1445009365Krash_Sample.pdf
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Krash (Sample)
Manual for Krash Why Krash?
Towards the end of preparation, a student has lost the time to revise all the chapters from his / her class notes / standard text books. This is the reason why K-Notes is specifically intended for Quick Revision and should not be considered as comprehensive study material. It’s very overwhelming for a student to even think about finishing 300-400 300-400 questions per subject when the clock is ticking at the last moment. This is the reason why Kuestion serves the purpose of being the bare minimum set of questions to be solved from each chapter during revision. What is Krash?
Krash is a combination of K-Notes and Kuestion which effectively becomes a great tool for revision. A 50 page or less notebook for each subject which contains all concepts covered in GATE Curriculum in a concise manner to aid a student in final stages of his/her preparation. A set of 100 questions or less for each subject covering almost every type which has been previously asked in GATE. Along with the Solved examples to refer from, a student can try similar unsolved questions to improve his/her problem solving skills. When do I start using Krash?
It is highly recommended to use Krash in the last 2-3 months before GATE Exam. How do I use Krash?
Once you finish the entire enti re K-Notes for a particular subject, you should practice the respective Subject Test / Mixed Question Bag containing questions from all the Chapters to make best use of it. Kuestion should be used as a tool to improve your speed and accuracy subjectwise. It should be treated as a supplement to our K-Notes and should be attempted once you are comfortable with the concepts and basic problem solving ability of the subject. You should refer K-Notes K-Notes before solving any “Type” problems from Kuestion.
© Kreatryx. All Rights Reserved. Reproduction or re-transmission of this material, in whole or in part, in any form, or by means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written consent of Kreatryx, is a violation of copyright law. Any sale/resale of this material is punishable under law. Subject to Ghaziabad Jurisdiction only. © Kreatryx. All Rights Reserved.
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Krash (Sample)
Krash - Sample Engineering Maths (Linear Algebra & Differential Equation)
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Krash (Sample)
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Krash (Sample)
LINEAR ALGEBRA MATRICES
A matrix is a rectangular array of numbers (or functions) enclosed in brackets. These numbers (or function) are called entries of elements of the matrix. 2 0.4 8 Example: order = 2 x 3, 2 = number of rows, 3 = number of columns 5 -32 0 Special Type of Matrices 1. Square Matrix
A m x n matrix is called as a square matrix if m = n i.e. number of rows = number of columns The elements aij when i = j 1 Example: 4
a11a22 .........
are called called diagonal elements
2
5
2. Diagonal Matrix
A square matrix in which all non-diagonal elements are zero and diagonal elements may or may not be zero.
1 0 Example: 0 5 Properties
a. diag [x, y, z] + diag [p, q, r] = diag [x + p, y + q, z + r] b. diag [x, y, z] × diag [p, q, r] = diag [xp, yq, zr] c.
diag x, y, z
1
diag 1 , 1 , 1 y z x
t
d. diag x, y, z = diag [x, y, z] e.
diag
n
x, y, z diag xn , y n , zn
f. Eigen value of diag [x, y, z] = x, y & z g. Determinant of diag [x, y, z] = xyz 3. Scalar Matrix
A diagonal matrix in which all diagonal elements are equal.
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Krash (Sample)
4. Identity Matrix
A diagonal matrix whose all diagonal elements are 1. Denoted by I Properties:
a. AI = IA = A b.
n
I
1
I
I c. I d. det(I) = 1
5. Null matrix
An m x n matrix whose all elements are zero. Denoted by O. Properties:
a. A + O = O + A = A b. A + (- A) = O 6. Upper Triangular Matrix
A square matrix whose lower off diagonal d iagonal elements are zero.
3 Example: 0 0 7.
4
5
6
7
9
0
Lower Triangular Matrix
A square matrix whose upper off diagonal elements are zero.
3 Example: 4 5 8.
0
0
6
0
9
7
Idempotent Matrix 2
A matrix is called Idempotent if 1 Example: 0 9.
A
A
0
1
Involutary Matrix
A matrix is called Involutary if
2
A
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I.
