n 1.3
-16; 3 2 -10 - 3 ];
6
/, For problem 2. ; 5 28 -26 - 8 ] ;
5 4;
6
1 3 20];
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
MATLAB
4.7
» »
% F o r p ro b le m 3. A = [ 3 5 1 0 ; 4 2 -8 0; 8 3 -1 8 0 ] ; » C = A ( :,[1 :3 ]); » ra n k (A ),ra n k (C ) ans =
2 ans
=
2 » A = [ 9 27 3 3 12; 9 27 10 1 19; 1 3 5 9 6 ] ; » C = A ( :,[ 1 :4 ]); » ra n k (A ),ra n k (C ) ans = 3 ans = 3
If the rank of A and C are the same, then the system has a solution. If the augmented m atrix has a higher rank, then the system has no solution. 16. (a) » » » ans
m 3 = m a gic(3 ) ;m 4 = m a gic(4 ) ;m 5 = m agic(5 ) ;m 6 = m a gic(6 ) ; . . . m 6 = m a gic(6 ) ;m 7 = m a gic(7 ) ;m 8 = m agic(8 ) ;m 9 = m a g ic (9 ); C ra n k(m 3 ) ra n k (m 4 ) ra n k (m 5 ) ra n k (m 6 ) ra n k (ra 7 ) ra n k (m 8 ) ra n k (m 9 )] = 3 3 5 5 7 3 9
To generate other magic squares we can take mi *, as transposing will only interchange row and column sums. Or we could interchange two row, say via m i[2 1 3 : i , :] and still have a magic square. Or any combination of transposes and row (or column) interdianges. As to patterns in the ranks, all of the other magic squares constructed from the mi will keep the same rank. So the sequence above will always be the same. Observe the odd i have rank (mi) = i, i.e. rank(m5) = 5 . The even i may not seem to have any clear pattern. We experiment some more, continuing to use the same notation. » m l0 = m a g ic (1 0 );m ll= m a g ic (ll);m l2 = m a g ic (1 2 );m l3 = m a g ic (1 3 );m l4 = m a g ic (1 4 ); » [r a n k (m lO ) r a n k ( m l l) ra n k (m l2 ) r a n k (m !3 ) r a n k ( m l4 ) ] ans = 7 11 3 13 9
Combined with the previous work these confirm the pattern for the odd order matrices and show » [ ra n k (m 4 ) ra n k (m 6 ) ra n k (m 8 ) r a n k (m lO ) ra n k (m l2 ) r a n k ( m l4 ) ] ans = 3 5 3 7 3 9
Thus it appears that ra n k (m a g ic ( 4 k ) ) = 3 while ra n k (m a g ic (4 k+ 2 )) = 2k+3. (b) » A = [1 2 3; 4 5 6; 7 8 9 ] ; » ra n k (A ) ans =
2
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» A = [1 2 3 4; 5 » rank(A) ans = 2
6
7
8
; 9 10 11 12; 13 14 15 16];
Each of the matrices will have rank 2. Since the difference between any two consecutive rows is n * [1 1 .. . 1], the first row and [11 • • • 1] form a basis for the row space. Henee rank= 2, always. (c) rank(u * v7) will always be one, provided u, v are non-zero. In fact the j ’th column of u * v 7 is Vju, múltiple of u. Henee u is a basis for the column space of u * v7 (provided some Vj / 0, and u * 0 ) . So dimensión of the column space is 1. » u = 2 * ra n d (4 ,1 )-1 ; v = 2 * ra n d (4 ,1 )-1 ; » A = u*v1; » rank(A) ans = 1 These matrices will always have rank 1, since all columns are multiplies of u. (Provided u ^ 0 and v ^ 0.) 17. (a)
» » »
n = 4; A = round( 1 0 * (2 * ra n d (n )-l)); m = 5; B = ro u n d (1 0 * (2 * ra n d (n ,m )-l)); B (3 ,:) = B ( l , : ) - B ( 2 ,: ) ; B (4 ,:) = B( 2 , : ) ; */• reduce th e rank of B.
Note that to reduce the rank of a matrix with fewer rows than columns we must make some rows equal to linear combinations of other rows, rather than columns. » rank(A ), ra n k (B ), rank(A*B) ans = 4 ans = 2 ans = 2 If A is invertible, and B has rank k ) then A B has rank k. This relates to problem 10 in MAT LAB 4.5 because the rank of B is the number of linearly independent columns in B, and the conclusión of Problem 10 says multiplication by an A preserves independence of a collection of columns. 0>)
» n = 6 ; A = ro u n d(10*(2*rand(n)-T )); » A (3 ,:) = A (l, : )~A(2, : ) ; A (4 ,:) = A(2, : ) ; */, reduce th e rank of A. » m = 5; B = ro u n d (1 0 * (2 * ra n d (n ,m )-l)); » rank(A ), ra n k (B ), rank(A*B) ans = 4 ans = 5 ans = 4
T h e Rank, Nullity, Row Space and Column Space o f a M a trix
MATLAB
4.7
(c) Form a 5 x 7 random m atrix A and make two rows by linear combinations of other rows. A reasonable conjecture is the rank of A B is the minumum of the ranks of A and B. (d) » A = [1 -1 0; 2 0 2; 3 1 4 ]; » B = [ 1 -3 2; 1 -3 2; -1 3 - 2 ]; » rank(A ), ran k (B ), rank(A*B) ans = 2 ans = 1 ans = 0 A refined (correct) conjecture is rank(AR) < min(rank(A), rank(jB)).
349
350
C hapter 4 V ector Spaces
In stru cto r^ Manual
S e c tio n 4.8
Note most Solutions compute transition matrix by computing C ~ l . An alternative is to find x# = C ~ lx by reduction (C | x) —> (I | C - 1 ^ ) for x = (xi • • •xn)t .
i.e. (x + y ) / 2 ( } ) + ( * - y )/ 21 ( _ * ) = ( * ) • o r _ (
2 3 \ r - i _ 1 r - 2 - 3 V U - 2 - 3 \ M _ /(-2 x -3 j/)/5 \ V -3 -2 ;’ _ 5 \ 3 2 y ’ 5 \ 3 2 , / ^ “ ^ (3x + 2y)/5 j
/2, —1/\/2)}. 3 . (i) if a = 6 = 0 , {( 1 , 0) , (0 , 1 )}; (ii) if a / 0 and 6 = 0 , {(0 , 1 )}; (iii) if a = 0 and 6 / 0 , {( 1 , 0)}; (iv) if a / 0 and 6 / 0 , {(b/y/a2 + 62, —a/y/a2 + 62)}
4. Let vi = ( ^b) an 6c
' a/y/a2 + 62 + c2 b/y/ a 2 + 62 + c2 c/y/a2 -b 62 -f c2
Henee
11. L is spanned by the vector
6/V 7Ó \ 5/V'70 0 0 \ —3 /\/7 0 /
\-3 /> /S > / 3/v/2l0 14/V210
and U3 =
and
is an orthonormal basis for L.
form a basis for H . Then u i =
and
12. The vectors
\2 /\/5 /
’- b/y/ a 2 + b2 ' a/y/a2 + 62 0,
is a basis for 7r. Then u i =
10. As abe / 0, then a ^ 0. Thus
/° \ 0 1 0
/W 5 \ o o o \ 2/y/b/ Q/V to \ h/y/70
/1 /V 5 \ 0 0 0
_2 _ 7 !
0 0
+
2/y/EJ
\ 1/
- 3 /\/7 0 /
To find 114 we compute V4 = v 4 — (v 4 * ui)ui — (v 4 • u 2)u 2
0
\
1/V2IÜ/ /1 /V 5 \
/ 0\ 0 0 1
(v 4 • U3)u 3 =
0 0 0 0
_8_
\4 /
\2 /v /5 /
+
12 770
6/> /^0\ 5/V70 0 0
V2T0
/-2/y/2Í0\ 3/1/2T0 14/-s/210 0
1/V 2 1 0 /
V -3 /V 7 Ó /
8/15 ^ 4/5 -4 /1 5
and
1
4 /1 5 /
/-8 /^ 6 5 \ 12/V465 henee u 4 = -4 /V 4 6 5 15/v^65 4 /V 4 6 5 / 13. A basis for the solution space is
and henee
forms a basis for
14. The set of vectors <
l /i - 1/2
( i/V5\ 0 1/ V 2 V 0/
1/2
1 2
1 1/2 V—1/2 .
7 ^ /6 6 is an orthonormal basis. l/i/66 -4/V66 ) )
l/\/Í2 \ 3 /\/l2 -i/V ñ l/y/ñ /
:rf -
O rth o n o rm al Bases and Projections in Mn
Section 4.9
1/VÍ2\ /-1 /3 3 /\/Í2 - l/v /Í 2 “ l/y/ñ / 2/3 j 15.
=
/ 2/3 1 / 3 - 2 / 3 \ 1/3 2/3 2/3 and \ 2/3 - 2 / 3 1/3 /
= I.
16. SinceP Q (P Q Y = P Q Q ' P * = P I P t = PP* = I then PQ is orthogonal.
18. As Q is symemtric and orthogonal, then QQX == QQ = Q2 = I. 19. Since Q is orthogonal then QQX = /. Henee, detQQ* = det Q = ±1.
det Q det Qx = ( det Q)2=
1, whichimplies
20. AA* = A2 = ( S*n * °OSJ N\ = f í for any real number t. \ c o s t —sin t J 1J 21. If v¿ = 0, then let c,- = 1 and Cj = 0 for j ^ Then we will have civi -f C2V2 + •••-+■ cnv n = 0 with d / 0, which implies the set of vectors is linearly dependent. 22. (a) From problem 2, {(l/%/2, —\/y/2)} is an orthonormal
(b)
basis for H . Thus
H 1 = {x G M2 : x - ( l / \ / 2 , —1/V2) = 0} = {(x,y) € M2 : x = y} = {(*,*) : x G IR}. So an orthonormal basis for H L is { (l/\/2 , l/\/2 )} .
(c) 23. (a) If a = b = 0 then{(l, 0), (0,1)} is an orthonormal basis for H . So in this case proj# v = v by theorem 4. If either a / 0 or b ^ 0 then {(b/y/a2 + 62, —a/y/a2 -f 62)} is an orthonormal basis for H ) and proj# v = 0. (b)For the case a = b = 0, H L = {0} by part (iii) of theorem 6. If a ^ 0 or b / 0, then Í71 = {x £ M2 : x • (6, —a) = 0} = {t(a,6) : t £ M}. Thus an orthonormal basis for H L is {{a/y/a2 -f 62, b/y/a2 + 62)}. (c) If a = 6 = 0, then v = v -f 0. If a / 0 or 6 ^ 0, then v = 0 A v. í / -b /y/ a * + b*\ ( —ac/y/ ( a 2 + b2)(a2 + b2 + c2) ' 24. (a) We may assume a ^ 0. By problem 10, < í a/y/a2 + b2 1 , I —bc/y/(a2 + b2){a2 A b2 + c2) l \ °/ \ ( a 2 A 62) / v/(a 2 -f6 2)(a2 + 62 + c2)> is an orthonormal basis for i/. So proj# v = 0 + 0 = 0. (b) Upon solving ther system Ui • x = 0, 112 • x = 0, we find {(a, 6, c)} is a basis for H L , and henee {(a/y/a2 -f 62 -f c2, b/y/a2 -f b2 -f c2, c/y/a2 + b2 + c2)} is an orthonormal basis. (c) v = 0 -f v. f/-2/V5\ 25. (a) From problem 6 we have < I 0 I 1 V 1 /V 5 /
/ 2 / V ^ / 3 5 \ '| , I 3\/5/7 I > for an orthonormal basis of 77, and henee V 4 V 5 /3 5 / J
/-2 /v /5 \ 5 A / 2 x/ 5 / 3 5 \ /-1 8 6 /4 9 \ projj/ v = 2\/5 I 0 + -^ 3x/5/7 = 75/49 . \ 1/V 5 / 7 \4 V ^ /3 5 / \ 118/49 /
370
C hapter 4 V ecto r Spaces
In structor’s M anual
(b) Solving the system u i • x = 0 ,112 • x = 0 gives {(3, —2,6)} for a basis of H L . So an orthonormal basis for H is {(3/7, —2/7,6/7)}. (c) v =
/ —186/49 \ 75/49 +
V 118/49/
/ / 3 /7 \ - 2 /7
•
/ —3 \ \ 1
/
3 /7 \ - 2 /7
V\ 6/7/ V 4/ / V 6/7/
=
/-1 8 6 /4 9 \ / 3 9 /4 9 \ 75/49 + -2 6 /4 9
\ 118/49/ V 78/49/
26. (a) An orthonormal basis for H is { (2 /\/2 9 ,3 //2 9 ,4 /\/2 9 )} . So proj# v = (18/29,27/29,36/29). í/-3 \ /-2 \) 0 gives < I 2 I , I 0 I> for a basis of H L . Henee 0/ v 1 /J
(b) Solving the system u j -x =
IV
-3/v/T3\ / -8//577\ ) 2 //Í3 , 0/
/18/29\
- 1 2 //3 7 7 > is an orthonormal basis for H L . 13/^/377/ J
V
1
- / — 8/v/377\
(~Z/s/Ü\
(c )T = i S ^ J " 7 1 í (
2/^ j - ^ w
/18/29\ / ll/29\
( i - 1I 2/ ' ^ 7 7l / = lV27/29 2 / 2 9/ l 3 //3 36/29 /l + l\ —7/29
—2 / \ / l 0 \ ^ i/V To \ ►forms an orthonor27. (a) From problem 9, the set of vectors 0 ’ 2 /v/ÍO 1 /v /í0 / J V - i /V S ó / 1 /5 ' 2 /V S o \ / —2 /-x/lO \ /1 /V 5 \ 5/V30 4 i 1 //Í0 I _ o -3 /5 mal basis for H . Henee proj# v = o 4/5 J o r v io 2/vw \/30 \ 2 / v /5 / l/v /ÍO / 17/5/ V -1 /V 3 Ó / ^ (b) Upon solving the system u, -x = 0, we find {(—2/v/Í5, l /\ /l 5 , —3 /\/l5 , l/\/Í5 )} is an orthonor mal basis for . /-2 /7 Í5 \ 4/5 \ l/5 \ 1/5 \ 6 - 2 /5 - 3 /5 l/x/15 | - 3 /5 (c) v = 4/5 4/5 + 6/5 - 3 /V Í 5 1 7 /5 / 1 / vT 5 / V 1 7 /5 / V—2 / 5 / 1/V5 0 o \2 //5
/0 \ 0 28. (a) The set of vectors < 1 < Vo/
/° \ 0 forms a basis for H ) and henee < 1
i 0
>
2 /V 3 0 \ 5/VSo I
Vo/
\3/
=
0 gives <
l V (c) v =
V 12/11/
V i
f-Z/y/22\
0 0/ 4 /1 1 ^ 4/11 3 +
-3/>/33
0 2 /v /2 2 / J —3/2
0/
is an orthonormal basis for H .
-
o V 3 //ÍI/ J
for a basis of H . So
V o/
1 //2
l/VTT
4/ll\ 4/11 3 12/11/
4 /H \ /° \ 4/11 _ o orthonormal basis. So proj# v = + 3 0 I V12 /1 1 / \0 / (b) Solving the system u¿ • x
/i/V T I^
3/22 3/22 0
4/11' 4/11
1/11
V 12/11/
- 1 5 /1 1 ' + , - 1/ 11,
is an
O rth o n o rm al Bases and Projections ¡n R "
29.
|ui -
u 2 |2 =
(ui -
u 2) • ( u i -
u 2) =
|ui|2 -
2ux -u2 +
| u 2 |2 =
1+ 0 + 1 =
2.
So
|ux -
Section 4.9
u 2| =
371
V2.
30. Use induction on n. If n = 1, then |u x|2 = 1. Suppose |u x + u 2 + • • • + u „ _ x|2 = n — 1. Then |u x + u 2 + • • • + u „ |2 = (u x + u 2 H 1- u„) • (u x + u 2 + f- u„) = u„ • ( u x + u 2 H 1- u„) + (u x + u 2 + • • • + u n_ x) • (u x + u 2 + • • • + u„) = 2u„ • (u x + u 2 + • • • + u „ . x) + u„ • u„ 4(u x + u 2 + • • • + u n_ x) • (u x + u 2 + • • • + u „ _ x) = 0 + l + n — l = n. By induction, this proves the result. 31. For linear independence we want a 2 + 62 ^ 0. For the vectors to form an orthonormal basis we need a2 + b2 = 1. 32. Suppose
^^
} *s an orthonormal basis for M2. Then a2+ 62 = c2 + d2 = 1, and ac + bd= 0.
We may assume a ^ 0. So c = —bd/a. Substituting this into c2+ d2 = 1 and solving for d gives
¿=±.,Th«., ( ') = (_ *)« ( ') = (-*). 33. Suppose |u 4 v| = |u| 4 |v|. Then |u -f v |2 = (u 4 v) • (u 4 v) = |u |2 4 2(u • v) 4 |v |2 = |u |2 4 2 |u||v| 4 |v |2. Thus u - v = |u ||v |, which implies u = Au for some scalar A, and henee u and v are linearly dependent. 34. |u 4 v |2 = |u |2 + 2(u* v) -f |v |2 < |u |2 4* 2|u||v| 4 |v |2 = (|u| 4- M )2. Taking square roots, we obtain |u 4- v| < |u| 4- |v|. 35. Use induction on n. For the case n = 1, we have Xi # 0, so dim span {xi} = 1. Assume the result is true for k = n. Suppose that |xi 4 x 2 4 • • - 4 x n+i| = |x i| 4 |x2| 4 • • - 4 |xn+i|.
(*)
We want to show |xi 4 x 2 4 • • • 4 xn | = |x i| 4 |x21H b |xn |. If we do not have equality, then by the triangle inequality, |x i 4 x 2 H h x n | < |xi | 4 |x 21H b |x n|. But adding |xn+i| to both sides would give |xi 4 X2 4 • • • 4 x„^_i| < |xi | 4 |x 214 • • • 4 |x„+i|, which contradicts (*). Thus, we have equality and henee, dim span {xi,X 2 , .. .,x n} = 1 by the induction hypothesis. As |xi 4 X2 4 h x n+i| = |xi 4 x 2 4 • • • 4 x n | 4 |xn+i|, then by problem #33, we have x n+i = A(xi 4 X2 4 • • • 4 x n). So x n+i G span {xi, X2 , . . . , x n}, and henee dim span {xi, X2 , ... ,x n+i} = 1, which proves the result. 36. By theorem 4 we have v = (v • u i)u i 4 (v • U2)u 2 4 |v |2 = v • v = [(v • u i)u i 4
h (v • u n)u n] • [(v • u i)u i 4
= (v • u i) 2(ui • Ui) 4 (v • u 2) 2( u 2 • u 2) 4 = | v - u i | 2 4 |v • u 212 H
b ( v * u n)un . Henee, h (v • u „ )u n]
h (v • u n)2(un • u n), since u¿ • u j = 0, i # j
h |v - u n|2, since u, -u, = 1.
37. Let v G H . Then for every k G H L, we have v*k = 0. Thus v G (H L)L, which shows H C (H ±)L . Suppose v G ( í f 1 )1 . Then v « k = 0 for every k G H L. By theorem 7, there exists h G H and p G H L such th at v = h 4 p. Thus v p = 0 = h ; v2 = [3 -5 0 0 5] 1; v3 = [ 2 1 4 1 3 ] » ; u l = v i / norm (vl) */, Normalize v i. = 0
-0.4170 -0.6255 -0.6255 0.2085 » u 2 = v2 u2 = 3.0000 -3.6957 1.9565 1.9565 4.3478
(v 2 ** u l)* u l '/, o rth o g o n a liz e .
» u 2 = u 2 / norm(u 2 ) u2 = 0.4276 -0.5268 0.2789 0.2789 0.6197
*/• norm alize.
1
.
O rth o n o rm al Bases and Projections in M n
MATLAB
» u3 = u3 - (v3 ’+uD +ul - (v 3 , *u2)*u2 u3 = 0.4682 1.6696 1.1749 -1.8251 1.3887 » u3 = u3 / norro(u3) u3 = 0.1507 0.5376 0.3783 -0.5876 0.4471 » »
A = [u l u2 u3] ; */• This i s th e m atrix o í v e c to r s . A**A */• Dot each column in A w ith every o th e r column in A. */, I f t h i s i s th e i d e n t it y , th en th e v e c to rs a re orthonorm al. ans = 1.0000 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 0.0000 1.0000 Check r r e f ( [A v i v 2 v 3 ]) is
[I e l c 2 c3] to verify v i ’s are linear combinations of u i ’s.
(c) » » » ul
v i = [ - 1 ; 2 ; 0 ; 1 ] ; v2 = [ 1 ; - 1 ; 2 ; 2 ] ; v3 = [1; -2 ; 3; 1]; v4 = [-1 ; 2; -1 ; 4] ; u l = v i / norm (vl) */. Normalize u l . = -0.4082 0.8165 0
0.4082 » u 2 = v2 u2 = 0.8333 -0.6667
(v 2 **ul)*ul '/, o rth o g o n a liz e .
2.0000
2.1667 » u 2 = u 2 / norm(u 2 ) u2 = 0.2657 -0.2126 0.6378 0.6909
'/• norm alize.
» u3 = v3 - (v 3 '* u l)* u l - (v3'*u2)*u2 u3 = -0.5424 0.0339 0.8983 -0.6102
4.9
373
374
In stru cto r^ Manual
C hapter 4 V ecto r Spaces
» u3 = u3 / norm(u3) u3 = -0.4466 0.0279 0.7398 -0.5025 » u4 = v4 - (v 4 , * u l)* u l - (v 4 , *u2)*u2 - (v4**u3)*u3 u4 = -0.8851 -0.6322 -0.2529 0.3793 » u4 = u4 / norra(u4) u4 = -0.7505 -0.5361 -0.2144 0.3216 » » ans
A = [ul u2 u3 u4] ; */• This is th e m atrix of v e c to r s . A'*A '/• Dot each column in A w ith every o th e r column in A. */• I f t h i s i s th e i d e n t it y , th en th e v e c to rs a re orthonorm al. =
1.0000 0.0000 0.0000 0.0000
0.0000 1.0000 0.0000 0.0000
0.0000 0.0000 1.0000 0.0000
0.0000 0.0000 0.0000 1.0000
(d) Repeat above with 4 random vectors. 2. The solution set for A x = 0 has x = l y will be: » » » » ul
vi =[1 ; v2 = [-3 ; v3 = [-1 ; ul = vi / = 0.7071 0.7071 0 0
» u 2 = v2 u2 = -1.5000 1.5000 1.0000 0
; 0 ; 0]; 0; 1; 0 ]; 0; 0; 1]; norm (vl)
1
(v 2 , * u l)* u l
» u 2 = u 2 / norm(u 2 ) u2 = -0.6396 0.6396 0.4264 0
—3z — l w with the
other variables arbitrary.
•/. y = 1 , z = w = 0 ; */. z = l , y = w = 0 ; •/. w = 1 , y = z = 0 ; '/, Perform Gram-Schmidt.
Soabasis for H
O rtho no rm al Bases and Projections in M n
MATLAB
4.9
» u3 = v3 - (vS ’+uD +ul - (v 3 , *u2)*u2 u3 = -0.0909 0.0909 -0.2727 1.0000 » u3 = u3 / norra(u3) u3 = -0.0870 0.0870 -0.2611 0.9574 » A = [u l u2 u3] A= 0.7071 -0.6396 0.7071 0.6396 0 0.4264 0 0
*/, The f i n a l b a s is . -0.0870 0.0870 -0.2611 0.9574
3. (a) Let / = y/a2 + ó2, which is the |v| and |z|. Then vi • = (—ab + 6a)//2 = 0. Also, |vi| = |v |// = /// = 1, and |v 2 1 = |z|/ / = l/l = 1. Henee the set is orthonormal. Since this is an orthogonal set of nozero vectors, it is linearly independent. Any set of two linearly independent vectors is a basis for M2. (b) » V = [1; 2] ; » v i = v / norm (v), vi = 0.4472 0.8944 v2 = -0.8944 0.4472 » w = [-3 ; 4] ; » p l = (w **vl)*vl pl = 1 2
v 2 = [ - 2 ; l]/n o rm (v )
*/, Not ic e t h a t v l ’*vl = 1.
» p2 = ( w ** v2 )* v2 '/, N otice t h a t v 2 , *v2 = 1. p2 = -4 2 » p r j t n ( w , v l ) ; p r in t -deps fig 4 9 3 b l.e p s » p r jtn (w ,v 2 ); p r in t -deps f ig493b2. eps
375
376
Instructor’s Manual
C h ap ter 4 V e c to r Spaces P is the prqjecüon of U onto V
P is the prqjection of U onto V
(C) » p l + p2 */• This should be w. ans = -3 4 » lin c o m b (v l,v 2 ,w ); p r in t -deps fig 4 9 3 c .e p s ; u = [0.447214:0.094427] v = [-0.894427:0.447214] w = [-3;4]
Note that in the p r j t n graphs the projection onto a vector is formed by dropping a perpendicu lar onto the vector. Then in the lincomb graph the parallelogram formed is a rectangle since v i , v 2 are perpendicular.
» w = [4; 2 ]; » p l = (w, * v l)* v l pl = 1.6000 3.2000 » p2 = (w, *v2)*v2 p2 = 2.4000 -
1.2000
'/• N otice t h a t v l '* v l = 1.
'/• N otice th a t v l ^ v l =
1
.
O rth o n o rm al Bases and Projections in Mn
MATLAB
4.9
*/, This should be w.
» p l + p2 ans = 4.0000 2.0000
(e) Your choice. (f) Using H as the span of
{v
=
a v i} ,
p x is in H and p 2 is in H L since
v 2 •v i =
0, and w = P Í + P 2-
4. (a) » v = [2 ; v = 0.8944 0.4472
1
] / norm ([ 2 ; l ] )
» w = [3; 5] ; » p = (w’*v)*v P = 4.4000 2.20 0 0
» norm(w-p) ans = 3.1305
'/. th e d is ta n c e from w to p.
(b) » » »
c = 2. 5; z = c*v; norm(w-z)
*/• This should be la r g e r th an |w-p|.
ans = 3.9564 (c)
» w =[-3 ; 2] ; » p =(w'*v)*v P = -1.6000 -0.8000 » norm(w-p) ans = 3.1305 » » »
c =1.5; z =c*v; norm(w-z)
'/, th e d is ta n c e from w to p.
*/, This should be la r g e r th an |w-p|.
ans = 4.5405 (e) Label p at foot of perpendicular dropped from w to line through v, w —p is this perpendicular, and w —z is the dashed hypotenuse. The diagrams show that if z = a v in H = span{v} is not proj# w = p then w —z is the hy potenuse of a right triangle with one side w —p. Henee |w —p| < |w —z|.
377
378
Instructor's Manual
C h ap ter 4 V ecto r Spaces
5. (a) » vi = [-1 ; 2; 3 ]; v2 = [0; 1; 2 ]; » zl = v l/n o rm (v l) zl = -0.2673 0.5345 0.8018 » z2 = v 2 - (v 2 * * z l)* z l; » z2 = z 2 /norm (z 2 ) z2 = 0.8729 -0.2182 0.4364 (b) » z = [-1 ; -2 ; » z**vl , z **v2 ans = 0 ans = 0
1
];
Since H is a two dimensional subspace is a nonzero vector in H L, it will form on the basis {z}.
of M3, H L will be one dimensional(3 — 2 = 1). Sincez a basis. The vector n is the resultof using Gram-Schmidt
(c) » n = z /n o rm (z ); » w = [1; 0; 0 ]; '/• Choose a v e c to r. » whl = (w **zl)*zl + (w, *z2)*z2 */, Use method 1. whl = 0.8333 -0.3333 0.1667 » wh2 = w - (w’+nj+n wh2 = 0.8333 -0.3333 0.1667
'/,Use method 2. (should be
th e same as 1 .)
(d) To get h drop a perpendicular from w to line through n. w —h can be represented as the arrow from h to w, which is the side of a parallelogram opposite to p, the projection of w. 6. (a) » » » B
u l = [0; -2 ; -3 ; -3 ; 1]; u2 = [3; -5 ; 0; 0; 5 ]; u3 = [2; 1; 4; 1; 3]; A = [u l u2 u 3 ] ; B = orth(A ) = 0.3906 0.2298 0.0157 -0.6509 0.4563 0.3293 0 0.7747 -0.1098 0 0.1937 -0.8814 0.6509 0.3184 0.3199
O rth o n o rm al Bases and Projections in R n
» B'*B ans = 1.0000 0.0000 0.0000
»
MATLAB
4.9
'/♦ V erify th a t th e columns a re orthonorm al. 0.0000 1.0000 0.0000
0.0000 0.0000 1.0000
x = r o u n d (1 0 * (2 * ra n d (3 ,l)-l));
Recall that A x is a linear combination of the columns of A, so it is in the range of A. To verify Theorem 4, we use the columns of B as the orthonormal basis. » w = A*x w=
'/, This and th e next d is p la y should be th e same
-11
-27 -43 -13 -14 » (w’* B ( : , l ) ) * B ( : , l ) + (w' *B(: , 2 ) ) *B( : , 2) +(w’*B(: , 3 ) ) *B( : , 3) ans = -11.0000 -27.0000 -43.0000 -13.0000 -14.0000
» A = 1 0 * (2 * ra n d (6 ,4 )-l); » B = orth(A ) B = -0.6351 0.1746 0.2858 -0.0476 0.4925 0.4137 -0.0451 0.0167 -0.5723 0.3622 -0.4933 0.5789 -0.6763 -0.5076 0.0353 -0.0611 0.5451 0.1089
» w = 1 0 * ( 2 * ran d ( 6 , 1 ) - 1 ) w= -4.4584 8.2763 0.5949 -0.7111 8.8196 -8.9983
0.2796 0.4788 0.4736 0.0574 -0.2394 -0.6385
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In stru cto r^ Manual
C h apter 4 V ecto r Spaces
» c = w'*B c = -6.3757 » z = c* z = -6.3757 1.3810 -3.2613 6.5911
1.3810
-3.2613
» p = B*z P = 3.1961 1.3539 2.3712 3.1414 3.9128 -7.3338
*/• The dot product o í w w ith each column in B. •/. (w’*ul w'*u2 w’*u3 w’*u4) 6.5911 */, So z=(w, *ul w**u2 w**u3 w, * u 4 ), =B, *w
% The p r o je c tio n o í w onto H i s */• (w .ul)ul+(w .u2)u2+(w .u3)u3+(w .u4)u4 */, Note p=B*B, *w combining p rev io u s ste p s
0>) » x = 1 0 * (2 * ra n d (4 ,1 )-1 ); » h = A*x h = 49.1183 19.2271 -38.5432 80.7957 80.5036 -54.4691 » norm(w-h), norm(w-p) ans = 135.5423 ans = 12.3026
'/• h i s in H.
'/• Compare,
The distance from w to proj# w is always smaller (or equal to) the distance from w to h, any arbitrary vector in H . (c) Since V4 is a linear combination of v i, V3 and z, it may be replaced by z in the basis for H. » z = A * [ 2 ; 0; -3 ; 1]; » C = [ A(:,[1:3]) z ] ; D » w = 1 0 * ( 2 * ran d ( 6 , 1 ) - 1 ) ; » p l = B*B’*w; » p2 = D*D **w; » p l - p2 ans = 1.0e-14 * 0.6661 -0.0888 -0.1776 0.4441 -0.1776 -0.3109
= o rth (C ); */, Use b a s is B. We le a re d in (a) t h i s is p r o je c tio n '/• Use b a s is D. '/, Compare. Get zero up to round o f í e r r o r .
O rth on orm al Bases and Projections in Mn
MATLAB
4.9
The projection of w onto H does not depend on which basis you choose. Here, p l —p2 has some small round off error. (d) The entries in B*w are the coefficients in definition 4 since u¿ • w = u¿ * w. To form the linear combination of the columns of B with these as multipliers, as in definition 4, form B ( B t w). (a) If a vector v is in the nuil space of A *, then A fv = 0. By taking the transpose of this we may replace it by v xA = 0. Since any thing in H can be written as A x for some x, we have that v x(Ax) = (vtA)x = Ox = 0. This means that v is orthoganal to anything in H or that v E H L. Conversely, if v E H L , we have that for any vector y E H y v*y = 0. For any vector x in the domain of A, A x is in H , so we have vM x = 0, or by taking the transpose x t(i4tv) = 0. This means that A tv is orthogonal to every vector x. This can happen only when A iv = 0, henee v is in the nuil space of A t . (b) » A = ro u n d (1 0 * (2 * ra n d (7 ,4 )-l)); » B = o rth (A ); C = n u ll( A ’ ); */, V erify t h a t columns of C a re orthonorm al. » C'+C ans = 1.0000 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 0.0000 1.0000 (c)
» w = r o u n d (1 0 * (2 * ra n d (7 ,l)-l)); » h = p = O C ’+w; */, This should be zero (up to round o ff e r r o r ) . » w - (h+p) ans = 1.0e-14 * 0.1776 -0.4441 0.5329 0.0444 -0.1776 0.0888 -0.0888 » h ’*p ans = - 5 . 3291e-15
'/, h ,p should be ( e s s e n ti a ll y ) o rth o g o n al.
» B*B * + C*C ) ans =
'/, This should be I .
(d)
1.0 0 0 0
0.0000
0.0000
0 .0000
0.0000
0.0000
0.0000
0.0000
1.0 00 0
0.0 00 0
0.0000
0 .0 00 0
0.0000
0.0000
0.0 00 0
0.0000
1.0000
0.0000
0.0000
0.0 00 0
0.0000
0.0 00 0
0.0000
0.0000
1.0 00 0
0 .0 00 0
0.0 00 0
0.0000
0.0000
0.0000
0.0000
0.0000
1.0 00 0
0.0 00 0
0
0.0000
0.0000
0.0 00 0
0 .0000
0 .0 00 0
1.0 00 0
0.0000
0.0000
0.0000
0 .0 00 0
0 .0000
0
0.0 00 0
1.0 00 0
(e) Since h = B B tyr and p = C C Xw, and since w = h + p, we have that w = ( B B X-f C C X)w. Since this is true for every vector w, it follows that I = B B X-f C C X.
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9. (a) The ith coordínate of v in this basis is u|v. Since uj is the ith row of B *, the vector of coordi nates is just B iw. (b) Since the norm of both u, and w are one, cos(0¿) = u , • w, where is the angle that u¿ makes with w. (c) (i) » deg = 180 /p i; » w = [1 ; l]/norra([l;l]) » v i = [ 1 ; 0 ] ; v2 = [0 ; 1 ] ; » acos( w, *vl)*deg ans = 45.0000
'/• The angle between w and v i in d eg rees.
'/, The angle between w and v2 in d e g re e s .
» acos( w, *v 2 )*deg ans = 45.0000 (¡i) » w = [- 1 ; 0]; » v i = [ 1 ; l]/n o rm ( [ 1 ; 1 ] ) ; » v 2 = [ - 1 ; l]/n o rm ( [ - 1 ; 1 ] ) ; » acos( w '*vl)*deg ans = 135
'/. The angle between w and v i in d eg rees.
*/, The angle between w and v2 in d eg rees.
» acos( w' *v 2 )*deg ans = 45.0000 (d) » B = [2 » B' * B ans = 1 0 0
2 -1 ; 2 -1 2; -1 2 2 ]/3 */, This should be th e i d e n t it y . 0 1 0
» s = [1 ; 1 ; 1 ]; » w = s/n o rm (s); » c = w' * B; » a c o s(c ) * deg ans = 54.7356 54.7356
'/, The co sin es of th e a n g le s. '/, E n trie s agree w ith acos ( (s **B( : , i) )/n o rm (s )) 54.7356
O rth on orm al Bases and Projections in K "
10. (a) » B = 1 /s q r t (2) * [1 1; 1 - 1 ]; » B* * B '/• This should be th e i d e n t it y . ans = 1.0000 0 0 1.0000 (b) » Bl = 1/14 * [-4 - 6 12; 6 -12 -4 ; 12 4 6 ] ; '/• This should be th e i d e n t i t y . » B l’ * Bl ans = 0 0.0000 1.0000 0.0000 1.0000 0.0000 1.0000 0 0.0000 (c)
» B2 = 1/39 * [-13 14 -34; -26 -29 -2 ; -26 22 19]; '/, This should be th e i d e n t it y . » B2* * B2 ans = 0 1 0 0 1 0 0 0 1 (d) » B3 = o r th ( r a n d ( 3 ) ); » B3> * B3 ans =
•/, This should be 0.0 00 0
1.0 00 0
0 .0000
0 .0000
1.0 00 0
0
0 .0000
0
1.0000
(e)
» » » » » B4
v i = [ - 1 ; 2; 3 ]; v2 = [0; 1; 1]; v3 = [ - 1 ; 2 ; u l = v l/n o r m ( v l) ; u 2 = v 2 - ( u l , *v 2 )* u l; v 2 = u 2 / norm(u 2 ) » u3 = v3 - ( u l **v3)*ul - (u 2 ’*v3)*u2;; u3 = u3 B4 = [ul u2 u3] = -0.2673 0.7715 0.5774 0.5345 -0.5774 0.6172 0.8018 0.5774 -0.1543
» B4> * B4 ans =
•/. This should be
1.0 0 0 0
0.0000
0.0000
0.0000
1.0 00 0
0.0000
0 .0000
0.0000
1.0000
MATLAB
4.9
383
384
C h ap ter 4 V e c to r Spaces
In structor’s M anual
11. (a) » B = B1*B2 B = -0.1905 0.6190 -0.7619 » B' * B ans = 1.0000 0.0000 0
0.6996 0.6300 0.3370
'/• This should be th e i d e n t it y . 0.0000 1.0000 0.0000
» B = B1*B3 B = -0.2959 0.3625 -0.4714 -0.8601 0.8308 -0.3589 » B’ * B ans = 1.0000 0.0000 0.0000
0.6886
-0.4689 -0.5531
0 0.0000
1.0000
0.8837 0.1950 0.4254 '/, This should be th e i d e n t it y .
0.0000 1.0000 0.0000
0.0000 0.0000 1.0000
» B = B2*B4 B = -0.4180 0.0989 -0.2604 -0.9654 0.8703 -0.2413
-0.9030 0.0148 -0.4293
» B1 * B ans = 1.0000 0.0000 0.0000
*/. This should be th e i d e n t it y . 0.0000 1.0000 0.0000
0.0000 0.0000 1.0000
» B = B3+B4 B = -0.4188 -0.0527 -0.2893 0.9540 0.8607 0.2950
0.9065 -0.0782 0.4148
» B} * B ans = 1.0000 0.0000 0.0000
*/• This should be th e i d e n t it y . 0.0000 1.0 00 0 0.0 00 0
0.0000 0.0000 1.0000
(b) See the solution to Problem 16 above.
O rth o n o rm al Bases and Projections in IR"
MATLAB
4 .9
(a) » B = l / s q r t ( 2 ) * [1 1; 1 -1] ; » A = inv(B) A= 0.7071 0.7071 0.7071 -0.7071
*/, B from MATLAB Problem 10 changed by Problem 11.
» A* * A ans = 1.0000 0
#/# This should be th e i d e n t it y . 0 1.0000
» A = in v (B l) A= 0.4286 -0.2857 -0.4286 -0.8571 0.8571 -0.2857 » A» * A ans = 1.0000 0.0000 0.0000
0
0.8571 0.2857 0.4286 */, This should be th e i d e n t it y .
0.0000 1.0000 0.0000
» A = inv(B2) A= -0.3333 -0.6667 0.3590 -0.7436 -0.8718 -0.0513 » A» * A ans = 1.0000 0.0000
'/, M atrices B l, B2, B3, B4 p re v io u s ly e n te re d */, in MATLAB Problem 10, and not changed.
0.0000 0.0000 1.0 00 0
-0.6667 0.5641 0.4872 */• This should be th e i d e n t it y .
0.0000 1.0000 0.0000
0 0.0 00 0 1.0000
» A = inv(B 3); » A* * A ans =
*/, This should be th e i d e n t it y .
