Cls Jeead-15-16 Xii Mat Target-5 Set-2 Chapter-4
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Chapter
4
Determinants Solutions SECTION - A Objective Type Questions
1.
⎡ cos sin ⎤ Let A = ⎢ ⎥ , then |2A| is equal to ⎣ sin cos ⎦
(1) 4cos2
(2)
1
(3)
2
(4)
4
(4)
2
Sol. Answer (4) |A| = cos2 + sin2 = 1, |2A| = 22.|A| = 4
2.
2
1
If is non-real complex cube root of unity, then
2
1 is equal to
2
1
(1) 0
(2)
1
(3)
3
Sol. Answer (1)
1 A 2
3.
2
2 1
1
1 2
2
2 1 2 1 2
1
0
2
0 2 1
1
0
0
1
If A is any of square matrix of order n, then A(adjA) is equal to (1) 1
(2)
|A|In
(3)
0
(4)
|A|n
(3)
(x – y)(y – z)(z – x)
(4)
xy + yz + zx
Sol. Answer (2) Clearly option (2) is correct. 1 x
4.
Value of determinant 1 y 1 z
(1) 0
x2 y 2 is equal to z2
(2)
xyz
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Determinants
Solution of Assignment (Set-2)
Sol. Answer (3)
1 x
x2
1 y
y2
1 z
z2
R1 R1 – R3, R2 R2 – R3
0 xy
x2 y 2
0 y z
y 2 z2
1
z2
z
= (x – y)(y – z)(z – x) 1 x
5.
The value of the determinant 1 y 1 z
x3 y 3 is equal to z3
(1) (x – y)(y – z)(z – x)
(2)
(x – y)(y – z)(z – x)(x + y + z)
(3) (x + y + z)
(4)
(x – y)(y – z)(z – x)(xy + yz + zx)
(3)
4
Sol. Answer (2)
1 x
x3
1 y
y3
1 z
z3
R1 R1 – R2, R2 R2 – R3
0 x y ( x y )( x 2 xy y 2 ) 0 y z 1
( y z )( y 2 yz z 2 )
z
z3
= (x – y)(y – z)(z – x)(x + y + z) 6.
If A is 3 × 3 matrix and |A| = 4, then |A–1| is equal to (1)
1 4
(2)
1 16
(4)
2
(4)
81
Sol. Answer (1) A 1
7.
1 1 A 4
If A is a square matrix of order 3, |A| = 3, then |adj adjA| is equal to (1) 35
(2)
37
(3)
9
Sol. Answer (4) |adjA| = 32 = 9 |adj(adjA)| = 92 = 81 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
8.
Determinants
215
If A, B and C are three square matrices of the same order such that A = B + C, then det A is equal to (1) det A + det B
(2)
det B
(3)
det (A)
(4)
det (B + C)
Sol. Answer (4) A=B+C For addition there is not specific rule.
a1
9.
b1
If = a2 a3
b2 b3
2a1 3b1 4c1
c1
c2 , then value of 2a2 3b2 4c 2 c3 2a3 3b3 4c3
(1) 2
(2)
4
b1
c1
b2 b3
c 2 is equal to c3 (3)
(4)
2
(3)
HP
(4)
AGP
(4)
4
Sol. Answer (4) 10. If the system of equations x + 4ay + az = 0 x + 3by + bz = 0 x + 2cy + cz = 0 have a non-zero solution, then a, b, c ( 0) are in (1) AP
(2)
GP
Sol. Answer (3) 1 4a a 1 3b b 0 1 2c
c
R2 R2 – R1, R3 R3 – R1 1 4a a 0 3b 4a b a 0 0 2c 4a c a
ab + bc = 2ac
2 1 1 b a c
a, b, c are in HP
11. Let
ax3
+
bx2
3x x 1 x 1 + cx + d = x 3 2 x x 2 , then the value of d is x 3 x 4 5x
(1) 5
(2)
0
(3)
–6
Sol. Answer (3) Put x = 0 0 d 3
1 1 0 2
3 4
0
d = –6 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Determinants
Solution of Assignment (Set-2)
12. Let x + y + z = 6, 4x + y – z = 0, 3x + 2y – 4z = –5. The value of for which given system of equations does not have a unique solution is (1) 0