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Krash (Sample)
Matrix Equality
Two matrices Am n and Bp q are equal if
m=p; n=q
i.e., both have same size
aij = bij for all values of i & j. Addition of Matrices
For addition to be performed, the size of both matrices should be same. If [C] = [A] + [B] Then cij aij bij i.e., elements in same position in the two matrices are added. Subtraction of Matrices
[C] = [A] – [B] = [A] + [– [–B] Difference is obtained by subtraction of all elements of B from elements of A. Hence here also, same size matrices should be there. Scalar Multiplication
The product of any m × n matrix A a jk and any scalar c, written as cA, is the m × n matrix cA = ca jk obtained by multiplying each entry in A by c. Multiplication of two matrices
Let Am
n
and Bp q be two matrices and C = AB, then for multiplication, [n = p] should
hold. Then, n
cik aij b jk j1
Properties:
If AB exists then BA does not necessarily exists.
Example: A 3
4
, B4
5
, then AB exits but BA does not exists as 5 ≠ 3
So, matrix multiplication is not commutative.
Matrix multiplication is not associative. A(BC) ≠ (AB)C .
Matrix Multiplication is distributive with respect to matrix addition A(B + C) = AB +AC
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Krash (Sample)
If AB = AC
B = C (if A is non-singular)
BA = CA B = C (if A is non-singular) Transpose of a matrix
If we interchange the rows by columns of a matrix and vice versa we obtain transpose of a matrix.
1 eg., A = 2 6
3
5
;
4
A
T
1 3
2
6
4
5
Conjugate of a matrix
The matrix obtained by replacing each element of matrix by its complex conjugate. Properties
a.
b.
A
c.
A
B
KA
d.
A
AB
A
B
KA AB
Transposed conjugate of a matrix
The transpose of conjugate of a matrix is called transposed conjugate. It is represented by A
a.
.
A
A
b. A B c. KA
d. AB
A B
KA
B A
Trace of matrix
Trace of a matrix is sum of all diagonal elements of the matrix. Classification of Real Matrix T
a. Symmetric Matrix : A A T
b. Skew symmetric matrix : A A c. Orthogonal Matrix :
A
T
A
1
T
; AA =I =I
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Krash (Sample)
Note:
a. If A & B are symmetric, then (A + B) & (A – B) are also symmetric b. For any matrix
T
AA
is always symmetric.
A + AT c. For any matrix, is symmetric & 2 d. For orthogonal matrices,
A AT is skew symmetric. 2
A 1
Classification of complex Matrices
a. Hermitian matrix :
A
A
b. Skew – Hermitian matrix : A
c. Unitary Matrix : A
A
1
A
; AA 1
Determinants
Determinants are only defined for square matrices. For a 2 × 2 matrix
a 11 a21
= a22 a12
a11 a22
a12 a21
Minors & co-factor a11
If
a12
a13
21
a22
a23
a31
a32
a33
a
Minor of element a21 : M21 Co-factor of an element aij
a12
a13
a32
a33
1 i j Mij
To design cofactor matrix, we replace each element by its co-factor Determinant
Suppose, we need to calculate a 3 × 3 determinant
3
3
3
j 1
j 1
j 1
a1 jcof a1 j a2 jcof a2 j a3 j cof a3 j
We can calculate determinant along any row or column of the matrix.
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Krash (Sample)
Properties A
T
A
Value of determinant is invariant under row & column interchange i.e.,
If any row or column is completely zero, then A
If two rows or columns are interchanged, then value of determinant is multiplied by -1.
If one row or column of a matrix is multiplied by ‘k’, then determinant also becomes k
0
times.
If A is a matrix of order n × n , then KA
K
n
A
Value of determinant is invariant under row or column transformation
AB A * B
A
A
n
A
1
n
1
A
Adjoint of a Square Matrix T Adj(A) = cof A Inverse of a matrix
Inverse of a matrix only exists for square matrices
A 1
Adj Adj A
A
Properties 1
a.
AA
b.
AB
c.
ABC
d.
A
T
1
A
1
I
B1 A1 1
1
A
C 1B 1 A
A 1
1
T
e. The inverse of a 2 × 2 matrix should be remembered
a c
b
d
1
d b ad bc c a 1
I. Divide by determinant. II. Interchange diagonal element. III. Take negative of off-diagonal element. © Kreatryx. All Rights Reserved.
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Krash (Sample)
Rank of a Matrix
a. Rank is defined for all matrices, not necessarily a square matrix. b. If A is a matrix of order m × n, then Rank (A) ≤ min (m, n) c. A number r is said to be rank of matrix A, if and only if
There is at least one square sub-matrix sub-matrix of A of order ‘r’ whose determinant is non-zero. non-zero.
If there is a sub-matrix of order (r + 1), then determinant of such sub-matrix should be 0.
Linearly Independent and Dependent
Let X1 and X2 be the non-zero vectors
If X1=kX2 or X2=kX1 then X1,X2 are said to be Linearly Dependent vectors.