1.0 00 0
0.0 00 0
0.0000 0.0000
1.0000 0.0000
0.0000 0.0000 1.0000
» A = inv(B4) A= -0.2673 0.5345 0.7715 0.6172 0.5774 -0.5774
0.8018 -0.1543 0.5774
» A' * A ans = 1.0000 0.0000 0.0000
*/, This should be th e i d e n t it y . 0.0000 1.0000 0.0000
0.0000 0.0000 1.0000
(b) The m atrix B is orthogonal if B %B = I. Since this is the definition of the inverse of B ) we see that A = B %is in fact B ” 1. Since A t = B y we then have A*A = B B ~ X — /, i.e. B ~ l is orthogo-
386
In stru cto r^ M anual
C hapter 4 V ecto r Spaces
13. (a) » det(B ) ans = -1.0000 » d e t(B l) ans = 1
» det(B 2) ans = 1.0000 » det(B 3) ans = 1 » det(B 4) ans = -1.0000 (b) Take the determinant of B 1,B = /, and use det (B %) = det (B ). We get det (B )2 = det (/) = 1, so det (H) = ±1. (c) Since the volume of the parallelepiped formed by Qu, Qv and Qw is that of the parallelepiped formed by u, v and w multiplied by | det (Q)|, and since det (Q) = ±1, they will have the same volumes. 14. (a) » Q = Bl; deg = 180/p i; » v = 2*rcm d(3,1 )—1; » w = 2 * ra n d (3 ,1 )-1 ; » norm (v), norm(Q*v) ans = 0.4455 ans = 0.4455
'/• compare th e le n g th s .
» acosCv'+w /(norm (v)*norm (w ))) * deg */, The angle between v and w ( in d egrees) ans = 75.1669 » qv = Q*v; qw = Q*w; » acosíqv'+qw /(norra(qv)*norm (qw ))) * deg */, The single between Qv and Qw ans = 75.1669 For any v and w, multiplication by Q will preserve both lengths and angles. (b) » Q = o r th ( 2 * r a n d ( 6 ) - l) ; » q ’*Q ans = 0.0000 1.0000 0.0000 0.0000 1.0000 0.0000 1.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0.0000 0.0000 0.0000
*/, This should be th e i d e n t it y . 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000
0.0000 0 0.0000 0.0000 1.0000 0.0000
0.0000 0.0000 0.0000 0.0000 0.0000 1.0000
O rth o n o rm al Bases and Projections in M n
» v = 2 * ran d ( 6 , 1 ) - 1 ; » w = 2 * ran d ( 6 , l ) - l ; » norm(v) , norm(Q*v) ans = 1.4703 ans = 1.4703
MATLAB
4.9
'/, compare th e le n g th s .
» acos(v**w /(norm (v)*norm (w ))) * deg */• The angle between v and w ( in d egrees) ans = 103.8352 » qv = Q*v; qw = Q*w; » acos(qv**qw /(norm (qv)*norm (qw ))) * deg */, The angle between Qv and Qw ans = 103.8352 Orthogonal matrices preserve angles and lengths.
(c) » » »
Q = o r th ( 2 * r a n d ( 6 ) - l) ; x = 2 * ran d ( 6 , 1 ) - 1 ; z = 2 *rand( 6 , 1 ) —1 ; xx = inv(Q )*x; zz = inv(Q )*z; */• Q i s th e t r a n s i t i o n m atrix from Q to S, */, so inv(Q) i s th e t r a n s i t i o n from S to Q. » norm (x-z), norm (xx-zz) '/, Compare, ans = 2.5292 ans = 2.5292
The distance between two vectors can be computed using coordinates with respect to any or thonormal basis. (d) Since multiplication by Q or by Q-1 does not change lengths, we expect that changing from one orthogonal basis to another will not increase any errors in the original representation. For example, if x is the true vector, and z is the computed vector, then the error would be |x —z|. In the new basis, this would be |xx —zz| which has the same size as |x —z|. (e) From the identity Q*Q = /, we get Q v Qw = (Qv)*Qw = v t(QtQ)w = v*w = v w. |QV| = y/Qv • Qv = y/w • v = |v|, and so acos(Qv • Q w /|Q v| |Qw|) = acos(v • w /(|v | |w|)). (f) |Qx —Qz| = |Q(x —z)| = |x —z|, as Q preserves lengths. Henee Q preserves distances between points. 15. (a) Choose an angle, say pi/ 4 and form V as in MATLAB 4.8.9(b) » V**V ans = 1 0
*/, This i s I , so th e r o ta t io n V i s o rth o g o n a l, f o r any an g le. 0 1
For the same angle form P , R ) and Y as in MATLAB 4.8.10(a). » P**P ans = 1.0000 0 0
*/, These w ill be I f o r any cingle so P,R,Y othogonal: 0 1.0000 0
0 0 1.0000
387
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Instructoras Manual
C hapter 4 V ecto r Spaces
» R'*R ans = 1 0 0 » Y’*Y ans = 1 0 0 (b) The standard basis is orthonormal, rotation preserveslengths and angles,and the columns of a rotation m atrix are just the rotations of the standard basis. Henee these ncolumns form an or thonormal set which must be a basis since the set has n independent vectors in the n dimensional space Mn. (c) The product of orthogonal matrices is also orthogonal, so the attitude m atrix will be orthogonal. (d) Since A is the transition m atrix from the satallite’s basis to the standard coordinates, A ~ 1 will be the transition m atrix back. » » » A
v i = [.7017; -.7 0 1 7 ; 0 ]; v2 = [.2130; .2130; .9093]; v3 = [ .1025; -.4 1 2 5 ; .0726]; */, Solve V = inv(A) * I f o r A. A = in v ( [vi v2 v 3 ]) = -0.4998 1.7793 0.3542 0.9910 0.2321 0.2321 1.3618 -2.9069 -2.9069
» A9*A ans = 11.6696 9.1339 -4.6179
'/, This should be I . 9.1339 8.6292 -3.9057
-4.6179 -3.9057 3.0865
Since A*A was not the identity, A is not orthogonal. This indicates an error. We could have checked [v i v 2 v3] '« [ v i v 2 v3] / I directly without finding A. (e) » » » » » » » A
ph = p i/4 ; P = [ cos(ph) 0 s in (p h ); 0 1 0 ; -s in (p h ) 0 cos(ph)] a l = - p i/ 3 ; R = [ 1 0 0 ; 0 c o s (a l) - s i n ( a l ) ; 0 s in ( a l) c o s (a l)] th = p i / 6 ; Y = [ c o s (th ) - s i n ( t h ) 0 ; s in ( th ) c o s (th ) 0; 0 0 1 ] A = Y*R*P*eye(3) = 0.9186 -0.2500 0.3062 0.8839 -0.1768 0.4330 0.3536 -0.3536 -0.8660
*/, P itc h . •/. R o ll. */, The yaw m a trix .
The ij th element of A* I will be the dot product of the zth column of A with the j th column of /. Since the columns have unit length, we can take the arccosine to compute the angles between them. »
acos( A* * eye(3) ) * deg
ans = 23.2837 104.4775 72.1705
100.1821 64.3411 27.8856
110.7048 150.0000 69.2952
•/. th e a n g le s, in d e g re e s. co l */, c o l. 2 f o r ( 0 , 1 , 0 ) , . . .
1
fo r ( 1 ,
0
,
0 ) 1
O rth o n o rm al Bases and Projections in Mn
MATLAB
4.9
(a) » x = 2 * r a n d ( 3 ,l ) - l; v = x / norm (x); » H = eye(3) - 2*v * v ' H= -0.0169 -0.9999 -0.0023 -0.9999 0.0169 -0.0023 -0.0023 -0.0023 1.0000 » H' * H ans = 1.0000 0.0000 0.0000
*/♦ This should be th e i d e n t it y . 0.0000 1.0000 0
0.0000 0 1.0000
(b) » n = 7; x = 2 * r a n d ( n ,l) - l; v = x / norm (x); */, P a rt (b) . » H = eye(n) - 2*v * v*; » norra( eye(n) - H’+H ) '/• This should be z ero : H'+H should be I . ans = 3 .4777e-16 (c) Compute H %H\ H f H = (P - (2vví )t )(7 - 2vvt) = ( II - 2vvt - 2vvt + ( -2 v v í ) (- 2 v v í ) = I - 4vvf+ 4vvív v t Since v*v = 1, the last term, 4vvtv v t is the same as 4vví , the second term. HeneeH %H = I. (d) This is a sample of the resulting plot, using » v v = [ 1 ; 1 ] ; x = [—1 ; 2 ] ; and plotting the reflected vector using line type
(e) Since H represents the reflection across a line, H — H 1. This can be proved by noticing that H = H t , and since H is orthogonal, H* = H " 1.
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C h ap ter 4 V ecto r Spaces
S e c tio n 4.10
An efficient altérnate route to the Solutions in Problems 4-8 is to compute A ty ) and solve A*Au = A*y by elimination, skipping (AtA )~ 1. . Then u
1. A =
^
68/21Y -19/42 J ’
y = (136 - 19x)/42 2 ^ = ( í l)-y= (55 - 2x)/7 Z1 1
3. A =
1\ 4
1 -2
( l ) - A'A =
( i 2 5 ) ' >
12 13 11 14
4 10 30\ - / 310 -270 50' [ 10 30 100 , (A*A)-1 = — -2 7 0 258 -5 0 | . Then 30 100 354 / 40 I 50 -5 0 10,
9 /2 \ u = | - 2 7 /5 I ; y = (45 - 54x + 10x2)/10
5. A
=
9508 -4182 -790 \ /4 7 /1 8 \ -4182 2169 393 . Then u = 25/12 , y = (94 + 75x + l l x 2)/36 -790 393 73/ Y H /36 /
1 2592
6. A
(i:!)- - (i)....
3 -4 54 \ -4 54 -334 , (A1A) -1 54 -334 2418)
=
1 21728
(l 1 1\ /-i\ -6 5 13 93' 1 3 9 2 , A* A = 13 93 469 | , (A*A)-1 1 5 25 . y = 1 93 469 3189 1 -3 9 4/ Vi 7 4 9 / 9577 270 —319 \ /-7 4 3 5 /2 7 1 6 \ 270 912 -142 I . Then u = -863/1358 1, y = (-7435 - 1726x + 641x2)/2716 -3 1 9 -1 4 2 37, 641/2716/
7. As with the linear approximation on page 420-421, Au = proju y. Then, Au_L(y — Au) => u = (AtA)- 1 A‘y(1 1 0
3 9 27 \ 0
0
1-11-1 1 2 4 8 \l 11 1 / ( 1944 1 -6 0 (A* A )" 1 = 2520 -1440 420 84x3)/252.
8. A
=
y
-6 0 1000 -150 -7 0
/ - 2\ / 5 3 5 4 , A* A = = 15 -2 \ 35 2/ -1440 420 \ -150 -7 0 . Then u = 1755 -525 252 -525 175/
5 15 35 99
15 35 99 275
35\ 99 275 795/
900 \ -426 , y = (900 - 426x - 27x2 + -2 7 84/
Least Squares Approximation
Section 4 .1 0
9. This is a generalization of problem 7. The same reasoning applies. 1 1 1\ / a \ 10. (a) Let y = a + bx + ex2. Then I 1 —1 1 \1 3 9 / \ c ) Note that 8.55 —5(—2) + 5a: + 1.97 a:2. /I 1 X\ -1 1 1 ,u= (b) A = 1 39 Vi - 2 4 /
11. A =
1.97(—2)2 = 26.43. So all four points lie on the same parabola y = 8.55 —
10 100 > 30 900 50 2500 , y = 1 100 10000
( 5.52' 15.52 I . Then a = 8.55, b = —5 and c = 1.97. 11.28 /
/ 8.55 \
( -5 V 1.97 /
!5 °\ 260 325 500
/ 5.52 \ 15.52 y = 11.28 . Then 26.43/
A* A = |
.Au| = 0.
0 => |y —
5 365 44125 \ /108.715' 365 44125 6512375 . Then u = 4.906 44125 6512375 1044960625/ \ -0 .0 1 ,
i G*7f\ / Vi 175 30625/ c = 108.715 + 4.906a: —O.Olx2, using c for cost.
(\ \\ 1 1.5 2.25 . Then u = 12. A y= 1 2.5 6.25 \1 16/ (a) «o = 10.898 ft. (b) vo = 60.947 ft./sec. (c) g/2 = -15.318 ^ g = -30.636 ft./sec.2
ylu= y => y —.Au=
10.898' 60.947 -15.318,
391
392 Chapter 4
Vector Spaces
Instructor’s Manual
CALCULATOR SOLUTIONS 4.10 Problems 13-16 ask you to find the linear regression line for some x-y data pairs and then problems 17-20 request the quadratic regression curve for the same data; each type is to be calculated to eight significant digits. To have the results listed to the requested accuracy we place the calculator in scientific mode with 8 digits to the right of the dec imal place by using the MODE menú or by keying in S C l:F lx 8 [ENTER] . Next we carryout the desired regres sion calculation using the functions LINR and P2REG from the STAT menú, both of which have x-list and y-list arguments. Two different approaches to entering the data and carrying out the calculations can be followed. Either we can do the (x,y) pair data entry together from the STAT menú and then perform the regression calculations as suggested in the text: 1.
We enter the STAT menú and prepare to enter the lists via: iSTATl (F2)
2.
We ñame the lists (with the problem number since we’ll have to reuse each one) via: X41013 [ENTER] Y41013 [ENTER]
3.
We then enter the (x,y) points in order by 57 [ENTER] 84 [ENTER| 43 [ENTER] 91 [ENTER] 71 [ENTER] 36 [ENTER] 83 [ENTER] 24 [ENTER[ 108 [ENTER] 15 [ENTER 1 141 |ENTER| 8 [ENTER] . (Wecan type [EXIT[ at this point to indícate we are done entering data, althogh this is not necessary.)
4.
Then we calcúlate the linear regression equation: (a) We go to the STAT CALC menú by [2nd[ [CALCl (if we stopped the previous step without an lEXITl ) or by (STÁfl (Fl) (if we used [EXIT[ ). (b)
[ENTER] [ENTERl (to accept the x-list and y-list entered previously; or insert a different x-list and/or y-list ñame before each [ENTERl .
(c)
[F2] to calcúlate and display the regression coefficents for the line y = a bx, or [MOREl (fT) to calcúlate and display the regression coefficients {c2, Ci, c0) for the 2nd degree polynomial y = c2x2 + c ix + c0 (yes the cofficients for the polynomial come out in this (reversed) order ).
+
Or we can enter the x-data into a list and the y-data (say for Problem 13) into a list from the [2nd] [LIST] menú by { 5 7 ,4 3 ,7 1 ,8 3 ,1 0 8 ,1 4 1 } [STO ] X41013 [ENTERl and { 8 4 ,9 1 ,3 6 ,2 4 ,1 5 8} [ST O ) Y41013 [ENTER] . Then compute the linear regression equation by literally keying in L I N R (X41013, Y41013) [ENTERl To see the coefficients of the linear regression line y = a + bxw c must enter SHWST [ENTER] which displays the last computed STAT CALC result. For the 2’nd degree polynomial regression equation from this data (i.e. for Problem 17) instead of LINR we key in P2REG (X41 0 1 3 , Y41 0 1 3 ): PRegC [ENTERl which calculates and displays the coefficients for the quadratic regression curve.
.
,
Although you are not asked to get the graphs of the regression curves, it is very easy to do and very helpful in assesing the reasonableness of the least squares fitting that has been done. After you perform the regression step above, you can look at the graphs by doing the following steps: G l. Goto the GRAPH RANGE menú by entering [GRAPH] [F2| and establish a reasonable graphing range which endoses the min and max of the x-list and the y-list. For Problem 13 and 17 this might be done by entering valúes xMin=0 @ xMax=150 Q yMin=0 @ yMax=100. The valúes you enter here should be nice valúes slightly outside the x-list and y-list limits. (If you want to have axes with tick marks you will have to set L abelO n on the GRAPH FORMT menú, and you should set xS cl= 10 (or 25) and vScl=10(or25) during the RANGE setting operation. The choices for scales should be chosen to mesh
Least Squares Approximation
Calculator4.10 393
nicely with the range valúes you choose and not generate too many tick marks.) G2. Now that the range and format are set, you can graph the data with [STAT] (F3| [F2l . (Or from the Home screen you could enter SCAT (X 41013,Y 41013).) G3. To draw the regression curve you enter [F4] from the STAT DRAW menú. (Note that this will draw the regression line if the last CALC was LINR and will draw the quadratic regression curve if the last CALC was P2REG.) 13. Once the data is in lists, l in r (X41013, Y41013) :SHWST lENTERl shows the regression line is y = a + bx witha = 1 1 6 .7 1 7 6 6 1 andb = - . 87933592. The scatter plot and regression line for this data look like:
14. Once the data is in lists, l in r (X 41014,Y 41014) : shwst [ENTER 1 shows the regression line is y = a + bx witha = 3 5 .9 3 5 6 7 5 5 4 6 andb = -8 3 .4 2 9 5 6 5 6 3 4 7 . The scatter plot and regression line for this data look like:
15. Once the data isin lists, LINR (X41015, Y41015) :SHWST [ENTER 1 shows the regression line is y = a + bx witha = -1 .1 1 9 3 0 5 7 6E+2 andb = 2 .1 3 3 2 6 5 2 7 . The scatter plot and regression line for this data look like:
394 Chapter 4
Vector Spaces
Instructors
16. Once the data is in lists, l i n r ( X 4 1 0 1 6 , Y 4 1 0 1 6 ) : S H W S T [ENTER 1 shows the regression line is y = a + bx with a = 1 9 4 2 1 5 7 5 6 andb = 1 . 1 9 2 0 6 5 0 7 . The scatter plot and regression line for this data look like:
17.
Once the Problem 1 3 data isin lists, P 2 R E G ( X 4 1 0 1 3 , Y 4 1 0 1 3 ) regression equation is y = c 2 x 2 + c i x + c 0 with {c2,C i,c0} =
:PR egC
[ENTER 1 shows the quadratic
{ 1 .3 5 7 3 9 1 6 0 E - 2 ,- 3 . 3 9 1 3 5 8 8 2 , 2 . 17421277E 2}.
The scatter plot and regression curve for this data look like:
18. Once the Problem 14dataisin lists, P 2 R E G ( X 4 1 0 1 4 , Y 4 1 0 1 4 ) regression equation is y = c2x2 + ct x + c0 with
:P R egC
[ENTER] shows the quadratic
{ C2 , C U C0 } = { - 5 . 9 5 4 8 1 0 7 3 e 1 , - 5 . 1 2 5 7 7 2 5 4 e 1 , 4 . 1 5 7 9 8 1 1 5 e 1 } .
The scatter plot and regression curve for this data look like:
19. Once the Problem 15 datais in lists, P 2 R E G ( X 4 1 0 1 5 , Y 4 1 0 1 5 ) regression equation is y = c2x2 + ci x + cq with
:P R egC
[ENTER 1 shows the quadratic
Least Squares Approximation
Calculator 4.10 395
(^2* Cj, Co) { 4 . 0 5643705E -5, 2 . 0 1 7 9 8 2 0 2 7 6 2 , - 5 . 38851804E1}.
The scatter plot and regression curve for this data look like:
20. Once the Problem 20 data is in lists, P2REG(X41020,Y41020) :PRegC [ENTER] shows the quadratic regression equation is y = c2x2 + cxx + c0 with [c2, c u c 0] =
{ - 5 . 7 7 5 1 4 1 1 9 , - . 7 0 7 8 3 6 0 6 , - 4 . 23378036E -2}.
The scatter plot and regression curve for this data look like:
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Instructor’s Manual
C h ap ter 4 V ecto r Spaces
M ATLAB 4.10 1. (a)
» » »
x y A
=[ 1 2 - 1 3 . 5 2 . 2 4 ] ’; = [2 . 5 4 - 1 .4 - 2 ] ’ ; = [o n es( 6 , l ) x] ;
V, See form ulas 3, 4 in th e t e x t .
(b) » u = inv(A**A) * A* * y u = 2.9535 -1.1813 » v = A\y v = 2.9535 -1.1813 (c) (Problem should say “and compare with
| y
—
A u | ” )
» norm( y - A*u) ans = 0.4066 » w = u + [ .1 ; - . 5 ] ; » norm(y - A*w) ans = 2.9712 The sum of squares of coordínate differences in y-coordinates between the data points and the least squares line is smaller then than for any other line.
(d) » B = orth(A ) B = 0.1599 -0.4433 0.3199 -0.2540 -0.1599 -0.8221 0.5598 0.0301 0.3519 -0.2161 0.6398 0.1248 » h = B*B1>*y h = 1.7722 0.5909 4.1347 -1.1810 0.3547 -1.7716
'/, This i s th e p r o je c tio n of y onto H.
Least Squares Approximation
MATLAB
4 .1 0
*/, Compare w ith h.
» A*u ans = 1.7722 0.5909 4.1347 -1.1810 0.3547 -1.7716
(e) Here is the plot generated by the code in the text:
The line seems to be a good fit to the data. (f) » [ 1 2.9] * u ans = -0.4722
2. (a)
» » » A
x = [10 30 50 100 175]’ ; y = [150 260 325 500 6 7 0 ]’ A = [ o n e s ( 5 ,l) x x.~2] =
*/. See form ulas (1 1 )-(1 2 ) in th e t e x t . '/• r e c a l l x . ~ 2 squares th e elem ents in x.
10
100
30 50
900 2500
100
10000
175
30625
(b )
» u = inv(A ’*A) * A'* y u = 108.7146 4.9064 -0.0097
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C hapter 4 V ecto r Spaces
Instructoras Manual
» v = A\y v = 108.7146 4.9064 -0.0097 » norm( y - A*u) ans = 15.4359 » w = u + [ .1 ; - . 2 ; - .0 5 ] ; » norm(y - A*w) ans
*/, Compare w ith u.
=
1 .6566e+03 As in problem 1, | y —Au\ is the minimum. » B = orth(A ) B = 0.0031 0.1576 0.0278 0.4100 0.0773 0.5786 0.3093 0.6334 0.9474 -0.2666 » h = B*B**y h = 156.8051 247.1472 329.7042 502.0374 669.3062 » A*u ans
0.8226 0.3711 0.0206 -0.4134 0.1197 Vt This i s th e p r o je c tio n of y onto H.
'/, Compare w ith h.
=
156.8051 247.1472 329.7042 502.0374 669.3062 Here is a graph of the data: » » » »
u = A \y; s = mi n ( x ) :( m ax (x )-m in (x ))/ 1 0 0 :m ax(x); f i t = u ( l ) + u (2 )* s + u ( 3 ) * ( s . ~ 2 ) ; p l o t ( x , y , ' w**, s , f i t ) 700
600
500
400
300
200
100 0
20
40
60
80
100
120
140
160
180
Least Squares Approximation
» Cl 75 75“2] * u ans = 421.9528 » [1 200 200 “2] * u ans = 700.7343
M A T L A B 4.10
*/. For x = 75.
'/. For x = 200
3. Problem 12. */, the data. » x = [1 1.5 2.5 4]* ; » y = [ 5 7 67 6 8 9 . 5 ] * ; '/, See formulas (11)-(12) in the text. » A = [ones(4,l) x x.~2] A = 1.0000 1.0000 1.0000 1.0000 1.5000 2.2500 6.2500 1.0000 2.5000 16.0000 1.0000 4.0000 » u = A\y u = 10.8977 60.9470 -15.3182
'/, See problem 1.
»
'/• needed for part (c).
g = u(3)*2
g = -30.6364
Even the estimates s(t) = u (l) -f u(2)t + u ( 3) t 2 = so + vqí \ g t 2 we get (a) the height is sq = 10.9, (b) the initial velocity is i>o = 60.9, and (c) gravity is g = —30.6. 4. (a)
» » »
r = 1.5; t = -3.8 xx = [x; r] ; yy = [y; t] ; A = [ones(7,l) xx]; uu = A\yy;
(i); (ii) Use the modified command p lo t( x , y , *, r , t , ’wo*, s , f i t , *-r *,s ,f i t l , *: b }) to get solid and dotted lines for visibility in black and white print outs. Here is a graph of the two different lines. The solid line is from problem 1, the dotted line is found using the additional data point, located near the middle bottom of the plot. Note the new point lies far away from the approximately linear plot of the original data. Thus it is an “outlier” .
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Chapter 4 Vector Spaces
Instructor^ Manual
(iii) The outlier moves the entire least squares line toward it and slightly away from fitting the other data. Since the original line so closely matches all but one point it seems like a better fit. (b) A different modification of Problem 1: » r=4.9;t=4.5; » xx = [x; r] ; yy = [y; t] ; » A = [ones(7,1) xx]; nu = A\yy uu = 2.1482 -0.3998 » ss = min(xx):(max(xx)-min(xx))/100:max(xx); » fit2 = uu(l)+uu(2)*ss; » plot(x,y,'w*1,r,t,’w o * ' - r ' , s s , f i t 2 , *:b*); » print -deps fig4104b.eps
Again the new point, idetified via a “o” on the plot, has a y valué far away from the generally expected position based on the original data and the original (solid) least squares approximation. This “outlier” causes the least squares approximation to rotate significantly (to the dotted line) and fail to fit the general trend of most of the data.
Least Squares Approximation
M A T L A B 4.10
5. (a)
» x = [-0.0162 -0.0515 0.0216 0.0628 0.0855 0.1163 0.1316 -0.4416]'; » y = [ -0.0315 -0.0813 -0.0339 -0.0616 -0.0919 -0.2105 -0.3002 -0.8519]'; » 1 = length(x); A = [ ones(l,l) x] ; */. For the least squares line. » u = A\y u = -0.1942 1.1921 » B = [ ones(l,l) x x.~2]; » v = B\y v = -0.0423 -0.7078 -5.7751
'/• For the least squaresquadratic.
» norm(y-A*u) ans = 0.4419
'/• The linear error,
» norm( y-B*v) ans = 0.1171 » » »
t
'/. The quadratic error,
s = min(x):( max(x)-min(x) )/100:max(x); fit = u(l) + u(2)*s; f it2 = v (1) + v(2)*s + v(3)* (s.~2) ;
We shall plot the lines using ’r-* and *b: * to make them solid and dotted. distinction on the black-and-white page. »
plot(x,y, ’w** ,s,íit,}x - } ,s,f it2, *b: *) ;
This will show the
*/, }x - }, 'b: * for printing.
The quadratic fit is much better; the norm is smaller and the *’s are much closer to the quadratic. The original data has a parabolic shape, not a linear shape. In fact it appears that the lower left point might even be an “outlier” for the quadratic fit. (Try refitting omitting this point).
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Chapter 4 Vector Spaces
(b) » x = [.32 -0.29 .58 0.71 0.44 0.88]'; » y = [14.16 51.3 -13.4 -29.8 19.6 -46.5]’; » 1 = length(x); A = [ ones(l.l) x] ; */. For the least squares line. » u = A\y u = 35.9357 -83.4296 » B = [ ones(l,l) x x.~2]; » v = B\y v = 41.5798 -51.2577 -59.5481
'/, For the least squares quadratic.
» norm(y-A*u) ans = 25.3326
'/, The linear error.
» norm( y-B*v) ans = 15.2469
'/, The quadratic error. *
» » » »
s = min(x):( max(x)-min(x))/100:max(x); fit = u(l) + u(2)*s; fit2 = v(l) + v(2)*s + v(3)* (s.~2) ; plot(x,y,}w* *,s,fit,*w-} ,s,fit2,*w:9);
The quadratic fit is slightly better; the norm is smaller and the *’s seem closer. 6.
(a) » x = [0:8]*; '/• Integers from 0 to 8. » y = [15.5 15.9 16.7 17.1 17.8 18.2 18.3 19.2 20.0]'; » 1 = length(x); A = [ ones(l,l) x] ; */• For the least squares line. » u = A\y u = 15.4867 0.5367
Least Squares Approximation » » »
M A T L A B 4.10
s = min(x):( max(x)-min(x))/100:max(x); fit = u(1) + u(2)*s; p l o t ( x , y , ’w-1);
(b) Solve 15.5 -f .5367# = 25, to get 17.7 which is in 1997. 7. » x = [600 600 700 700 700 900 950 950]»; » y = [40 44 48 46 50 48 46 45]'; » 1 = length(x); A = [ ones(l,l) x]; '/•For » u = A\y u = 40.8537 0.0066 » B = [ ones(l,l) x x.~2]; » v = B\y v = -78.0000 0.3200 -
» » » »
the least squares line.
'/• For the least squares quadratic.
0.0002
s = min(x):( max(x)-min(x))/100:max(x); fit = u(l) + u(2)*s; fit2 = v(1) + v(2)*s + v(3)* (s.~2) ; p l o t ( x , y , , s , f i t , *w— 1,s,fit2,'w: *);
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Chapter 4 Vector Spaces
(Note the two points with x = 600 are obscured by labelling). The quadratic curve seems to fit better since the data seem to have a distinct upward then downward pattern. So we may recommend that the temperature be choosen at the máximum valué of the quadratic, » t = -v(2)/( 2*v(3)) t = 800.0000
'/• y=at~2+bt+c=a(t+b/2a)~2+(c-b~2/4a) . =a(t+(b/2a))2+(c-(b~2/4a)).
(Rewriting a quadratic y = at2 + bt + c = a(t + provided a > 0 ).
¿ ) 2
+ (c — f“ ), shows the máximum is at t = —6 / 2 a,
8. » mile » 1 = length(xm); A = [ ones(l,l) xm] ; */. For the least squares line. » u = A\ym u = 290.7737 -0.3424 » » »
s = min(xm):( max(xm)-"min(xm))/100:max(xm); fit = u(l) + u(2)*s; plot(xm,ym,’w**,s,fit,,w-));
(b) The slope is —0.3424, so the record has decreased by .3 seconds per year on average. (c) Solve 290.8 - .3424* = 3 * 60: » x = (3*60 - u(l) )/u(2) x = 323.4915
This is the year 2123. 9. (a) » x » p » y » 1 » u u =
= [0:10]»; = [5.3 7.29.6 12.9 17.1 23.231.4 38.6 50.262.9 76.2]’; = log(p); = length(x); A = [ones(l,l) x] ; '/,Fortheleast squares line. = A\y
1.7322 0.2706
Least Squares Approximation » » » »
M A T L A B 4.10
s = min(x):( max(x)-min(x))/100:max(x); fit = u(l) + u(2)*s; fite = exp(fit); plot(x,p, ,s,fite, 'w-O;
The fit appears reasonable, although the recent trend (since about 1850) seems to grow less quickly than the fitted exponential). (ii)
» exp( u(l) + 15 * u(2)) ans = 327.1814
'/.The population in 1950.
(b) (i) The predicted population is higher than the actual population, in fact all the new population data lie well below the fitted graph in (a). (ii) We continué the x scale from (a). » x = [11:18]’; » p2 = [92.2 106.0 123.2 132.2 151.3 179.3 203.3 226.3]’; » y = log(p2); » 1 = length(x); A = [ ones(l,l) x] ;'/, For the least squares line. » u = A\y '/•If x=l:8 used, u(l) would be increased u = */•by 10*u(2), i.e. to 3.1141+1.286=4.4001 3.1141 0.1286 » » » »
s = min(x):( max(x)-min(x))/100:max(x); fit = u(l) + u(2)*s; fite = exp(fit); plot(x,p2,’w*’,s,fite,*w-’);
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Chapter 4 Vector Spaces
Instructor^ Manual
The population still looks exponential. The growth rate is the coefñcient in front of the x in the exponential. From part (a) this was .27 and in part (b) it is .13, which is significantly less. (iv) Using u from part (ii): »
exp( u (1) + 20*u(2))
ans
=
294.7384
» x = [.1164 .0121 .0562 .0931 .0664 .1728 .1793 .1443 .1824]»; » y =[.12128 .17185 .13365 .1485 .12637 .10406 .10703 .1189 .09952]»; » 1 =length(x); A = [ ones(l,l) x] ; */, Forthe leastsquares line. » u =A\y u = 0.1645 -0.3418 » s =min(x):( max(x)-min(x))/100:max(x); » fit = u(l) + u(2)*s; » plot(x,y,*w*»,s,fit,*w-»);
The least squares linear equation: Fe — Mg = .1645 — .3418Ca. This seems to fit the general pattern of the data, although the large deviations for smaller Ca valúes might raise some doubts.
Inner Product Spaces and Projections
S e c tio n 4.11 1. (i) (A, A ) = a h + a ¡ 2 +
> 0. (ii) (A , A) = 0 implies
h
n
if .A.
n
g -f* c¿¿) ~ ^
-f- ^ j iacg = (A, 5 ) + (A, C*).
*=i
t=i
(iv) Similarly, (A + J9, C) = (A, C) + (5 , C). (v) As
407
= 0 for each i, so A = 0. Conversely, n
0 then (A, A)— 0. (iii) (A, J3-T C ) — ^
Section 4.11
¿= i
= &uaix-, then (A, 5 ) = (5 , A) = (B , A), (vi)
( a A ,£ ) = ^ ( a a ¿ ¿) 6 ¿t- = a ( y ^ a „ 6 t, j = a (A ,£ ). (vii) (A ,a £ ) = (ai?, A) = a ( B , A ) = a(A, £ ) = w'=i
*=i
á (A ,£ ). 2. Suppose ||A|| = 1. Then y /(A,A) = + a¡2 H a2n = 1. Conversely, if (A, A) = 1, then ||A|| = 1.
+ a \2 H
h a \ n - 1. As (A, A) > 0, then
h
3. Let Ei be the n x n m atrix with 1 in the ¿, i position and 0 everywhere else. Then {E\, an orthonormal basis for Dn .
, . . . , En} is
4. As |A| = %/5, .he»
1/v| )
U,
= ( 2^
B' = B -(B ,U > W >
(" ” ) + ^
=
( VÍ
=
( ~ U / 0 22/“ )' “ d he"“ " I = ( ‘ U / '/5^ 22/^005)
5. u i = ( l/\/2 , i / \ / 2) and v 2 = (2 —«, 3 + 2z) —[(2 —i, 3 + 2¿) • u i]u i = (i, 1). So u 2 = («’/ v ^ , I/a/2). 6.
Start with the standard basis { l,x ,x 2 ,x 3}. From example 8 , ui = 1, u 2 = \/3(2x — 1), and 113 = ti 1 fi 3^ /3 \/5(6x 2 —6 x + 1). We have (v 4 ,u i) = / x 3 dx = - , ( v 4 ,u 2) — / (x 3 )[\/3(2x — 1)] dx = — —, and Jo
(v4, u 3) = J
(x 3 )[>/5(6x2 -
6x + 1)] = x 3 - ^ x 2 + | x -
6x
4
Thus v 4 = x 3 - \ - ^ [ V 5 ( 2 x - 1 )] - |^ [ \ / 5 ( 6 x 2 4 20
-f 1 )] dx = and \\v'4\\ =
7. Start with the standard basis { l,x ,x 2}. As J J
x/V2dx =
0,
llv 3 ll = [ /
8.
f 1í 3 3 2 3 1V , / ( x -----x H— x ------- ) dx
Jo V
2
l 2 dx =
2
then v 2 = v 2. As ||v2|| =
We have (v 3 ,u i) = J
, then
x2 d x |
111
(*a ~ l/3 ) 2dx j
1/2
20)
5
20V 7
= l / \ / 2 - Since
= ^/§ =
x 2 / \ / 2 dx = v^2/3, and (v 3 ,u 2) = J
/
Ja
(v2 ,ui)
(6
—
then u 2 = \ j ^ x -
] J ^ x 3 dx = 0. So V3 = x2 — 1/3
l 2 dx
= 6 —a, then ui = l \/6 —a.
2V3 x- a) 3/ 2
-(6
As
2
______
/ x¡\Jb — a dx = 2^ b ~ h t ’ then v '2 = x ~ \ ^ b + Ja ( f e - a ) 3/ 2 Henee u 2 = 2VZ
. Henee
= | y | , and henee u 3 = \ j \ { ^ 2 ~ !)•
We start with the standard basis { l,x ,x 2}. Since fb
Jo
and
= { /
+ a) . We have (V3 ,
Ui) =
[* _
+ °)
^ i/2
dx >
/ x 2/y/b — a dx = Ja
(v2 ,ui) = —
fe3 —a 3 ____ 3y6 —a
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Chapter 4 Vector Spaces
and (v 3 , u 2) =
2\ / 3 x2
Instructor’s Manual
x - ^ ( b + a) /( b - a)3/2dx =
(2 «
b ) + ^ ( a — 26)
^ ^
_
1 (b —a ) 5 / 2 V3 — (v 3 ,u i) u i - (v 3 ,U 2 )u 2 = x 2 — (a + 6 )x + - ( a 2 + 4a6 + 6 2) and Uv^U = — -^-j=— . Henee
u3 =
6-^5 (6
- a ) 5/ 2
x 2 —(a + 6 )x + —(o 2 + 4a6 + 6 2)
9. If A = (a¿j) and B = (bij), then (AB*)ij = ^T^a¿*6
so that tr (.AB*) = E E M
,
t= l j = 1
jfc=l
(i) ( A i 4 ) = t r ( ^ * ) = ¿ ¿ a ? j > 0 . *=1 i=i (ii) If tr (AA*) = 0 , then a'f- = 0 for all i and j, and henee A = 0. Conversely, if A = 0, then (A, A) = 0. (iii) (A, B + O) = tr (A(B + C)* = tr(A (B ‘ + C*)) = tr(AB* + AC*) = ti (AB*) + tr (AC*) = ( A , B ) + (A,C). (iv) A + B , C ) — tr ((A b )C*) = tr (AC* + B C ‘) = tr (AC*) + tr (BC*) = (A, C) + (B, C). (v) ( A , B ) = tr (AB*) =
= tr(£A *) = (B, A). *= 1 j = l
n
*= 1 J = 1
n
n
n
(vi) (aA ,j3) = tr ( ( a A ) 5 í ) = J 2 j 2 aan bij = a Y , Y , dijbij = a t r ^ l ? * ) = a(A,£?). 1=1
j-i
*= i y = i
(vii) Similarly, ( A ,a B ) = a( A,B). 10. ||A ||2 = tr(A A ‘) = ¿ ¿ a ? - . »= 1 j = 1
11
f 1(M
( 01) ( ° ° ) ( ° ° )
Vo ° y ’ \ ° °y ’ V 1 ° / ’ Vo v
1 2 . (i) (z, z) = a2 -j- b2 >
0. (ii) Suppose (z , 2:) = 0, then a 2 -f b2 = 0 so that a = b = 0. If J2r = 0 then (z, z) — 0. (iii) (z, W! + w 2) = a(ci + c2) + 6(di -f d2) = aci + 6c?i + ac 2 4- bd2 = (z, ici) 4- (z, w2). (iv) Similarly, (zi + ^ 2 ,u;) = (zi,w) + (z2}w). (v) (z, u>) — ac -f- bd = ba + db = (u;, z). _______ (vi) (az, tu ) = aac 4 - a 6 d = a(ac 4 - 6 c?) = a(z, tu ) . _____ (vii) Similarly, (z, au;) = a(z, w). Finally, ||z|| = yj{z~z) — y/a2 + b2.
13. (a) (i) (p,p) = p((a ) 2 4 - p( 6 ) 2 4- p(c ) 2 > 0.
(ii) If (p,p) = 0, then p(a) = p( 6 ) = p(c) = 0. Since a quadratic equation can have at most 2 roots, then p{x) = 0. Conversely, if p(x) = 0 then (p,p) = 0. (iii) (p,g 4- r) = p(a)[g(a) 4- r(a)] + p( 6 )[g(6 ) 4- r( 6 )] 4- p(c)[g(c) 4- r(c)] = p(a)q(a) 4- p( 6 )g(6 ) 4p(c)g(c) 4 - p(a)r(a) + p( 6 )r( 6 ) 4 - p(c)r(c) = (p, g) 4 - (p, r). (iv) Similarly, (p 4- q, r) = (p, r) 4- (g, r). (v) (P,V) = p(a)g(a) + p ( 6 )g(6 ) 4 -p(c)g(c) = g(a)p(a) + g(6 )p( 6 ) 4 - g(c)p(c) = (g,p). (vi) (ap, g) = ap(a)g(a) 4 - otp(b)q(b) + ap(c)q(c) = a[p(a)g(a) + p( 6 )g(6 ) 4 - p(c)g(c)] = a(p, q). (vii) Similarly, (p, ag) = a(p, g). (b) No, since (ii) does not hold. For example, let a = 1, 6 = —1, and p(x) — (x 4- l)(a? — 1) — x 2 — 1. Then (p,p) = 0, but p / 0.