(2)
1
(3)
2
(4)
(1) 0
(2)
1
(3) 2sinB tanAcosC
(4)
2sinAsinBsinC
3
Sol. Answer (4) To not have unique solution, 1 1 1 4 0 3 2 4
1(2 – 0) – 1( – 3) + 1(2 –2) = 0 –3=0 =3 13. If A + B + C = , then value of
sin( A B C )
sin B
sin B cos( A B )
0 tan A
cos C tan A is 0
Sol. Answer (1)
sin sin B
sin B 0
cos C tan A
cos( C ) tan A 0 sin B cos C
0
sin B 0
cos C tan A
tan A
0
= 0 (Skew symmetric matrix of odd order)
14. The value of lying between 0 and and satisfying the equation 2
(1)
5 3 , 24 24
(2)
7 5 , 24 24
(3)
1 sin2
cos 2
4 sin 4
sin2
1 cos 2
4 sin 4
2
sin
7 11 , 24 24
2
cos
= 0 are
1 4 sin 4
(4)
11 , 24 24
Sol. Answer (3) R2 R2 – R1, R3 R3 – R1 1 sin2 cos2 4 sin 4
1 1
1 0
0 1
0
C1 C1 + C2 + C3 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Determinants
217
2 4 sin 4 cos 2 4 sin 4 0 1 0 0 0
0
1
4sin4 = –2 sin4 =
1 ⎛ ⎞ sin ⎜ ⎟ 2 ⎝ 6⎠
7 11 , 24 24
x
6
15. The number of real roots of the equation 2 3 x 3 2 x (1) 0
(2)
1
1 x 3 = 0 is x2 (3)
2
(4)
3
(3)
628
(4)
–4
(3)
x = –7
(4)
x = –9
Sol. Answer (4) x 2
6 3 x
1 x 3 0
3
2x
x2
–5 (x – 2)(x + 3)(x – 1) = 0 x = 2, –3, 1
4 4 0 c = 0, then a + b + c is equal to 16. If a b 4 a b c4 (1) 41
(2)
116
Sol. Answer (4) C1 C1 + C2 + C3 0 4 abc 4 b4 abc 4
b
0 c
0
c4
(a + b + c + 4) = 0 a + b + c = –4
x 2 3 1 17. The equation 4 x 2 10 4 = 0 is satisfied by 2x 1 5 1 (1) x = –2
(2)
x = –5
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Determinants
Solution of Assignment (Set-2)
Sol. Answer (3) x 2 3 1 4 x 2 10 4 0 2x 1
5
1
(x – 2)(10 – 20) – (4x – 2)(3 – 5) + (2x – 1)(12 – 10) = 0 = –10(x – 2) + 2(4x – 2) + 2(2x – 1) = 0 = –10x + 20 + 8x – 4 + 4x – 2 = 0 2x + 14 = 0 x = –7
18.
x
4 y z
y z
4 z x is equal to 4 xy
(1) 4
(2)
x+y+z
(3)
xyz
(4)
0
(1) a + b + c + abc
(2)
a2 + b2 + c2 + ab + bc + ca
(3) 3abc – a3 – b3 – c3
(4)
a3 + b3 + c3 – 3abc
Sol. Answer (4) R1 R1 – R2, R2 R2 – R1 xy 0 y x y z 0 zy z
4 xy
= –4((x – y)(z – y) – (y – z)(y – x)) = 0
19.
a b c b c a is equal to c a b
Sol. Answer (3) C1 C1 + C2 + C3 abc b c abc c a abc a b 1 b c (a b c ) 1 c a 1 a b 0 bc c a (a b c ) 0 c a a b 1
a
b
= (a + b + c)(ab + bc + ca – a2 – b2 – c2) = –(a3 + b3 + c3 – 3abc) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Determinants
x 1
(1) 1, 2
1
x 1 1 = 0 are 1 x 1
1 1
20. The roots of the equation
1
219
(2)
–1, 2
(3)
1, –2
(4)
–1, –2
Sol. Answer (2) C1 C1 + C2 + C3 x 1 1 1 x 1
1
1 1 x 1
1
x 1 1 x 1 x 1 x 1
1 1 x 1
1
1 1 ( x 1) 1 x 1 1
1 1 x 1
1
0 x 2 0 ( x 1) 0 x 2 2 x 1
x 1
1
= –(x + 1)(x – 2)2 Roots are –1, 2, 2
21. The value of the determinant (1) 1
1 logb a is equal to loga b 1
(2)
logab
(3)
logba
(4)
0
0
(3)
3
(4)
a+b+c
Sol. Answer (4)
= 1.1 – logba × logab =1–1=0
22.