If X 1kX2 or X2 kX1 then X1,X2 are said to be Linearly Independent vectors.
Note:
Let X1,X2 ……………. Xn be n vectors of matrix A
If rank(A)=no of vectors then vector X1,X2………. Xn are Linearly Independent
If rank(A) 0 will be
1 n pv 1 n q 1 n pv 1 n q
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dv 1 n p v 1 n q dt dv 1 n pv 1 n q (D) dt
(B)
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Krash (Sample)
d2 y
dy 4 3y 3e2x , the particular integral is dx dx2
05. For
(A) (C)
1
2x
e
15 2x
3e
(B)
1 2x e 5
(D) C1ex
06. Solution of the differential equation 3y
dy dx
2x
(A) ellipses (C) parabolas
C
2
e
3x
0 represents a family of
(B) circles (D) hyperbolas
07. Which one of the following differential equation has a solution given by the function
y 5 si sin 3x 3x 3 dy 5 cos 3x (A) dx 3
(C)
d2 y d2 x
0
(B)
9y 0
dy dx
(D)
5 cos 3x 0 3
d2 y dx2
9y
0
3
d3 y
dy 2 0 are 4 y 08. The order and degree of a differential equation dx 3 dx respectively (A) 3 and 2 (C) 3 and 3
(B) 2 and 3 (D) 3 and 1
..
09. The maximum value of the solution y (t) of the differential equation y(t)+ y (t) = 0 with
.
initial conditions y 0 1 and y(0) = 1, for t ≥ 0 is (A) 1 (C)
(B) 2
(D)
2
10. It is given that y 2y y 0 , y(0) = 0 & y(1) = 0. What is y(0.5) ?
(A) 0 (C) 0.82
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(B) 0.37 (D) 1.13
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Krash (Sample)
11. A body originally at 60° cools down to 40 0 in 15 minutes when kept in air at a
temperature of 25°c. What will be the temperature of the body at the end of 30 minutes? (A) 35.2° C (C) 28.7°C
12. The solution
(B) 31.5°C (D) 15°C
d2 y 2
dx
2
dy dy 0 in the range 17 y 0 ; y 0 1, dx dx x 4
0
x
4
is given by
4x (A) ex cos 4x
1 sin 4x 4
1 (C) e4x cos 4 x sin x
4
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(B) ex cos 4x 4x sin 4x 4 (D)
26
4x cos 4x 1 sin 4x
e
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Krash (Sample)
Solutions Type 1: Linear Algebra 01. Ans: (A) Solution: Rank=no. of rows=no. of columns A
0 As
all rows & columns are linearly independent
A=Non-Singular matrix
02. Ans: (B) Solution: I
2
1
1
1
y
x2
1 1
0
x
By just looking at matrix I we can conclude that:
One solution of above matrix is (0,0) as row3’s all elements are zero hence |I|=0
Second solution is (1,2) as R1=R3 Linearly dependent rows
Third solution is (-1,1) as R2=R3 Linearly dependent rows
03. Ans: (B) 2
Solution: X
I
X
X I Multiplying by
X
1
1
X I
Or
1 0 a 1 1a 1 X 1 I X 2 2 0 1 a a 1 1 a a a 1 a 04. Ans: (A) T
Solution: X 4 3 , Y4
Matrix operation X Y
T
T
Order 4 3
T
T
3 4
X Y
3
1 PY 1 X T 1 PT T T & P2 3 and P X y P
33
X Y
1
33
P X T Y 1
23
P X T Y1 PT
22
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Krash (Sample)
05. Ans: (C) Solution: R1 R1 R3 , Replace
0 1 We get A 1 A
0
1 1 B 2
0
0 1
0
Rank=2
06. Ans: (B) Solution: Eigen values=-3, 3, 5
1 Eigen vector corresponding to 5 is 2 1 For
3
M
Eigen values -27, 27, 125
But Eigen vector will remains the same
1
1
2
T
07. Ans: (B)
1 1 1 6 Solution: A | B 1 4 6 20 1 4 u For no solution
A A | B n
A n A
0
1 4 24 1 6 1 4 4 0 4 24 24 6 0
3 18 6
But A | B 0
1 4u 80 8 0 1 20 u 6 4 4 0 => u 20 © Kreatryx. All Rights Reserved.