Inner Product Spaces and Projections
Section 4.11
14. (i) (x ,x ) = * 2 + 3 * 2 > 0(ii) Suppose (x ,x ) = 0, then x\ + 3*| = 0, which implies * x = * 2 = 0. Conversely, if x = 0, then (x ,x ) = 0. ( i i i ) ( x ,y + z) = * 1 ( 3/1 + zi) + 3 * 2 (j/2 + z 2 ) = * 12/1 + 3 * 2 j/2 + * 1 * 1 + 3* 2 z2 = (x ,y ) + (x,z). (iv) Similarly, (x + y, z) = (x, z) + (y, z). (v) (x, y) = xiyi + 3* 2 y2 = 3/1 * 1 + 3 p2 * 2 = (y, x). (vi) (ax , y) = a*ij/i + 3a* 2 y2 = « (* 13/1 + 3 * 23/2 ) = « (x ,y ). (vii) Similarly, (x ,o y ) = a (x ,y ). 15.
( -
3)
= ^/22 + 3(—3)2 = v/3l.
16. no; (i) (( 0 , 1 ), (0 , 1 )) =
0
-
1
=
- 1
<
0;
(ii) (( 1 , 1 ), ( 1 , 1 )) =
0;
(v) (x, y) = - ( y , x).
17. Let A be any real number. Then 0 < ((Au + (u, v )v ),(Au + (u ,v )v ) = (Au, Au) + (Au, (u , v)v) + ((u,v)v,A u) + ((u ,v )v ,(u ,v )v ) = (u,u)A 2 + A (u,v)(u,v) + A (u,v)(v,u) + (u, v )(u ,v )(v , v) = ||u || 2 A2 + 2 |(u, v )|2A + |(u, v )| 2 ||v ||2. The last line is a quadratic equation in A. If we have aA2 + 6 A + c > 0 then aA2 + 6 A + c = 0 has at most one real root, and henee b2 —4ac < 0. Thus, 4|(u, v ) | 4 — 4 ||u || 2 |(u, v )| 2 ||v ||2 < 0. So |(u, v ) | 2 < ||u || 2 ||v||2. Taking square roots, we obtain |(u, v ) |< ||u ||||v ||. 18. ||u + v ||2 = ( u + v ) - ( u + v) = (u ,u ) + (u, v) + (u, v) + (v, v) = ||u ||2 + 2 Re(u, v) + ||v||2. By the CauchySchwarz inequality, we have \/R e(u , v ) 2 + Im(u, v ) 2 < |u| |v|, so that Re(u, v) < IMIIMI. Henee ||u + v ||2 < ||u || 2 + 2 ||u || ||v|| + ||v ||2 = (||u|| + ||v||)2. Taking square roots, we obtain ||u + v || < ||u|| + ||v||. 19. By definition, H x = {p £ íhtO, 1] : (p, h) = 0 for every h £ H}. Let p(*) = ax3 + bx2+ ex + d £ P 3 [0 , 1], We want to find conditions on a, 6 , c, and d such that p(*) £ H L . We have (p, *2) = / (a * 3 + bx2 + Jo
ex + d)x2dx = § + ^ + ^ + ^ = 0 , and (p, 1 ) = / (ax 3 + bx2 + ex + d)dx = 7 + ^ + ^ + d = 0 . O 5 4 o J o 4 o ¿ Upon solving this system of equations, we obtain {(0, —15,16, —3), (20, —30,12, —1)} for a basis of the solution space. Henee, H L — span {—15x2 + 16x —3, 20z 3 —30x2 + 12x — 1 }. 20. To show the orthogonal complement of the even functions, H = { / E C[—1,1] : f { —x) = /(# )}, is ^ e set of odd functions, {g £ C[—1,1] : g(—x) = —flf(x)}, observe that for any odd g(x) f!_1g(x)dx = 0. This follows by applying the change of variables, x = —z to the integral and applying the definition of oddness. Now for any odd g and even / , / ( —x)g(—x) = /(x ) ( —g(x)) = —(f(x)g(x)), i.e. fg is odd. Henee (/, g) = f * 1(fg)(x)dx = 0. Thus odds C H 1 . Conversely, suppose g £ H L . Write g(x) = | ( g(x) + g{—x)) -f | ( g{x) - g{—x)) = ge(x) + g0(x), where ge is even, i.e in H , and g0 is odd. Then since g £ H ± i (g>ge) = 0, which implies 0 = (ge + 9o, 9 e ) = Í 9 e y 9 e ) + ( g o , 9 e ) = (9 e , 9 e ) as the odd function g 0 £ H 1 . But this says ||0e||2 = 0, so ge = 0. Henee g = g0l i.e. g is odd. So H L C odds and we are done. (It is much cleaner to write out all of this using the inner product notation rather than putting in the definite integráis.) 21. We have {\,y/Z(2x — l),\/5 (6 x 2 — 6 x + 1)} for an orthonormal basis of H . As (v ,u i) =
í
(1
+
Jo
2x + 3x 2 - x 3)dx =
i
4
( v ,u 2) = / (1 + 2x -f 3x 2 - x 3 )\/3(2x - l)dx = i i ^ , and (v ,u 3) = Jo 60
(l + 2x + 3x 2 —x 3 ) \ / 5 ( 6 x 2 —6 x + l)dx =
then proj^ v =
is orthogonal to i /, so an orthonormal basis for H L is
L
+
D
Z\J
Note that proj^ v —v
410
Instructor's Manual
Chapter 4 Vector Spaces
2x + 3x2— x3)20\/7 (x3 - ^x2 + ^x — 4y) dx =
~ ' / 7 nn / = f 3 w
20^
3 2 3 - r + r
henee l+ 2 x + 3 x 2— x3 = ( ^ x 2 + - ^ x + ^ ) +
1\ /3 , 13 19\ / , 3 2 3 - 2ó) = ( r + y 1 + s o J + { - ’ + r ~ r
22. We want to calcúlate projp 2[0 ^ sin
1\ + ñ )
Since {1, \/3(2x — 1), \/5(6x 2 — 6 x + 1)} is an orthonormal
basis for P 2 [0 , 1]. We have projp 2[0 ^ sin ^ x = ^sin —x, 1^ -f ^sin —x, \/3(2x — 1)^ \/3(2x — 1) 4^sin ^ x , \/5(6x 2
— 6 x 4-
1)^
a /5 ( 6 x 2 —6 x 4 -1 ) .
^sin l x ) V/3(2x—1 ) dx =
J
Since ^sin —x,
= j
sin ^-x dx
^sin — x, \/3(2x
=
4), using integration by parts and ^sin ^-x, \/5(6x 2 —6 x
— 1 )^ =
4- 1 )^ =
í (sin ~ x ) V^5(6x2 —6 x + 1) dx = ^ ^ ( t t 2 + 127r —48), then projp 2[0 )i] sin ^ x = -^-[(107r2 + 1207T — Jq V Z / 7T ^ 7T 480)x2 + (—12ít2 - 112tt + 480)x + 3tt2 + 16tr - 80]. 23. We want to calcúlate projp 3j01] eos ^-x. Since {1, V^(2x —1),-\/5(6x2 —6 x + 1)} is an orthonormal basis A* for P 2 [0 , 1 ], we have ^cos ^ x , 1 ^ = J
eos ^ x dx =
(eos |-x, \/3(2x —1 )) = J
^cos —x^ \/3(2x —
1) dx = —^ ( 7r —4), and icos ^-x, \/5(6x 2 —6 x 4- 1)^ = í icos ^ x ) \/5(6x 2 —6x4-1) dx = 47T V J / Jo ^ 7T 6 127T —48). Henee, projp2^01] eos ^ x = — [(107r2-M207r —480)x24-(—Stt 2 —1 2 8 7 r 4 - 4 8 0 )x 4 - 7r2 4 - 2 4 7 r —80].
«
*•=( i+ J r-í)
™ -a ’ = { -
w
-
w
v
£ $ ) - ’aa’ = i-
26. Suppose a is the ¿th column of A, i = 1 ,2 ,..., n. We have A*A = B where 6 ty = a¿ • ay ■= (a¿, ay). Then A is unitary if and ony if B = 7, i.e. 6¿y = 1 if i = j and 6¿y = 0 if ¿ / j. Henee, A is unitary if and only if the columns of A constitute an orthonormal basis for C*. 27. (i) ( / , / ) = í /(x ) /( x ) dx > 0 since /(x ) /( x ) > 0. (ii) Suppose ( / , / ) Ja
= í /(x ) /( x ) dx
= 0. As
Ja
/( x ) /( x ) = / 2 (x) -f f%{x) > 0 for all x € [a, b] and / E CV[a, 6], then /(x ) = 0 on [a, 6]. Conversely, if /(x ) =
0
on [a, 6], then ( /, / ) =
0.
(iii) (/, g+h) = í f(x)[g(x) + fc(x)] dx = í Ja
f(x)[g(x)+h(x)] dx =
Ja
í f(x)g(x) dx + f f(x)h(x) dx = (f,g) + (f, h). (iv) Similarly, ( f + g,h) = (f, h) + (g, h). (v) (f , g ) = Ja
Ja
pb
/ f( x ) g (x ) d x = v/a
p b ______________
pb
/ f(x)g(x) dx =
/ g(x)f(x) dx = (g , f ). (vi) ( a /, g) =
«/a
pb
Ja
/ af{x)g(x) dx
=
Ja
*/, Note z = */, and p = B*z from the formula for p -0.7047 + 0.7876Í -0.7063 + 0.9850Í 0.8179 + 0.2196Í -0.0638 - 0.0596Í » B**w # /. This will ans = -0.7047 + 0.7876Í -0.7063 + 0.9850Í 0.8179 + 0.2196Í -0.0638 - 0.0596Í » B*(B’*w) */. Si ans = -0.5014 + 0.6019Í 0.2835 + 0.1884Í 0.9411 + 0.3912Í -0.0977 + 0.2916Í 0.5543 - 0.7369Í 0.5884 + 0.5017Í (b) » x = 2*rand(4,1)-1 + i*(2*rand(4,1)-1); » h = A*x '/• h is in H. h = 1.2117 - 0.2441Í 0.1575 + 0.1130Í -0.9003 - 0.9038Í 0.2202 - 1.0490Í -0.5889 - 0.1895Í -0.5081 - 1.5998Í » norm(w-h), norm(w-p) ans = 4.2915 ans = 0.7440
*/. Compare,
The projection of w on H is the vector in H closest to w.
in the complex case
413
414
Chapter 4 Vector Spaces
Instructor’s Manual
(c) Since V4 is a linear combination of Vi, V3 and z, it may be replaced by z in the basis. » » » » » »
z = A * [ 2; 0; -3; 1]; C = [ A(:, [1:3]) z]; D = orth(C); w = 10*(2*rand(6,1)-1); pl = '/, Use basis B. '/. Use basis D. p2 = D*D'*w; '/. Compare pl - p2
ans = l.Oe-13 * 0.2132 - 0.2021Í -0.0444 + 0.0200Í -0.1599 + 0.1354Í -0.0797 + 0.2792Í 0.0622 - 0.0311Í 0.1954 - 0.0133Í
The projection of w onto H should not depend on which basis you choose. Here, p l —p2 is zero up to round-off error. 4. (a) See MATLAB 4.9 problem 8 , replacing * by ' throughout. 0>) » A = (2*rand(7,4)“1) + i*(2*rand(7,4)-l); » B = orth(A); C = nullCA*); » C ’*C '/. Verify that columns of C are orthonormal. ans = 1.0000 0.0000 - 0.0000Í 0.0000 - 0.0000Í 0.0000 +0.0000Í 1.0000 0.0000 - 0.0000Í 0.0000 +0.0000Í 0.0000 + 0.0000Í 1.0000
»
w =
(2*rand(7,1)-1) + i*(2*rand(7,1)-1);
Use proj# w = B * B ' w from Problem 3(a). Also note that columns of C are an orthonormal basis for Í71 , by part (a) of this problem. So » h = B*B>*w; p = C*C’*w; » w - (h+p) ans = 1.0e-15 * 0.1110 + 0.6106Í -0.3331 -0.4441 + O.lllOi 0.2220 - O.lllOi 0.1110 - 0.1665Í -0.2220 + O.lllOi -0.0555 + O.lllOi
*/• Using projection formula in Problem 3(a) '/# This should be zero (up to round off).
» h^p ans = -1.7347e-16-
'/• h, p essentially orthogonal 2.2204e-16i
Inner Product Spaces and Projections
M A T L A B 4.11
(a)
» Al = (2*rand(4)-l) + i*(2*rand(4)-l); » A2 = (2*rand(4)-l) + i*(2*rand(4)-l); » Q1 = orth(Al); Q2 = orth(A2); » normíQl’+Ql - eye(4)) */, This should be zero: ans = 5.7132e-16 » norm(Q2’*Q2 - eye(4)) '/• This also should be zero. ans = 6.8807e-16
Since the matrices are unitary, their columns are orthonormal, and henee linearly independent. Since there are four columns, they must form a basis for C4. (b) » A = inv(Ql); » normíA^A - eye(4)) '/. This should be zero. ans = 4.7682e-16 » A = inv(Q2); » normCA'+A - eye(4)) '/, This should be zero. ans = 7.2160e-16 (c) » A = Q1*Q2; » normíA^A - eye(4)) '/• This should be zero. ans = 9.2361e-16 (d) » v = (2*rand(4,1)-1) + i*(2*rand(4,1)-1); */, Part (d)> » norm(v) - norm(Ql*v) '/, This should be zero. ans = 6.6613e-16 » norm(v) - norm(Q2*v) '/• This should be zero. ans = 8.8818e-16
(e) Repeat above for two random 6 x 6 .
416
Chapter 4 Vector Spaces
Instructor’s Manual
S e c tio n 4 .12 1
. Let 5 be a set of linearly independent elements of V. If 5 is maximal, S forms a basis by Theorem 1 and we are done. If not, then there exists vi £ V such that Vi £ span S. Then if S U {vi} is maximal, we are finished by Theorem 1. Otherwise, we continué the process. The process must stop with a maximal set of linearly independent elements S U { v i , . . . , v m}. This gives a basis for V.
2. Let 5 be a set such that V C span 5. Choose vi £ S. If {vi} is not a basis of V , we can find V2 £ S such that V2 ^ span {vi}. Then vi and V2 are independent and if {vi, V2 } is maximal, we are finished by Theorem 1. Otherwise, we continué the process. The process must stop with a maximal set of linearly independent elements { v i , . . . , v m} C S. This gives us a basis for V. 3. Because T is a chain, either A\ C A 2 or A 2 C A\. So the result is true if n = 2. Suppose the result is true for n — 1 sets in a chain T. Then for A \ , . . . , A n- 1 , one of the sets, say Ak, contains all of the others. Then consider the n sets A \ , .. . , A n- i , A n. Then either Ak C A n or A n C Ak. If Ak C A n) then A n contains all of the other sets. If A n C Ak, then Ak contains all of the other sets. Then, by mathematical induction, given n sets in a chain T, one of the sets contains all the others.
Review Exercises for Chapter 4
R eview Exercises for C hapter 4 í í ~ 2\
1. yes; Basis:
1 ) ’l °
2. no; if (x, y , z) satisfies x + 2y - z < 0 then ( - x , - y , - z ) satisfies x + 2y —2 > 0. /-
3. yes; Basis: <
/ _1\ / - 1 0 1 1 °l 0 0 ’ 1 ’ 1 \ 0/ 0/
>; dimensión = 3.
4. no; ( 1 , —4 ,3 ) satisfies the equation but (—1 , 4 , —3) does not.
5. yes; Basis: (Eij : j > ¿), where E¡j is the matrix with 1 in the = n(n + l)/2 .
position and 0 elsewhere; dimensión
6. yes; Basis: {1, x, x2, x3, x4, x5}; dimensión = 6. 7. no; (x5 + 1) + (—x5 -f 1) = 2; not closed under addition. 8.
; dimensión = 5.
yes; Basis:
not closed under addition. 10. yes; infinite dimensional. 11.
13.
15.
17.
19.
2 4 = —24 => linearly independent. 3 -6 130 100 = 2 10
14.
100 0 0 100 = 1 => linearly dependent. 00 10 000 1 12 3 0 0 0 1 -8 = —7 => linearly independent. 00 0 7 01 0 0 1 10 0 1 -1 0 0 = 4 => linearly independent. 0 01 1 0 0 1 -1 1 3 -5 a) 5 0 5 24 6 (b)
2 3 -1 1 -2 -4 4 6 -2
12.
-180 => linearly independent.
= 0 =í* linearly dependent.
16.
18.
24 = 0 => linearly dependent. 36 1 0 2 - 4 2 -1 0 = 0 2 -1 5 1 2 3 0
linearly dependent.
0 0-1-8 = 0 => linearly dependent. 0 - 1 0 7 0 0 0 0 110 0 -110 0 = —4 => linearly independent. 00 1 1 0 0 1-1
Instructor^ Manual
Chapter 4 Vector Spaces
—3/2 21. x = 2z — 3y/2 Basis:
► ; dimensión = 2.
1 0
22. y = 2x/3 Basis:
; dimensión =
1
. °\ o -1
23. y — Zx — z -\- w Basis: <
dimensión = 3.
V i/ 24. Basis: {x, x2, x3}; dimensión = 3. /I 0 25. Basis: < 0 \0
0 0 0 0
0 0 0 0
0\ /0 0 1 0 0 J ’ 0 0/ \0
00 10 00 00
'0 0 0 0 ' 0000
0\ 0 0 0/
'0 0 0 0\ 0000 0000
0 0 10
; dimensión = 4.
,0 0 0 1 / J
,0 0 0 0 /
dimensión =
26. Basis:
27.
1 -2
-2
4
o
o ) ;^
= span{ ( ? ) } ’^
) = 1,Range
6
¿ = span { ( _ J ) | . p(A ) = i-
s p » { ( " i ) } , , ( 4 = 1, Range A —
p{A) = 2.
1 0 0\
0 ; N a = {0}, v(A) = 0 , 0 0 1/ 0 1
P(A) = 3.
Range A = span
30.
2
4 -2
-1 -2
1
-
Range A = span |
fV o1 o2 _ 1o V/ ’ N a
~ 1*
=
span
419
Review Exercises for Chapter 4
2 -1 32. 3 1 -2 \2 -3 5 /
1 - 1
0
1
/I
3\ 0 3 6/
R ange A — span
33' C =
(2
0
1 3\
0 1 - 1 0 0 0
0 0
V oo
0 0
,P(A)
2 ) :
o\
/l/^\ 0
0'
1/V5>
l /\/2 -l/y/2 0/
(c)P=
o
0
-1A /2/ /l/2 \ 1/2 _ , !/2 1 ~ \ 1/2/
44. Start with the basis {1, x, x2}. r
v '2 = x - 1. ||v'2|| = U
1/2 \
( l\
: i / 2 ) ¡ the» 1/ 2 /
í
Jo
0 0
/!/2 \ 1/2 1/2 + V1/2/
1/2 -1/2 -1/2 1/2
l 2dx = 2. Let ui = 1/V2. (v2,ui) =
í (x/y/2)dx
Jo
i 1/ 2
,2
(x - l ) 2 dxj
=
2/V2.
r2
= \/2/3. u 2 = \ft y2 {x ~ !)• (v s .u j) =
(x 2 / - / 2 )d x -
8/3V2. (V3 , U2 ) = \/3 /2 J (x 3 — x2)dx = 4/V6. V3 = x 2 — (2x — 2) — 8/3 — x 2 — 2 x — 2/3. I K ||
r /2 i 1/2 [jf (x 2 - 2 x - 2 / 3 ) 2dx
=
2^14/15.
Orthonormal basis: {l/-v/2, \/3 /2 (x —1), \/Í5 /2 \/l4 (x 2 —2x —2/3)}. 45. proj ex = (e* ,u i)u i
+
(ex, u 2 )u 2 + (e®, 113)113
P 2[0,2]
= (e 2 - l)/2 + 3(2)(x - l)/2 + 15(-2e2/3 - 10/3)(x 2 - 2x - 2/3)/56 = (—5e2/28 - 25/28)x2 + (5e2/14 + 67/14)* + (13e2/21 - 61/21)
Review Exercises for Chapter 4
=
( J 1 ) 0
(ty * *
- ¡ I ) ( l )
=
( » > < ^ > "
=
n ( - S 1 >
= ( v 0 - = W 2 - ‘'
( i ) ......
/
6 —1 0 \ 6 9 -6 -1 0 - 6 7
V
22
/l l l W 2 -1 1 11
4
5\ - / —16\ -3 = 9 I ; y = (—16 + 9* + 7ar2) / 6 7/
0
6V
421
422
Notes
Instructors
Definitions and Exampies
Section 5.1
C hapter 5. Linear Transform ations Section 5.1 Xi + X2 1. linea,; r) = (;;) + ( £ )
( z ) =
„(*)=„rx
4. linear; If*. + «) = T (»■ + » ) = ( B + ¡^) = („“) + ( » ) = T *' + T *=i T *=1
( ax\' ax2
n = T x + T r> T (a x ) = T
*= 1
\ OCXfi t
423
424
Chapter 5 Linear Transformations
11. linear; T( x + y) =
Instructor's Manual
/ x\ x
x+y
m y
= T(x) + T(y); T(ax) =
+
Vx/
\x + y/
Vy
( ax \ ax
(x\
\ax /
/
X
—a
= aT(x)
\x)
Xl + x 2 (X i + X2) +
(¿ 1 + * 2) ^
12. linear; T ( x t + x 2) = T I yi + 2/2 I 1 W\ + w2 ■ ( (2/1 +J/2) + (u>i + ^ 2)A
^ X! +
Zx ^ +
2/1 + m
^
X2 +
zi + z 2
Txx
T (« x ) =
f
“* + ) +
( * + 1)
= a
^2/ + » /
Z2
2/2 + ™2
)
Tx 1 +
= «T *
13. not linear; if a ^ 0 or 1, xyzw ± 0 then T (ax ) =
j) = a2 ( ^ ) ^ a ( ^ ) =
14. linear; T(A + A') = (A + A')B = A B + A 'B = T(A) + T ( A ’); T(aA) = (a A ) B = a{AB) = a T ( A ) 15. not linear; T( A + B ) = (A + B)*(A + B) = ( A ' + B ^ A + B) = A U + A ^ + B U + B ^ ± ^ M + 5 ‘5 = T(A) + T(B) unless A*B + B*A = 0 16. linear; same solution as problem 14 17. not linear; i f a ^ O o r l
then T(aD) = a 2D 2 ^ a D 2 = aT (D )
18. not linear; T (aD ) = I 4- a D ^ a(7 + D) = aT (D ) unless a = 1 19. linear; T[(ao“l_aix4“a2x2)4-(¿>o“ffriíC“l-^2íE2)] ^ (ao4~i>o)"b(ai4“&i)* = ao4-aix4"6o4~6i* — T(ao4~aix4“ a2x2) + T(60 + bix + &2*2); T [a(a0 4- aix 4- a2x2)] = a a o + a a ix = a (a 0 4- aix) = a T (a 0 4* a i* 4- a2x2) 20. linear; T[(a04-aix4-a2x2)4-(&o4-&i*4-&2*2)] = (ai4-&i)4*(a24-&2)* == a i4 -a 2x4-i>i4-&2* = T (a04-aix4a2x2)4-T(604-&i*4-&2*2); T [a(ajlo 4 -aix 4 -a2x2)] = aai-\-aa2x = a(ai + a 2x) = aT(a 0+ aix + a 2x 2) h ax” 4- b 4- bx 4- bx 2421. linear; T(a4-&) = (a 4- 6) 4- (a 4- 6)* 4- (a 4- b)x 2 4------h(a4-&)*n = a4-ax4-ax24 h 6xn = T(a) 4- T(b); T(aa) = a a 4- aax 4- a a x 2 4 h a a x n = a (a 4- ax 4- ax2 4------- haxn) = aT(a) 22. not linear; T(ap(x)) = a 2\p(x )]2 ^ a[p(x)]2 = aT(p(x)) unless a = 0 or 1, or p = 0. 23. not linear; if a
0 or 1 then T ( a /) = a 2/ 2(x) ^ a / 2(x) = a T /
24. not linear; T ( a f ) = a f ( x ) 4-1 ^ a (/(x ) 4-1) = a T / unless a = 1 25. linear; T ( f i + f 2) = í ( f x(x) + f 2(x))g(x) dx = í f 1 (x)g(x)dx + í f 2(x)g(x) dx = T / i 4 - T / 2; Jo Jo Jo T ( a f ) = f a f (x ) g (x ) d x = a í f(x)g(x) dx = a T / jo
jo
26. linear; T (/i4 -/2 ) = [ ( h + f ^ g ] ' = (/i4 -/2 ),í/ + ( / i 4- / 2) ^ = /í^ + Z ^ + Z i^ + ^ ^ + í/z^ y = T f i + T f 2; T ( a f ) = [(cr/)y]/ = * (/* )' = a T / 27. linear; T (/(x ) + y(ar)) = /(* - 1) + g(x - 1) = T /(x ) + T(/(x); T (a /(x )) = a f ( x - 1) = a T /(x ) 28. linear; T ( / + (/) = /(1 /2 ) + ^(1/2) = T f + T g ; T ( a /) = a / ( l / 2 ) = a T f 29. not linear; if a ^ 0 or 1, and det (j1) / 0, then T(aA) = det (aA) = a n det
a det A = aT(A)
30. Geometrically, T rotates a vector in the xy-plane through an angle of 180 degrees, or, equivalently, T reflects a vector through the origin.
Definitions and Examples
Section 5.1
31. ( , ) t ( 2 ) = 2 [t ( J ) + 2 t ( ; ) ] = 2
(b) T ( ~ ? )
= - 3T(o) + 7T( i ) = - 3 ( ¡ )
32. ( .) * =
+ 7 T (
(b) * ( - ’ ) = ( $
°) =
+
33. T rotates a vector counterclockwise around the z-axis through an angle 0 in a plañe parallel to the xy-plane. 34. T rotates a vector counterclockwise around the y-axis through an angle 0 in a plañe parallel to the xz-plane. 35. Suppose a < 0. We have T[(a —a)x] = T'(Ox) = OTx = 0, so that 0 = T (a x —ax ) = T (ax ) T (—ax ) = T (ax ) —aT x . Thus, T (ax) = a T x if a < 0.
36. Define T(Á) = T Í Z Z Z ]
4-
= ( ^ ^ ) . Since T(A + B) = ( ^ + £ £ J £ ) =
\ 031 «32 O33 J
( an ai2) + (
x
J 11 l 12) = T(A) + T(B), and T(aA) \ h l i>22/ then T is linear. Many other examples possible.
\ a21 «22 /
= ( “ au
a fl1 2 )
\ a a 2i a a 22y
=
a
( '“ n
012 )
\ 021 }; ¿'(T) = 2; Range T = M2; />(T) = 2
5' ^
= (* ")•
= ( ' í ' + ")• KerT = { ( o o ) } : ^ =
r =
p(T) = 4. 6.
Ker T = {0}; i/(T) = 0; Range T = {a + ax + ax 2 + ax 3 : a € M};/>(T) = 1.
7.
K erT = {A : A* = —yl}; i/(T) = (ra2 —n)/2; Range T = {A : A issymmetric};p(T) = (ra2 + ra)/2.
8. K erT = { / G C [0 ,1] : / = constant}; í/(T) = 1; Range T = { / G C [0,1]} by fundamental theorem of calculus; Range T is infinite dimensional. 9. K erT = { / E C [0,1] : /(1 /2 ) == 0}; K erT is infinite dimensional; Range T = R; p(T) = 1. 10. Ker T = {(0,0)}; i/(T) = 0; Range T = M2; p(T) = 2. 11. For any v £ F , v = a iv i + a 2 V2 +
h
for some (ai, a 2 , • • •, an). Then
Tv = T (a iv i + a2v 2 + = aiTvx + a2T v 2 H = ai •0
b a„v„) 1- a„T vn
a2 •0 -f- * * Q>n *0 = 0.
Thus T is the zero transformation. 12. For any v E V, v = aiVi + a 2 V2 H
h anv n for some (ai, a2, . . . , an). Then
Tv = T (a iv i + a2v 2 + = a iT v i + a2T v 2 + = a iv i + fl2 V2 +
h flnV„) h flnTvn h anvn = v.
Thus T is the identity operator. 13. Range T is a subspace of M3 which contains the origin. Thus by Example 4.6.9 Range T is either a) {0}, b) a line through the origin, c) a plañe through the origin or d) M3. 14. K erT is a subspace of M3 which contains the origin. So as in Problem 13 K erT is either a) {0}, b) a line through the origin, c) a plañe through the origin or d) M3. 15. Tx = A x where A = ^ j , a , 4 , c G K . i.e. T ^
16. T x = A x where A = ^
d j / a ) ’ w^ere'c>^ ^
^
, T ^ ^ is arbitrary.
Properties of a Linear Transformation: Range and Kernel
18. Note that a basis for Range T is
)}
. Then for any
Section 5.2
, we want T
< 19. (a) If A £ kerT then A —A* = O. So A = A*. That is, A is symmetric. Conversely, if A = A*, then A —Al = O, so A £ ker T. (b) If A £ RangeT then there exists a m atrix B such that A = B — B %. Then A 1 = (B —B t)t = B t — B = —A . That is, A is skew-symmetric. Conversely if A i = —A, then T Á) = = A. 20. K erT = { / £ C^fO, 1] : x f' ( x) = 0 for # £ [0,1]}. Then we must have f '( x ) constant if / £ kerT. Range T = {xf(x ) : f (x ) £ C [0,1]}
= 0 all x. Then f( x )
21. Choose bases { u i , . . . , u n} for V', { w i , . . . , wm} for W. Then let (u¿) = w¿ and l y (u*) = 0 if k ^ i. These form a basis for L(V, W) since any v* is a linear combination of w s o T with T(ujfe) = vjk is a linear combination of T y . Specifically, if v* = Ec*¿w¿, k = 1, . . . , n, then T = EjycjyTjy. Independence of Tij follows from E ay 7 y (u /) = Eay wy = 0 => ay = 0. Therefore, dim L(V, W) — n m . 22. (a) Suppose T i,T 2 £ U. Then for every h £ H, (Ti + T2)h = T ih + T2I1 = 0 + 0 = 0 , and (aT i)h = a(T ih) = a • 0 = 0 . Then U is a subspace of L(V, V”). (b) d im í/ = n(n — fc). In fact extending { u i , . . . , u*}, a basis of H , to {ui, • • •, u n} a basis of Va, then T is in U if and only if T (u i) = • • • = T(u¿) = 0. In particular T (u*.+i),. .. ,T (u n) are n — k arbitrary vectors in the n dimensional space V. So if Ty (u¿) = \ij, Tij (u/) = 0, / i for k < i < n, 1 < j < n, then Ty are a basis for U. (See solution to previous problem.) 23. No. Let S =
anc^ ^ =
( n n ) ^ zero transformation.
^0
o)* ^ en ^
= ^0
0
)
= Zer° ^rans^orma^ on»an^
=
432
Chapter 5 Linear Transformations
Instructor’s Manual
S ectio n 5.3
1. As T
^
and T
then A t —
^
Since det A t = —1 / 0, then
kerT = {0}, range T = M2, z/(T) = 0, and p(T) = 2. í l\ J = 1 1 1 and
then
( o\
/ A
2. Since T
p(At)
/ x\ ( 1 = I “ M ) ^ e n A t = I 1 —1 1 . As columns not collinear,
= p(T) = 2, and henee v{T) = 2 — p(T) = 0. Thus kerT = {0} and range T =
span ^ | 1 j , í - 1
3. We have T
As At
(»)
-*■
= ( - 0 ' r ( ;)
q q^, then
= ( 1 ) - “ d r ( ? ) = ( - O - 80 l hat ^
= p(T) = 1, v{T) = 3 -
p(A)
' 1° \I \>, from Tx
ker T = span
4. A s T ^ J ^ = ^ ^
an(^
p{T)
= 2, range T =
=
span
(-2 '2
- O -
j (_ 2 ^}> and
= 0 only if x = íI x\ —zX1 3
= ( d ) ’ ^ en
= ( c d ) * ^ a = ^ = c = : ^ : = 0> ^ en ^
the zero
transformation, and henee p(T) = 0, i'(T) = 2, kerT = M2, and range T = {0}. If ad — cb ^ 0, then p(T) = 2, z/(T) = 0, kerT = {0}, and range T = M2. Suppose ad — bc = 0, and suppose at least one of a, 6, c, or d are nonzero. We may assume a / 0. Then p(T) = 1, ¿'(T) = 1, kerT = span | ^ a ) and range T = span | ^ c ) } ‘ 10 3 /2 N 0 1 —1/2 | then p(T) = 2 = # pivots, i/(T) = 1, and range
1 -1 2 >
5. At = ( 3 1 4 | . Since At
00
5-18,
T = span j í y ’ í
5.
At =
-i
2
r
2—4—2 ).
|
-3
6
Since
3,
and range T = span
7. A t
/ í
0 6 6
.
^
i -2 - r 0 0 J , then p(T) = 1, v(T) = 2, kernel T = span
0
0
0
0,
i
’l - 1 2 3' = ( 0 1 4 3 ). As A T 1
^ =
^ S° ^
At
0,
1 0 6 6'
0 1 4 3 | , then p(T) = 2, t'(T) = 2, and ker T = span 0 0 0 0.
r Also since pivots in columns 1 and 2, range T = span
-i’
-6 \ -3
0 1/
/-6\
I "4 \
0^
T h e M atrix Representation of a Linear Transform aron 1 -1 2
/
4 I' M A t
00
0
, then p(T) = 2, i/(T) = 2, and kerT = span <
0
2 -1 1 -1.
range T = span
o
A t = (j^J
= ( : ! )
4r = (
3 ^ /7
r ( í )
( - Í)
■
■ O
)
■ =
?
( . ; ) * ! ©
= y
( 1 )
1 ( s ) - and T
-
) - As det AT =
11^
( - i ) +
l
( 0 -
^
*
(5) = (!») = t
0, then p(T) = 2, v(T)
= (o
= ( 0
-
( " ¡ ) + f
-
( 5 ) - H “ ce-
0, ker T = {0}, and range T = M2.
= K - í ) + l ( ! ) - “ *d i ’ ( | )
' £ "£)■
*
(o
t
=
% ) - (o ° %)■
p(T) = 2, v(T) = 3 —2 = 1 , range T = M2, and (k e rT ) b x — span
2 /o\ -
/ I
-i\
/
1
/ 10\
-1\
2 . Thus A t = 14/5 11/10 . Since 14/5 11/10 \5 / V 1/5 2/5 / \ 1/5 2 / 5 /
-*■
0 1 , then p(T) = 2, V0 0 /
v(T) = 0, kerT = {0}, and (range T ) b 2 — span I [ 14/5 J , ( 11/10 ( 1 /5 / 2 /5 ;
V
V
0
13. T (l) = x3, T(x) = 1 — x, and T (x2) = 0. Henee A t =
1 0'
o ” o o I ’ p{T) = 2> v{T) = l ' range
Vi o o, T = span {x3, —x + 1}, and ker T = span {x2}. /!' 14. A t
i
, p(T) = 1, i/(T) = 0, range T = span{1 + x + x 2 + x3}, and kerT = {0}.
15. A t = (0 0 1 0 ) , p(T) = 1, i/(T) = 3, range T = M, and kerT = sp an { 1 , a?,a;3}. 16. T (l) = 0, T(x) = x, T (x2) = - 1 , and T(x3) = x. Thus range T = Pi, and kerT = span {1,x 3 —x}.
2
3
1 ) 0 ^0/ U
1\ -1 1 2/
= ( J ) - , ( - i ) + o G > r ( i)
= T
3
*3/4) * As det ÁT = 5 ^ 0’then I* ? ) = 2>V(T ) = °> kerT = {0}, and range T = M 2.
io- T ( ' ¡ )
u
433
1>
0 1-3-3 8' Á T = I " l - 2 5
Section 5.3
= ( ° J " J } ) • So p(T) = 2, v{T) = 2,
\ ►
434
Instructor’s Manual
Chapter 5 Linear Transformations
/ I -1 2 3 \ 17. T( 1) = 1 + x2, T{x) = - 1 + x, T ( x 2) = 2 + 4x + 6x2, and T ( x 3) = 3 + 3x + 5x2. So A t I 0 1 4 3 . As \l 065/ 1 —1 2 4 \ 0 143 1 065/
-»
/1 0 6 0\ 0 1 4 0 , then p(T) \0001/
3, v{T)
=
=
1, range T
=
P2,
kerT = span {6 + 4x — x 2}.
18
= ( 1 _ 2 ) ' r ( o o ) = ( = 2 - 0 ' r ( ? o ) = ( s í ) ' “ d r ( o O = ( 4 - 0 - Hen“ ,
T
1-12 , -1 0 2
1' 2 .
I
4
1 -2 5
1-12 .
1
2 — 1 1 — 1,
span |
—
10 0 — 2\
2
0 10-3
1 -2 5
4
001
0
.0 0 0
0/
2-11-1/
Since PÍvo^s in columns 1-3, Range T = span
“ •< ¡¡0 A?
1^
- 1 0 2
= (ü )-K S i) 1 1 1\ 1110 u o o
■ G
" M
" )
, then p(T) = 3, í'(T) = 1, kerT =
2
) * ( —2 —1 ) ’ ^5 l ) } ’
= (J O -^ K S !)
■ ( í o ) Thus
. So p(T) = 4, u(T) = 0, range T = M 22, and kerT = {0}.
\ l o o 0/ 20. T (l) = x = (x + 1) - 1, T(x) = x 2 = (x + l )2 - 2(x + 1) + 1, and T ( x 2) = x 3 = (x + l) 33(x + l ) 2 + 3(x 4- 1) — 1, and henee A t =
-1
í - l \
0
3 1 -3
0
0
1-2
. So p(T) = 3, v{T) = 0, kerT = {0}, and
1/
range T = span {x, x 2, x3}.
21.
/0 1 0 0 0\ 00200 1 D (l) = 0, £>(*) = 1, D (x2) = 2 x, D(x3) = 3x2, and D (x4) = 4x3. So A D = 0 0 0 3 0 I’
,, e’
\0 0 0 0 4 / p(D) — 4, v(D) = 1, ker D = P q} and range D = P 3 . f - l 0 0 0 0\ 00000
22. T (l) = - 1 , T(x) = 0, T (x2) = x2, T(x3) = 2x3, and T(x4) = 3x4. Thus, A r =
00100 00020
, and
00003/ henee, p(T) = 4, r'(T) = 1, kerT = span {x}, and range T = span {1, x 2, x3, x4}.
23. As D( x k) = k x k 4, then A d
/ o 1 0 0 • '° \ 0 0 2 0- • 0 — 0 0 0 3- • 0 , p(D) = n, v{D) = 1, range D = Pn- i , and \o 00 0 • • n )
ker D = P q.
T h e M atrix Representaron of a Linear Transform ation
Section 5.3
/O 0 2 0 0 \ 24. D (l) = 0, D(x) = 0, D (x2) = 2, £>(x3) = 6x, and D(x4) = 12x2.So A D = I 0 0 0 6 0 ,p(D) = 3,
V 000012/
v{D) = 2, ker D = Pi, and range D = span {2, 6 x, 12x2} = P k and is 0 if m < ib. Thus A d = (a,; ) is the (n —Jb + 1) x (n + 1) matrix, where for each l < i < n —Jb+1, a¿,fc+¿ = (ib + 1 — 1)*, and ay = 0 otherwise.So there is a pivot in each row. p(D) = ra —ib + 1, t/(D) = (n + 1) —p(D) = k, ker D = span {1, x, x2, . . . , x i- 1 }, and range D = span {1, x, x2, . . . , xn~k} = Pn-k-
( yk : /L k \
\
*
ib !