11 a c 1 bc 11 a d 1 bd is equal to 11 a e 1 be
(1) 1
(2)
Sol. Answer (2) 0 0 11 a
c d b(c d ) d e b(d e ) e
= (c – d)(d – e)
1 be 0 0
1 1
b b
11 a e 1 be
0
(as R1 R2 )
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Determinants
a 2
ab
23. If ab
b
2
ac
bc
Solution of Assignment (Set-2)
ac bc = ka 2b 2c 2, then k is equal to c 2
(1) 2
(2)
4
(3)
–4
(4)
8
(4)
–100
Sol. Answer (2) Put a = b = c = 1 1 1 1 1
1 1 k
1
1
1
k = –1(0) – 1(–2) + 1(2) = 4
1 24. Let f(x) =
x 1
x
2x x( x 1) ( x 1)x , then f(100) is equal to 3 x( x 1) x( x 1)( x 2) ( x 1)x ( x 1)
(1) 0
(2)
1
(3)
100
Sol. Answer (1) 1 2x
f ( x ) x ( x 1)
1 x 1
1 x
3 x ( x 1) ( x 1)( x 2) x ( x 1)
C1 C1 – C2, C2 C2 – C3 0 x 1
x ( x 1)
0 1
1 x
2( x 1)( x 1) 2( x 1) x ( x 1)
= x(x + 1)(–2(x2 – 1) + 2(x2 – 1)) = 0 f(100) = 0 25. If the system of the equations x – ky – z = 0, kx – y – z = 0, x + y – z = 0 has a non-zero solution, then the possible value of k are (1) –1, 2
(2)
1, 2
(3)
0, 1
(4)
–1, 1
Sol. Answer (4) To have non-zero solution 1 k 1 k 1 1 0 1
1
1
1(1 + 1) + k(–k + 1) – 1(k + 1) = 0 2 – k2 + k – k – 1 = 0 –k2 + 1 = 0 k = ±1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Determinants
sc a 26. If s = (a + b + c), then value of c sa c
(1) 2s 2
(2)
2s 3
a
b b
221
is
sb
(3)
s3
(4)
3s 3
Sol. Answer (2) sc a c sa c
a
b b sb
C1 C1 + C2 + C3 sabc a sabc sa sabc
a
b b sb
1 a b ( s a b c ) 1 s a b = 1 a sb 0 s 0 (2 s ) 0 s s = 1 a sb
= (2s)(s2 – c) = 2s3
sin x cos x cos x 27. The number of distinct real roots of cos x sin x cos x = 0 in the interval x is 4 4 cos x cos x sin x (1) 0
(2)
2
(3)
1
(4)
3
Sol. Answer (3) sin x cos x cos x cos x sin x cos x cos x cos x sin x
C1 C1 + C2 + C3 sin x 2cos x cos x cos x sin x 2cos x sin x cos x sin x 2cos x cos x
sin x
1 cos x cos x (sin x 2cos x ) 1 sin x cos x 1 cos x sin x Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Determinants
Solution of Assignment (Set-2)
0 cos x sin x 0 (sin x 2cos x ) 0 sin x cos x cos x sin x 1
cos x
sin x
0 (sin x 2cos x )(cos x sin x ) 0 2
1 1
0 1
1 cos x sin x
= (sinx + 2cosx)(cosx – sinx)2 = 0 tanx = –2 or sinx = cosx x=
is only solution 4
Number of solution = one
28.