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Krash (Sample)
08. Ans: (B) Solution: Orthogonal matrix,
P
T
P
1
.............. 1
Note: Orthogonal matrix presence the length of vector [Euclidean Norm]
X X1 X2 X
T
X 1 X 2
PX
2
T
PX PX PX
X P PX T
T
From (1) PX
2
X P T
1
PX
X T IX X T X X2
09. Ans: (D) Solution: Characteristic equation= 2
2
2 1 0 .
Every matrix satisfies its characteristic
equation [Cayley Hamiltonian theorem] 3
P
2
P
2P I
Px Multiplying
by
P
1
P
2
P 2I 2I
P
1
10. Ans: (C) Solution: This equation will be solvable even for non-square matrix A of B vector is linearly
independent of columns of A. 11. Ans: (D) Solution: Over determined system will have no solution as number of equations are more
than number of unknowns. 12. Ans: (C) Solution:
Eigen 2
1 2 2
&
&
E i g en 3 5
1 2 35
1 2 1 35 12 21 35 0 1 7 or 2 5
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Krash (Sample)
Type 2: Differential Equation 01. Ans: (B)
x, y dy g x, y dx 0 Solution: f x, N
M
For exact differential equation M
y
N
x
f x
g y
02. Ans: (D) Solution: For linear differential equation:dy dx
Py Q
0 P & Q must be function of x Or constant k f x
03. Ans: (B) Solution: dx
kx
2
dx dt
dt
dx x2 k dt
kx
2
0 , x
t 0
a
Variable separate type 1
x
kt c
At t=0 1
a
1 x
c 1
a
kt
04. Ans: (A) Solution:
dy dt
Py q. yn [ Bernoulli’s equation]………….(i)
1 n Let V y
dv n dy 1 n y dt dt
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Krash (Sample)
dy dt
yn
dv .............. ................... ..... 2 1 n dt
Equation (1) can be written as y
n
dy dt
Py
dv
1 n dt dv dt
1 n
q
Pv q
1 n pv 1 n q
05. Ans: (B) Solution: Let
d dx
D, f D
D
2
4D 3
0
3e2 x 3e2 x e2x P.I 15 5 f 2 06. Ans: (A)
[Variable separable]
Solution: 3ydy 2xdx 0
On integration y2
3y 2 x 2 K [Ellipse, x2 2
For circle, coefficient of
2
x
2 k 3
]
coefficient of y 2
07. Ans: (C)
5sin 3x ............. 1 On differentiating both sides Solution: y 5si 3 dy 5 3 cos 3x dx 3 Again differentiating both d2 y sin 3x 9 5 sin 3x 5 3 3 si 2 3 3 dx
d2 y From (1) dx2
9y 0
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Krash (Sample)
08. Ans: (A) 3
dy 2 Solution: 4 y 3 dx dx d3 y
On squaring both sides 2 dy 3 d3 y 2 3 16 y dx dx
Order=3,
Degree=2
09. Ans: (D) Solution:
2
m
1
0
m i
Solution y C1 cos x C2 ssiin x
y 0 1 => C1 1 and y 0 1 => C2
y cos x sin x 2 si sin x 450 y max
1
2 at
4
10. Ans: (A) Solution: y 2y y 0 y 0 0 & y 1 0
m2
m
1
2m 1 0 2
m=-1 (Replacing Roots)
0
Solution
ye
x
C
1
xC2
y 0 0 C1 y e
x
xC 2
y 1 0 C2
0
Solution of difference equation y 0
y 0.5
0
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Krash (Sample)
11. Ans: (B) Solution: By Newton’s law of cooling (0=Temp) Q 0 Q 1
d dt
Secondary temperature
Initial temperature
Q Q 0
k
dQ
k Q Q dt 0
kt
ln Q Q0 k t c Q Q0 e Q
Pe
kt
Q0
At t 0 P Q1 P
0
Q
60 60
0
25
kt
35e
kc
.e
Q 0
0
35
25
After, 15 minutes
40
25
35e
k 15
:
K
.0565
After, 30minutes Q
0.56530
35 35e
25 31 31.48
0
31 31.5 C
12. Ans: (A) Solution: Let
Then,
2
m
d dx
2m 1 17 7
my
0
m 1 4i
Solution,
y ex C1 cos 4x 4x C2 sin 4x 4x y 0 1 =>
1
C1
y ex cos 4x 4x C2 sin 4x 4x On differentiating both sides y ex cos 4x 4x C2 sin 4x ex 4 sin 4x 4C2 cos 4x 4x
At y 0 : 0 4C2 1 4 C2 1
4
sin4x Solution= y e x cos 4x 4 © Kreatryx. All Rights Reserved.
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