-m ) x k, and henee A T = diag(60,i i , • • - , M> where bk = .=0^
, = o ^ “ l> ) Thus, p(T) = n + 1, v(T) — 0, kerT = {0}, and range T = Pn.
% >-
í1 1 / x k dx = - — - for each 0 < k < n. So A j = ( 1 ,1 /2 ,1 /3 ,..., l /( n 4- 1)), and henee, Jo A?4" 1 />(J) = 1, í/(J) = 0, ker J = {0}, and range J = M.
28. J ( x k) =
/1°0\ 29. A t = 1 0 1 0 1, p(T) = 3, i/(T) = 0, range T = P2, and kerT = {0}.
\0 0 1 /
/o \
/
0\
/-1 \
ÍI\
fo
o -1 1 \
30. T (l) = 1 0 , T(x) = 1 , T (x2) = I 0 , and T(x3) = 1 . Thus A t = I 0 1 0 1 \ y V - 1/ V i/ Vo/ Vo - 1 '0 0 - 1 1 \ /O 1 o 0 \ 0 1 0 1 1 —*■ | 0 0 1 0 ), then p(T) = 3, í/(T) = 1, kerT = Po, and range T = M3. k0 - l 10/
. As 10/
V 0001 /
31. Let Ers 6 Mmn be the m x n m atrix with e,¿ = 1 if i = r, j = s, and 0 otherwise. Then T T r3 = E sr 6 Mnm, and Ay = (a,-¿) where 1,
if i = (¿ — l)m +
and j = (é — l)n + ib for Jb = 1 , 2, . . . , n and £ = l , 2 , . . . , m
aij —'
0,
32-
T {1 )
-
( - !)
otherwise
“ d T ( !) = ( 1 + í> * > * = (_í i + O
/o o o' 33. D (l) = 0, D (sinx) = cosx, D(cosx) = - s i n x . Thus, A d = I 0 0 —1 J , range D = span (sin x, eos x), Vo 1 0 . and ker D = IR.
435
436
Instructor’s Manual
Chapter 5 Linear Transformations
/ 1 1 °\
34. D(ex ) = er , D(xex) = ex + xex, and D(x2ex) = 2xejr + x 2ex . Henee A# = 1 0 1 2 1, range D = K, \0 0 1 / and kerD = {0}. 35. r ( J ) = proi„ ( J ) = ( ; / » ) , and T ( ? ) = p,ojH ( J ) = ( " $ ) . So
At =
( $
- $ ) .
36. (i) By theorem 1, y E range T if and only if y E range A?. So range T = range A t • (ii) This follows from (i). (iii) By theorem 1, Tx = 0 if and only if A t x = 0. Thus kerT = N a t (iv) By (iii) v(T) — dim kerT = dim ( N a t ) = v ( A t ) 37. (i) Let Bi and B 2 be bases for V and W, respectively. By theorem 3, (T v)#2 = A t ( v ) b x- If w E range T, then w = Tv for some v E V. Henee, (T v)#2 = (w )#2 = A t ( 'v) b1- S o ( w ) # 2 £ range A t . Similarly, range A t C (range T )#2, and henee, range A t = (range T )# 2. It follows that p ( A t) = P(T). (ii) We have T v = 0 if and only if (T v)#2 = (0)#2 if and only if A t { v ) b x = (0)#2. Henee, (kerT )#! = ker A t, and i/(T) = í'(A t). (iii) As p (A t ) + v (A t ) = n by theorem 4.7.7, we have p(T) -f i/(T) = n. 38. 40. 42. 44. 46.
expansión along the x-axis withc = 4 reflection about the x-axis shear along the x-axis with c = —3 shear along the y-axis with c = —5
(ii)
39. possible compression along the y-axis with c = 1/4 41. shear along the x-axis with c = 2 43. shear along the y-axis with c = 1/2 45. reflection about the line y = x «•
( 17“
)
(-0.75,4)
/NV (0,4)
(-0.75,0)
T h e M atrix Representaron o f a Linear Transform ation
50'
(-1/2 l)
*G-í)
*
Gí)
«■
G
5 3 '
( ’ oí)
Section 5.3
437
438
Instructor’s Manual
Chapter 5 Linear Transformations
54-
(¡o)
“ •
(10)
For 56-63 use inverses of elementary transformations which take A t to reduced echelon form.
«• (uxsíX í -OGOG-5'?) -
( í í ) ( « ) ( í _ ?)
«■ ( í ! ) ( - i ! ) ( í v S ) ( i 7/í )
«•
( ? : ) ( -
»•
(!!)(!?)
«■ ( ! i ) ( ”
) ( J
m
h
/°3 ) 0
2/? )
) ( i - ! ) ( i 5 ) ( í T/ ; )
T h e M atrix Representaron of a Linear Transform aron
M A T L A B 5.3
M A T L A B 5.3
The Solutions to 5.1 explain the addition of a line type argument to ,grafics\ 1. To recreate Figure 5.8(a) we use » » » »
pts=[[0 0]» [3 0]’ [3 2]» [0 2]’]; lns=[[l 2] 1 [2 3]’ [3 4] * [4 1] ’] ; grafios(pts,lns,,r ,,,* ,,5,,- ,) print -deps fig531.eps 5.----------------------------- .----------------------------4
3 2 .
« ---------------------------------------------
1• 0-
*----------------
-1• -2-3.
4.
-5‘--------------- 1 ---------------5
0
5
(a) » »
» » » »
A=[.5 0 ; 0 3]; '/# Expand y-axis by factor of 3, compress x-axis factor .5 grafics(pts,lns,,r l,’*>,10,>:>) hold on grafics(A*pts,lns,’b*,*+ *,10,* — *) hold off print -deps fig531a.eps
10 8 6 4
2 0 -2 -4
-6 -8 -1-10 0
-5
0
5
10
439
440
Instructor^ Manual
Chapter 5 Linear Transformations
(b) Recreate the shear transformations in Figures 5.8(b),(c). » » » »
» » » » » » » » »
A=[l 2 ; 0 1] ; */• Shear along x-axis, c=2 , Fig. 5.8(b) subplot(121); grafics(pts,lns,’r*, , 7 ,* :*) hold on grafíes(A*pts,lns,’b*,* + *,7,*— *) title(’Fig. 5.8. (a) and (b)>) hold off A= [1 -2 ; 0 1]; */• Shear along x-axis, c=-2 , Fig. 5.8(c) subplot(122); grafics(pts,lns,>r*,*** ,7,* :1) hold on grafics(A*pts,lns,,b >,* + *,7, 9— *) title^Fig. 5.8. (a) and (c)’) hold off print -deps fig531b.eps F ig . 5 .8 . ( a ) a n d (b )
F ig . 5 .8 . ( a ) a n d (c )
6
6
4
4
2
* .............#
0
-------3if
A
-
------
0
-2
-2
-4
-4
-6
-6 -5
0
- k ---------- h s . * .............# " v Is : ^ : ---------- ■%
2
5
-5
0
(c) A shear transformation along y-axis » » » »
» »
A=[í 0; -2 1]; */, Shear along y-axis, c=-2 graíics(pts,lns,'r*, , 8 , * : ’) hold on grafics(A*pts,lns,’b ’,*+ >,8,*— *) hold off print -deps fig531c.eps
4
" ^ 8 - 6 - 4
-2
0 2
4
6
8
5
T h e M atrix Representation of a Linear Transform aron
M A T L A B 5.3
441
2. (a) The m atrix R for rotaion counterclockwise by 7r/2 is given by R = (ií(e i) /Z(e2)) = í - i
since the x-axis goes to the y-axis and the y-axis goes to the negative x-axis. The m atrix E for expansión along the x-axis by a factor of 2 is
(o?)'
(b) » » » » » »
» »
» » » » » »
» » »
pts = [
0 3 3 8 8 11 11 15 15 11 8 8 0 10;... 0 0 3 3 0 0 7 7 10 10 12 7 7 9]; lns = [ 1:13 ; [ 2:13 1 ] ]; R = [ 0 -1; 1 0] ; 54 Rotation by pi/2 counterclockwise E =[ 2 0; 0 1]; '/, Expansión along the x-axis by a factor of 2 subplot(121); grafics(pts,lns,’r*,9* 9,30, 9: 9) ; hold on grafics(E*R*pts,lns,’b*, , 3 0 , 9— 9) ; title(Original - dotted*) xlabel('Rotated then expanded - dashed*) hold off subplot(122); grafics(pts,lns,9r 9,***,30, 9: 9); hold on grafics(R*E*pts,lns,9v 9, 9+9,30, 9— 9) ; xlabel(*Expanded then rotated - dashed*) hold off print -deps fig532b.eps Original - dotted
3. (a) T(x) = projv x = (v -x )v is linear since( v • ( x i+ X2))v and (v • a x )v = (c*(v • x))v = a (v • x)v. The m atrix Since (v • ej) = v j , this means P = (viv V2V . . . vnv). 0 >) (i) »
v
» » » »
P =[v(l)*v v(2)*v]; grafics(pts,lns,*r9,9o 9,20,9:9) hold on grafics(P*pts,lns,’w*,,+ >, 20 , 9— 9) print -deps fig533bi.eps
»
=[ 1 0 ] ’ ;
= ((v -x i) + (v*X2))v = (v -x i)v + (v *X for P forT has T(e¿) forits j th column.
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Chapter 5 Linear Transformations
Instructor's Manual
(ii) The kernel of P is {x : (v • x) = 0} which is just all vectors perpendicular to v, i.e. the yaxis. (That is just saying that every vector perpendicular to v projects to zero.) The range of P is just the set of all múltiples of v, since every column of P has this form. Alternatively projection on v gives múltiples of v.
» » »
» » »
» » »
» »
»
w = [1 1]’; v = (l/norm(w))*w; P = [v(l)*v v(2)*v]; grafics(pts,lns,’r* , , 2 0 ,* :*) hold on grafics(P*pts,lns,’w*,’+ *, 20,* — *) print -deps fig533c.eps
w = [-1 1]’; v = (l/norra(w))*w; P = [v(l)*v v(2)*v]; grafics(pts,lns,,r >,’o*,20,* : }) hold on grafics(P*pts,lns, ’w 1,,+ *, 2 0 / — *) print -deps fig533d.eps
T h e M atrix Representation of a Linear Transform aron
M A T L A B 5.3
443
(e) Repeat for your own figures. 4. (a) The diagonals of a rhombus with two sides x and F x (the reflection of x in the line through v) will be perpendicular bisectors. Thus the diagonals meet at projv x, and from the bisection property, 2projv x = x + Fx. Since this is true for all x, and since P is the m atrix for the projection, 2P = J + F o r F = 2 P - J .
» v = [10]*; P = [ v(l)*v v(2)*v ] ; % Use the projection matrix from Problem 3 » F = 2*P - eye(2) X Reflection matrix from results in (a) F = 1 0 0 -1 » grafics(pts,lns,>r* , , 2 0 , 9: 9) » hold on » grafics(F*pts,lns,,>+>,20, » print -deps *fig534b.eps’
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» i = [-1 1]’; ? = (l/norm(w))*w ; */. Form the unit vector along y= -x » P = [ v(l)*v v(2)*v ]; F = 2*P - eye(2) */• Reflection in line through v F = 0.0000 -1.0000 -1.0000 0.0000 » grafics(pts,lns,’r ’,’* ’,20, 9 : 9) » hold on » grafics(F*pts,lns,’w ’,’+ ’,20, 9— ’) » print -deps ’fig534c.eps9
6. (a) » vi = [1 0 3 -1] '; v2 = [2 -1 4 3]’; v3 = [3 2 0 -1] '; v4 = [ 4 2 1 1]*; » rref([vi v2 v3 v4]) */, If this is I then vi’s forra a basis. ans = 1 0 0 0
(b)
0
1
0
0
0 1
0
0
0
0
0
1
We form various matrices from the given data: » » » » » » » »
[2 0 6 -2]»; Tv3 = [ 1 - 1 1 4 ] ’; Tv4 = [5 1 17 -10] Tvl = [ 3 - 1 7 2]’; Tv2 V = [ vi v2 v3 v4 ]; •/. So T(cl*vl+. ..+c4*v4) = TV* [el ... c4] ’ TV = [Tvl Tv2 Tv3 Tv4] ; cel = V\[l 0 0 0]’; ce2 = V\[0 1 0 0]’; */♦ cei are coordinates of ei with ce3 = V\[0 0 1 0]’; ce4 = V\[0 0 0 1]’; % respect to vi when ei = V*cei •/• T(ei) = T(V*cei) = TV*cei Tel = TV*cel; Te2 = TV*ce2; Te3 = TV*ce3; Te4 = TV*ce4; C = [ Tel Te2 Te3 Te4 ] */. C = (T(el) T(e2) T(e3) T(e4))
c = -6,0000 -2.8000 -23.6000 2 0 .0 0 0 0
11.0000 4.6000 42.2000 -34,0000
-
4.0000
3,0000
1 .2 0 0 0 14.4000
1.8000 12.6000 - 1 2 .0 0 0 0
1 0 .0 0 0 0
T h e M atrix Representation of a Linear Transform aron
M A T L A B 5.3
(c) We’ll rename some of the quantities from (b) » A = V ; B = TV » B/A - C ans = 1.0e-13 -0.0178 0.0356 -0.0089 0.0089 -0.1066 0.1421 0.0711 -0.1421
V. This should be all zeros, up to round off
0.0133 0.0067 0.0711 -0.0533
0.0089 0.0044 0.0178 -0.0355
The comments in (b) help to explain why C = B A “ 1 is the m atrix for T. Specifically note the coordinates, ce,- of e¿ with respect to the basis vl,v2,v3,v4 solve the equation A*ce,- = e,- , so ce,- = A ~ 1e,-. But T(e,-) = J3(ce¿ from the definition of T. Putting these together in the definition of C, C = ( r ( e i ) T (e 2) T(e3) T (e 4)), yields C = B A " 1. (d) We identify a basis for the kernel and the range of T from rref (C). » r=rref(C) r = 1.0000 o o o » C(:,1:2) ans = -
6 .0 0 0 0
-2.8000 -23.6000 2 0 .0 0 0 0
0 .0000
0 0
1.6250 1.2500 0 0
-1.8750 -0.7500 0 0
11.0000 4.6000 42.2000 ■ ■34.0000
Since there are pivots in columns 1 and 2 of r, {C(:, 1), C(:, 2)} form a basis for Range C . »
kl = [-r(l,3) -r(2,3) 1 0]
*/. Kernel of C from r*x = 0:
First x3=l, x4=0
kl = -1.6250 -1.2500 1.0000 0 » k2 = C-r(l,4) -r(2,4) 0 1] */. k2 = 1.8750 0.7500 0 1.0000
Then
x3=0, x4=l
Thus a basis of K erT is {k^k^}. 7. Compute T as a product: T = i 2( 7r / 4) * Ey(3) * E x ( 2) * R ( —7t / 4), where R{ 6) is rotation counterclockwise by 0, and E x ( k ), Ey{k) are expansions in the x, respectively y, directions by the factor Note the order is correct since the right factor is performed first and the left factor is performed last. (a) » Rpi4 = [ [ cos(pi/4); sin(pi/4) ] [ -sin(pi/4); cos(pi/4)]]; » Ex2 * [ 2 0 ; 0 1 ] ; Ey3 = [ 1 0 ; 0 3] ; » T = Rpi4 * Ey3 * Ex2 * inv(Rpi4) */, clockwise by pi/4 is the inverse of Rpi4. T = 2.5000 -0.5000 -0.5000 2.5000
An alternative way to find the matrix for T would be to compute T (ei) in 4 steps via geometry, and similarly for T (e2)
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Instructor's Manual
» A = [ [1;1] [-1;1] ] ; % Form the transition matrix from basis B to Std. » TB = inv(A)*T*A */, T in basis B is: B to Std, then T, then Std to B. Theorem 5 TB = 2 0 0 3
(c) TB shows that for vectors in the direction (1,1)* T acts by expansión by a factor of 2, while for vectors in the direction (—1,1)* T acts by expansión by a factor of 3. Since these two directions are independent, the action of T on any vector can be deduced by expanding it in each of the two new basis directions and performing the relevent expansions and adding the results back together.
Isomorphisms
Section 5.4
Section 5.4 1. Since (aA)* = a A 1 and (.A + B )* = A*+J3*, T is linear. Also if A* = 0 then A = 0. Henee ker T = {0}. So T is 1-1. Since dim M mn = dim M nm, then by Theorem 2, T is onto. Thus T is an isomorphism. 2. A t is invertible if and only if v ( A t ) = 0. v { A t ) = 0 if and only if kerT = {0}. K erT = {0} if and only if T is 1-1. Thus A t is invertible if and only if T is 1-1. By Theorem 2, T is 1-1 if and only if T is onto. Therefore, A t is invertible if and only if T is an isomorphism. 3. Suppose T is an isomorphism. Then T x = A t (x )b 1 = 0 if and only if x = 0. Then det A t ^ 0. Conversely, suppose det A t ^ 0. Then Tx = A t (x) bi = 0 has only the trivial solution. Then T is 1-1. Since d im F = dim W = n, T is also onto by Theorem 2. Thus T is an isomorphism. /ai
O- -
°\ 0
/«i\
• an /
\an/
0 a2 •
4. Define T : Dn —►Mn by T
\ 0
a2
, T is easily seen to be linear, 1 — 1 and onto. So
Dn , Mn are isomorphic. 5. dim{A : A is n x n and symmetric} = n(n + l)/2 = m. 6. Let V = the set of n x n symmetric matrices and let W = the set of n x n upper triangular matri/ «11 «12 ••• «in \ a 1 2 a22 • • • a2n ces. Note that d im F = di mW = n(n -f l)/2 . Define T : V —+ W by T \ üln
/«ll «12
0 a22 \
0
«ln ^ «2n
Clearly kerT = {0}. Thus T is 1-1. Clearly T is also onto. Then T is an iso-
J
morphism and thus V ~ W. 7. Define T : V —►VT by Tp = also onto. Then V ~ W.
Then kerT = {0} and thus T is 1-1. Since d im F = dim lF
5, T is
8. Suppose Tp = p + p' = 0. Since p is a polynomial, p + p' = 0 implies that p = 0 (look at highest degree term). Then kerT = {0} and therefore T is 1-1. By Theorem 2, T is also onto. Then T is an isomorphism. 9. mn = pq, i.e. dim(Mmn) = dim(Mpg). /«i
10. Define T : Dn —►Pn- i by T
0
0 a2
°\ 0
= ai + «2^ +
V 0
h «n^n
Clearly kerT = {0}, so T is
«n /
1-1. Since dim D n = dim Pn_i = n, T is also onto. Then Dn ~ Pn-i* 11. Repeat the proof of Theorem 6 with the understanding that the scalars ci , c2, .. .,cn are complex numbers. 12. Suppose T f = Tg. Then f ( x —3) = g(x —3) for all x E [3,4]. That is, f{x) = g{x) for all x E [0,1]. Then T is 1-1. If f ( x ) E C [3,4] then f ( x + 3) E C [0,1]. Then T f ( x + 3) = /(* ). So T is onto. Therefore, T is an isomorphism. 13. T(Ai + A2) = (Ai + A 2)B = A \ B + A 2B = TAi + TA2. T(aA ) = a A P = aTA. So T is linear. Suppose TA = AJ5 = O. Then A = OP"-1 = O. So kerT = {O} and therefore T is 1-1. Since dim Mnm = nm < oo, T is also onto by Theorem 2. Thus T is an isomorphism.
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Chapter 5 Linear Transformations
14. Suppose Tp{x) = x p \ x ) = 0. Then p \ x ) = 0 => p(x) = constant. Then kerT = {p GPn • p{%) = c, c G M}. That is, kerT ^ {0}. Then T is not 1-1 and therefore not an isomorphism. 15. If h £ H then proj# h = h. So T is onto. If H = V then T will be 1-1. 16. Let {v,-} be a basis in V. Then {Tv,-} is a basis in W. Define S : W —> V by 5(Tv,-) = v¿. Then S (T v) == v for all v G V. 17. Problem 3 showed that A is invertible. We need T ” 1 (Tx) = T ~ 1(Ax) = x. Then T""1x = A ~ xx since A - 1(i4x) = x.
18. T~*(p) = p ( x ) / x , since any polynomialp with p(0) = 0 is divisible by x . 19. Define T : C —►R 2 by T(a 4 - i 6 ) = (a, 6 ). Let ¿i = ai 4 - ib\ and Z2 =
^2
+
Then
T(^i + Z2) = T((ai + 02) + i(b\ + 62)) = (ai + a j, &i + 62 ) = (ai,
6 1)
+ (« 2 , ^2 )
= T ( z i ) + T ( z2)
If a G M then T (az) = T(aa + ¿a 6 ) = (aa, a&) = a(a, 6 ) = aT(z). So T is linear. If T(z) = (0,0) then z = 0 + ¿0 = 0. So kerT = {0}. Then T is 1-1. Since dimC = dimM 2 = 2, T is also onto. Therefore, € - l 2.
20. Let ci = ai + ¿&i,...,cn = an + ibn . Then let T : » E 2n be defined by T( c i , . . . , cn) = (ai, 6 1 , . . . , an, 6 n). Let di = ei + i f i , . . . , dn = en 4 - ¿/n . Then T ( c i , . . . , cn) + ( di , . . . , dn)) = (ai + ei, 61 + / 1 , . . . , an 4* e ni bn 4 */n) — ( a i , 61, . . . , an, -|- (e i, , . . . , en, /n) = T( c i , . . . , cn) 4- T( di , . . . , dn) If a G M then T ( a ( c i , .. . ,cn) = ( a a i , a 6 i , .. . , a a n, a6n) = a ( a i , 61, . . . , an, 6n) = a T ( c i , . . . , c n) So T is linear. kerT = {0}, so T is 1-1. Since dim C f = dimM2n = 2n, T is also onto. Therefore, C£ - M2n.
Isomorphisms
MATLAB 5 .4
M A T L A B 5.4
1. T is to be definedby T (v,) = w¿ where » »
vi = [1 0 0 0] '; v2 = [2 1 0 0] '; v3 wl = [1 2 1 0]’; w2 = [2 5 3 0]'; w3
[-2 1 2 0]’; v 4 = [ 3 4 2 1]’; [ - 1 - 1 - 1 2 ] ' ; w4 = [ 0 3 7 7]';
(a)
»
rref([vl v2 v3 v4]) # /, This will be I. So vi’s form a basis. And T defined.
ans 1 0 0 0
(b)
0 1 0 0
We form various matrices from the given data: % Since this too is I, wi’s form a basis.
» rref([wl w2 w3 w4]) ans = 1 0 0 1 0 o 0 o
T will be an isomorphism. In fact p{T) = 4 (as the w, will form a basis for Range T) and v(T) = 4 —p(T) = 0. So T is onto M4 and 1-1, and thus an isomorphism. (c) As in the solution to MATLAB 5.3.6, the m atrix for T with respect to the standard basis can be found, efliciently, by: » »
= [ vi v2 v3 v4 ]; TV = [ wi w2 w3 w4]; A = TV/V */, This will be the matrix for T in the standard basis.
A =
»
1.0000
o
2.0000
1.0000
1.0000 1.0000 0 o R = rref(A)
0.5000 -4.0000 1.0000 -9.0000 o o 1.0000 5.0000 Use this to find bases for Range T and Ker T.
R = 1
0 0 0
0 1 0 0
o 0 1
0 0 0
o
1
Since rref (A) is /, the columns of A (corresponding to the four pivots) form a basis for Range T and Ker T = {0}. Thus Range T = M4. Henee T is 1-1 and onto, henee an isomorphism. (d) If 5(w,*) = v¿ then the matrix for S is » B = V/TV */• TV=[ wl w2 w3 w4 ] so reverse V, TV roles from (c) B = 0.8667 1.8667 -0.8667 -0.0667 -1.8667 0.8667 0.1333 0.0667 -0.6667 -0.6667 0.6667 0.6667 0.1333 -0.1333 0.1333 0.0667 » B*A •/. If I then B = inv(A) ans
1.0000 0.0000 0.0000 0.0000
0.0000 1.0000 0 0.0000
0.0000 0.0000 1.0000 0.0000
0 0.0000 0.0000 1.0000
449
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Instructor's Manual
Chapter 5 Linear Transformations S ectio n 5.5
sin 9 *X cos 0 \ (x\xi cos# —x sin0 I 4
2
2
x3)
/ x\ sin 6 + x 2 cos 9 \ •I xi eos# —x 2 sin# I= x 2 + x\ + x§ = x \
• x, so |Tx| = |x|, or check
*3)
A* A = I.
(x\ cos 9 —x
2
sin 9 \ í x\ cos 9 —X3 sin 9 \ x2 I • í x 2 I = x\ 4 - x\ 4 - x§ = x • x, and henee, |Tx| = |x|.
xi sin 9 + x 3 cos 0 /
\ xi sin 0 + X3 cos 0 )
3. Using theorem 1, we have T x • Tx = (A 5x) • (A £x) = x • [(AS)* ABx] = x • (BtA tA B x ) = x • x. Thus |Tx| = |x|. / 1 / 7 5 - 1 / 7 5 i/V 3 \ 4. Ay = I 1 /7 2 1 / 7 6 —1 /7 3 1, with respect to the bases V 0 2 /7 6 1 / 7 3 / Bx =
| ^
| . As A TA ^ = I, AT is
1/3 j , ^ 2 / 3 j , ^ - 2 / 3 j | and S 2 = | ^ 0 j , ^ 1 j ,
orthogonal. 5. By theorem 2, isometries preserve inner produets. 6.
Assume x and y are nonzero. Let 2 denote the angle X'V TX-TV • between T x and Ty. By theorem 3.2.2, we have cos^>i = ]x||y[ an(^ c o s ^ 2 = fTx[|TyT• Using theorem = ||r¡jyj = cos ^ 2 - As 0 < / 2 dx = _ q gQ
0,
(v 4 , u 2) = j
ai
U3
= - ^ ~ (3 x 2 — 1). To find
Z ^ /6 \ x 3 [ ~ 2 ~x J
ft
U 4,
we
>/6
~
and (v 4 ,u 3) =
— V4 _ (v 4 ,u 2 )u 2 = x 3 — | x , and (v^l = 2 ^ 2 / 5 ^ .
\/l Thus u 4 = — t=(5x 3 — 3x). Let { e i,e 2 ,e 3, e4} denote the standard basis for M4. Define Tu,* = e¿ 2y2
Isometries
HTUl j = ° 2 ^ = e 3 + ^ e 1j,.„dr(«3X¡' =
for each i. Then T (a 2X2) = a2T í 3y iQ U3
( 2V 2 VÜ \ TT m, 2 3X “3 1 5 ^ 64 + ~5~e2 ) ' Hence’T (a° + axX + ° 2X + “3 X )
_ / 2\/2 V6 \ ( h\/fU4 + ~1TU2 ) =
= ^>/2ao + - Y ’a2, ^ = a i + ^ a3> 3^f^a2’ 2«o + ^ ao«2 + o o
5
Section 5.5
+ r a i «3 + 5 7
‘ Check that Hr (a° + ° lX + 02x2 + “a*3)!!2 =
= ||ao + a i* + a 2x 2 + a 3a:3||2. Thus, T is an isometry.
12. Recall that M 22 is an inner product space with (A, B) =
tr (A Bi ). We have { u i, u 2, 113, U4}
= { ( 0 0 ) ’ ( 0 0 ) ’ ( l o ) ’ ( 0 1 ) } ^o ra n orthonormal basis of M22. Define Tu¿ = e¿ for each
i. Then T
( : i ) -
(a b c \d
. Thus T is an isometry.
13. We have { u i , u 2, u 3,u 4} = |
ü ) ’ ( o o ) ’ ( 1 o ) ’ ( o 1 ) } ^°r an or^ onorma^ basis °f ^ 22, and
{v ij V2, V3, V4} = 1 1 /^ / 2,
"" 1 )>
Define Tu¿ = v¿ for each i. Then
i)
~~
i ^or an orthonormal basis of / hf —1 , 1]. + -~ p c (3 x 2 — 1) +
= a/y/2 +
yfi
~~ ^X^ an(^
(:¡)
. Henee T is an isometry. = a 2 + b2 + c2 + d 2 = 0 14. Recall that Dn is an inner product space with (A, B) = tr(A i?). Let Ei denote the n x n matrix with 1 in i, i position and 0 every where else. Then the set {Ei : i = 1, 2, . . . , n) is an orthonormal 0 0 • • 0 a 22 0 • 0 0 a 33 • •
0\ 0 0
an
basis for Dn . Define TEi = e,, i = 1, 2, . . . , n. So TA = T
0
||TA ||2 =
0
0
/ «11 \
022 =
ann /
«33
As
V ann /
— ||A||2, T is an isometry. t=i
15•
^ “ ( - ¿ a e + a)
+ 2i 3 ‘ S6« ) = ' 16- j‘, = ( s3 +21
17. As A* = A, then a,-,- = a¡7, and hence, a,¿ is real. 18 AA* ~ ( ( 1+ O/2 \ ( cos(2a ) sin( 2a ) \ (e) If v = (cos(a) sin a * then F = 2 P - J = 0 , \ ! , * 0 • 2/ \ í I = ( • x x • v ' v n \ 2 cos(a) sin(a) 2 sin (a) — 1J \ sm(2a) —cos(2a) J This is exactly R X , i.e. R with its second column negated.
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Instructor^ Manual
2. T is to be defined by T (v,) = w,* where » »
vi = [2/3 1/3 -2/3]*; v2 = [1/3 2/3 2/3]’; v3 = [2/3 -2/3 1/3]*; wl = [l/sqrt(2) l/sqrt(2) 0] *; w2 = [-1 1 2]’/sqrt(6); w3 = [1 -1 l]Vsqrt(3);
As in the solution to MATLAB 5.3.6, the matrix for T with respect to the standard basis can be found, efficiently, by: » V = [ vi v2 v3 ]; W = [ wl w2 w3 ]; » A = W/V */, This will be the matrix for T in the standard basis. A = 0.7202 -0.4214 -0.5511 0.2226 0.8928 -0.3917 0.6571 0.1594 0.7368 » A ’*A Test for orthogonality. If I then A is orthogonal ans = 1.0000 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 0.0000 1.0000
Now to verify that T maps an orthonormal basis to an orthonormal basis, we just check that the v¿ and wi both form orthonormal bases: » V>*V % This is I, showing the vi form an orthonormal basis ans = 1 0 0 0 1 0 0 0 1 » W ’+tf 7, This is I, and shows the wi form an orthonormal basis ans = 1.0000 0 0 0 1.0000 0 0 0 1.0000
Any isometry preserves lengths, by definítion. Henee it will map sets of unit vectors to sets of unit . vectors. But it also preserves inner produets, by Theorem 6 and henee preserves orthogonality. Thus it will map orthonormal bases to orthonormal bases.
Review Exercises for Chapter 5
R eview Exercises for C hapter 5 1- T ( x i , y i ) + T (x2)y 2 ) = (0, - yi) + (0, - y 2) = (0, - ( y i + y2)) = T ( ( x x, yx) + (x2,y 2)) T (a(x, y)) = (0, - a y ) = a(0, - y ) = aT (x, y). Therefore T is a linear transformation. 2. T ( x 1, y l , z 1) + T
(
x
y2, z2) = (1, yx, zx) + (1, y2, z2)
2,
= (2,(yi + y2),(z i + z2)) ( 1 , ( 2/1 + í/ 2 ) ,
(^1
+ xr2 ) )
= T ( (x l t yi,zi) + (x2, y2, z2)) Therefore T is not a linear transformation. dP1
y
2
y
«i| 1
2
3 . -----1-----does not necessarily eq u a l------------. Therefore T is not a linear transformation. yi V2 Vi + 2/2 4. T(a + bx) + T(c + dx) — (ax + bx2) + (ex + dx2) = (a + c)x + (b + d)x2 = T((a + bx) + (c + dx)) T(a(a + 6x)) = a(ax + bx2) = aT(a + bx). Therefore T is a linear transformation. 5. T(pi) + T(p2) = (1 + pi) + (1 + p2) = 2 + pi + p2 1 + (pi + p 2) = T(pi + p 2) Therefore T is not a linear transformation. 6. T ( h ) + T ( /2) = / i ( l ) + / 2(1) = ( h + / 2)(1) = T ( h + h ) T ( a f ) = a f ( 1) = T (/); therefore T is a linear transformation. 7. Since
2 -1
4
/1 2 — 1\
8. 1 2 4
7
# 0, Ker T = {(0,0)}; v(T) = 0; Range T = M2; p(T) = 2. /I 2 — 1\
3 J —>- 1 0 0
\ 1 2 — 6/
V00
/1 2 0\
5 1 —+ ( 0 0 1 j ; K erT = {(x,y, z) : z = 0 and x = —2y}; v(T) = 1. Note 5/
\00°/
that 3 ñ i —R 2 — R 3 = (0,0,0). Then Range T = {(x, y, z) : z = 3x —y}; p(T) = 2. Or choose pivot columns in A t : Range T = span . I W
V-6/J
9. Ker T — {(x, y, z ) : x = 0 and y = 0}; v(T) = 1; Range T = IR2; p(T) = 2. 10. Ker T = {0}; j/(T) = 0; Range T = {a + bx + ex2 + dx 3 + ex4 : a = 0 and 6 = 0}; p(T) = 3. 11.LetA = ( * " ) . T h . „ A B = p(T) =
+
K erT = { ( ™ ) } ; n ( r ) = 0; Range r = M2!i
4.
12. K erT = { / G C [0,1]: /(1 ) = 0}; Ker T is infinite dimensional; Range T = M; p(T) = 1. 13. í 4t = ( ¡ ¡ _ J ) ; K erT = span |
j ; v(T) = 1, Range T = span { ( _ j ) }'- />C0 = 1-
455
456
Instructor’s Manual
Chapter 5 Linear Transformations
14. A t = ( q q j ) ; K erT = span | ^ 0 ^ | ; v(T) = 1; Range T = M2; p(T) = 2.
15. A t
_ / l 0 - 2 (A — \0 2
0 3 /’
K erT = span <
°\1 -3 |
f 2\ 0 1
1
\0/
’
; Range T = ®2; p(T) = 2; v(T) = 2.
0
2/ J
/ 0 0 0 0\ 10 0 0
16. A t =
; Ker T = {0}; v(T) = 0; Range T = span {x, x 2, x3, x4}; p(T) = 4; v{T) = 0.
0 10 0 0 0 10
\0 0 0 1/ /-ll
17. .At =
00'
0 0 - 1 1 I; KerT = {0}; "(T) = 0; Range T = M22] ^ 00
= 4; ^
= °-
0 2;
18. T (l, 1) = (0,5); T (l, 2) = (-1 ,8 ); (0,5)Ba = (20/13,5/13); ( - l , 8 ) Ba = (33/13,5/13); AT = ¿ ( r a s )
K " T = { 0 } ; ‘' ( T ) = 0 ;< R “ g e T ) ' i- = SI>“ * { ( 25 / í l ) ■( 35 / í l ) } - = 2 -
19. Expansión along the x-axis with c = 3. 21. Shear along the y-axis with c = —2.
20. Compression along the y-axis with c = 1/3. 22. Shear along the ar-axis with c = —5.
* (5 !)
24
í 1
°)
^0 \ / z )
Review Exercises for Chapter 5
457
25. ( - 3. 4)
(- 3 .2 )
27. The row operations to transform to the identity m atrix are: 1. R 2 + 2iíi; 2. R 2/&’, 3. R\ —37^2 ( 28.
22)
- ( -
21)
(os) (
01)
The row operations to transform to the identity m atrix are: 1. - f l i / 3 ; 3 . R 2/ 5; 4. E i + 2 J ? 2 / 3 (
0 5 \
_
\-3 2 y -
/O
l \ / - 3 0 \
U ° A
( í
O W l
ov Vo sy Vo
- 2 /3 \
V
29. The row operations to transform to the identity m atrix are: 1. Ri~£tR2\ 2. R 2 4" 3. i?2/22; 4. i?i —3i?2
( 5)- c?i)(_i?)(¿ (á;) 1
30. The row operations to transform to the identity m atrix are: 1. R i m ^ - R 2\ 2. R 2 ~ 2iíi; 3. —7^2/9j 4. i?i —57^2
31. T(ao + aix + a2x 2)
do ax , a2á | of M2 and the orthonormalirasis {l / y / 2 , >/3/2 • x} of P i[—1,1].
32. Use the standard basis | L e tT ^
= 1 /^ 2 and T
^
= s / z j 2 - x . Then T ^
y = ^ ¿ 2 ) ' T ^en (x ,y ) = ° í a 2 + h h - (T x,T y) = aiü 2 + bib2. Thus T is an isometry.
j
= a/y/2 + y/Z/2 ■bx. Let x =
and
(a1a2/2 + ( a 1&2+a2¿>i):c-v'^/2-|-3M2a:2/ 2) dx =
458
Notes
Instructors
Eigenvalues and Eigenvectors
Section 6.1
C h a p t e r 6 . E ig e n v a lu e s , E ig e n v e c t o r s a n d C a n o n ic a l F o r m s Section 6.1 - 2 - A -2 = A2 + A —12 = (A + 4)(A —3); Eigenvalues: —4,3 1. —5 1 —A A + 41
= ( - 5 ” 5 ) =■
A —31
2.
= (is l | )
{ ( ”
5) }
-1 2 -A 7 = A2 + 10A + 25 = (A + 5)2; Eigenvalues: —5, —5 -7 2 - A => E - 5 = span ( ( J ^
A + 51 3.
= span { ( l ) }
2 —A -1 5 - 2 -A
Geometric multiplicity of —5 is 1.
= A2 + 1; Eigenvalues: i, —i.
A ~ lI= (
= > E t = s p a n | ( 2 + * ) } = span
5 -2 - i)
i ) } ’ Then taking conjugates,
E - i — span { ( 2 ~ t-) }• 4.
^
O 3
^|
^_
^*Senva^ues: - 3 , - 3
A + 31 = ( 0 0 ) ^ E - 2, = ffi2. Geometric multiplicity of —3 is 2. 5.
^
q _3 _ ^ = (—^ —-^)2j Eigenvalues: —3, —3
>1 + 3 /:
6.
( " ) ^ E - 3 = span ^ ^ 0 )
Geometric multiplicity of —3 is 1.
3 —A 2 = A2 —4A + 13; Eigenvalues: 2 + 3i, 2 —3i. -5 1 - A Á -
I =
( :1 ' - 5
-1
Then E 2 - 3 Í = span ( ( ^
-
3 ¡)
^0 }’
=*■ E í + 3 i =
sp “
{ ( :1 + - 5 ) }
=
sp “
{ ( (S i -
l)/2 ) } ■
conj ugatcs-
In 7-20, to find one root of P ( A), try división by (A — r) with ± r a factor of the constant term. Once di visión yields one root repeat with the quotient. Also see MATLAB 6.1.3 Solutions for 6.1.8 for an illustration of how to use row operations to get a factored form of the characteristic polynomial.
7.
1 - A -1 0 - 1 2 - A - 1 = A(1 —A)(A —3); Eigenvalues: 0,1,3. 0 —1 1 —A 1 -1 0\ / 1 —1 0\ /1 0 -1 ' A - 0 1 = | - 1 2 —1 1 —*■ ( 0 1 - 1 -► 0 1 - 1 0 -1 1/ \ 0 —1 l) \ 0 0 0, 0 -1 0 \ f /- r A — I = | —1 1 —1 j = > /? != span < í 0
E q = span
459
460
Instructor’s Manual
Chapter 6 Eigenvalues, Eigenvectors and Canonical Forms
Ez = span
A — ZI
8.