(a x a x )2
(a x a x )2 1
( b x b x )2
(b x b x )2 1 is equal to
(c x c x ) 2
(c x c x )2 1
(1) 0
(2)
2abc
(3)
a 2b 2c 2
(4)
abc
(3)
and
(4)
Neither nor
Sol. Answer (1) C1 C1 – C2
4 (a x a x )2 1 4 ( b x b x )2 1 4 (c x c x )2 1 = 0 (as C1 and C3 are proportional) 29. The determinant D=
cos( ) sin( ) cos 2 sin cos sin is independent of cos
(1)
sin
cos
(2)
Sol. Answer (1) D = cos( + )(cos cos – sinsin) + sin( + )(sin cos + cos sin) + cos2(sin2 + cos2) = cos( + ) cos( + ) + sin( + ) sin( + ) + cos2.1 = cos2 1 4 30. The roots of the equation 1 2
20 5 0 are
1 2x 5 x 2
(1) –1, –2
(2)
–1, 2
(3)
1, –2
(4)
1, 2
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Solution of Assignment (Set-2)
Determinants
223
Sol. Answer (2) 1
4
20
1 2
5
1 2x 5 x 2
0
6
15
0 2 2 x 5(1 x 2 ) 1
5x 2
2x
= 30(1 – x2) + 30(1 + x) = 0 1 – x2 + 1 + x = 0 x2 – x – 2 = 0 x = 2, –1
1 a b 31. If a, b, c are sides of the ABC, 1 c a 0 , then value of sin2A + sin2B + sin2C is equal to 1 b c
(1)
4 9
(2)
3 2
(3)
9 4
(4)
3 3 2
Sol. Answer (3) 1 a b 1 c a 0 1 b c
0 ac ba 0 c b ac 0 1
b
c
(a – c)2 – (b – a)(c – b) = 0 a2 + c2 – 2ac – (bc – b2 – ac + ab) a2 + b2 + c2 – ab – bc – ca = 0 a=b=c sin2A + sin2B + sin2C =
3 3 3 9 = 4 4 4 4
a a2 32. The parameter, on which the value of the determinant cos( p d )x cos px cos( p d )x does not depend sin( p d )x sin px sin( p d )x 1
upon, is (1) a
(2)
p
(3)
d
(4)
x
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Determinants
Solution of Assignment (Set-2)
Sol. Answer (2) 1 a a2 cos( p d )x cos px cos( p d )x sin( p d )x
sin px
sin( p d )x
= (cospx .sin(p + d)x – sinpx . cos(p + d)x) – a(cos(p – d)x . sin(p + d)x – sin(p – d)x . cos (p + d)x) 4 + a2(cos(p – d)x . sinpx –sin(p – d)x . cospx) = sindx – asin2dx + a2sindx ⎡ 11 33. If A = ⎢ ⎢⎣ – 13 ⎡17 (1) ⎢ ⎢⎣13
7⎤ ⎥ , then adj(adj A) is 17 ⎥⎦
– 7⎤ ⎥ 11 ⎥⎦
⎡ 11 ⎢ ⎢⎣ – 13
(2)
7⎤ ⎥ 17 ⎥⎦
(3)
⎡ – 17 ⎢ ⎢⎣ 13
(3)
(3)
7 ⎤ ⎥ – 11⎥⎦
(4)
⎡ – 11 ⎢ ⎢⎣ – 13
1
(4)
–1
adj A
(4)
A2
7⎤ ⎥ 17 ⎥⎦
Sol. Answer (2) For a square matrix of order 2 × 2, adj (adj A) = A ⎡1 34. If A2 = 8A + kI where A = ⎢ ⎣ 1
(1) 7
0⎤ ⎥ , then k is 7⎦
(2)
–7
Sol. Answer (2) ⎡ 1 0⎤ ⎡ 1 0⎤ ⎡ 1 0 ⎤ A2 ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ 1 7 ⎦ ⎣ 1 7 ⎦ ⎣ 8 49 ⎦ 0 ⎤ ⎡ k 0 ⎤ ⎡8 k ⎡8 8 A kI ⎢ ⎥ ⎢ 0 k ⎥ ⎢ 8 8 56 ⎣ ⎦ ⎣ ⎦ ⎣
0 ⎤ 56 k ⎥⎦
Clearly, k = –7
⎡sin cos ⎢ 35. If A = ⎢cos sin ⎢⎣ 0 0 (1) –AT
0⎤ ⎥ 0⎥ , then A–1 is equal to 1⎥⎦ (2)
A
Sol. Answer (3) |A| = sin2 + cos2 = 1 ( 0 ). 1 Hence A–1 exists. Since, A
1 adj( A) and for the given question |A| = 1, hence A–1 = adj(A) |A|
36. If A is a matrix of order 3 and |A| = 2, then |adj A| is (1) 1
(2)
2
(3)
8
(4)
4
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Solution of Assignment (Set-2)
Determinants
225
Sol. Answer (4) det (adj A) = (det A)n – 1, where n is the order of matrix A, therefore |adj A| = |A|3–1 = 22 = 4 5
C0 5 37. The value of the determinant C1 5
(1) 0
(2)
C2
5 5 5
C3 14 C 4 1 is C5
1
– (6!)