1 -A 1 -2 -1 2 -A 1 = —A3 + 2A + A —2 = —(A — 1)(A —2)(A + 1); Eigenvalues: 1,2, —1. (See the 0 1 - 1 -A Solutions to MATLAB 6.1.3(a) for a way to use row operations to find this characteristic polynomial in factored form.) 0 1-2\ / 1 - 1 -1\ / 1 0 —3 \ f/3' A —I — - 1 1 1 -► 0 1 - 2 — 0 1 -2 =* Ei = span \ 2
0 1 —2 /
9.
VO1 - 2 /
V0 0 0 /
A -21 =
1 0 -1 0 1 —3 | => E 2 = span 00 0
A+I =
10 -1 \ 0 1 0 1 =>• E - 1 = span 0 0 0/
; Geom. multiplicity of 1 is 2.
E 1 0 = span
A - 107 =
1- A 2 2 02 -A 1 = —A3 + 5A2 —8A + 4 = —(A —1)(A —2)2; Eigenvalues: 1,2,2 -1 2 2 —A A —I =
10 3\ 0 11 ) =>■ Ei = span 000/
1 —2 —2 \ / 1 —2 0 0 0 1 1 —*■ I 0 0 1 0 0 2 / V0 0 0 Geometric multiplicity of 2 is 1.
A -21 =
11.
1'
5 —A 4 2 4 5 —A 2 = -A 3 + 12A2 - 21A + 10 = -(A - 1)2(A - 10); Eigenvalues: 1,1,10 2 2 2 —A ( 4 4 2\ 4 4 2 j => Ei = span 2 2 1/
10.
I VI
E 2 = span
-A 1 0 0 -A 1 = —A3 + 3A2 —3A + 1 = —(A — l) 3; Eigenvalues: 1,1,1 1 -3 3 - A /-I 10\ / 1 —1 0 A —I = \ 0 -1 1 ) -f ( 0 1 -1 \ 1 -3 2 / V0 - 2 2 Geometric multiplicity of 1 is 1.
1 0 -1 0 1 —1 | => Ei = span 00 0
Eigenvalues and Eigenvectors
12.
-3 - A -7 24 -A 1
-5 3 = —A3 + 3A2 —3A + 1 = —(A —l) 3; Eigenvalues: 1,1,1
2 2 -A
/ —4 —7 —5 \
A-I = [
/1 2
2 3 3 i-+
1
[ 0 1—1
V 1 2 1/ \ 0 1 —1 Geometric multiplicity of 1is 1. 13.
Section 6.1
10
3
—3
0 1 —1 | => E i = span
00
0
A -1 -1 1 - 1 -A 0 = —(A3 + A2 + A + 1) = —(A + 1)(A2 + 1); Eigenvalues: -1, i, —i (See the 1 0 - 1 -A Solutions to MATLAB 6.1.3(a) for a way to use row operations to find this characteristic polynomial in factored form.) 2 -1 - 1 \ 0 j => E -1 = span 1 0 0/
A+ I = j 1 0
E¡ = span
A —i l =
Then £L¿ = span
14.
7 - A -2 -4 3 -A - 2 = -A3 + 4A2 - 5A + 2 = -(A - 1)2(A - 2); Eigenvalues: 1,1,2 6 -2 -3 - A /6-2-4\ /1-1/3-2/3 3 —1 - 2 — 0 0 0 \ 6 —2 —2 / \0 0 2 Geometric multiplicity of 1 is 2. 5 —2 —4 \ / I - 0 .4 - 0 .8 A - 27 = | 3 - 2 - 2 — 0 -0 .8 0.4 6 -2 -5 / \ 0 0.4 -0 .2 A —/ =
=>• Ei = span
10
-1 \
0 1 —0.5 1 =>• E2 = span 00
0/
= —A3 + 5A2 —8A + 4 = —(A —1)(A —2)2; Eigenvalues: 1,2,2
15.
1 0 4/3 \ A —I =
00
0 1 =$■ E i = span
0 1 1 /3 /
1 0 3/2 \ A -21 =
0 1 1/2 1 => E 2 = span
00
0/
Geometric multiplicity of 2 is 1.
4 —A 1 0 1 2 3 —A 0 1 16. = (2 - A)(48 - 44A + 12A2 - A3) = (A - 2)2(A - 4)(A - 6); -2 12 - A - 3 2 -1 05 - A Eigenvalues: 2,2,4,6
462
Instructor’s Manual
Chapter 6 Eigenvalues, Eigenvectors and Canonical Forms 1 1 / 2 0 1/ 2 '
A - 21
0
=
0 0
00 0 2 0 -2 2 0 -2
100
1'
0 0 0
0
0 1 0 -1 ,0 0 0 0-
E2
=
span Geometric multiplicity of 2 is 2. A -41 =
0 1 0 1\ 2 -1 0 1 ] 2 1 -2 - 3 ^ 2 -1 0 1/
1 - 1/2 0 l / 2 \ 0 10 1 0 1 1 0 Vo 0 0 0/
/ I - 1 /2 0 1/2 \ 0 -10 2 | 0 04 4 I V0 0 0 0/
A -61 =
/ I 0 0 1\ I 0 10 1 100 11 Vo 0 0 0 /
/ - 1\ => E 4 = span <
0000 I a —A 0 0 0 00 / 0 0 a —A 0 0 = (a —A)4; Eigenvalues: a, a, a, a 0 Oa-A 0 0 0a-A
/0 6 0 0\ 0000 1 „ A —a l = 0000 I a ~ S1 Vo 0 0 0 / Geometric multiplicity of a is 3.
19.
r / i \ 0 0 0 1 i ) 0 0 1 ’ k. V o/ v o / V i/ >
a -A 6 0 0 0 a —A c 0 = (a —A)4; Eigenvalues: a, a, a, a 0 Oa-A 0 0 0 0 a —A /0 6 0 0\
OOc O oooo I a~ s Voooo/ Geometric multiplicity of a is 2.
A -al
20.
<
í ol \] í °0\ 0 ’ 0 vo/ \í/ J
a -A 6 0 0 0 a —A c 0 = (a —A) ; Eigenvalues: a, a, a, a 0 Oa-A d 0 0 Oa-A
/0 6 0 0\ A - a l = i Ü Ü f °, => E a = span I 000d Vo 0 o 0 /
q
Vo,
'
-2 => f?6 = span < 1 0/
a -A 0 0 0 0a-A 0 0 17. = (a —A)4; Eigenvalues: a, a, a, a 0 Oa-A 0 0 0 Oa-A ( 0 0 0 0\ 0000 ] A -a l. =$■ Ea = IR4. Geometric multiplicity of a is 4.
18.
y
\)
(\ 0 0 -\¡2 \
0 10 001 \0 0 0
-1 -1
I >, Geometric multiplicity of a is 1.
(
j
l\
4 -2 2/ >
Eigenvalues and Eigenvectors
21.
Section 6.1
a -A b = (a — A)2 + b2; Eigenvalues: a + bi, a — bi -b a - A
(-» < ) ( -') = (o) Thus ( ^ j an(^
i ) are eigenvectors of A =
22. Note that det (A* —XI) = det (A —X/)* = det (A —A/). Then the characteristic equations for A and A t are the same and thus A and A t will have the same eigenvalues. 23. For each A¿, 1 < i < k, there exists x,- ^ 0 such that Axf- = A¿x¿. Then aAx{ = aX,x¿. Thus aAt-, 1 < i < k, are eigenvalues for a A. Conversely if a ^ 0 and is an eigenvalue for a A with eigenvector x, then (aA)x = vx, so Ax = Thus ~ = A¿ or z/ = aX,• some 24. Note that A ~ x exists if and only if A x = 0 only for x = 0, i.e. if and only if 0 is not an eigenvalue for A . This gives the result. 25. Using the same notation as in problem 23, we have A xí = A¿x¿. Then x¿ = í4” 1A,*Xí. Then ~-x¿ = At A“‘1x¿. Thus, — , 1 < * < k, are eigenvalues for A” 1, and reversing the roles of A, A” 1, we get 1/A* A* are all the eigenvalues. 26. Using the same notation as in problem 23, we have (A —al )xi = Ax¡ —ax,- = (A,* —a)xj. Thus A¿ —a , 1 < i < k, are eigenvalues of A —al. Conversely, if (A —a /) x = Ax, then A x = (A + a )x , so A4-a = A¿ i.e. A = A¿ —a. 27. Using the same notation as in problem 23, we have A 2xí = A(Ax¿) = AA¿x» = AjAxt- = A2x¿. Thus A?, 1 < « < are eigenvalues of A 2. Conversely, if A 2x = v x then (A 2 —v l ) x = 0 or (A + y/vI)(A — y/v l)x = 0. Henee either (A — y/vl)x = 0 or y = (A —\/^0x 7^ 0 and (A + V ^ I ) y = 0. These say one of ± v ^ is an eigenvalue for A, i.e. ± v ^ = some h ° r v = A¿. 28. Assume x¿ is an eigenvector for A for the eigenvalue A,*. Then A nXi = A n~1(Ax¡) = i4n" 1(AiX,*) = Af(i4n“ 1x,*). Now repeating this argument n — 1 more times yields Anx,- = A” x¿. Thus each A” is an eigenvalue for A n . Conversely, suppose v is a (complex) eigenvalue for A n with an associated eigen vector x and A = y/v is one complex n ’th root of v. Then A n — v i = n¿= iC ^ ““ e2irtJ^nXI). Let y0 = x and y m = IlyLiC^ —e2Kl^ nXI)x for m — 1 ,..., n. Since y n = (A n —v l ) x = 0, there exists a smallest k > 0 with y* = 0. Since Ar > 0, yib_1 ^ 0 and (A —e2irtk^nXI)yk_ 1 = y* = 0. This says e2irtk/nX is an eigenvalue for A, so A/ = e2*tk/nX, for some /. But A" = (e2,r**/nA)n = An = v, i.e. v = AJ1. This shows every eigenvalue of An is the n ’th power of some eigenvalue of A and finishes the solution. 29. p(A)v = (a0I + aiA + --- h anA n)v = aov 4- ai Av + h an Anv = a0v + ai Av 4----- h anAnv = (ao 4- aiA 4 -------h anAn)v = p(A)v. 30. Using the same notation as in problem 23 and the result of problem 29, we have p(A)x¿ = p(A¿)x¿, for 1 < i < k. Then p(At), 1 < i < k, are eigenvalues ofp(A ). (Note it takes a lot more work to show there are no other eigenvalues for p(A).)
463
464
Chapter 6 Eigenvalues, Eigenvectors and Canonical Forms
31. Let A =
det
/ a n 0 ••• 0 \ 0 022 O O
Instructor^ Manual
A is an eigenvalue of A if and only if det (A —XI) =
\ O • • • O ann / / , n = 2 2 , . . . , 2 5 ans = 0.9909
1.0582
1.0005
1.0496
A Model of Population Growth
M A T L A B 6.2
497
So both the ratios Tn/T n_i and P j , n / P a , n are still quite variable in the years 21 to 25 ( the first changes nearly 13 per cent a year and the second nearly 5 per cent a year). » p=zeros(2,5) ;p(: ,l)=A~46*pO; */. Now put all yeaxs 46-50 into one 2 rowed matrix » for k=2:5,p(:,k)=A*p(:,k-l); end » p(l,:)./p(2,:) */, Ratio of juveniles to adults for years 46,..., 50 ans = 2.9184 2.9254 2.9194 2.9245 2.9201 » T=sum(p) ; » T(2:5)./T(l:4) 7. Ratio of Tn to T{n-1>, n=47,...,50 ans = 1.0273 1.0260 1.0272 1.0262
For the years 46 to 50 the variation is down to at most .2 per cent a year. (c) The ratio of the absolute valúes of the smallest to largest eigenvalue in MATLAB problem one was .8358/1.4358 = .5818, while for the present problem the ratio is .8766/1.0266 = .8539. The convergence to a stable distribution is governed by equation (8) and requires that (A2/A i)n —* 0. This approach to zero is very slow for a number like .8539, near 1, much more rapid for numbers like .58 which are closer to zero. 3. »
A = [ 0 1 ; .6 .8]; p0=[100 200]*; */. Growth matrix and initial deer population
(a) Compute the (largest) eigenvalue for A: » CV,D]=eig(A) V = -0.9044 -0.6181 0.4267 -0.7861 D = -0.4718 0 0 1.2718
The m atrix has largest eigenvalue = 1.2718, and the other eigenvalue is strictly less in magnitude. Also there is an eigenvector for the largest eigenvalue with all components positive. Under these conditions we have seen that Tn/T n- \ approaches the largest eigenvalue, i.e. the long term growth rate is 1.27. (b) The only change in the model is that the adult population in the following year will be decreased by those adults from the previous year killed by hunting which is just h times the adult popula tion. This simply modifies the matrix A by subtracting h from the adult survival rate, A{2,2). (c) » AH=A; AH(2,2)=A(2,2)-.6; » [AH~10*p0 AH~20*p0 AH~30*p0 AH~40*p0 AH~50*p0] ans = 46.8229 13.5931 3.8386 1.0818 0.3048 43.6535 12.0274 3.3830 0.9531 0.2685 » [V,D]=eig(AH) V = -0.8265 -0.7503 0.5629 -0.6611 D = -0.6810 0 0 0.8810
498
Chapter 6 Eigenvalues, Eigenvectors and Canonical Forms
Instructor^ Manual
The components of the vectors A H npo get smaller and smaller, decreasing to zero. (By year 50, after rounding to integer valúes, there are no deer left.) Alternatively, note that the largest eigen value, representing the total population growth rate, is .8810, less than one, so eventually the to tal population goes down by about 12% a year; obviouly this leads to eventual elimination.
» AH(2,2)=A(2,2)-.3; */. Modiíy AH(2,2) for h=.3 » [AH~10*p0 AH~20*p0 AH~30*p0 AH~40*p0] ans = 1.0e+03 * 0.2925 0.5440 1.0111 1.8793 0.3115 0.5788 1.0758 1.9994 » max(max(eig(AH))) V. Largest eigenvalue > 1 , so explosive growth ans = 1.0639 » AH(2,2) =A(2,2)-.5; 7. Try h=.5 » [AH~10*p0 AH~20*p0 AH~30*p0 AH~40*p0] ans = 88.3481 47.4730 25.2999 13.4808 84.1617 44.5902 23.7564 12.6583 » max(max(eig(AH))) 7%Largest eigenvalue, still < 1, explaining decay ans = 0.9390 » AH(2,2)=A(2,2)-.4; » [AH~10*p0 AH~20*p0 AH~30*p0 AH~40*p0] ans = 162.1221 162.4977 162.5000 162.5000 162.7267 162.5014 162.5000 162.5000 » max(max(eig(AH))) */, Largest eigenvalue = 1, so eventual steady State, ans =
1 For an h to determine a steady state, the largest eigenvalue must be one. Equation 10 in the text can be adapted to test for equality of the largest eigenvalue to 1 and becomes ¿ = (1 —/?)/) » p30=A~30*p0 % Distribution after 30 years. p30 = 1.0e+04 * 9.0060 4.2565 2.9328 » v=zeros(3,6);v ( :,l)=p30; X Columns of v will be v30,v31,etc. » for k=2:6,v(:,k)=A~(29+k)*pO;end X Calcúlate the populations at 31-35 » T=sum(v) ;T(2:6) ./T(l:5) X sum the columns of v and take ratios ans = 1.2705 1.2701 1.2704 1.2702 1.2704 » W=v*diag(ones(l,6)./T) X Columns of W are w30,w31,w32,etc. W = 0.5561 0.5563 0.5561 0.5562 0.5562 0.5562 0.2628 0.2626 0.2628 0.2627 0.2627 0.2627 0.1811 0.1811 0.1811 0.1811 0.1811 0.1811
Since entries in v n are all nonnegative, the entries in w n, vn(i)/su m (v n), are exactly the proportion of vn in the ¿’th entry. limn-^oo Tn/ T n- i = 1.2705 from the results for years 31-35; thus it appears the total population is growing at about 27 per cent per year. Also limn_oo w n = (.5562, .2627, .1811)* and the entries give the eventual proportions of juveniles,1 to 5 year olds, and over 5 year olds in the population. Thus even though the populations are growing, the pro portions are not changing. oo » CV,D]=eig(A) X D(2,2) is the largest eigenvalue, it is greater than 1 V = 0.8281 -0.8674 -0.0730 -0.5133 -0.4097 -0.4489 0.2252 -0.2825 0.8906 D = -0.9679 0 0 0 1.2703 0 0 0 0.0976 » z=-V(:,2) X An eigenvector for D(2,2) with all positive elements z = 0.8674 0.4097 0.2825 » zz=z/sum(z) zz = 0.5562 0.2627 0.1811 » T(6)/T(5) - D(2,2) , zz-W(:,6) ans = 7.4247e-05 ans = 1.0e-04 * -0.2471 0.3067 -0.0596
X Differences cióse to zero as expected
499
500
Chapter 6 Eigenvalues, Eigenvectors and Canonical Forms
Instructoras Manual
The conclusión is that the vector of eventual proportions of each age class agrees with the proportions of the entries in the eigenvector for the largest eigenvalue. (e) Again equation 9 justifies the result about T(n)/T(n-1) and also the conclusión that the direc tion of v n should coincide with the direction of the positive eigenvector for the largest eigenvalue. Specifically, p „ / [ l 1 l]p n « a 1Anv 1/( a iA n [l 1 l]v „ ) = v i / [ l 1 1]v í . 5. »
P=[.8 .2 .05; .05 .75 .05; .15 .05 .9];
(a) (See MATLAB 1.6 Solutions to problem 14 for details. Here are interpretations) The ith component of P nx represents the number of households buying product i after n months. As n gets larger, P nx seems to be getting closer and closer to a fixed vector, (900 500 1600)*, implying that the market share of each product stabilizes over time. (b) » CV,D]=eig(P) */. This has D(l,l) = 1 as largest eigenvalue, V(:,l) all positive V = 0.4730 0.7071 0.8018 0.2628 0.0000 -0.2673 0.8409 -0.7071 -0.5345 D = 1.0000 0 0 0 0.7500 0 0 0.7000 0
Any initial starting vector x with all positive entries will have a nonzero component in the direc tion of the eigenvector for the largest eigenvalue, since 0 < (1 1 l)x = (1 1 l ) ( a i r (:, 1) + a27 (:, 2) + a3V(:, 3)) = ax + 0 + 0 for any such vector. Now extend the discussion leading to equation (9) to M3, to conclude that when Ai = 1 is the largest magnitude of an eigenvalue, then P nx will approach some fixed múlti ple of the eigenvector, v i. In the present case this will approach the limit y = a iv n = ((1 1 l)x )v n 3000vi since x = (1000,1000,1000)*. (In general if the eigenvector vi had not satisfied (1 1 l)v i = 1, then it would have to be replaced by z = v i/s u m (v i), an eigenvector normalized so that the components sum to 1). The limit vector has components which represent the long term distribution of households which will be buying a given product each month. (taking conjugates, eigenvector for A = 1 —i:
( 2 + 6i ) - Th“ c = G - i 2 + ‘ ) 5. By problem 6, section 6.1, the eigenvalues of A are 2 + 3¿, 2 —3i. Since the eigenvalues of A are dis tinct, A is diagonalizable. Eigenvector for A = 2 + 3i: ^
i + 3f ^ » taking conjugates, eigenvector for
A= 2- 3i; (-l-3 ¡) Th“ C= ( - l +3¡-l-3¡)' 6. By problem 7, section 6.1, the eigenvalues of A are 0,1,3. Since the eigenvalues of A are distinct, A is diagonalizable. Eigenvector for A
( =l \ W
■0:
( ~ l for \ A= [ 1 1 ; Eigenvector V i/
1: I
0 I ; Eigenvector for
A /l- l 1' - 2 . Then C = 1 0 -2 1/
\1
1
1,
7. By problem 8, section 6.1, the eigenvalues of A are —1,1,2. Since the eigenvalues of A are distinct, A is diagonalizable. Eigenvector for A = 1: [( 2* 1\ ; Eigenvector for A = 2: I( 3x\I . Eigenvector for W \ij (l\ /311\ A = - 1 : I 0 1; Then C = I 2 3 0 I .
Similar Matrices and Diagonalization
; Eo = span . Then A is diagonalizable and C = II 111 1 0 1 ]. 1 /1 V0 1 1
11. By problem 14, section 6.1, the eigenvalues of A are 1,1,2. Ei = span
f?2 —
f/2 \} /I 02\ span < I 1 1 >. Then A is diagonalizable and C = [ 3 —2 1 1 . I\2/J Vo 1 2 ; 12. By problem 15, section 6.1, A is not diagonalizable since the eigenvalue of 2 has algebraic multiplicity two and geometric multiplicity one. 13.
14.
-3 - A -7 -5 '- 3 ' 2 4 —A 3 = 1 — 3A + 3A2 — A3; Eigenvalues: 1,1,1. E 3 = span ^ j 1 j A is not 1 22 -A diagonalizable since the algebraic multiplicity of 1 is three and the geometric multiplicity of 1 is one. -2 - A -2 0 0 -5 1- A 0 0 = A4 + A3 — 11 A2 + A — 12; Eigenvalues: 3, —4, i, —i. Since the eigen0 02 —A -1 0 0 5- 2 - A /" 2 \ 5 ; Eigenvector for A = —4: 1 1/ 0> 0\ 0 0 . . Then ; Eigenvector for the conjúgate A = —i: 2 —i 2+ i 5 5/
valúes are distinct, A is diagonalizable. Eigenvector for A = 3:
/1 \ 1 ; Eigenvector for A = i: 0 \0 / -2 1
C —
0
0'
51 0 0 10 2+ i 2- i r 10 5 5>
506
Instructor^ Manual
Chapter 6 Eigenvalues, Eigenvectors and Canonical Forms
0 15. By problem 15, section 6.1, the eigenvalues of A are 2,2,4,6. Ei = span < 1 i \0 / span <
1 - 1/ )
—
/O - 1 1 i \ 0 1 1 1 Then C 1 0 1 -1
l\ 1
(~l\ 1 ► ; £4 0 1/ >
>] E q = span
Vo
1-1
1/
16. Since A is similar to £ , B = D l AD for some invertible m atrix D. Since B is similar to (7, C = E ~ l B E for some invertible matrix E. Then C = E ~ 1D ~ 1A D E = (DE)~l A(DE). Thus A is similar to C. 17. Since A is similar to B, B = C ~ XA C for some invertible matrix ) » A = [3 9.5 -2 -10.5; -10 -42.5 10 44.5; 6 23.5 -5 -24.5; -10 -43 10 45] » d=eig(A), dd=d.~20; % Take the 2 0 ’th power of the eigenvalues of A. d = -3.0000
2 .0 0 0 0
1.0000 0.5000 » E=diag(dd) E = 1.0e+09 * 3.4868
0 0 0
0
0.0010 0 0
0
0 0.0000 0
0
0 0 0.0000
511
512
Chapter 6 Eigenvalues, Eigenvectors and Canonical Forms » CV,D]=eig(A); » E-D~20 ans = 1.0e-05 * 0.4768
0 0 0
Instructor^ Manual
54 Find the eigenvectors for A. 54 Zero u p to round-off, s o E=D~20.
0 0 0 0
0
0 0 0
» A~20-V*E*inv(V) ans =
0 0 0 0
*/. Zero up to round-off, so A~20=V*E*inv(V).
0.0000
-0.0001
0.0000
0.0001
-0.0014 0.0007 -0.0014
-0.0056 0.0028 -0.0056
0.0014 -0.0007 0.0014
0.0056 -0.0028 0.0056
(c) From C ~ XDC = A we get A = C D C " 1. Then A n = (C - 1D C ) ( C - 1DC )...(C-1D C ) ( C - 1DC), n factors of C D C ~ l . The interior adjoining C C ~ l all give I, leaving n factors of D,i.e. A n = C D nC - 1. (Compare with 6.3, problem 17 or 21.) 5. (a) A x = Ax for A > 0 says that A expands or compresses x, depending on whether A > 1 or A < 1. (b) A expands or compresses each eigenvector by a factor given by the associated eigenvalue. If A is diagonalizable there is a basis of eigenvectors. Since any vector is a linear combination of the eigenvectors in this basis, the effect of multiplication by A on any vector can be described as a linear combination of the expansions or compressions along the directions of the basis vectors.
(c) » A=[5/2 1/2;1/2 5/2]; » [V,D]=eig(A) 54 Since D has distinct entries, A is diagonalizable V = 0.7071 0.7071 -0.7071 0.7071 D =
2
0
0
3
» V*diag([l l]./min(V)) ans =
-1 1
54 This divides each eigenvector in V by its* minimum
1 1
The last result shows the eigenvectors are in the directions (1,-1)* and (1, l ) t and the eigenval ues in D show multiplication by A expands in the direction (1 -1)* by a factor of 2 and expands the direction (1 1)* by a factor of 3. To sketch the image of the rectangle, whose corners are at eigenvectors, take the diagonal running from the (-1,-1) córner to the (1,1) córner and stretch by a factor of 3 in each direction; take the other diagonal and stretch by a factor of 2 in each direc tion. Your sketch should yield a rhombus.
Similar Matrices and Diagonalization
M A T L A B 6.3
(d) (i) » »
A [15 -31 17; 20.5 -44 24.5; 26.5 -58 32.5]; Distinct diagonal entries in D>0 so A is diagonalizable CV, D]=eig(A)
V =
-0 . 4243 -0 . 5657 -0 . 7071
-0.2453 -0.5518 -0.7971
0.5774 0.5774 0.5774
0
0 0 1.0000
D =
2 . 0000 0 0 » inv(V)*A*V ans =
0.5000
0
*/• This veriíies A is diagonal izable
2.0000
0 .0 0 0 0
0.0000
0.5000
0.0000
0.0000
0.0000 0.0000 1.0000
» V*diag([l 1 1] ./min(abs(V))) ans =
-1.0000
-1.0000
1.0000
-1.3333 -1.6667
-2.2500 -3.2500
1.0000 1.0000
» ans*diag([3 4 1 ] ) ans = -3.0000 -4.0000 -4.0000 -9.0000 -5.0000 -13.0000
*/. Give eigenvectors with nice entries
y.
Still nicer eigenvectors
1.0000 1.0000 1.0000
The last m atrix computed has three independent eigenvectors for columns, and shows that the geometry of A is given by an expansión by a factor of 2 in the direction (3 4 5)*, compression by a factor of 1/2 in the direction of (4 9 13) and expansión by a factor of 1 in the direction (111)*. (ii) » B=10*rand(3)-5*ones(3,3); A=B,*B A = 31.6145 -26.8123 -23.8618 -26.8123 23.4677 20.1571 -23.8618 20.1571 32.6538 » [V,D]=eig(A) % D has 3 distinct positive entries, % so A is diagonalizable V = -0.6203 0.6599 0.4240 0.5305 0.7511 -0.3930 0.5778 0.8160 0.0188 D = 0 0.4161 0 0 10.5457 0 76.7743 0 0
The three independent eigenvector columns of V give directions of expansion/compression for multiplication by A.
513
514
Chapter 6 Eigenvalues, Eigenvectors and Canonical Forms
Instructor’s Manual
6. (a) » A=[22 -10; 50 -23]; e = eig(A) , d = det(A) e =
2 -3 d =
-6 » A=[8 3; .5 5.5]; e = eig(A) , d = det(A) e = 8.5000 5.0000 d = 42.5000 » A= [5 -11 7; -2 1 2 ; -6 7 0]; e = eig(A) , d = det(A) e =
2 .0 0 0 0
1.0000 3.0000 d =
6 » A = [26 -68 40;19 -56 35;15 -50 33]; e = eig(A) , d = det(A) e = -
2.0000
2 .0 0 0 0 3.0000 d =
-12
For each A the eigenvalues in e are distinct, so a set consisting of one eigenvector for each eigen value will be a basis, the matrix C with this basis as columns will be invertible and C ~ XA C will be diagonal with the eigenvalues on the diagonal; this says A is diagonalizable. For each A the product of the eigenvalues is equal to det (A).
» A = [ 38 -95 55; 35 -92 55; 35 -95 58]; V. Matrix from MATLAB 6.1, 1 » det(A), (-2)*3*3 % This verifies det(A) = the product of the eigenvalues ans = -18 ans = -18 » A = [ 1 1 .5 -1 ; -2 1 -1 0; 0 2 0 2; 2 1 -1.5 2] A = 1.0000 1.0000 0.5000 -1.0000
-2.0000 0
1.0000 2.0000
-1.0000 0
o 2.0000
2.0000
1.0000
-1.5000
2.0000
» det (A), (l+2*i)~2*(l-2*i)~2 */. This verifies det(A) = product of eigenvalues ans = 25 ans = 25.0000
S im ilar M atrices and D iag o n alizatio n
M A T L A B 6 .3
(c) If A is diagonalizable, then det (A) is the product of the eigenvalues for A (counting algebraic multiplicity). Proof: Since A = C ~ 1D C where D has the eigenvalues for A along its diagonal, det (A) = (1 / det (C)) det (D ) det (C ) = det (D ) = d iid 22-*-dnn* The equalities come from the product rule for determinants, the fact that det (C ” 1) = 1 / det (C) and the fact that the determinant of a diagonal matrix is the product of its diagonal entries.
515
516
In stru cto r’s M anual
C h a p ter 6 Eigenvalues, Eigenvectors and Canonical Forms
S e c t io n 6 .4
1. The eigenvalues o f A are Ai = 5 and A2 = —5. The corresponding eigenvectors are v i = ^ ^
and
y» = ( 4 ) |v i| = |v 2| = y/E. Thus Q =
= >h ( l —2 ) and
(^i _ 2 ^ ’
_ 1 /2 5 0 \ _ /5 OV ~ 5 V 0 -2 5 / \0 - 5 J
2. The eigenvalues of A are Ai = 1 and A2 = 3. The corresponding eigenvectors are v i =
an(^
v» = ( 0 |v ,l = |v , | =
Thus Q = ^ -
= j . ( l ” ¡) “ d
= K 3 S)“ (iS)3. The eigenvalues of A are Ai = 0 and A2 = 2. The corresponding eigenvectors are v i =
y¡ = ( . } ) • W = N
anc^
= V i Thus)/2, A4 = (1 — y/%)/2.
517
518
Instructoras M anual
C h a p ter 6 Eigenvalues, Eigenvectors and Canonical Forms
1—y/Ei\ The corresponding eigenvectors are v i =
|vi| =
1
, |v2| =
1
, |v3| = y l 2 + ( i~ ^ )
0 Vi0-2n/5 Vl0+2>/5 ^
oo 10 \01
9.
Vl0-2\/5 V10+2V5 o o o 0/
,v2 =
2
,v 3 =
0 0/
v4 =
= ¿ \ / l 0 —2\/5, |v4| = | \ / l 0 + 2\/5.
/0 0 0 02 0 , and Q tA Q — 0 0 1^1
V0 0
0\ 0 o
0^
/
Let u be an eigenvector corresponding to A with |u| = 1. Then 1 = |/u | = \Q 1Qu| |AQ‘u | = |AQu| ( since Q is symmetric) = |A2u| = A2. Henee A = ±1.
|AQ- ■i„i
-
10. We have B = Q*AQ and C = R tB R 1 where Q and R are orthogonal. As B = R R tB R R t = R C R } = l +
2
= 36;
1 — a/2
a/
a /
1-VIO
and (4 _ ^ ( x ') 2 + (4 + V Í Ó W
D = ^4 - v/ÍO ^ ^
)2
3
3
5 ± 3x/2
\/4-
©4
2 a/ 2 , | v 2 | =
+
2\/2; Q =
« ■
W (
J
i n :
J ) ; | V1|
13(*')2
=
V 2 6 ,|v 2|
=
v ^ 6;Q
W )!_
2
2
= - 7 . Hyperbola;
6
,2
5/2
5/2 — 6
169
—A = ^ 2 - X =
-± =
1 + V2
-1
1 1 =
£> =
1
6
= 292.5°. /5 ^
= 0 = ^ A = ± 1 3 /2 ; V l= ( , l j ¿)
- 0
\
- 2a/2 V 4 + 2 V 2 /
^ and .C5- + o3v^ .(^ )2 + Í5— l ^ l ( j / ) 2 = 1 ; Ellipse; and 2 V ^ (a:' )2 + 6- A
/-i + V5\
y / 4 - 2 V 2 \ / 4 + 2\/2 a/ 4
= -7 ;
- 2VT0 /
;v,=(
- 1 + V2 =
a/20
= 36; Ellipse; 0 » 35.78°.
12 r ^ ^ V ' v r ^ - i - i ~ x ~ v 2 = A2 -5 A + 7 /4 = 0=*A = l ¿ - V —3 /2 4) U ; U y ’ -3 /2 4 - A
|v i|
\
a /2 0 + 2 a /I0 > /2 0 - 2 a /Í0 V \/20 + 2a/10
( 5 + 3V1 5 - 3V^
D =
3 - A -3 = A2 — 8A + 6 = 0 = > A = 4 ± VTO; v i = ( 1 + -3 5 - A
> lv il = \/20 + 2\ / l 0, |v2| = \/20 - 2 a/Í 0; Q =
and v 2 — ^ +
\
2 /2
4 -
1 + y/IÜ
and v 2 = ^
+
+ ^ ^ ^ V )2 = 4; Ellipse; 0 = 22.5°.
and
3 -3 ^ ( x 3 5/ U
=
A/ 4 + 2 V 2 © t - 2 % / 2 1 1
^ 2 , |v2| = V 4 - 2V5; Q =
VV4 + 2VI ( 3 + ^ 0 33 - J )
1
, an d v 2
=
(^ J ^ a n d
ftt 11.31°.
14. Two straight lines, a single straight line, or a single point. 15.
(
i"1"1
-1
1 -1
V-l-1
1
-( x ')2 + 2(í/)2 + 2(z ')2-
1
—A
- 1
-1
-11-A -1 -1 -11-A
-1 0 0\
= - 4 + 3 A2 -A 3 = 0 => A = - 1 ,2 ,2 . |
020 0 0 2/
;
Q u a d ra tic Form s and Conic Sections
-1-A
2
2
2 = 21 + 3 A - A2 —A3 = 0 =►A = - 3 ,
2 - 1 -A 2
^ ( -3(x')2 +
2
3 —A
525
2±Vff
2 1 -A
±
^
2
2
22 - A
0
2
Section 6.5
)2+(
2
^
(z,)2
0 0 0'
= —18A + 9 A2 —A3 = 0 =» A = 0,3,6; ( 0 3 0
0 4 —A
0 0 6,
+ 6(z')2. 1 —A
1 -1 0 18. | - 1 2-1 0 -1 1
-1
0
-12-A -1 0 -11-A
= -3A + 4A2 - A3 = 0 => A = 0,1,3;
(y,)2 + 3(2')2.
1
19.
1 2 7/ 2 ' 1 13-1 2 33 0 ,7 /2 - 1 0 1,
20.
3 -7 /2
I
1 0 1/2 0> 0 1 0 -1/2 1/2 0 0 0 0-1/2 0 1>
-
7 /2
-2
-
1/2 - 1
3 /2 \
1/2 1 /2
0
1/2 - 1 / 2
21.
-1
3/2
1 /2
3 - 2 - 5 /2 -2 -6
0 - 5 / 2 1/2
1 /2
-1 /
22. Note that det A < 0 since we have a hyperbola. Then det A < 0 regardless of the valué of d. Thus the equation represents a hyperbola for any nonzero valué of d by Theorem 2, ». 23 ■ ( s í ■ £ : ) ( ; )
■
( i z u i z i )
=
T.™
+ * « y + c(^
-
a(x eos 0 —y sin 6) 2 -f b(x eos 0 —y sin 0)(x sin 0 + y eos 0) + c(x sin 0 + 2/ cos # ) 2 • Then the coefficient of the xy-teim is (—2a sin 0 cos 0 + 6 eos2 0 —6 sin 2 0 + 2csin 0 cos 0) = (c —a )(2 sin 0 cos 0) + 6(cos2 0 —sin 2 0). This is 0 if (c —a) sin 20 + 6 cos 20 = 0, or 6 cos 20 = (a — c) sin 20. So cot 20 = (a — c ) / 6. 24.
If a = c then cot 20 = 0 which implies that 20 = ±w/2. Then 0 = ± 7r/ 4 .
25. Let A =
an(^ ^
= ^ 6'/2 ^
^ en ^ ere exists a unique orthogonal matrix Q such
that A = Q A ! Q *. Then A and A' are similar. Thus they have the same characteristic polynomials. But det (A — AJ) = A 2 — (a + c)A + (ac —62/4 ) = A2 — (a' + c')A + (a V — 6/2/4 ). a) So a + c = a' + cf from equality of A terms. (b) 62 —4ac = 6/2 — 4ac from equality of constant terms. 26.
F ( x ) = Ax-x. = .Dx'-x'. D = diag(Ai, A2, . . . , An). But if Dx'-x' is to be greater than or equal to zero for all x' £ Mn, then A¿ > 0, 1 < i < n. If D x 1 • x' > 0 for all x' ^ 0, then De 1 • e i = Ai > 0, De 2 • e 2 = A2 > 0 , .. .,D e n en = An > 0. If At* > 0, 1 < i < n, then D x '-x ' = Ai(a:/1)2H hAn(x ^)2 = F(x) > 0, for all x £ Mn and F(x) = 0 if and only if x = 0. Thus F(x) is positive definite if and only if A has positive eigenvalues.
27. In problem 26, it is shown that F(x) > 0 => Ai > 0, 1 < i < n. If A,* > 0 then F(x) = D x 1 •x! = A i(x i )2 H h An(xJj)2. Then F(x) > 0 for all x £ Mn. Thus F(x) is positive semidefinite if and only if the eigenvalues of A are nonnegative. A =
^0
2
) ; ^ = 2’3; A is positive definite.
526
C h a p ter 6 Eigenvalues, Eigenvectors and Canonical Forms
jj _ 3 ) ¡ ^ = “ 3 , —3; A is negative definite.
29 . A = (
30 . A = ( q
31.
^=
(_ } “ j);A
33 . A = ^
34
^ *s indefinite.
^ = (3 ± v/ 5) / 2 ; A is positive definite.
A = ( j
32
In s tru c to r’s M anual
= (3 ± V S )/2 ; A is positive definite.
g ^ > A = 2 ± y/5; A is indefinite.
^ _3 ^ ; A = —2 db \/5; A is indefinite.
35. A =
^
j 2 ) ’^= (~ 3 ± a/5)/2; A is negativedefinite.
36. det> 1/ = ± ^ / —X i / ^ x * . This represents two straight lines through the origin with slope ± \ / —A1 /A 2. If Ai and A2 have the same sign then the only solution is x' = 0 and yf = 0. These are the equations of a point. If det A = d = 0 then either Ai = 0 or A2 = 0. If Ai = 0 then we have A2(y*)2 = 0 => yf = 0, which is a straight line. If A2 = 0 then we have x ; = 0, which is a straight line. Note both Ai = A2 = 0, does not correspond to a true curve; we have all x, y. 38. (a) Equation (1) is a hyperbola if det A = Ai A2 < 0. (This uses det (A) = product of eigenvalues). (b) Equation (1) is an ellipse, circle or degenerate conic (possibly empty or a single point) section if det A = A1 A2 > 0.
Q u a d ra tic Forms and Conic Sections
M A T L A B 6.5
M ATLAB 6.5 1.
For problem 10: 2x2 + xy + y 2 = 4 (a) Using the formula from Theorem 2 we see
^
»
A= [2 1/2; 1/2 1];
»
[Q,D]=eig(A)
Q * -0.9239 -0.3827 D = 2.2071 0
(c)
So (Av)*v=4, v=(x y)' is the given equation.
0.3827 -0.9239 0 0.7929
» det(Q) ans = 1.0000 » theta=atan2(Q(2,l),Q(l,l)) theta = -2.7489 » theta*180/pi ans = -157.5000 » ans+360 ans = 202.5000
Y% det(Q) is one, so just find angle of column 1
*/, atan2(y,x) gives polar angle of (x,y)
% Convert to degrees
# /.To give a rotation angle in [0,360)
(d) The equation in the new coordinates is 2.2071a/2 -1- .7929¡/2 = 4. It is an ellipse. » det(A) ans = 1.7500
Y. det(A)>0
for this ellipse,
Y%so Theorem 2 is verified.
(e) The picture below shows the ellipse with major and minor axes along the eigenvectors (with angles 6 = 292.5° and 6 = 202.5 and y/(dj\i),i = 1 , 2 as the lengths of the semi-major and semiminor axes. The foci are at x 1 = ± \ j d ( Ax 1 —A2 l ),yf = 0.
y
a » sqrt{4/.7829} b a sqrt{4/2.2071} - foci at Y - ±sqrt{4/.7829-4/2.2071)
527
528
In s tru c to r^ M anual
C h ap ter 6 Eigenvalues, Eigenvectors and Canonical Forms
2. For problem 8: x 2 + 4xy + 4y 2 — 6 = 0 (a) Using the formula from Theorem 2 we see »
A=[l 2; 2 4];
»
CQ,D]=eig(A)
*/• So (Av),v=6, v=(x y)* is the given equation.
(b)
Q =
0.8944 -0.4472 D = 0 0 0 5
0.4472 0.8944
(c) » det(Q) det(Q) is one, so just find angle of column 1 ans = 1.0000 » theta=atan2(Q(2,1) ,Q(1,1)) */• atan2(y,x) gives polar angle of (x,y) theta = -0.4636 » theta* 180/pi */• Convert to degrees ans = -26.5651 » ans+360 5ÍTo give a rotation angle in [0,360) ans = 333.4349
(d) The equation in the new coordinates is 5?/2 = 6 , which represents two straight lines. » det (A) ans = 0
*/.det (A)=0, with d=6, '/•so case iii of Theorem
2 applies
(e) A picture would show two straight lines, at \j = ± i / 6 / 5 . The distance of each of these lines from the origin is \f{d¡ A2), where A2 = 5 is the one nonzero eigenvalue. The inclination of these lines (angle between the x-axis and the lines) is —26.5651°, and the normal to these lines is given by U2 = Q (:, 2), the eigenvector for A2. 3. For problem 4: xy — 1 (a) Using the formula from Theorem 2 we see »
A=[0 1/2; 1/2 0];
»
[Q,D]=eig(A)
(b)
Q = 0.7071 -0.7071 D = -0.5000 0
0.7071 0.7071 0 0.5000
So (Av),v=l, v=(x
y)9
is the given equation.
Q u a d ra tic Form s and Conic Sections
M A T L A B 6.5
( ib,then A m = A k ■A m~k = 0 • A m' k = 0. I
24. We will show that N £
r” 1 I « ••• 0 | „
0 0
0
\ ••• 0
r— 1 \0 0
0
using induction on r. If r = 1, then the
••• 0/
result is true. Suppose the result is true for r = £. Let Nk = (n,j) and Nj¿ = (a¡; ). Then N %+ 1 = (bij)
Jordán Canonical Form
where 6¿¿ =
Section 6 .6
533
If j < i + 1 or i > k — 1, then 6¿¿ = 0. If j > i 4 - 1 and i < k — 1, then 5= 1
• _ _ í 1, if (¿, j) = (a, a + r + 1 ) where a = 1 , 2 , . . i —r — 1 which means ij — n*\*+la*+ l,i ~ at + l,j — i o otherwise
^0
0 ••• 0 | ,„
\o
. Hence N t has index of nilpotency k.
o
0 o
^ +1
\
Nk-i
ó/
o
/-a 0 0 0 \ / a 0 0 0 \ /a 0 0 0 \ /a 1 0 0 \ /alOO' 06001 06001 I 06101 (OalOj { Oa OO 25. OOc O ’ 0 0 c 1 ’ I 0 0 6 1 l ’ l 0 0 a 1 I ’ I 0 0 6 1 V o o o d / V o o o c / \ 0 0 0 6 / V O O O a / \0 0 0 6 , a, b, c, and d are not necessarily distinct, and the blocks may be permuted along the diagonal.
0 0 0>
(-1
-1
26.
10 0'
/-I 0 0 0> 0-100 0 0 2 1 0 0 0 2;
0-100 0 02 0
0-100 0 0 2 0 0 0 0 2;
0
0 0 2;
'-1 1 0 0 ' 0-100
0
0 2 1
0
0 0 2;
; the blocks may be permuted along
the diagonal. -4 0 0 0
27.
28.
3 0 0 0
0 3 0 0
0 0 3 0
00 30 03 00
0\ 0 0 3/
— 4 0 0 0'
0' 0 0 3;
(
'— 4 0 0 0\
0 3 10 ; the blocks may be permuted along the diagonal. 003 1 0 0 0 3/
0 300 003 1 0 0 0 3; '3 0 03 00 ,0 0
/3 0 0 Vo
0 \ 00 3 1 03/
0 3 0 0
0 0\ ( i 1 0 0\ 10 1 0 3 10 ; the blocks may be permuted along the diagonal. 3 1 I’ 0 0 3 1 0 3/ \0 0 0 3 /
/-3 0 0 0 0\ / —3 1 00 0 \ / —3 1 0 0 0 \ 0 -3 0 0 0 0 -3 0 0 0 0 -3 0 0 0 29. 0 04 00 > 0 0 40 0 J 0 04 0 0 0 0 040 0 004 1 0 0 04 0 0 0 0 0 4/ 0 0 00 4 / 0 0004/ the blocks may be permuted along the diagonal.
V
V
0 0\
/6
0 -7 0 0 0 30. 0 0 -7 0 0 0 00 - 7 0 V0 0 0 0- 7 /
0 0 0
/6
0 0
/6
0 0
0 0
0
0 -7
0
0
0 0\
0 0\
/6
0 0
-7 0 0 0 0- 7 0 0 0 0 -7 1 0 0 0 -7 /
0
-7 0
0
0 - 7 1 0
0 0
0 V0
0 0
0 0 -7
1
0 00 - 7 /
^
0 - 7 1 0 0 0 0 - 7 1 0
0 Vo
V0
/ —3 0 0 0 0 \ / —3 1 0 0 0 \ / —3 0 0 -3 0 0 0 0 -3 0 0 0 0 0 04 10 > 0 04 0 0 > 0 004 1 0 0 0 04 1 0 0004/ 0 0 0 0 4/ V o
1
0 -7 /
; the blocks may be permuted along the diagonal.
004 1 0 0 0 4/
534
C h a p ter 6 Eigenvalues, Eigenvectors and Canonical Forms
( —1 31.
0
0
0
0\
0 -7 0 0 0 0 0 -7 0 0 0 0 0 -7 0 0 0 0 0 -7 / /—7
0
0
0
0\
0 - 7 1 0 0
(-1
0
0
0
V 0
0
0
0 0 1 0 -7/
/-7
1
0
0
0
0 - 7 1 0
0 - 7 1 0 0 0 0 - 7 1 0
0 0
0 0
0 0
0 -7 1 0 0 -7/
0\
0 -7 0 0 0 0 -7 0 0 0 0 -7
0 0
( —1
0 0
Instructo r's M anual
0 0\
0 -7 0 00 0 0 -7 01 0 00 - 7 1 0 0 0 0 -7 /
0\ ; the blocks may be permuted along the diagonal.
0 -7 1 0 0 -7/
32. As C ~ XA C = J , then det (iC~ XA C ) = det (C ) " 1 det ( A) det (C) = det (A) = det (J ) = AiA2 • • • A„.
Jordán Canonical Form
M A T L A B 6.6
535
M A T L A B 6.6 1.
(a) Let » J=diag( [2 2 3 3]); */• First enter the diagonal entries of J » J(3,4)=l */.Next enter the one off diagonal 1 J = 2 0 0 0 0 2 0 0 0 0 3 1 0 0 0 3 » cl=[l 1 2 l]1; c2= [2 3 4 3]'; c3=[2 5 3 3 ] ’; c4=[-l 3 06 ] ’; » C=[cl c2 c3 c4] ; A = C*J/C A = -14.0000 5.0000 8.0000 -5.0000 -62.0000 20.0000 31.0000 -18.0000 -30.0000 9.0000 17.0000 -9.0000 -54.0000 15.0000 27.0000 -13.0000
(i) We show the columns el and c2 are eigenvectors for A for the eigenvalue A = 2, by showing ( A - 21)a = 0 : » (A-2*eye(4))*cl */, If (essentially) 0 then el is eigenvector for A ans = 1.0e-14 * -0.1776 0.3553 0 -0.3553 » (A-2*eye(4))*c2 ans = 1.0e-14 * -0.5329 0 -0.3553 0 » */• Preceding is (essentially) 0, so c2 an eigenvector for A
(ii) Now we investígate the eigenvector properties of c3 and c4. » (A-3*eye(4))*c3 ans = 1.0e-13 * -0.0178 0 0.0355 0.1421
'/. (Essentially) 0 so c3 is an eigenvector for A
» (A-3*eye(4))*c4 ans =
'/• Not 0 (not cióse) so c4 is NOT an eigenvector for A
2.0000
»
5.0000 3.0000 3.0000 */• Note (A-3I)c4 = c3, which is an eigenvector, i.e.
536
C h ap ter 6 Eigenvalues, Eigenvectors and Canonical Forms
In s tru c to r^ M anual
» (A-3*eye(4))*((A-3*eye(4))*c4) ans = 1.0e-12 * -0.1865 -0.6821 -0.3304 -0.5969 » */• (Essentially) 0 so c4 is a generalizad eigenvector for A
(iii) Repeat the calculations with any other choice of an invertible C . (iv) To see that A = C J C ~ X has A = 2 as an eigenvalue of algebraic and geometric multiplicity 2 and /¿ = 3 as an eigenvalue of algebraic multiplicity 2 and geometric multiplicity 3 (for any invertible C ) first note that since A and J are similar, they have the same characteristic polynomials(Theorem 6.3.1). But d et(J —t i ) = (2 —í) 2(3 —t ) 2 as J — t i is upper triangular, so both 2 and 3 are eigenvalues with algebraic multiplicity 2 . As to geometric multiplicity, it is immediate that if E ( \ ) is an eigenspace for J, then C E ( A) is the corresponding eigenspace for A = C J C ~ l , and both have the same dimensión since C is invertible. But J —21 is already in echelon form from which we see that v ( J — 27) = 2 since xi,X 2 will be free variables in any solution to (J — 2 /)x = 0. Thus the geometric multiplicity of 2 as an eigenvalue for A is 2. For the eigenvalue 3, J —3 / is also in echelon form, but with 3 non-zero rows, so v ( J — 31) = 4 — 3 = 1. Thus the geometric multiplicity of 3 as an eigenvalue for A is 1 . (b) Form a new J and a new A: »
J = diag([3 3 3 3]); J(l,2)=l ; J(2,3)=l
» A = C*J/C A = 24.0000 -5.0000 15.0000 0.0000 42.0000 -10.0000 15.0000 -3.0000
-11.0000 -8.0000 -19.0000 -8.0000
6.0000 4.0000 12.0000 7.0000
(i) » (A-3*eye(4))*C(:,4) % Get the easy case over: c4 is eigenvector for A ans = 0 0 0 0 » for j=l:3, (A-3*eye(4))~j*C(:,j) , end ans = 1.0e-14 * 0 0.0888 -0.1776 0
ans 1.0e-12 * -0.0853 0.1137 -0.0853 -0.0426
Jordán Canonical Form
M A T L A B 6.6
537
ans = 1.0e-ll * -0.0917 -0.0568 -0.1819 -0.0568 » */• Each was (essentially) zero, so cl,c2,c3 are generalized eigenvectors
This shows that el and c4 are actually eigenvectors for A. To verify that c2 and c3 are only generalized eigenvectors we must check that (A — 3J)c 2 / 0 and (A — 3 / ) 2C3 ^ 0. » (A-3*eye(4))*c2 ans = 1.0000 1.0000
% Not
zero so c2 only a generalized eigenvector for A
% Not
zero so c3 only a generalized eigenvector for A
2 .0 0 0 0 »
1.0000 */. Note (A-3I)c2 = el
» (A-3*eye(4))~2*c3 ans = 1.0000 1.0000
2 .0 0 0 0 »
1.0000 Note (A-3I)~2 c3 = el
V.
(ii) R ep eat th e above w ith an o th er invertible C of th e sam e size. T h e p a tte rn of zero an d non zero p ro d u e ts and th e conclusions a b o u t which colum ns of C are eigenvectors or generalized eigenvectors should be th e sam e as in (i). (iii) As in (a)(iv ), det (A — t i ) = det ( J — t i ) = (3 — t ) 4 so 3 is an eigenvalue of algebraic m ul tip licity 4. Also, J — 37 is in echelon form , and shows there are 2 free variables in Solutions to ( J — 3 /) z = 0. A pplying C to th e nuil space of J — 3 / yields th e nuil space of A — 37; in p a rtic u la r, a basis of tw o in dependent Solutions to (J — 3/)z,- = 0 becom es x¿ = Cz¿, a basis o f tw o ind ep en d en t Solutions to 0 = ( A — 3 1)x = C ( J — 3 /)C '~ 1(C'z). T h u s th e geom etric m u ltip licity o f th e eigenvalue 3 is 2. (c) F orm a new J an d A w ith »
J=diag([2 2 3 3 ] ) ;
» A=C*J/C A = 19.0000 -29.0000 36.0000 -
2 1 .0 0 0 0
-4.0000 11.0000 -9.0000 6.0000
J(l,2)=l ; J(3,4)=l ;
-9.0000 14.0000 -17.0000 10.0000
5.0000 -8.0000 11.0000 -3.0000
(i) We expect columns el, c3 to be eigenvectors and c2, c4 to be generalized eigenvectors. » for j=l:4, (A-J(j,j)*eye(4))*C(:,j) , end ans = 1.0e-14 * -0.0888 0.1776 -0.1776 0
538
C h ap ter 6 Eigenvalues, Eigenvectors and Canonical Forms
In s tru c to r^ M anual
ans =
1.0000 1.0000 2 .0000 1.0000 ans = 1.0e-13 * -0.0711 0 -0.1421 0 ans =
2.0000 5.0000 3.0000 3.0000
Since (A —27)ci = 0 and (A —37 )c 3 = 0 , e l , c3 are eigenvectors for A for the eigenvalues 2,3 respectively. Also, note that the second product above says (A — 27)c2 = c i(^ 0 ), so c 2 is not an eigenvector for A but is a generalized eigenvector for A (as (A —2I ) 2C2 = (A —27)ci = 0 ). Similarly, the fourth product above says (A — 37 )c 4 = C3 and shows c4 is a generalized eigenvector for A. (ii) A repetition with different invertible C will yield exactly the same patterns of eigenvectors and generalized eigenvectors. (iii) The algebraic multiplicities of 2 and 3 are both 2, since det (A — t i ) = det («7 — ¿7) = (2 —í) 2(3 —t ) 2 has (t — 2) and (¿ —3) as factors of order 2. However, J — 21 has 3 pivots, after moving the second row to the bottom, so v ( J — 27) = 4 — 3 = 1, and thus the geometric mul tiplicity for the eigenvalue 2 for A is 1. (As usual the eigenspace for J maps under C to the eigenspace for A.) Similarly J — 37 has 3 pivots and the same reasoning shows the geometric multiplicity for the eigenvalue 3 of A is 1. 2. First we form an invertible 5 x 5 matrix, C : » C = round(10*rand(5)-3*ones(5,5)) C =
-
1
1
-3 4
4
2
2 5
-
1 4 -3
3
1
2
4 3
6
-2 4
1
6 - 2 - 2 5 4 » det(C) X If non-zero then invertible ans = -673 » X To make cl,c2 (generalized) eigenvectors associated with 2 » X and c3,c4,c5 (generalized) eigenvectors associated with 4 » X Form the following Jordán matrix » J = diag([2 2 4 4 4]); J(l,2)=l ; J(3,4)=l ; J(4,5)=l J =
2 0
1 2
0 0 0
0 0 0
0 0
0 0
0 0
4 0 0
1 4 0
0 1 4
Jordán Canonical Form
M A T L A B 6.6
Since this J has e i, e 2 as (generalized) eigenvectors associated with 2 and e 3, e 4, es as (generalized) eigenvectors associated with 4, then the similar matrix, A = £7*7(7“ 1 will have Ce i = c i, Ce^ = C2 as (generalized) eigenvectors associated with 2 (and also the last 3 columns of C will be (generalized) eigenvectors for A associated with 4). Also, for «7 ,henee for A , the algebraic multiplicity of 2 will be 2 and the algebraic multiplicity of 4 will be 3. (Look: det (A —t i ) = det ( C ( J — t I ) C ~ l ) = det («7—t i ) = (2 — ¿)2(4 — O3) Also the geometric multiplicity of 2 and of 4 will be 1 , for both «7 and A, since muítiplication by C will take the nuil space of «7 isomorphically onto the nuil space of A . The following calculates A and verifies these faets: » A = C*J/C A = 4.0134 0.8276 -0.6686 0.1040 -0.2823
0.0698 3.2110 4.5082 4.3210 5.4146
-0.4086 -0.5111 -0.5691 -1.6226 -2.5958
-0.1055 0.9153 -9.7251 -3.9316 -10.3284
0.7132 0.4740 10.3388 7.2140 13.2764
» (A-2*eye(5))*C(:,1) ans = l.Oe-13 * 0 0 -0.1421 0 0
•/•(Essentially) zero so el is an eigenvector, eigenvalue 2
» (A-2*eye(5))*C(:,2) ans = -1.0000 -3.0000 4.0000 4.0000 6.0000
•/•This gives C(:,l), so c2 is a generalized eigenvector.
» (A-4*eye(5))*C(:,3) ans = l.Oe-14 * -0.0444 0.0222 0.3553 -0.3553 0
•/(Essentially) zero so c3 is an eigenvector, eigenvalue 4
» (A-4*eye(5))*C(:,4) ans
•/•This gives C(:,3), so c4 is a generalized eigenvector.
2.0000 4.0000 -3.0000 1.0000 -
2 .0 0 0 0
» (A-4*eye(5))~2*C(:,5) /Also gives C(:,3), so c5 is a generalized eigenvector. ans =
2 .0 0 0 0 4.0000 -3.0000 1.0000 -
2.0000
540
Instructor's M anual
C h a p te r 6 Eigenvalues, Eigenvectors and Canonical Forms
Section 6.7 -2 - A
-2 —5 1 —A = A’ + J - 1 2 ¡ A = - 4 , 3 . C =
1.
3 —A - 1 -2 4 - A
= A’ -7A+10;A = 2.5-C = ( | Í ) : C - =
Ce“ C~'
-
- 1
°
2 - A
(-J").
_ 1 / 5 e - 4< + 2e3t 2 e - 4í - 2 e 3t\ 7 V 5e-4t - 5c3< 2e~4í + 5e3* J '
eAt = CeJtC ~ 1
2.
¡);J=
=
(
( J |)i
5
J
= (o “)-
2e2< - e" - 2e2‘ " 2e5‘
~ 3 V2e2t + 2e5‘ - e 2‘ + 4e5‘ / -
-1
5 - 2 -A ( 2 siní + cosí —sin í A 5 sin í —2 sin í + cosí J ’
eAt = CeJtC ~ 1 =
4. | 3 ' Í - X - 1 | = A2 - 2A + 2;A = 1 ± ¡ . c = ( 2 _ 3 2 + ' ) ; C -
f í+i
5. | _ 3 °
—
* I - A2 + 1 ; A = „ 5 2 -A
-
-3 ,-3 . C =
=
c .
( , ; J,
;
C-. - 1 ( J
;;
, .
( • .» ) .
' cos 5 — 2 sin í —sin í A 5 sin í cos í + 2 sin í / ’ c
-1 2 -A 7 = A -7 2 - A eAt = CeJtC ~ 1 = e~ ! 1 -2 -1 2- A 1 0 1 - 1 -A
/ 1 — 7í 7í A V —7í 7í + 1 ) '
1 —A 8.
7=
9.
C -
=—
-2-A
|
eAt = CeJtC~
7.
A2 + 6A + 9; A =
)—
4
— 5 sin í A sin í —2 sin í + cosí J ‘
V
7 4 —~A I = ( 7
6.
=
pAt _ rv'tr'-i _ S /'2 s in í + cosí
0 1-iy
V
= ¿ ( _ 2í : 1 ) ^
¡l
=
j -1
4 0
2-22 /
eAt = CeJtC ~ 1 = |
4 -
1
= 4
6
7\
0 -3 2
-e-* + 9e‘ - 2e2‘ - 2e~‘ + 2e2t - 7 e " ‘ - 9e‘ + 2e2‘ A 6e* —6e2t 6e2t —6e* + 6e2t I — e~%+ 3e‘ - 2e2t - 2e-* + 2e2t - 7 e _ * - 3e‘ + 2e2t)
8A + 5A2 -
A3; A =
1,2,2. C
=
/ 1 0 0'
;J =
3 -2
4
3
0'
1
1
0
—3 - 2 1/2
1 —3 0 \
C~l =
i/-l-2
\ l l l j
/ —1 0 o \ 0 1 0 . eAt = CeJtC ~ \ 0 0 2/
4 -A 6 6 13 - A 2 -1 -5 -2 -A
3 A
= —2 + A + 2A2 — A3;A = —1,1 ,2 . C = I 0 2 3 1; C ~ l = ^ I
02 1 \0 0 2 y
4e* - 3e2* + 6íe2t -12e* + 12e2‘ - 6íe2t 6íe2* A e* - e2t + 2íe 2t - 3e* + 4e2t - 2íe2‘ 2íe2t k-3e* + 3e2t - 4íe2í 9e* - 9e2t + 4íe2* - 4 í e 21 + e2< /
2
;
An Im p o rta n t A pplication: M a trix D ifferential Equations
Section 6 .7
541
10. x(“xo = - ^ ( _ ¿ + 2e«-2e*ÍÍ) (fc¡) = C 2a. Then x(t) = ( 2 j“ + ¿ y /3 + ( 6 - l a je4* /!} )' N° te that SÍn°e 6 > 2a’ 2a — 6 < 0. Then there exists t > 0 such that (a -f b)e*/3 + (2a — 6)e4*/3 < 0. That is, the first population will be eliminated.
12.,«)=e» ( 1- ;! ; | ) ( : ; $
) =
1
3 $ ) . ..to -«».(0)+,1(0))=o=>
t = x i ( 0) / ( x i ( 0) + X2( 0)). 13.
Let x i(t) = kg. of salt in tank 1 and x 2 (t) = kg. of salt in tank 2. Then ( * * ) x0 =
( 10°q^ ; | _ ° ° 3q"q3 _ 0 03° 10{ I=
v O T 0 3 and j
_ ”
0
( a \Q0J
= -0 .0 3 -V O Ü 0 3 . C PAt
r e jtr - 1
_ ~
(
30
30 ) ’
* 2 + 0.06A + 0.0006; A = -0 .0 3 ± \/O 0 0 3 . Let a = -0 .0 3 +
=
°-03
\ _ ~
=
( ;C - 0 .0 3 y ’
=
j ( 0.06\/0.0003
J’J® 5 0.03 V 0 .0 0 0 3 /’
( (e « + e^ )/2 -V Ó ^ Ü Ü 3 (e“ - e ^ ) /2 A \0 .0 1 5 (e“ —e^)/\/ÓÍ0003 ( - e ° + e^ ) / 2 ) ’
rf0\ - ( eAtx(
500(ea + eP) \ W ~ ^ —500v^L0003(e“ + eP) ) '
/ 1 (aebt —a)/b 0 ' 14. Note that if A = | 0 6 0 | then eAt = í 0 ebt 0 y 0 (cebt — c)/b 1 0 - a ¡ c i( 0) 0 \ 0 a x i( 0) - 0 0 0 0 0 / / 1 (—
/ xi(t)\ x 2 (t) \ x 3 (t)J +a¡ Bi ( 0))/(o x x ( 0) —/?) 0 \ / ®i(0) \ e(ar,(o)-/J)to a;2( 0) (/fc(«».(o)-/>)*- 0 )/(axi(O) - 0 ) l j W 0) )
x = eAtx { 0 ) = 1 0 \0
( * i( 0) + i 2(0)(-a;a:i( 0)e(QfXl(0)- ^)< + axx( 0))/(aarx(0) —0 )' x 2 ( 0 )e(-aXl(°'>-^t \ * 8( 0) + * 2( 0)(/k - « < - 0 ) / ( a x i ( 0) - 0 ) t (a) If a*x(0) < 0 then x 2 (t) < 0 which implies that no epidemiccan buildup. (c) If axx(0) > 0 then x'2 (t) > 0 and an epidemic will occur. =
15. (a) x'iit) = x'(t) = x 2 (t). x 2 (t) = x"(t) = —ax'(t) —bx(t) = — ax2 (t) — bxi(t). 0
(b) | - b
_a_
f
1 1 = A2 + aA + 6 = °-
16. A2 + 5A + 6 = 0 => A = -2,-3. C = (_* _ j); C ' 1 = eAt = C e J t c - i = (
3e-2< —2e-3< f x 1 (t)\ 3t —2e 2t + 3e 3ty V ^ Í O /
6e 2t + 6e
Then x = 3e-2t —2e“ 3*.
(® j); J = ( " 2 _°); At f l ) \® /
( Ze~™-2e~*\ \ —6e 2t + 6e 3 t J '
542
Instructor's M anual
C h a p ter 6 Eigenvalues, Eigenvectors and Canonical Forms
17. A2 + 6 A + 9 =
0
=>• A = - 3 , - 3 . C =
=
J = ( q - j)5^
( 3
= CeJtC~1 =
;eA< = C tJ%C~x =
_ 13 e-2t + 8e5 • ) ] . ( ; * / “ - ; / » ) .
/; (b) p ( A )
=
(~ l _°)
+
(
3. (a) p( A) = A 3 - 4A 2 + 3A; (b) p(A) =
5 -9 4' - 9 18 - 1 9 V 4 -9 5,
=
( H )
(J S )¡
2 -3 1 - 4 [ -3 6 -3 1 + 3 1 -3 2
/o o o\
I 0 0 0 I ; (c) A is not invertible, as constant term is 0. \0 0 0 /
4. (a) p(A) = A 3 —5 A 2 + 8A - 47; (b) p{A)
—9 34 2 4 \ / —110 8 \ = ( - 5 18 12 - 5 -1 6 4 + 8 -7 14 8 / \ —3 6 4 / '-1 10 8\ / -164 +5
-1
(c) A - 1 =
-3 5.
64/
0 0 0\ 000 ; 0 0 0/
1/2 0 - 1 / 2 ' -1/4 1 - 1 / 4 1/2-1 1/2 y
1 2 2V 021
\ —1 2 2 /
(a) p(A) = A 3 — 3A 2 + 3A — 7; /I -3 3 \ /0 0 1 \ /0 10 3 -8 6 - 3 1 -3 3 + 3 0 0 1 |\6 —15 10 / \ 3 —8 6/ V 1 - 33
(b) p(A) =
/
/0
(c) A - 1 = -
-
0 l\ /0 1 0 \ / I 0 0 V 1 -3 3 + 3 0 0 1- 3 0 10 \ 3 —8 6/ \ 1 — 33/ V001/
100\
0 1 0 001
=
/000\
0 0 0
V000/
;
3 - 3 1' = I1 0
00 1 0,
6 . (a) p(A) = A 3 - 3A 2 + 3 A - I ;
(b) p(A) = (
-20 - 3 0 - 3 3 \ / -1 0 -1 7 -1 6 ' 9 13 15 - 3 5 8 8 6 9 10/ \ 3 5 5, -10 -1 7 -1 6 5 8 8| +3
(c) A - 1 = -
3
5
5
0 0 0\ 000 ; 0 0 0/
+ 3
- 3 - 7 —5 \ /I 0 0 2 4 3 - 3 0 1 0
1
2
2/
\0 0 1
7. (a) p(A) = A 3 - 6A 2 - 18vi - 97; / (b) p(A) =
(c) A " 1 =
63 54 108 \ / 3 12 9 \ 180 189 324 - 6 18 27 36 \ 168 204 315 / y 25 19 42 /
-1
3 12 9 \ (2 -1 3 \ / 1 0 0' 18 27 36 + 6 4 1 6 + 1 8 0 1 0 \ 25 19 42 / \1 53/ \ 0 0 1^ /
-
/1 0 0
9
0 10 \0 0 1
0 0 0\ 000 ; 0 0 0/
-3 2 - l\ = | -2 /3 1/3 0 1 9 /9 -1 1 /9 2 /3 /
A D iffe re n t Perspective: T h e T h eo rem s o f C ayley-H am ilton and Gershgorin
Section 6 .8
8. (a) p{A) = A 4 - A 2 - A - 91;
10 0 2 1 \
(
10 110 2 17
(c)
/O
0 1 1 3 0 -2 3 1 - 2 0 \7 -l 4 1 - 1 1\ / 1 0 10 10 6 | I 2 - 1 02 3 0 I +I -1 0 0 1 6 2/ V 4 1 —1 0 8
0 1
11 0 3 1 0 / 3 1 6 -5 = -g 7-1 5 2
/O 6 0 0\ 4 9.
(a) p(A) = ( A - a l) 4-, (b) p(A) = I í 0n0n0nd^ I I \0 0 0 0/
1
inverse. If a ^ 0, then A 1 = — a4
( a b 0 0\ 0a c0 I ,,3
10.
11.
/ I 00 0 0100
-9
/I o o o 0 10 0 0010
Vo
0 0 0 0\ _ í0000 1
001
oooo r 0 0 0 0/
f 1/9 1 / 9 - 2 / 9 1/9 ^ 4 /9 - 5 / 9 10/9 4 /9 8/9 - 1 / 9 2 /9 - 1 / 9 1/9 1/9 7 /9 1 / 9 /
/0 0 0 0\ =
an 0n0n0n0n II ; (c) If a = °> then A does not have \0000/
° 2 2a 6 be a3 3a 26 3abe bcd\ °\ 0 a2 2 ac cd 0 a3 Za2c Zacd 1 | +4a 0 a3 3a2d 1 0 0 a2 2ad 0 0 0 a3) 0 0 0 a2) V 0 1 /a — b/a2 cb/a3 — bcd/á4' 0 1 /a —c /a 2 cd/a3 0 0 1 /a — d/a2 0 0 0 l/a>
0 0a d ] I 0 01 0 a n = 2,0 a0220 =a/5, 033 = \0 6 , ri0 0= 11 , r-¡ = 1 , and and Re A > 1.
= 1 ; |A —2 | < 1, |A — 5| < 1, or |A — 6 | < 1; |A| < 7
a n = 3, 022 = 6 , 033 = 5, 044 = 4, ri = 5/6, = 1 ,r3 = 1 , and r4 = 1; |A — 3| < 5/6, |A — 6 | < 1, |A —5| < 1, ór |A —4| < 1; |A| < 7 and ReA > 13/6.
545
546
C h a p ter 6 Eigenvalues, Eigenvectors and Canonical Forms
In stru cto r’s M anual
12.
a n = 1 , a 22 = 5, a33 = 6 , 044 = 4, ri = 8, r2 = 9, r3 = 5, and r4 = 5; |A — 1| < 8 , |A — 5| < 9, |A - 6 | < 5, or |A — 4| < 5; |A| < 14 and ReA > - 7 .
13.
a n = - 7 , a 22 = -1 0 , a 33 = 5, a 44 = 4, n. = 4/5, r 2 = 1/2, r3 = 3/4, and r4 = 1; |A + 7| < 4/5, |A + 10| < 1/2, |A — 5| < 3/4, or |A - 4| < 1; |A| < 21/2 and - 21/2 < Re A < 23/4.
(x+10)2 + y* - (1/2)*
s
-
e
y - lm z
(X”4)* + y* - 1*
x -R e z
(x+7)2 + y8 - (4/5)*
14.
a n = 3, a 22 = 5, a 33 = 4, a 44 - 3, a5s = 2, a 66 = 0, rx = 1 , r2 = 2, r3 = 6/5, r4 = 1 , r5 = 3 /2 , and r 6 = 1; |A - 3| < 1, |A — 5| < 2, |A - 4| < 6/5, |A + 3| < 1, |A - 2 | < 3/2, or |A| = 1 ; |A¡ < 7 and Re A > —4.
15. As A is symmetric, the eigenvalues are real. By Gershgorin’s theorem, we have A = R e A > 4 — (2 + 1 + 1/4) = 3/4. 16. As A is symmetric, the eigenvalues are real, and by Gershgorin’s theorem, —6 — (1 + 2 + 1) = —10 < Re A = A < - 4 - (1 + 1 + 1) = - 1 .
A D iffe ren t Perspective: T h e T h eo rem s o f C ayley-H am ilton and Gershgorin
Section 6 .8
17. (a) F ( X ) = (S 0 + S iA )(C 0 + CiA) = B 0 C 0 + (S 0Ci + S iC 0)A + S iC iA 2; (b) P ( A ) Q ( A ) = BoCo + BíACo + BoCíA + B U C i A and F ( A ) = S 0Co + (So> E 3+.V 2 = SPan S
-¡ V l \iy/ 2 / .
V o/ J 6.
f 0\ 1
( y, and E 3_ is^ = span <
°v
0
►
i
< \* V 2 /
I i 1'
p(A) = (A + 2)3; the matrix has —2 as an eigenvalue; E - 2 = span < I 0
7. A has eigenvalues 2 and —3, with corresponding eigenvectors ^
^
and ^
j ^ . Henee
C
8. A has eigenvalues 1 and —1/2, with corresponding eigenvectors ^
and ^
2) *^°
C = ( - J - J ) and C - U C = ( j _ 1 /« ) .
( 9.
°\
/-l-*"
A has eigenvalues —1, i, and —i, with corresponding eigenvectors f —1 1, I
V 1/ V ( -1 + A 1 j . Thus C = 1/
/
0 — 1 — i — 1 + ¿\
-1
1
1
V i
i
11
/—10
and C ~ l A C =
0 i \
0\
0
o o -* 7
.
1 | , and
1
0 \0>
Review Exercises fo r C h ap ter 6
10.
The eigenvalues of A are 2, 6 , and —3,with corresponding eigenvectors ( 1 0
V
' - 1 / V 2 1/V2 0 \ Q = [ l/y/2 1 / / 2 0 and Q*AQ = a
0
01/
( 2 0 0\ 06
0
.
\00 —3/
11.
The matrix A has 1 as an eigenvalue, with E\ = span < I 1 I >. So A is not diagonalizable.
12.
The eigenvalues of A are 16, —2, and —10, with corresponding eigenvectors I 0
1 I , and Il ~ 20 '
\2
/3 /V ^ 3 0 —2 /a /1 3 \ /1 6 0 0\ Henee, Q = ( 0 1 0 and Q*AQ = 0 -2 0 . \ 2 / \ / Í 3 0 3 /v T 3 / \ 0 0 - 1 0/
and E - a —
span
14. The eigenvalues of A are 2, 4, and 6, with E 2 = span
and Ee = span
► , E 4 = span <
/—10 0 1 10-10 . So C = 0 1 00 \ 0 1 11 f°\
15. A has 3 and —1 as eigenvalues, with E 3 = span
Henee C = | jj J |
16.
We have A = ^
(!)
mi
J I and C ~ l A C =
^q) •
/3 0 03
> and E - i — span 0 0
f - x\ ) >
11 ,
i/> í
0
0' 0
0 0-1 0 \00 0 -1.
e*8enva^ues
are 1/2 and —1/2, with corresponding eigenvectors
( - i ) * Hence> with ( ^ ) = ( - 1 / v l 1/vl) ( y ) ’then we haveT
an equation of a hyperbola.
1 1 Vo/
“ T
= 1, which
IS
558
C h ap te r 6 Eigenvalues, Eigenvectors and Canonical Form s
17.
In stru cto r’s M anual
As A = { t l ) ' A h a s i 4- V5 and 3 — y/2 as eigenvalues, with corresponding eigenvectors and
^ (1 +
Thus, with
(1 ~
^
^
where a = / 4 4- 2v ^ and
/----------a?'2 r/2 — = 1 , which is an equation of an ellipse. P = v 4 —2>/2 , then — -------^=- + — 8/(3 4- v 2) 8/(3 — v 2 ) 18. As A =
^ ^ias
^ 3v^2)/2 and (5 — 3 \/2 )/2 as eigenvalues, with ^ j __
and
( l + \ / ^ ) ^ corresP°nding eígenvecl ° rs* Let a = y/4 — 2\/2 and /? = \ A 4 - 2\ / 2 . Henee, with
( ? ) = ( < l - V 2 ) / « 0 + V 5 ) / 2 ) ( » ) ■ then 2/ ( 5 + 3V 5) + 2 / ( 5 - l¡ ^ ) = ’ ■Which is M eq"ati° n of an ellipse.
19. As A =
1
3 ^ ’ ^ ^iaS ^
Vl3)/2 and (3 — \/Í 3 )/2 as eigenvalues, with corresponding eigen
and ^3 —
vectors
Let a = \/2 6 4- 6 \/l3 and /? = \/2 6 —6>/l3- With
( ( 3 + V Ü ¡ ) / “ (3 - V Í 5 w ) ( l } ■ the” 1 0 / ( 3 + V I» ) + 1 0 / ( 3 - V i » ) =
" “ Ch U “
=
eq“ ti0" ° f
a hyperbola. 20.
Wehave A = ^ _ 2
4 ^ • The matrix has 0 and 5as eigenvalues, with
sponding eigenvectors. Let
^
2 l ) ( j / ) ’ ^ en **X‘^ =
and ^ 2 ) 33 corre~ which is a degenerate conic
section. 21 .
°\
/ M
\ corresponding A = ( 2 2 I 2 2 0 1has0, 4, and —3as eigenvalues,with I —1 1, I 1( lj \, and I (0 ° 1as \0 0 -3 / V 0) \oJ Vi/
/x '\ eigenvectors. Let
W
l /\/2 1 / / 2 0 \ / a : \ I — l/y /2 l / \ / 2 0 1 I y I , then we have A / 2 — 3z/2.
/
j yf 1= \
o
o
1/
\zj Solving (A — 7 )v 2 = v i for V2, we
22. The matrix A has 1 as aneigenvalue, with E\ = span | obtain V2 = ^ _ 2 ^> and henee, C = ^g
and ^ = ^0 l ) *
23. The matrix A has —2 as an eigenvalue, with £ L 2 = span ^ ^ 1 ^ obtain
v2= ( - J ) .
So C
=(\
and J =
.
Upon solving ( A 4- 27)v2 = v i, we
Review Exercises fo r C h ap ter 6
24. The m atrix A has —1 as an eigenvalue, with E \ = span
—3 ] f . Upon solving (vi + J)v 2 = v i
'-2'
2\
—1 ) and V3 =
and (vi + 7)v 3 = V2, we obtain V2 =
1
-
3,
/-I J = I
1
0 -1
2
( —5 —2
. Then C = j
2'
-3 -1 1 ) and V 7 3-2,
0\
1
.
V o 0-1/ 25. The eigenvalues of A are 1 and —1 , with ^
( j ; ) - d v
-
(*_;).**—
“
=
and ( ^ j 35 corresponding eigenvectors. With C =
-
( ¡ O
C
^ X I - ? )
-
/ — e* + 2e- í 2e* —2e- t \ ^ —e ^ e - * 2et — e~t J ' 26. From problem 23, we have that C =
^
and J =
^
^
So eAt = CeJtC ~1 =
Í 2 - l \ ( e~2t te~2t\ ( 0 1 \ _ 2f / 1 — 2< 4A (l O /V 0 e" 2í) \ - l 2 J ~ e \ —t l + 2 t ) ' 27. The eigenvalues of the matrix are —1 + 2i and —1 — 2i, with corresponding eigenvectors ^ “ •* ( _ , + ¡ ) -
(_!_(_!+*)
= ( ' - 1 + 2,;
^
^
then e - = C J C - =
( eos 21 — sin 21 —2 sin 21 \ \ sin 21 eos 21 + sin 21 J ‘ 28. As p(A) = A 3 - 7A 2 + 19A - 237, then A " 1 = ^ l ( - i4a + 7vl - 1 9 1 ) -1
23
23
- |
-1 8 6\ ( 2 3 1\ -3 -2 -1 + 7 -1 10 -1 1 -1 1 1 4 / \-2 -1 4 /
/100’ -1 9
010 Vo 0 1
'4 - 1 3 - 1 ' 4 10 - 1 3 - 4 5,
29. We have |A - 3| < 1/2 + 1/2 = 1 , |A - 4| < 1/3 + 1/3 = 2/3, |A - 2| < 1 + 1 = 2, and |A + 3| < 1/2 + 1/2 + 1 = 2. Henee | A| < 5 and - 5 < Re A < 14/3.
559
Notes
Instructors
M a th e m a tic a l Induction
A ppendix 1
A ppendix 1. M athem atical Induction 1 . For n = 1, the equation holds. Assume that the equation holds for n = k. Then 2 + 4 + 2(Jb + 1 ) = k(k + 1 ) + 2(fc + 1 ) = (k + 1 )(¿ + 2), which completes the proof.
2. For n = 1, the formula holds. Assume the equation holds for n = k. Then 1 + 4 + 7 + [3(4 I I )
| 1 = 3 P + 5¿ + 2 =
2] — * (3fc ~ 1} | 2
2
h 2k + b (3k — 2) +
+ 1)(3fc + 2) = (* + 1)[3(fc + l) ~ 1]. 2
2
3. For n = 1, the equation holds. Assume the formula holds for n = ¿. Then 2+5+8H------h [3 (¿ + l)—1] = 3*a + 7* + 4 = ( t + l ) ( 3 t + 4) _ (fc + l)[3(fc + !) + !]
2
2
2
2
4. As theequation holds for n = 1 , we assumethat it holds for n = k.Then [3(Jk + 1) - 1] = jfc2 + 2¿fc + 1 = (fc + l ) 2.
1 + 3 + 5+ b (2k — 1) + -
*(3fc + l)
5. ( | ) 1 = | < j = 1 , so inequality is true for n = 1. Now assume it is true for n = k. That is, ( |) * < £. Then Q )* +1 = \ ( 5 )* < H í ) = SF < I+T since 24 > 4 + 1 if * > 1 . 6 . As the inequality holds for n = 4, assume that it holds for n = k. Then 2 fc+1 = 2 • 2* < 2 • n! <
(*+l)Jfe! = (Jb + l)L 7. The formula holds for k = 1. Assume that the formula holds for n = k. Then 1 4- 2 + 4 + • • • + 2 *+1 = 2*+! - 1 + 2fc+1 = 2*+2 - 1 . 8. The equation holds for n = 1. Suppose that the equation holds for n = k. Then 1 + 3 + 9H
3t+ 1 - l 2
h3 *+1 =
. ot+i _ 3 t+2 — 1 +á " 2 • 1 , assumethat it holdsfor n= k.Then 1 + ^
9. As theequation holds for n =
^ + ------ 1-
=
1 1 2 - 2* ¿ +' 2fc+ 1 = 2 2t+ 1 ’
10.
For
n
= 1, we have 1 = 1 . Now assume that the equation holds for
(_>)*** - ! V
11.
u
r
-
i -
n
= 4. Then 1 — ^ u
r
For n = 1, we have 1 = 1. Suppose that the formula holds for n = k. Then l 3 + 23 H fc2(fe + 1 ) 2
I
(k
^ — ------ 1y
h (k + l )3 =
I 1)3 _ ( * + l ) a(* + 2)a -
12.
As the equation holds for n = 1, assume that it holds for n = &. Then 1*2 + 2*34 k(k + 1 )(¿ + 2) ( t + l ) ( * + 2)(fc + 3) + (k + 1 )(k + 2) =
13.
For n = 1 , we have 2 = 2. Suppose that the equation holds for n = fe. Then 1*2 + 3 * 4 4 1) - 1][2(4 + 1)] = k( k + í )(4 k ~ í ) + 2(2Jfe+ 1)(4 + 1 ) = í fc + 1 Kfc + 2)(4fc + 3) .
b(& + l)(Ar-f-2) =
h [2(Jfe +
561
562
Instructor's M anual
A ppendix 1 M a th e m a tic a l Induction
14.
For n — í the equation holds. So suppose that the equation holds for n = k. Then 1
22 — 1
1
1 7 + • • • + t i—.
"h ñ ó
32 — 1
3
1
1
r — 7 — 7771— 7 - 7 7
(Jb + 2 )2 — 1
1
7771— — 777 +
4
2(k + 1) 2(k + 2)
3 “ 4
1 1 2(k + 2) 2 (k + 3)
k2 + 4 k + l
k is even. Then (fc + l )2+ (k + 1) =
15. When n = 1, n2 + n = 2, which is even. Now suppose that k2 + k2 + 3Ar + 2 = k2 + k + 2(k + 1 ),which is even.
16. For n = 12, the inequality holds. Suppose that the inequality holds for n = k, k > 11. Then k + 1 < ^
+ 2 + 1 = Jfe+ 1 ). 17. If n = 1, then n(n 2 + 5) = 6 . Suppose 6 divides k(k2 + 5). Then ( k + 1)[(k + l )2 + 5] = (fc + 1)[(k 2 + 5 )- f(2 fc + 1)] = k(k 2 + 5 )-f 3(Jb2 + Jb+ 2). Using problem 15, we see that 6 divides (k + l)[(fc-f l )2 + 5]. 18. For n = 1, 3n5-f 5n 3+7n = 15. Suppose 15 divides 3&5 + 5k3 +7k. Then 3 ( ¿ + l) 5 + 5 (fc + l) 3+7(fc+ 1) = 3JSr5 + 15¿4 + 35Jk3 + 45ik2 + 37Ar + 15 = (3Jfc5 + 5Jk3 + 7k) + (15ik4 + 30Ar3 + 45Ar2 + 30Jfe + 15), which is divisible by 15. 19. We will show that (x — 1)(1 + x 4- x 2 H h a?*1” 1) = xn — 1. For n = 1 , we have equality, so suppose the equation holds for n = Then (x — 1)(1 + x + x 2 + h xk) = xk — 1 + (x — l ) x k = xk+l — 1. Thus xn — 1 is divisible by x — 1. n—1 xxyn~%~1 = xn — yn. For n = 1, we have equality, so suppose the equation
20. We will show (x — y) *=o
holds for n = k — 1. Then
k (x - y)
^ 2
k —l x'yk~l = ( x - y)xk + (x - y) ^
i=0
x*yk~’
»=0 A:— 1
= (x - y)xk + (x - y)y ^
xV
1 -’
•=o
= (x —y)xk + y(xk —yk) =
by induction
xk+l - i/*+1.
21. If n = 1, then (ab) 1 = ab. Now suppose that (ab)k = a* 6*. Then (ab)k+l = (ab)(ab)k = (ab)(akbk) = aakbbk = a*+ 16fc+1. 22. Proof by induction on n. Let / be a polynomial of degree 1, i.e. f ( x ) = a x + 6,a ^ 0. Then ax+b = 0 implies a: = — b/a. Therefore / has exactly one root. This proves the desired proposition for n = 1. Now suppose that it is true for n = fc. Let y be a polynomial of degree k + 1. g has at least one complex root, so let a be a root of g. Then g(x) = (x — a)h(x) by división where h is a polynomial of degree k. h(x) has exactly k roots by the induction hypothesis so g(x) has exactly k + 1 roots since g(x) == 0 only if (x — a) = 0 or h(x) = 0. This completes the proof. 23. Suppose det (A 1 A 2 • • •Am_ i) = det A\ det A i • • • det Am_ i. Then det ( A 1 A 2 • • - A m) = det (A 1A 2 • • •Am_ i)d e t (Am) = det A i det A 2 • • • det A m
M a th e m a tic a l Induction
24. Suppose (v4i + A 2 -f • • • + A k - i f = A\ 4- A\ 4- • • • + A \_ 1 . Then (A\ + A 2 + ( Ai + A i H h A k - i ) %+ A\ = A\ + A *2 + hA\ .
A ppendix 1
+ Aje)* =
25. If 5 is a set with 0 elements, then S is the empty set. The proposition states that S must have exactly 2° = 1 subset. But the empty set does have exactly one subset, namely, the empty set itself. So suppose that sets with h elements have exactly 2* subsets. Let S = {« 1 , * 2»• • • >^Jfe+i} be any set of size k + 1 . Then we may classify all subsets of S into two groups: those subsets which contain £*+1 and those that do not. Note that the number of subsets of S that contain ar*+1 is the same as the number of subsets that do not contain x*+i. Now, the subsets of S that do not contain are precisely the subsets of T = {a?i, By hypothesis, T has exactly 2* subsets. Hence there are 2 • 2* = 2*+1 subsets of S. 26. Suppose 2fc — 1 is even. Since 2 divides (2k — 1) + 2, then 2(k + 1) — 1 is even. To conclude something by induction, we would need to show that 2k — 1 is even for some integer k. But 2k — 1 is odd. 27. For the induction step to be valid, the intersection of Si and S2 would have to be nonempty to link the two sets of equalities. But if h = 2, S\ 0^2 is empty.
563
564
Notes
Instructors
Complex Numbers
Appendix 2
A ppendix 2. C om plex N um bers 1.
(2 - 3¿) + (7 - 4¿) = 9 —7i
3 . ( i + ¿)(i _ ¿) = 2 6 . 5i = 5e '*/2
9. 12. 15. 18. 20. 23. 26. 29. 32.
3 - 3i = 3 \/2 • e- ^ 4 1 - V3¿ = 2e “ w /3 - 1 - \/3 í = 2e “ 2**/3 (e3'*/4)/2 = - \ / 2 / 4 + y/2i/A 6e '*/6 = 3>/3 + 3¿ 3e " 2'*/3 = - 3 / 2 - 3V & /2 3 + Ai 7i 4e “ 3'*/5
2. 3(4 + i) - 5 ( - 3 + 6¿) = 12 + 3¿ + 15 - 30¿ = 27 - 27¿ 4. 7. 10. 13. 16. 19. 21. 24. 27. 30. 33.
(2 - 3*)(4 + 70 = 29 + 2* 5 + 5i = 5^2 • e ™/4 2 + 2^/3¿ = 4e**'/3 4>/3 - Ai = ge - '* / 6 e3'* = - 1 (e“ 3" / 4)/2 = - V 5 / 4 - v^¿/4 4e 5x,‘/6 = - 2 ^ 3 + 2s v ^ 23'* 74 = V 6/2 - >/5*/2 4 - 6i 16 3e47r*/n
5. ( - 3 + 2s)(7 + 3¿) = - 2 7 + 5¿ 8. - 2 - 2i = 2>/2 • e ” 3**/4 11. 3a/3 + 3i = 6e ” /6 14. -6 > /5 - 6* = 12e“ 5TÍ/6 17. 2e~ 7,r¿ = - 2
22. 4e “ 5,rí/6 = - 2 \ / 3 - 2t 25. é = 0.5403 + 0.8415* 28. - 3 - 8t 31. 2e “ x*/7 34. e- ° 012í
35. Suppose z = ol + i¡3 is real. Then /? = 0. Then z = a = z. Next, suppose z = z. Then a + i/3 = a — i/3Then i¡3 = — ifi => ¡3 = —¡3 => /? = 0. Then z is real. 36. Suppose z = a + */? is puré imaginary. Then a = 0. Then z = —i/? = —z. Next, suppose z = —z. Then a + i/3 = —a + *'/? Then a = —a =>►a = 0. Then z is puré imaginary. 37. Let z = a + */?. Then zz = (a + *’/?)( |z| = 1. Thus the unit circle is the set of points in the complex plañe that satisfies |z| = 1 . 39. The circle of radius a centered at zo. 40. The circle and interior of the circle of radius a centered at zq. 41. First note that (z)n = zn. Then p{z) = p(z), since coefficients are real. So if p(z) = 0, then p(z) = 0 = 0.
42. eos AB + i sin AB = (eos 6 + i sin B) 4 = (eos4 8 + sin4 8 —6 eos2 8 sin 2 8 ) + (4 eos3 8 sin 8 —4 eos 8 sin 3 8 )i Then eos A8 = eos4 8 + sin 4 0 —6 eos2 0 sin 2 0 = 1 —8 eos2 0 sin 2 0 and sin 40 = 4 eos3 0 sin 0 —4 eos 0 sin 3 0. 43. For n = 1, (cos0 + is in 0 )1 = cos0 + isin 0. Thus DeMoivre’s formula is true for n = 1. Suppose it is true for n = Jb, that is, (cos0 + isin0)* = cos(¿0) + isin(fc0). Then consider n = fc + 1. (eos 0 + i sin 0)*+1 = (eos 0 + i sin 0)* (eos 8 + i sin 0) = (cos(fc0) + i sin(&0))(cos 0 + i sin 0) = (cos(fc0) eos 0 —sin(¿ 0) sin 0) + (cos(¿ 0) sin 0 + sin(fc0) eos 8 )i = cos((fc + 1 ) 0) + isin((¿ + 1 ) 0) Thus DeMoivre’s formula holds for n = 1 ,2 ,.
566
Notes
Instructors
T h e Error in Num erical C o m p utations and C o m p u tatio n al C om plexity
A ppendix 3
567
A ppendix 3. The Error in N um erical C om putations and C om putational C om plexity 1. 4. 7. 10.
0.33333333 0.77777778 0.77272727 0.23705962
x x x x
10° 10° 10 1 109
3. 6. 9. 12.
2. 0.875 x 10° 5. 0.77777777 x 10° 8 . -0.18833333 x 102 11. 0.23705963 x 109
-0 .3 5 x 10" 4 0.47142857 x 101 -0.18833333 x 102 -0 .2 3 7 x 1017
13. 0.83742 x 10“ 2° 14. ca = |0.49 x 101 - 5| = 0.1; er = 0.1/5 = 0.02 15. ea = |0.4999 x 103 - 500| = 0.1; er = 0.1/500 = 0.0002 16. ea = |0.3704 x 104 - 3720| = 16; er = 16/3720 = 0.0043 17. €a = |0.12 x 10° - 1/8| = 0.005; er = 0.005 • 8 = 0.04 18. ca = |0.12 x 10" 2 - 1/800| = 0.00005; er = 0.00005 • 800 = 0.04 19. f a = | - 0.583 x 101 + 5 f | = 0.0033333...; er « 0.57143 x 10~3 20. ea = |0.70466 x 10° - 0.70465| = 0.1 x 10"4; er = 0.70465 x 10" 5 21. ea = |0.70466 x 105 - 70465| = 1; eB « 0.1419144 x 10~4 22. There are three different operations: (1) divide row i by au in colums i + 1 to n + 1, (2) multiply row i by aj¡ in columns í + 1 to n + 1, and subtract it from row j for j > i; (3) multiply 6,- by o,-,-, j < i, and n^
subtract it from bj. Operation ( 1 ) requires
ir =
n(n +
1)
— - multiplications, ir = n 4 - 1 —i. Operation
k=l i(t + 1) = “¿ ‘t’ + " ¿ ‘k
(2) «quipes
k= 1
k= l
=
C» - 1)K3>» - D +
=
muWplications
Jb= l
n~1 n 2 — n and additions. Operation (3) requires ^ k = — - — multiplications and additions. Adding these Jb=i fractions together gives the formulas in row 2 of Table 1.
23. There are three operations: (1) dividing row i by an in columns i + 1 to n + 1, (2) multiplying row i by aji in columns 1 to n -f 1 and subtracting it from row j for j > (3) the back substitution. Op/*\
•
n
»
eration ( 1 ) requires y . & = *=i
/ Tlifl
. 1\ “I" 1 )
tí
.
^
multiplications. Operation (2) requires y , fc=i n —1
multiplications and additions. Operation (3) requires
/?
TI3
—
TI
+ 1) = — t—
2
k = — - — multiplications and additions.
*=i Adding these fractions together gives the desired results. 24. Let (A\I) = (bij). At the kth step, we divide bkj by 6** for ir + 1 < j < n + k as the rest of row ir is zero, and then multiply bkj by a,-*, for i ^ ir. This gives n + n(n — 1) = n 2 multiplications at each step. As there are n steps, then there are n 3 total multiplications. As for the number of additions, note that at each step ir, we perform (n — ir)(n — 1) -h (ir — l)(n — 1) additions. Henee, there are n —1
n
£ ( » — k)(n — 1 ) +
fc=i and subtractions.
n
— !)(»* — 1 ) = (n — l)^ ^ (n — 1 ) = (n — l ) 2n = n 3 — 2n 2 + n total additions
t= 2
k=l
568
A ppendix 3 T h e Error in Num erical C om putations and C o m p u tatio n al C om plexity In s tru c to r^ Manual
25. For the operation of dividing a row by an in columns i + 1 to n, we have n(n — l) /2 divisions. At each step k, k = 1 , 2 , . . n — 1 , we multiply row ife by ay* in columns k + 1 to n, for j > fc, and n—i subtract it from row j. So this accounts for
fi(n l ) ( 2 n k2 = -----------
1}
additions and multiplications.
Jb=l Finally, keeping track of the diagonal elements and multiplying them together at the end requires n —1
3
^
3
2
multiplications. Adding these fractions gives — + - 5— 1 multiplications and —------— + — additions. o o o ¿ D 26. As it would require 4, 200 multiplications and 3, 990 additions, the average time would be 4,200 • (2 x 10"6) + 3,990 • (0.5 x 10~6) = 0.104 seconds. 27. Since it would require 3, 060 multiplications and 2, 850 additions, the average time would be 3,060 • (2 x 10~6) + 2,850 • (0.5 x 10~6) = 7.545 x 10" 3 seconds. 28. 50 x 50; 0.31 seconds 200 x 200; 19.96 seconds 10,000 x 10,000; 2.5 x 106 seconds n 29. A B = (cij) where aj = ]j>^a¡kbkj. As i = 1, 2, . . . , m and j = 1, 2, . . . , g, there are m • q • n multiplica-
Jb=l tions and m • q • (n — 1 ) additions.
Gaussian Elim ination w ith Pivoting
A p p e n d ix 4
A p p e n d ix 4 . G a u ssia n E lim in a tio n w ith P iv o tin g 2 -1 1.
-4 3 -2 2 -1 1 3 -8 3
1
I -4 3 -2 3 -8 3 1 -0.75 0.5 0 1 -0.260870 k0 0 0.130435 Then
-0.75 0.5 0.5 0 -5.75 1.5
0 .3 5 \ 0.165217 -0.482609/
-0.75 0.5 1 -0.260870 0
1
0.35 0.35 \ / 1 -0.75 0.5 -0.4 — -5.75 1.5 -0.95 0 -0.4 -0.95 / 0.5 0 \o 0.35 \ 0.165217 . -3.700000 /
x 3 = -3 .7 x 2 = 0.1653217 + 0.260870(—3.7) = -0.800002 x i = 0.35 + 0.75(—0.800002) - 0.5(—3.7) = 1.60000
4.7
2. 12.3
1.81 2.6 0.25 1.1 0.06 0.77
-5.047 \ /1 2 .3 0.06 0.77 11.495 — 1 -3.4 -0.25 1.1 7.9684 / l 4.7 1.81 2.6
’l 0.00487805 0.0626016 0 1.7807 2.30577 0 -0.233413 1.31285
0.647837' - 8, 09183 13.6976,
7.9684 \ 11.495 -5.047 /
/ 1 0.00487805 0.0626016 | -0.233413 1.31285 0 1.7807 2.30577 \o 1 0.00487805 0.0626016 0.647837' 0 1 1.29025 -4.52799 0 0 1.61401 12.6407 j
1 0.00487805 0.0626016 1 1.29025
0.647837' -4.52799 1 7.83186,
0
Then
x 3 = 7.83186 x 2 = - 4 .5 2 7 9 9 - 1.29025(7.83186) = -14.6330 xi = 0 .647837- 0.00487805(-14.6330)- 0.0626016(7.83186) = 0.228931
3.61 8.04 -2.7 -0.891 6.63 -4.38
3.
25.1499 \ 3.2157 -36.1383/
-0.222039 -0.073273 1.96690 7.49778 5.71520 -4.68188 0.222039 -0.073273 1 -0.819199 0 9.109069 Then
-►
12.16 -2.7 -0.891 -7.4 3.61 8.04 -4.12 6.63 -4.38
0.264449 \ 27.1068 -35.0488 / 0.264449\ -6.132556 39.168996/
3.2157\ 25.1499 -36.1383/
1 -0.222039 -0.073273 0 5.715197 -4.681885 . 0 1.966908 7.497780
0.264449' -35.048770 27.106823,
0.264449' -6.132556 4.300000. 1
1 -0.222039 -0.073273 0 1 -0.819199 0
0
x 3 = 4.3 x 2 = -6.132556 + 0.819199(4.30000) = -2 .6 1 x i = 0.264449 + 0.222039(-2.61000) + 0.073273(4.3) = 0.0
0.647837 13.6976 8.09183.
569
570
A ppendix 4 Gaussian Elim ination w ith Pivoting
4.1 -0.7 8.3 3.9 5.3 2.6 8.1 0.64 - 0.8 2.6 -5.3 - 0.2 7.4 -0.55 4.1 0.8 -1.3 3.6 1.6 0.8 0.377358 -1.39623 4.85472 \ 0.103774 /I 8.00189 4.27020 -1.06981 24.8297 | 1 0 -0.8547168 14.0245 3.47453 -24.1244 1 Vo0 -1.33019 4.71698 1.51698 -11.5838/ 4.85472 \ / I 0.377358 -1.39623 0.103774 0 1 0.533649 ■•0.133695 3.10298 j 0 0 1 0.232053 -1.48283 I Vo 0 0 0.0798278 0.590816/
4.
In stru cto r’s M anual
0.2 7.4 -0.55 8.1 0.64 -0.8 3.9 -0.7 8.3 1.6 -1.3 3.6 0.377358 -1.39623 0.103774 /I 1 0.533649 -0.133695 1 0 | 0 14.4806 3.36026 0 1.33914 Vo 0 5.42683 0.103774 / I 0.377358 -1.39623 1 0.533649 -0.133695 0 0 1 0.232053 0 -
Vo
0
1
0
4.85472 \ 3.10298 -21.4722 -7.45625/ 4.85472 \ 3.10298 -1.48283 7.40113/
Thenx 4 = 7.40113 x 3 = -1.48283-0.232053(7.40113) = -3.20028 x 2 = 3 .1 0 2 9 8 - 0.533649(-3.20028) + 0.133695(7.40113) = 5.80030 * i = 4.85472-0.377358(5.80030) + 1.39623(-3.20028) - 0.103774(7.40113) = -2.57044
5. Gaussian elimination with partial pivoting: '
0.1 0.05 0.2
25 8
/ 12 12 -3 8 15 -7 0.2 0.05 \ 0-1
1 .3 \
-3 15
-*•
10
2/
10\ 2
1 .3 /
1 2.08 -0.25 0.833\ / l 2.08 -0.25 0 10.588 0.346 0 1 0.588 ^0 0 0.318 1 .2 7 / \0 0 1 Gaussian elimination without partial pivoting: 0.1 0.05 0.2 12 25 -3 -7 8 15
1 .3 \
—►
0.833' 0.346 3.99,
/1 0
2.08 -0.25 13.3 Vo -0.158 0.225
Then x 3 = 3.99, x 2 = —2.00, x x = 5.9!
1 0.5 2 1 -1.42 0 0 0 45.3
1 0.5 2 0 19 -27 0 11.5 29
10 2/
22.6
13 -7.68 181
/ I 0.5 2¡ 13 0 1 -1.42 -7.68 Then x 3 = 4.00, x 2 = —2.00, x i = 6.00. Vo 0 1 4.00 Exact solution: x\ — 6 , x 2 = —1 , x 3 = 4. Relative error with partial pivoting: x x : 0.00167, x 2 : 0, x 3 : 0.0025 Relative error without partial pivoting: x\ : 0, x 2 : 0, x 3 : 0. 6 . Gaussian elimination with partial pivoting: '
0.02 0.03 -0.04 2 4 16 50 10 8
1 0.2 0 1
^0
1.2
0 -0.012
0
/ -
6/ 0. 12 \ 1.6
0.16 -
-0.04 \
50
10 2
8
4 16 \ 0.02 0.03 -0.04
6\
0
0.2 0.16 - 1.2 1.44 0 0.026 -0.0432 Vo
(l -
-0.04 /
Then xz — 7 , x 2 = 10, x x = —3.
-
-0 .0 4 2 4 /
Gaussian elimination without partial pivoting: 0.02 0.03 -0.04 2 4 16 50 10 8 -2 1 -1.64 0 1
1.5
1.5 -22
-2
36 -65 108
-2 \ 32 106/
/I -+
-2
-1.45 8.43
Then x 3 = 8.43, x 2 = 12.4, x x = —3.74.
0 Vo
-2 1 -1.64 0 1.4
1.5
-2
-1.45 11.8
0.12' -1.92 -0.0424,
Gaussian E lim ination w ith P ivoting
A ppendix 4
Exact solution: x\ = —3, «2 = 10, x3 = 7. Relative error with partial pivoting: x\ : 0, a?2 : 0, x3 : 0. Relative error without partial pivoting: x\ : 0.247, x^ : 0.24, x3 : 0.204. 7. Exact solution: x\ = 15650/13, x 0, and y > 0. The córner points are (0,0), (380,0), (0,275), and (120,195). A máximum profit of $1,900 is earned when 380 chairs and no tables are produced. 4. no chairs; 275 tables; Profit = $2,200 5. 380 chairs; no tables; Profit = $3,040 6 . We have ax -f ay < c, bx + by < d, x > 0, and y > 0. If c/a < d/6, then the córner points are (0,0), (0, c/a) and (c/a,0). If d/b < c/a, then the córner points are (0,0), (0, d/6), and (d/b,0). In either case, as P = 3x + 4y, the owner will maximize profits by producing tables only. 7. The comer points are (0,10), (5,5), and (5,10). So the cost is minimized at $4,000 per day if x = 5 mgd and y — 5 mgd. 8.
comer points: (0,0), (0,4), (5/2,0), (1,3) /(0,4) = 16, /(5/2,0) = 15/2, /(1,3) = 15 / is maximized at (0,4) 9. córner points: (0,0), (0,4), (5/2,0), (1,3) /(0,4) = 12, /(5/2,0) = 10, /(1,3) = 13 / is maximized at (1,3) 10. comer points: (0,0), (0,3), (4,0) /(0 ,3) = 3, /(4 , 0) = 4 / is maximized at (4,0) 11. córner points: (0,0), (0,10/3), (3,0), (2,2) / (0 ,10/3) = 10, /(3 ,0) = 6, /(2,2) = 10 any point on the line 2x + 3y = 10 between (0,10/3) and (2,2) will yield the máximum valué / = 10 12.
comer points: (0, 0), ( 0, 1), ( 1, 0), ( 10/ 11, 10/ 11) / (0 ,1) = 5, /(1,0) = 3, /(10/11,10/11) = 80/11 / is maximized at ( 10/ 11, 10/ 11)
13. / (0 ,1) = 3, /(1,0) = 5, /(10/11,10/11) = 80/11 / is maximized at ( 10/ 11, 10/ 11) 14. / (0 ,1) = 1, /(1,0) = 12, /(10/11,10/11) = 130/11 / is maximized at ( 1, 0) 15. / (0 ,1) = 12, /(1,0) = 1, /(10/11,10/11) = 130/11 / is maximized at (0, 1) 16. córner points: (0,4), (4,0) (7(0,4) = 20, (7(4,0) = 16 g is minimized at (4,0) 17. córner points: (0,3), (4,0), (2,1) 0(0,3) = 15, 0(4,0) = 16, 0(2,1) = 13 0 is minimized at ( 2 , 1) 18. comer points: (0,1), (1/3,0), (1/5,1/5) 0(0,1) = 8, 0(1/3,0) = 4, 0(1/5,1/5) = 4 any point on the line 12a; + 8y = 4 between (1/5,1/5) and (1/3,0) will yield the minimum valué 0 = 4
582
Application 1 Linear Programming
Instructor’s Manual
19. comer points: (0,1), (1,0), (1/13,8/13) (/(0,1) = 7, g{1,0) = 3, 0(1/13,8/13) = 59/13 g is minimized at (1,0) 20. comer points: (2,0), (0,5/2) 0(2,0) = 6, 0(0,5/2) = 5 g is minimized at (0,5/2) 21. Let x and y denote the amount of the first and second food groups, respectively. We want to minimize / = 0.5z + y subject to the constraints: 0.9# + 0.6y > 2, O.lx + QAy > 1, x > 0, and y > 0. The córner points are (10,0), (0,10/3), and (2/3,7/3). Upon computing / at the córner points, we find that 2/3 Ib of Food I and 7/3 Ib of Food II provides the diet requirements at minimum cost. The cost per Ib is $ 8/9 « 89 cents. 22. Let x and y denote the number of regular and super deluxe pizzas, respectively. We want to maximize P = 0.5x + 0.75y subject to the constraints: x + y < 150, 4x + 8y < 800, 0 < x < 125, and 0 < y < 75. The córner points are (0,0), (0,75), (50,75), (100,50), (125,25), and (125,0). Upon evaluating P at the córner points, we find that when Art makes 100 regular pizzas and 50 super deluxe pizzas, his profit is maximized at $87.50. 23. Let x and y denote the number of species I and II, respectively. We want to minimize E = 3x + 2y subject to the constraints 5x+ y > 10, 2x + 2y > 12, x + Ay > 12, x > 0, and y > 0. The córner points are (1,5), (4,2), (0,10), and (12,0). Upon evaluating E at the córner points, we find that if x = 1 and y = 5, then the energy expended will be minimized at 13 units. 24. (a)
(b)
(c) By sketching lines, observe that / is maximized at (20/7,6/7) with / = 58/9, and g is minimized at (1/7,4/7) with g = 58/7.
Linear Programming: T h e Córner Point Method
Application 1.2
25. (a) yr
(b) The first two inequalities ae satisfied for all (xi, X2, £3) as long as X1 +X3 > 5 and X2 > Xi4 x3—8. Henee the problem is unbounded. 26. (a)
(b) From the first two inequalities we have that 6x1 + 4 x2 + 3^3 < 14, but this contradicts the third inequality. 27. (a) Minimize C = 2xi 4 2.5x2 -f 0.8x3 subject to the constraints: x\ 4 X2 -f IOX3 > 1, lOOxi 4- 10x2 4lOx + 3ye50, 10xi + 100x2 4 IOX3 > 10, x\ > 0, X2 > 0, and X3 > 0. (b) ( 0, 0, 0); ( 0, 0, 1/ 10); (0, 0,5); (0, 0, 1); (0, 1, 0); (0,5 , 0); ( 0, 1/ 10, 0); ( 1, 0, 0); ( 1/ 2, 0, 0); ( 1, 0, 0); (0,49/9, —4/9); (0,1/11,1/11); (0, -4/9,49/9); (49/99,0,5/99); (1,0,0); (4/9,0,5/9); (4/9,5/9,0); (1,0,0); (49/99,5/99,0); (53/108,5/108,5/108) (c) (0,0,5); (0,5,0); (1,0,0); (1,0,0); (1,0,0); (4/9,0,5/9); (4/9,5/9,0); (1,0,0); (53/108,5/108,5/108) Feasible solution Cost (in dollars) 4.00 (0, 0, 5) (0, 5, 0) 12.50 2.00 ( 1, 0 , 0) 1.33 (4/9, 0, 5/9) (4/9, 5/9, 0) 2.28 1.13 (53/108, 5/108, 5/108) (e) Cost is $245/216 « $1.13 when x\ = 53/108 « 0.49 gallón of milk, X2 = 5/108 « 0.046 pound of bef, and X3 = 5/108 « 0.046 dozen eggs consumed daily. 28. Let x, y, and z denote operations I, II, and III, respectively. We want to maximize P = 100x4 150y 4 200z and N = x + y 4- z subject to the constraints: 0.5x 4 V2 Z < 80, x > 0, y > 0. The córner points are (0,0,0), (160,0,0), (0,80,0), and (0,0,40). Both P and N are maximized at (160,0,0). Henee, to maximize the revenue and the total number of operations, 160 of type I, 0 of type II, and 0 of type III should be performed.
583
584
Application 1 Linear Programming
Instructor’s Manual
29. Let xi,X2,and X3 denote the number of cans of mixtures 1, 2, and 3, respectively. We want to maximize P = 0.3a:i + 0.4*2 4- 0.5x3 subject to the constraints: ^xi + ^X2 < 10,000, ^xi + 1 x2 + 2
\x$ < 12,000, 2
-^2 o
o
2
o
+ \x% < 8,000, x\ > 0, X2 > 0, and x3 > 0. The córner points are (0,0,0), 2
(8000,18000,4000), (20000,0,4000), (4000,24000,0), (8000,0,16000), (0,24000,0), and (20000,0,0). Upon evaluating P at the comer points, we find that if (xi)X2 )xs) = (8000,18000,4000), then P is maximized with P = $11,600.
Slack Variables
Application 1.3
A pplication 1.3 *1 + *2 +81 = 3 1. 2xi + %2 +52 = 7 *1**2 > 0
3xi + x2 —^3 +«i =4 2. 2a?i + x2 + x3 H-52 = 6
2xi + X2 +Si = 10 3xi + 2x2 H ~ 52 = 30 3. 4xi + 7x2 +53 = 20
3xi + X2 + 2x3 +$i 2xi + 3x2 + 7x3 4. 4xi + 8 x 2 + 5x3
Sl,S2 > 0
si,s2>s3 >
2x2 +51 6. (a) 3xi + 7x2 X\ +
—
8
= 12 +53 = 9 Si, 52, 53 > 0
+52
5
=
+52 = 20 51,52 > 0
( . (1 2 1 0 I 5\ fí 2 1 0 5 \ / I 0 7 -2 |- 5 \ 1^0 1 -3 1 5J Vo 1 -3 1 | 5 ) ( J \3 7 0 1 | 20 J Then x\ = —5 —75i + 252; ^2 = 5 + 35i —52. /l 1 2 0 I 5\ / 1 1 2 0 |5 \ A 0 7/3 1/3 120/3 '• \3 0 7 1 | 20/ ” * VO -3 1 1 I 5 / VO 1 -1/3 -1/3 | -5/3 J Then xi = (20 —7x2 — s2)/3; si = (—5 + x2 + s2)/3. A 0 1/7 -2/7 -5/7\ 2 1 0 I 5\ 1^0 7 3 1 | 20y “* ^0 1 3/7 1/7 20/7 ) Then x2 = (20 —3xi 5s2)/7; si = (-5 - xx+ 2s2)/7. A
2xi + 5x2 +«i = 12 9. (a) 4xi + 9x2 +s2 = 20 «1,52 >
0
(2 5 1 0 | 1 2 \ fl 5/2 1/2 0 W 1,4 9 0 1 | 20^ “ * 1^0 1 2 - 1 Then x\ — (— 8 + 9si —5s2)/2; x2 = 4
(í
0
Vo 1 251 +
-9/2 5/2 1-4' 2
- 1 4
52.
2 0 5 1 ( í 00 5 /2 1/2 4 19 0 Vo i Then x% = (12 —5x2 — si)/2; s2 = —4 + x2 + 2«i.
10. (
xi + 2x2 + 2:3 +«! = 8 11. (a) 2xi + 5x2 + 5x3 +s2 = 35 «1,«2 >
0
2 1 1 0 I 8^ /I 2 1 1 0 8 \ A 0 -5 5 -2 I-30 5 5 0 1 135) Vo 1 3 -2 1 19) 1^0 1 3 - 2 1 19 ) Then xi = —30 + 5x3 —5«i + 2s2; x2 = 19 —3x3 + 2«i —s2. 12. (a) Same as lia). 2 1 0 A 0 -1/2 -3/2 -1/2 -19/2 \ ( J 22 5 5 1 VO 1 5/2 5/2 1/2 35/2 )
= 12 +53
= 8
S l , S 2,S 3 >
0
7xi + X 2 + 3x3 + X4 +51 3xi + 2x2 + 5x3 + 12x4 2xi + 5x2+8x3 + 2x4
= 15 +52
(2
Then xi = (35 —5x2 —5x 3 —s2)/2\ si = (—19 -I- x2 + 3x 3 + s2)/2.
0
585
586
Instructor’s Manual
Application 1 Linear Programming
13. (a) Same as lia). r.v (2 1 1 1 0 8 \ fl 1/2 1/2 1/2 0 4 \ fl 0 3/5 1 -1/5 W V5 5 2 0 1 35J \0 5/2 - 1/2 -5/2 1 15/ -f Vo 1 _1/5 -1 2/5 Then 2:2 = (5 - 3xi - 5si + S2)/ 5; *3 = (30 + xi + 5si - 2s2)/5. 14. (a) Same as lia). (b) si = 8 - xi — x2 —2x3; s2 = 35 —2xi —5x2 —5x3. xj + 3x2 +si 2xx 7x2 15. (a) 3 x i + + 8x2
+ s2
= 5 20 +S3 = 40 =
S l,S 2 ,S 3 >
13
(b)
10
o
3 10 0
0
2 7 0 1 0
1-210 0-1-301
. 3 8 0 0 1
10 7-30 0 1-2 1 0
0 0 -5
11
’l 0 0 -8/5 7/5 0 1 0 3/5 -2/5 k0 0 1 -1/5 -1/5 Then xi = (120 + 8s2 —7ss)/5; x2 = (-20 —3s2 + 2s3)/5; si = (—35 + s2 + s3)/5. 16. (a) Same as 15a). '3 0 0 1 1 1 0 0 1/3 1/3 5/3 \ (b) [ 7 1 0 2 0 0 1 0 -1/3 -7/3 25/3 8 0 13 0 0 0 1 1/3 -8/3 80/3 / Then x2 = (5 —x\ —si)/3; S2 = (25 + xi + 7si)/3; S3 = (80 - xi + 8si)/3. 2xi + 4x2 + 17
8x3 +si
, x 2 xi + 5x 2 + 12x3
3xj + 6x2 + 13x3
12
+S 2
= 25
+S3 = 60 s l . s2 , s3 >
o
si = 12—2xi —4x2 — 8x3 (b) s2 = 25—2xi — 5x2 —12x3 s3 = 60—3xi — 6x2 —13x3
18. Same as 17a) '2 4 8 1 0 0 1 2 4 1/2 0 0 0 14 -110 (b) 2 5 12 0 1 0 0 0 1 -3/2 0 1 3 6 13 0 0 1 1 0 - 4 5/2 -2 0 -20\ fl 0 0 -7/2 -2 4 0 1 4 -110 13 — 0 1 0 5 1 -4 \ 0 0 1 -3/2 0 1 0 0 1 -3/2 0 1 4 2/ Then xi = 148 + 7si/2 + 2s2 —4s3;X2 = —155 —5si —S2 + 4s3; x3 = 24 + 3si/2 —s3. 19. (a) Same as 17a). fl 0 0 -1 -4 -1 -13\ 1 2 0 4 8 0 12\ (b) j 0 2 0 5 12 1 25 I —> [ 0 1 0 5/2 6 1/2 25/2 .0 3 1 6 13 0 60/ Vo 0 1 -3/2 *5 -3/2 45/2 / Then xi = (25 —5x2 —12x3 —S2)/ 2; si = —13+ x2 + 4x3 + s2; s3 = (45 + 3x2 + 10x3 + 3s2)/2. 20. (a) Same as 17a). 10-4 -1 - 1 0 1 2 8 4 0 0 0 1 6 5 / 2 1/2 0 (b) ( 0 2 12 5 1 0 0 0 - 5 -3/2 -3/2 1 .0 3 13 6 0 1 1 0 0 1/5 1/5 -4/5 -31 \ 0 1 0 7/10 -13/10 6/5 79/2 I Then xi = (395 — 7x2 + 13s2 1 2 s 3 ) / 1 0 ; 0 0 1 3/10 3/10 -1/5 -9/2/ X3 = (-4 5 - 3x 2 - 3s 2 + 2s3)/10; si = (-155 - x 2 - «2 + 4s3)/5.
T h e Simplex Method I: T h e Standard Maximizing Problem
Application 1.4
A pplication 1.4
Note: pivots are in parentheses. 1. 1 0 (2) -1 (2) 0 1 -2 0 3 1 1 0 0 1 2 3 1 -1
1 2 (3) 2
3.
5.
2 (5) 1
7.
1 (2) 3
9.
1 1 (2) 1 Xi
X2
0 1
0
0
1
1
-3
^3
3 0 2
2 4 0
Xi
X2
-1
-1/3 X2
X3
5 6 / 0 1 0 0
0 0 1 0
(1) 0 1 1 (5) 1
(1) (2) 2
6.
1 (3) 5 4
8.
1 -1 (2) 2
1 0 0
(1) 1 3 1 0 0 1 0 0 0
(2) 2 3 3 -1 (1) -1 1
-1 (2) 1 3
0 1 0
s2 1 1/3 0 1/3 0 -1/3
0 1 0 0
0 0 1 0
5 7 14 /
1 0 0 0
0 0 1 0 0 1 0 0
5 6 7 /
5 6 4 / 5/3 2/3 / 2/3 Xl
1 0 0 1 0 0
2 2 /
si *2
%2
#3
1 3/2 1 0 3/2 5 0 1/2 -1
si «2
Si
1/2 1/2 -1/2
S2
0 1 0
1 3 /-I
S1 «2
X3
1/3 0 -2/3 Si
1/2 1 -1/2 «i «2 1 0 -1 1 -3 0
0 1
0
«2 0
2/3 2 / - 4/3
1/2 3 / - 1/2
1
0 1 1
X3 S2
X2 S2
Xl
«2
1 2 / 1 2 /
0 1 0
«i
00 1
x2 *3 1 1 -1 0 -2 0
0 1 0
S2
1 2/3 4 0 0 ■■4/3
1 -1/2 1 0 - 1 5 0 3/2 0 Xl 1 -2 -1
S i
4.
7 4 /
1 0 0 0
x4 *3 14/3 25/3 2/3 7/3 -1/6 -13/3
(1) -1 2
5 3 1 /
0 1 0 1 0 0
1 2
-1
%2
2/3
Xl
1 (3) 4
(1) -2
1
2 -1 1
1 0 0
3 8 -1
3
H
1 0 0 0 1 0 0 0 1 0 0 0
3 1 (2) 3
(1) 1 2
1 2 /
The solution is unbounded.
Xi S2
587
588
Instructor's Manual
Application 1 Linear Programming Xi
14.
15
1
S2
S3
0
1
0
-3 /2
7 /2
s2
0
0
1
-1 /2
5 /2
S2
1 /2
X3
X2
X3
-7 /2
1 /2
1 /2
5 /2
5
3 /2
1 /2
1
0
0
1 /2
-5 /2
-5 /2
0
0
0
-3 /2
Xl
X2
1
s
«2
1
/ -
X2
«1
«2
0 1
-1 /5 8 /5
1 0
-2 /5 1 /5
2 7 /5 4 /5
0
-13 /5
0
-1 /5
/- 4 /5
Xl
2 5
3 8
0
0 1
7 4
1
-1
0
0
/
«i «2
3 /2
«1 Xl
(4/5,0);/= 4/5 16.
Xl
X2
«1
«2
1 2
1 5
1 0
0 1
1 2
2
1
0
0
/
S1 «2
2
x2
Si
S
1 0
1 3
1 -2
0 1
0
-1
-2
0
1 0
Xl «2
/ - 2
(1,0);/: U
Xl
X2
«1
52
2
3 2
1
0 1
6
0
5
0
0
/
3 4
X2
Xl 0 1
1
0 Xl
18.
Si s2
5
5i 3 /5 -2 /5 -7 /5
0 0
1
5i
52
0 1
5 /3 2 /3
1 0
-2 /3 1 /3
8 /3
0
7 /3
0
-4 /3
/ - 2 0 /3
Si Xl
5 /3
52 -2 /5 3 /5 -2 /5
8 /5 3 /5 / — 5 2 /5
X2
53
0 1
0 0
2 1
Si S2
0
1
3
S3
0
0
/
0 0 1
5 6 4
0
x3
5
1
1
0
1
0 0
0
0
1
2
0
1
2
1
0
X3
5i
52
1
(3/5,8/5);/= 52/5
Xl
52
X2
1
X2
Xi
X2
X3
si
s2
S3
0 1
0 0
1
-1
0
0
1
0 0
0
0
1
0
-1 /2
1 /2
0
0
0
-1
-1 /2
-1 /2
Xl
X2
X3
«1
«2
«3
s2
0 0
2 7
*2 s2
1 /3
0 1 0
-1 /3 -1
-1
1 /3 2 2 /3
2 /3 1
S3
1 0 0
1 /3
3
Xl
0
0
-4
-1
0
0
Xl
1
1
Xl X2 X3
1 1 / - 4
( 1, 1, 1); / = 4 19.
2
Xl
X
1 1 2
1 -2 -1
1 2 1
0 0
0 1 0
1
1
-3
0
0
Si
/
/ -
5
(3,2,0); / = 5; or (0,5,0) or . . . . There are an infinite number of Solutions in the constraint set.
*2
*3
Si
1
-1
-1
-1
1
2 2
*1 20.
s2
S3
1
0
2
0
1
0 0
5 6
S2
-1
1
0
0
1
7
S3
1
3
0
0
0
( 1 3 ,1 9 ,0 ) ;/= 45
/
Si
2
Xl
X
*3
Si
S2
S3
0 0
0
1
1
1
0
1
5
0
2
1
11 19
si x2 Xl
1
0
3
0
1
1
13
0
0
-8
0
-4
-3
/ — 45
T h e Sim plex M eth o d I: T h e Standard M axim izing Problem
ari
x2
s
H
2
S3
«4
¡Ej
1
1
1
0
0
0
1
2
0
1
0
0
1
0
0
0
1
0
0
0
0
1
3 4 5 /2 3 /2
0 0
0
0
/
si
s2
s3
s4
2 0
0
5
1 8
X2
«1
«2
«3
A pplication 1 .4
S4
«x
0
1
-1
1
0
0
1
«2
0
0
-2
1
1
0
1/2
S 3 -*•
1
0
2
-1
0
0
2
S4
0
0
1
-1
0
1
-2
-3
0
0
1/2
0 0
589
x2 S3 xx s4
— 18
/
(2 ,1 );/= 1 8
22.
xi
x3
x2
X3
0 0 0 1
0 0
1 0
0 0
0 1
0
0
Xx
1 0 0 4 1 -1
0 1 0
1 0 0 1 0 0
0 0 1
0 0 0
0
0
1
0
0
0
1
1
5
1
3
0
0
0
0
/
3
«1 S2 S3 -*• S4
S1
S2
53
S4
1 /4 -1 /4
1
-1 /4
-1
0 0
1 /4 1 /4
0
0
0
0
-1 /4 1
0
0
-3 /2
-5
-3 /2
X3
S1
s2
1
sx
1 1 /4 1 /4 1 /4
X2
XX *3
1 / -
9 /2
( 1 / 4 , 1 / 4 , 1 ) ; / = 9 /2
23.
Xi
*2 *3
1 3 -1 1
3 2 2 2
Si
6 4 1 2
s
1
2
0 1 0 0
0 0 0
3
Xx
S
0 0 1
12 10 5
0
/
0 1 0 0
Sx s2 S3
x2
0 21/8 0 3 /4 1 7 /8 0 - 1/2
1 -5 /8 0 1 /4 0 1/8 0 - 1/2
S3
-7 /8 -1 /4 3 /8 -
1/2
11/8
sx Xx x2
5 /4 2 5 /8 / — 1 5 /2
(5 /4 ,2 5 /8 ,0 );/= 1 5 /2
24.
Xx
x2
X3
1
1
-1
1
-1
1
1
0 0
1
2
3
0
1
-1
S2
S1
1
0
1
S3
Xx
X2
X3
0 0
0 0
1
si
0
1
2
s2
1
3
S3
0
0 0
0
0
0
1
0
0
/
1
0
S1
S2
53
1/2 0 1/2 1/2 1/2 0 1/2 1/2 0 -3 /2 -2 - 5 / 2
2 3 /2 5 /2 / — 13
( 3 / 2 , 2 , 5 / 2 ) ; / = 13 25.
^1 ^2 2?3 1 1 2 2 1 1 2 -1 3 1 2 5 1
-1
xx
26.
s1 1
0 0 0 0
1
x2
x3
0 2 1 1 0 -1 0 6 0 -1
0 0 1 0 0
2
sx
3 4 0
3 5
5
7
15
0 1
s2
5 7
s x s2
1
1
0 2 3
0 0 0
0 0
6
0
0
0 1
Si s2
8
S3
9
S4
/
s3
s4
1 -1 /4 -3 /4 0 3 /4 -1 /4 0 - 1/2 1/2 0 7 /4 -9 /4 0 -1 /4 -1 /4
x4
x2
1 1 1 1
«2 «3 «4 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0
0 0 0 1 0
3 /4 1 3 /4
Si Xi
1/2
x
1 3 /4
s
/ — 1 5 /4
S3
s4
0 0 1
1
Si
2 3 4
S2
0
0 0 0 1
0
0
/
3
S
S4
3 4
(1 3 /4 ,0 ,1 /2 ) ; / = 15 /4
x2 x\ X3
590
Instructor's Manual
Application 1 Linear Programming Xl
*2
*3
xa
S2
S3
«4
1
0
0
9 /8
5 /4
0
-5 /8
3 /8
7 /8
0
0
0
-2 1 /1 6
-5 /8
1
-7 /1 6
1 /1 6
5 /1 6
«2
0
1
0
-1 /1 6
-1 /8
0
5 /1 6
-3 /1 6
1 /1 6
Xl
5 /8
*3
«i
0
0
1
3 /8
-1 /4
0
1 /8
1 /8
0
0
0
-7 7 /1 6
-1 3 /8
0
-1 5 /1 6
-3 9 /1 6
/
-
Xl
2 2 7 /1 6
.(7/8,1/16,5/8,0);/= 227/16 27. Let a, 6, and x denote the amount of each grade of plywood to be produced. We want to maximize P = 40a+ 306+ 30x subject to the constraints: 2a+ 56+10x < 900, 2a+56+3x < 400, 4a+26+2x < 600, a > 0, 6 > 0, and x > 0. Upon writing the information as a simplex tableau and solving, we obtain a b X a b X Si S2 S3 Si S2 S3 2
5
10
1
0
0
900
Si
0
0
7
1
-1
0
500
2
5
3
0
1
0
400
s2
0
1
1 /2
0
1 /4
-1 /8
25
4
2
2
0
0
1
600
S3
1
0
1 /4
0
-1 /8
5 /1 6
2 7 5 /2
Si
b a
-5 /2 -3 5 /4 P - 6250 P 0 0 -5 0 So P is maximized at (137.5,25,0) with P = $6250. 28. (a) Let xi, X2, and X3 denote the amount of Heidelberg Sweet, Heidelberg Regular, and Deutschland Extra Dry to be produced, respectively. We want to maximize P = x\ + 1.2x2 + 2x3 subject to the constraints: x\ + 2x2 < 150, x\ + 2x3 < 150, 2xi + X2 < 80, 2xi + 3x2 + X3 < 225, x\ > 0, X2 > 0, and X3 > 0. Upon writing the information in a simplex tableau and solving, we obtain s4 Xl Si x2 * 3 S2 S3 40
30
20
0
0
0
1
2
0
1
0
0
0
150
1
0
2
0
1
0
0
150
S2
2
1
0
0
0
1
0
80
S3 s4
2
3
1
0
0
0
1
225
1
1 .2
2
0
0
0
0
P
X\
X2
X3
si
s2
Ss
Si
S4
50 1/3 0 -2/3 Si 4/9 -1/3 1/9 65 X3 1 20 Xl 0 1/9 2/3 -2/9 0 0 -2/9 -1/3 4/9 40 x2 0 0 -11/15 2/5 -8/15 P —198 Henee P is maximized at (20,40,65) with P = $198. Note that we used all of the resources except 50 bushels of the Grade A grapes (si = 50, and $2 = «3 = S4 = 0). Henee we would want an increase in Grade B grapes, sugar, and labor to improve the company’s profit. 29. Let c, v, and 6 denote the amount of chocolate, vanilla, and banana ice cream to be produced, respec tively. We want to maximize P = c + 0.9v + 0.956 subject to the constraints: 0.45c+ 0.5i/ + 0.46 < 200, 0.5c+Q.4t/ + 0.46 < 150, 0.1c+0.15u + 0.26 < 60, c > 0, v > 0, and 6 > 0. Upon writing the information in a simplex tableau and solving, we obtain c V b Si S2 S3 0.45 0.5 0.4 1 0 0 200 0.5 0.4 0.4 0 1 0 150 1 0.1 0.15 0.2 0 0 60 1 0.9 0.95 0 0 0 P c v 6 Si S2 s3 -0.35 0 0 1 -2 2 20 3 1 0 0 10 -20 300 -1.75 0 1 0 -7.5 20 75 -0.0375 0 0 0 -1.875 -1 P- 341.25 0 0
0 0 0 1 0
0 1
1 0 0 0 0
T h e Simplex Method I: T h e Standard Maximizing Problem
Application 1.4
Henee, if Kirkman makes 0 gallons of chocolate, 300 gallons of vanilla, and 75 gallons of banana, then the profit is maximized at $341.25. As s\ = 20, then 20 gallons of milk went unused. 30. Let /, s, and tdenote the fraction of an hour fast walking, leisurely strolling, and talking to voters, respectively. We want to maximize D = 3/ -f s subject to the constraints: /-f s < 3/4, / —s —¿ < 0, /+ < 1, / > 0, s > 0 ,and t> 0. Upon writing the information in a simplex tableau and solving, we obtain s t s t s2 Si S2 «3 «1 S3 f / 1 1/4 1 0 0 1 -1 0 t 1 1 1 0 0 1 Si 3/4 -1/2 1 -1/2 1/4 s 1 1 0 0 1 0 0 1 0 S2 1 1 0 0 1/2 0 1/2 1/2 1 -1 -1 0 0 0 S3 f P 0 0 0 0 0 0 -1 -1 -1 3 1 0 P - 1.75 Henee, if (/, s,t) = (1/2,1/4,1/4), then D is maximized and D = 1.75. 31. Let xi, X2, X3, and X4 denote the amount of syrup, cream, soda water, and ice cream, respectively. We want to maximize C = 75xi + +50x2 + 40x4 subject to the constraints: X4 < 4, xi —X2 < 1, x\ + X2 —X3 < 0, xi + X2 + X3 + X4 < 12, xi > 0, X2 > 0, X3 > 0, and X4 > 0. Then Xl X2 X2 x± s1 S 2 S 3 S 4 1 0 1 -1
1 0 1 1
1 0 -1 0 0
75 50 Xl
X2
1 1 0 0
1 0 0 0 0
40
0 1 0 0 0
0 0 1 0 0
12
Sí
4
S2
0 1
S3 Si
1/4
-1/4
C «3 1/4
0
1
0
*4
*3
0 0 0 1 0
«1
S2
s4
3/2 x2 x4 4 Xl 1/4 -1/4 1/4 5/2 1/2 - 1/2 -1/2 4 X3 -125/4 -35/4 -125/4 -25/2 C - 845/2 Thus if 2.5 oz of syrup, 1.5 oz of cream, 4 oz of soda water, and 4 oz of ice cream ar used, then the number of calories is maximized at 422.5. 0 0 1 0 0
1 0 0 0 0
0 0 0 1 0
0 1 0 0 0
-1/2 0 1/2 0
32. Let xi, X2, and X3 denote the amount of money invested in stocks, bonds, and a savings account, re spectively. We want to maximize P = 0.08xi+0.07x2+0.05x3 subject to the constraints: X1+X2+X3 < 10,000, xi —0.5x2 < 0, xi —X3 < 0, xi + X2 < 8,000, xi > 0, X2 > 0, and x3 > 0. Then X\
X2
X3
1 1 1 1 -0.5 0 1 0 -1 1 1 0 0.08 0.07 0.05 X\
0 1 0
0
X2
Xs
0 0 1 0 0
1 0 0 0 0
S1
S2
Ss
S4
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
s1
1 1 -1
10,000
0 0 8,000
Si «2 «3 «4
p
S2
Ss
S4
0 0 0 1 0
0 1 -1
-1 -1 2 2
2,000 2,000 6,000 1,000
X3
Xi X2
-1.5 -1.5 S2 0 P — 680 -0.006 -0.01 -0.01 Henee if (xi, X2, xs) = (2000,6000,2000), then P is maximized and P = $680.
Instructor^ Manual
Application 1 Linear Programming
33. Let r, e, p, and n denote the number of rings, earrings, pins, and necklaces, respectively. We want to maximize E = 50r + 80e + 25e -f 200n subject to the constraints: 0 < r < 10, 0 < e < 10, 0 < p < 15, 0 < n < 3, and 2r -f 2e -f p + 4n < 40. Then r e p n «i s2 «3 «4 S5 4 40 1 0 0 0 0 0 0 0 0 0
0 0 0 1
50 80 25 200
1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 $1
0 1 0 0 0 0
0 0 1 0 0 0
0
0.5 0 1
1 0 0 0 0 0
S2
0
0 0 0 0 1 0
0.5 0 0
0 0 0 1 0
10 10
S4
«5
0
0 0 0 1 0 0
1 -2 0 0 2 -100
1 0 1
«3 s4 «5
15 3 E
Ss
-1
«2
3 4
n r e s4 ss
10
15 -0.5 6 0 -25 £■-1,600 -30 Thus with (r, e,p, n) = (4,10,0,3), E is maximized and E = $1600. Note that the jeweler can make 2 pins instead of a ring, with the same profit, so there are more Solutions. -0.5
34. Using the same notation as in #29 of Application Section 1.2, we have Xi *2 X3 «i «2 S 3 1/2 1/3 1 0 0 0 10,000 «i 1/2 1/3 1/2 12,000 0 1 0 52 0 1 1/3 1/2 0 0 8,000 53 0.3 0.4 0.5 0 0 P 0
x\ x2 xs
Si
S2
S3
xx 4,000 *3 3 -3 3 18,000 X2 -0.2 -0.4 -0.6 P — 11,600 As before, (xi,ar2,x3) = (8000,18000,4000) and P = $11600. 35. Using the same notation as in #28 of Application Section 1.2, we have 1 0 0 0
Xl
0 0 1 0
X2
0 1 0 0
0 -2
Xs
si
0.5 1 2 100 150 200
2 2
8,000
Xx
80 P
1 0
X2
Xs
Si
0.5
1
2
1
1
1
1 0
Xx
-2 0
Si
X2
x3
1 2 4 0 -50 -200 Xx
X2
Xs
Si
2 -200
160 P - 16,000
Xx
Si
1
OO co
80 si 1 2 4 2 160 Xx N -2 N — 160 0 -1 -3 Henee with (#i, x2, £3) = (160,0,0), both P and N are maximized, P = $16000, and N = $160. 36. Note: See Application Section 1.5 for information on the method used to solve this problem. From problem 21, we wantto minimize g = 0.5pi + 1/2 subject to the constraints: 0.9pi + 0.6p2 > 2, O.lpi + 0.4y2 > 1, j/i > 0, and p2 > 0. The dual problem is to maximize / = 2xx + x2subject to the constraints: 0.9xi + 0.1x2 0, and x2 > 0. This gives Xx X2 Sx s2 *1 *2 «1 «2 0.9 0.1 1 0 1 0 4/3 -1/3 1/3 0.5 Xx 0.6 0.4 0 1 1 0 1 -2 3 0 S2 X2 2 0 0 1 0 0 -2/3 -7/3 / and
T h e Simplex Method I: T h e Standard Maximizing Problem
Application 1.4
Thus with 2/3 Ib of Food I and 7/3 Ib of Food II, the cost is minimized at 88.9
89 cents per pound.
37. Using the notation from problem 22, we have X Si S2 S3 «4 y i 1 0 0 0 150 1 Si 4 0 1 0 0 800 8 S2 125 1 0 0 0 1 0 S3 S4 1 0 0 0 1 75 0 P 0 0 0 0 0.5 0.75 x y si s2 s3 s4 -1 0.25 0 0 50 0 1 y -2 1 0 25 0 0 0.25 S3 X 1 0 2 100 -0.25 0 0 1 25 0 0 -0.25 0 1 S4 0 0 -0.25 -0.0625 0 0 P - 87.5 Thus with 100 regular pizzas and 50 super deluxe pizzas, the profit is maximized at $87.50. 38. Note: See Application Section 1.5 for information on the method used to solve this problem. From problem 23, we want to minimize E = 3yi + 2t/2 subject to the constraints: 5yi + 2/2 > 10, 2yi + 2j/2 > 12, t/i + 4t/2 > 12, 1/1 > 0, and t/2 > 0. The dual problem is to maximize / = 10xi + 12x2 + 12x3 subject to the constraints: 5xi + 2 x2 + x3 < 3, x\ + 2 x2 + 4^3 < 2, x\ > 0, X2 > 0, and x3 > 0. Then X\
X2
x3
5 1
2 2
1 4
10
12
12
S1
S2
1 0
0 1
3 2
0
0
/
Si S2
X\
X2
x3
1
0 1
-3 /4
1 /4
-1 /4
-19 /8
-1 /8
0
-9
-1
0 0
S\
S2
1 /4
Xi
5 /8
7 /8
X2
-5
/ — 13
Henee if the predator catches 1 of species I and 5 of species II, then the energy will be minimized at 13 units.
594
Instructor’s Manual
Application 1 Linear Programming
Application 1.5 i (2 - 1 M
1.
V 3 2 4J
f 22 1 A y —1 4 6 5 /
12 .
11.
Minimize y = 5j/i + 62/2 subject to yi + 3y2 > 4 - l y i - 2y2 > 3
Minimize g = 5yi + 7j/2 + 2/3 subject to Vi+ 3j/2 + í/3 > 2 2yi + 2y2 + y3 > 5 13.
y1.y2.y3 > o
yi»y2 > o
14. Maximize / = *1 + *2 + 3*3 subject to 2*i + *2 < 5 *1 + 2*2 + *3 < 3
Maximize / = * 1 + *2 subject to 2*i + *2 < 2 *1 + 2*2 < 3
15.
* 1, *2 > 0
g = 5yi + 6y2 ;t to iy2 > 1 12 > 1 Iy2 > 1 >0
Maximize / = 13*i + 21*2 + H *3 subject to *1 + 4*2 —3*3 < 2 2*i + *2 —*3 < 5 *1 + 2*2 + 4*3 < 3 *1,*2,*3 > 0
* 1,2:2,*3 > 0
16. Minimize g = 5yi + 6y2+ 7y3 subject to yi - y2 + 2y3 > 2 -yi + y2 - y3 > 8 —yi + 2y2 + y3 > 3 y - l , y 2,y3>0
Minimize g = 8yi + 6 y2 + 25y3 subject to 2yi + 4y2 + 8y3 > 4 3yi + y2 + 7y3 > - 1 yi + 2 y2 + 4y3 > 9 y1.y2.y3 > 0
T h e Simplex Method II: T h e Dual Mínimum Problem 20.
19.
2yi > 2
Maximize / = 10xi + 14x2 + 5^3 subject to xi + 2x2 + 5x3 + 2x4 < 3 Xi —X2 —8x3 —X4 < 1
3yi > - 1 4t/i > 5
Xi -}- 2x2 4- 3x3 4“3x4 K.12
Minimize g = Í2yi subject to 2/1
2/1
>1
>
x i + x 2 — 3x 3 — 5x4 < 5
* 1 ,* 2 ,* 3 ,* 4 < 0
0
X2 «1 1 2 1 1 2 0 1 1 0 Xl X2 1 0 0 1 0 0
*1
*2 *3 1 1 2 1 1 3
22 .
2 1 1 23.
Xl X2 X3 1 4 -3 2 1 -1 1 2 4 13
21
«2 0 1 0
2 3 /
Si s2
*2 1/2 3/2 1/2
«2 1 0 0 1 0 0
«1
1
*4
2 -4 -5 3 0 X3 5/2 -11/2 -11/2 -2 -30
1 2 /-I
*1
S2
*1
5 3 /
0 1 0 0
Si 1 0 0 0 0 x4
1 -3 -6 1 -14
Xl 2 5
0
/
S2 0 1 0 0 0
xz 0 1 0
*2
1 -1 1 2 -2 -5
51 52
0 0 1
Si 4/7 -9/7 -1/7 -41/7
g = 5/3 at (1/3,1/3).
X2
Si S2 S3 0 0 0
«i «2 1/2 0 -1/2 1 -1/2 0
Xl
1/3 4/3 / - 5/3
2/3 -1/3 -1/3 2/3 -1/3 -1/3
11
x2 *3 1 2 5 1 -1 -8 1 1 -3 1 2 3 10 14 5 Xl x2 1/2 1 3/2 0 1/2 0 0 0 3 0
*1
Xl 1 0 0
S2
«1
Xl X2 X3 1 22/7 0 0 -39/7 0 0 -2/7 1 0 -117/7 0 24.
Application 1.5
3
«2
«3
0
s2 0 3/7 1 -5/7 0 1/7 0 -50/7 S3 S4 0 0 0 0 1 0 0 1 0 0 Si S2 1/2 0 1/2 1 -1/2 0 -1 0 -7 0
X2 X3 4 -3 -7 5 -2 7 -31 5 0
1 0 0
«i
«1 «2 1 -1 0 1 0 -3
17/7 2/7 1/7 / - 232/7 3 1 5 12 / s3 0 0 1 0 0
2 3 / —9 1 -2
0 1
-1
0 0
-13
0 0 1 0
/
2 1
Xl
1
53
—26
52
Xl
S2 X3
g = 232/7 at (41/7,0,50/7).
53
54
3/2 5/2 7/2 9 / —21
g = 9 at (0,3).
s1 S2 S3
si 52
S4 0 0 0 1 0
«1
X3
x2 s2 sz s4
595
596
Instructor's Manual
Application 1 Linear Programming
Xi x2 £3 £4 0 1 13/3 2 1 0 -11/3 -2 0 0 -11/3 -5 -2 1 0 0 0 0 -19 -8
51
s2
1/3 -1/3 1/3 2/3 -2/3 -1/3 -1 0 -8 -2
s3
$4
0 0 1 0 0
0 0 0
2/3 5/3 8/3 9 / —26
1
0
*2 «3 «4
g = 26 at (8,2,0,0). 25. Let yi = lbs. of Food I and 2/2 = lbs. of Food II. Minimize Dual problem: Maximize / = 2a?i + £2 9 = 0.52/1 + 2/2 subject to subject to 0 . 92/1 + O . 62/2 O . I 2/1 + 0 . 42/2
>
0.9£i 4- 0.l£2 < 0.5
2
> 1
0.6£i + 0.4#2 < 1
/ , >0
£1, £2 > 0
2 1 2/2
Xl *2 «1 S2 Xl x2 Si S2 0.9 0.1 1 0 1 1/9 10/9 0 5/9 0.5 si 2/3 0.6 0.4 1 0 1/3 0 1 -2/3 1 S2 2 1 0 7/9 -20/9 0 0 0 / 10/9 / Xi *2 Si s2 Xl 1 0 4/3 -1/3 1/3 -2 2 0 1 3 x2 -2/3 -7/3 0 0 / -8/3 Then yx= 2/3 Ib. and y2 = 7/3 Ib. Cost per Ib. = ($8/3)/31b. = $0.89/lb 26.
£1
£2
0.9 0.1 0.6 0.4 0.3 0.7 2 1 £l
£2
Sl
«2
«3
1 0 0 1 0 0 0 0
0 0 1 0
*1
S2
£l
0.5 1 2 /
Si s2
-+
S3
1 0 0 0
£2
«1
«2
s3
1/9 1/3 2/3 7/9
10/9 -2/3 -1/3 -20/9
0 1 0 0
0 0 1 0
S2
5/9 2/3 11/6 / —10/9
Xl S2 S3
S3
1 0 0 0
Xl 1/3 0 4/3 -1/3 0 -2 2 1 3 0 *2 s- 3 1/2 1 -2/3 1 0 / - 8/3 0 -2/3 -7/3 0 Then 2/1 = 2/3 Ib., y2 = 7/3 Ib. and 2/3 = 0 Ib. Cost per Ib. = ($8/3)/3 Ib. = $0.89/lb. 27. Let 2/1 = lbs. of chemical 1, 2/2 = lbs. of chemical 2 and 2/3 = lbs. of chemical 3. Vi
Minimize g = 20j/i + 15y2 + 5y3 subject to
>
20
+^ +^ | yi,V2,y3>
0
Since 2/3 = 100 —2/1 —2/2, 9 = 500 + 152/i + 102/2- To minimize 20 > 100
Vi Vi +
2x2 < 10 *1,*2 > 0
2/1,2/2 > 0
X\
X2
597
subject to
subject to
Xi x2 1 1 0 2 20 100
A pplication 1.5
«i S2 1 0 0 1 0 0 Si
15 10 /
si «2
Xl 1 0 0
X2 1 2 80
Si 1 0 -20
*2 0 1 0
15 10 / —300
Xl s2
S2
1 0 1 -1/2 0 1 0 1/2 -20 -40 0 0 / Then yi = 20 lbs., t/2 = 40 lbs.
Xi 10 5 X2 —700 and therefore ys = 40 lbs. The minimized cost is $1200.
28. Let 2/1 : hrs. for jogging, 1/2= hrs. for bicycling and 2/3 = hrs. for swimming. Minimize Dual problem: Maximize / = 2 x\ + 3000^3 9 = 2/1 + 2/2 + 2/3 subject to subject to - 2 / 1 + 2/2 -
2/3 >
0
—a?i + 600a?3 < 1 X\
V3 > 2
6 O O 2/1
+ 3002/2 + 3002/3 > 3 0 0 0
—X \ + X 2 + 300^3 < 1
2/i, 2/2,2/3 > o
Xl x2 X 3 «2 S 3 -1 0 0 1 -1/600 0 1 51 1 1 0 1 3/2 0 0 52 -1 0 1 1 -1/2 1 0 53 0 0 0 2 0 5 / Sí s2 S3 1/600 0 0 1/600 *3 -1/2 1 0 1/2 s2 -1/2 0 1 1/2 X2 -4 0 -2 /-6 Xl X2 si s2 X3 S3 0 0 1 1/450 -1/900 0 1/900 X3 1 0 0 Xl -1/3 2/3 1/3 0 0 1 0 -2/3 1/3 1 2/3 X2 i 0 0 0 2 -4 -2 /-8 Then 2/1 = 2 , 2/2 = 4 and 2/3 = 2. The minimized time is 8 hours.
*1
x2 X 3 «i 0 600 1 0 300 0 1 300 0 2 3000 0 Xl *2 *3 -1/600 0 1 3/2 0 0 -1/2 1 0 6 0 0
+ 300^3 < 1
®1, x 2, ^3 > 0
«1 s2 1/600 0 -1/2 1 -1/2 0 -5 0
3
S
0 0 1 0
1/600 1/2 1/2 /-5
*3 «2
S3
598
Instructor’s Manual
Application 1 Linear Programming
Application 1.6 1. (a)
Xl 2 -3 1
Si S2 1 0 10 0 1 -9 0 0 / S1 S2 X2 0 1 3/5 2/5 1 0 1/5 -1/5 -2 -1 0 0 (19/5,12/5);/ =11. 2. (a) (b)
(b)
*2 1 1 3
Xl x2 1 -1 1 1 -2 1 2 5 Xl
Si
X2
0 0 0 1 1 0 0 0 ( 8 , 4 ) ; / = 36.
Si S2
12/5 19/5 /-ll
S2
S3
0 1
0 0
0 0
1
12 12 -1 2
0
/
1 0 0 0
$1
«2
4 3 /-3
si Xl
*2 Xi
Si
s2 —*■ S3
Xl
x2
0 0
-1 /2
1
-1 /2
0 0
0
6
0
«3
1 1/3 2/3 0 2/3 1/3 0 1/3 -1/3 0 -4 -1
si S2 1 2/3 0 -1/3 0 1/3
Xl x2 0 5/3 1 -1/3 0 10/3
8 4 8 / - 36
Si x2 Xi
3/2
si
1
S2
S3
0
1 /2 1 /2 -1 /2
6 6 6
1
/ — 12
1 0 0
si «2 «1
T h e Simplex Method III: Finding a Feasible Solution Xi
#2
£3
Si
S2
S3
1
1 -1
2
1
0
1 1 1
0 0 0
1
0 0
5 3
1
-1
1 -1 1
0 1
0 0
Xi
0 s2 0 1 0 0
«i «2 «3
/
X2
Xz
S1
S2
$3
0 1 0 -1 1 0 0 1
3 2
1
0
0 0 0
1
1 1 -1 1
-1
2
0 0
Application 1.6
4 2 1
«i S2 Xi
/- 1
£2 x3 s1 s3 4 1 1 0 1 3 1 2 6 0 -5 0 -1 1 1 0 -1 0 -1 -1 0 0 0 /-5 (1 ,4 ,0 );/= 5. 4. The dual problem is to maximize f = xi + 3a?2 subject to the constraints: 2xi + X2 < 10, —3xi+ X2 < —9, xi > 0, and X2 > 0. By problem 1, g is minimized at (2,1) and g = 11.
5. The dual problem is to maximize / = 2xi 4- X2 subject to the constraints: x\ — X2 < 12,xi + X2 < 12, —2xi + X2 < —12, xi > 0, X2 > 0, and x3 > 0. By problem 2, g = 36 at (0,4,1). X\
X2
X3
Si
«2
s3
4 1 -2 1
3 5 5 3
-2 1 -1 4
1 0 0 1 0 0 0 0
0 0 1 0
Xl
%2
Si
0 -3.25 0 9.125 1 -0.875 0 16.875 Xx
X2
2 3 -4 /
1 0 0 0
-0.25 0.125 0.125 0.875
X3
S1
0 15 1 0 73 0 1 -10 0 0 -47 0 (1 ,0 ,2 );/= 9.
0 1 0 0
52
S 2
«i S2
s3 S3
0 -0.5 1 0.75 0 -0.25 0 2.25
1.5 0.25 1.25 / - 7.25
S3
2 1 8 6 -1 -1 -7 -3
2 2 1 f-9
*3 Si XI
*3 S2 Xi
599
600
A pplicatio n
Instructor’s Manual
1 Linear P rogram m ing
Review Exercises for Application 1
1.
2
N '¡*x+3y>12
3.
-1--*-- H \* / X 4x+3y-12 \
5.
6.
Review Exercises for Application 1
8.
7.
9.
¡x|>2
\y
I>Í this is not strictly determined. S _- sq / q
30.
; The farmer should pick the tomatoes on August 25.
31. The company should take a lógica! approach and the unión should take a legal approach. 32. The University of Montana should use the fullback run and Montana State University should use the prevent defense against a short pass. 33. Since a,y and a*/ are both saddle points then a,¿ < au < a*/ a,¿ < a*/ and a¡j > a^j > a*/ => a¿¿ > au. Then ay = a*/.
Two-Person Games: Mixed Strategies
Application 2.2
Application 2.2 1. pAq‘ = p f 1? / ^ = 4 15/4 ) = “*
2. pAq'
3. pAq'* = P ( 3/ 2 ) = 3/2
4. pAq‘ = p ( * ) = :
5. pAq'‘ = P ( q) = 2
6. pAq* = p ^
/ 4\ 7. pAq* = p I 3 1 =8 / 3
8. pAq* = p
5 j =26/9
/8/5\ 5 = 31/10
/ 23/10\ / 5/4\ 33/10 = 149/40 = 3.725 10. pAq‘ = p 3/2 = 31/20 = 1.55 V 21/5/ V V p0 = (1/2 1/2); q0 = (1/2 1/2); 7, = 1/2 12. Po = (3/5 2/5); q0 =(4/5 1/5); u = 7/5 po = (4/5 1/4); q0 = (2/5 3/5); t; = 2/5 14. Po = (1/2 1/2); q0= (1/2 1/2); t; = 0; fair Po = (1 0); q„ = (1 0); v = 0; fair 16. p0 = (1/2 1/2); q0= (1/2 1/2); v= 0; fair Po = (1 0); q0 = (1/2 1/2) or (0 1); t>= -1 po = (1 0) or (0 1) or (1/2 1/2); q0 = (1 0); t, = 1/2 Po = (1 0); q0 = (1 0); v = 3 po = (1/2 1/2); q0 = (3/8 5/8); v = 0; fair
9. pAq* = p 11. 13. 15. 17. 18. 19. 20.
21. A' =( * * ); po = (3/5 2/5); q0 = (0 4/5 1/5); v= 18/5 22. A! =(2 g ); Po = ( 4/5 1/5 0); q0 = (0 4/5 1/5); v = 14/5 23. A' = ^ 4 6 9 ^ ; A" = ^
p0 = (5/6 0 1/6 0); q0 = (5/6 0 1/6); v = 29/6 ft¡ 4.833
24. E(p,q) = (pi ( i - J ¡ ) = (Pl 1 _ P l ) ( J o - ^ l ) - I f t h e Patient h a s t h e °P‘ eration, then p = (1 0), and ü?(p, q) = 39 —29q\. If the patient does nothave the operation,then p = (0 1), and E{p, q) = 40 —35gi. The patient should have the operation if 39 —29gi > 40 —35gi, so that qi > 1/6. 25. Let q\ be the probability that the patient has the disease. So q* = ^ Then £(p,q) = (pi 1 - p i ) (
“12 ) ( .
^
^ =
^ ~~^ ^
^ ) = ( = v. Conversely, suppose ü7(p,qo) < v for every p. Let p be the strategy x=i 1=1 with 1 in the kth position and 0 everywhere else. Then the fcth component of ^4qg = P-Aqo < v. Henee the components of Aql are less than or equal to v.
44. (a) The first component of p0i4 is [an(a 22 —^ 21) + «21(011 —ai2)]/(an + 0,22 —
1 (0.79 0.13\ 18. T, _ (0.21 0.46 0.13 0.20^ q _(0.46 0.20\ fí_ (0.21 0.13\ n _ “ V°.31 0-25 0-2l O-23)' ~ Y0.25 0.23 J’ Y0 3l °-2l 7 ’ “ 0.5838 V°-31 °-79/ ’ . 1 (0.3959 0.1879^ 0.5838 V0-3401 0.2077 / 19. T
(1/8 1/4 1/8 1/8 3/8 \ (1 /8 3/8 \ 1/7 2/7 1/7 2/7 1/7 ; S = 2/7 1/7 ; R V1/4 1/2 0 1/8 1/8/ \ 1/8 1/8/ í (1 4 4 70 28 \ . ( 83 135' 126 92 111 107.
-
=
(1 /8 1/4 1/8' 1/7 2/7 1/7 \ 1/4 1/2 0,
(0.17 0.23 0.15 0.32 0.13\ (023 . 0.32 0 •l3") ff_ (0 .1 7 0.15\ n _ 1 ( 1 0.15\ 21 \0.15 0 Y 0.8075 Vo.15 0.83)' u —V°-15 0-21 0 0.38 0.26 Y ” Vo-0.38 0 .26/ ._ 1 (0.2615 0.377 0.169 \ _ 0.8075 V0 2088 0.3634 0.2353 ) 21. Let E¡, i= 0,1,2,3,4, be the state such that the animal has received iunits of food. The state moves from Ei to £,+i, i = 0,1,2,3, with probability of 4/5 and stays in E¡ with probability of 1/5. E4 is 1/5 4/5 0 0 0 \ 0 1/5 4/5 0 0 an absorbing state. T = 0 1/5 4/5 0 0 0
0 1/5 4/5 0 0 1/
617
618
Instructor’s Manual
Application 2 Markov Chains and Game Theory
f 1/5 4/5 0 0 0' 0 1/5 4/5 0 0 22. V = R= 0 0 1/5 4/5 0 \ 0 0 0 1/5 4/5 > = 320.
1/5 4/5 0 0' 0 1/5 4/5 0 0 0 1/5 4/5 0 0 0 1/5/
Q=
'5 20 80 320' 0 5 20 80 0 0 5 20 ,0 0 0 5,
Expected number
23. (a). Let Ei, i= 0,1,..., 8, be the state such that Gi has idollars. State Ei, i= 1, 2, ..., 7, moves to state Ei+i with probability of 3/7 and to state £
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