(3)
80
(4)
–576
(4)
x=1
Sol. Answer (4) 5
C0
5
C1 5 C2
5
5
C3 14
C4 5 C5
1 10 14 1 5 5 1 1 10 1 1
16 16 16 5 10
5 1
R1 R1 R2 R3
1 1
1 1 1 16 5 5 1 10 1 1 0 0 1 16 0 4 1 9 0 1
C1 C1 C2 , C2 C2 C3
= 16(0 – 36) = – 16 × 36
10 4 3 4 x 5 3 38. If 1 17 7 4 , 2 7 x 12 4 such that 1 + 2 = 0, then 4 5 7 5 x 1 7 (1) x = 5
(2)
x has no real value
(3)
x=0
Sol. Answer (1)
10 4 3 1 17 7 4 4 5 7 4 10 3 7 17 4 5 4 7 If 1 + 2 = 0 then 1 = – 2
…(i) …(ii)
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Determinants
Solution of Assignment (Set-2)
By equation (i) and (ii)
x 5
4
3
4
10 3
7 x 12 4 7 17 4 5 x 1 7 5 4 7 x=5
x 39. The value of y
z
x 2 yz 1 y 2 zx 1 is z 2 xy 1
(1) 1
(2)
–1
(3)
0
(4)
–xyz
Sol. Answer (3) x y z x y z
x 2 yz 1 y 2 zx 1 can be resolved as sum by two determinants as, z 2 xy 1 yz 1 zx 1 xy 1
x2 1 x y2 1 y z2 1 z
= 1 + 2 (say)
x where, 2 y z
yz 1 zx 1 xy 1
x2 1 y2 xyz 2 z
xyz x xyz y xyz
z
(multiplying R1 by x, R2 by y and R3 by z)
x2 1 x
xyz 2 y 1 y (taking common –xyz from C2) xyz 2 z 1 z x2
x 1
2 = ( 1) y z2
y 1 z 1
x y ( 1)( 1) = z
x2 1 y2 1 z2 1
= –1 = 1 + 2 = 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
i m 1 i m4
im
m 5
40. The value of i i m 6
Determinants
227
i m2 i m 3 , when i 1 , is
i m 7
i m 8
(1) 1 if m is a multiple of 4
(2)
0 for all real m
(3) – i if m is a multiple of 3
(4)
2
(3)
2
Sol. Answer (2) im
m 5
= i i m 6
i m 1 i m4 i m 7
i m2 i m 3 i m 8
1
i
i2
i m .i m 3 .i m 6 i 2 1
i i
1 i2
=0
(∵ R1 & R3 are identical)
41. The value of the determinant
1
– (25 1)2
210 – 1
210 – 1
– (25 – 1)2
1
1
25 – 1
25 1
5
2 –1 1 –
(1) 0
(2)
25 1 1
is
(210 – 1)2 1
(4)
4
Sol. Answer (4) Taking 25 + 1 = a and 25 – 1 = b, then 210 – 1 = (25 + 1) (25 – 1) = ab, therefore the given determinant equals.
a 2
ab
ab
b 2
1 b
1 a
1 b 1 a 1 2 2 a b
Multiplying R1 with b, R2 with a and R3 with a2b2 ba 2 1 3 3 ba 2 a b ba 2
ab 2 ab 2 ab 2
1 1 1
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Determinants
0 1 3 3 ba 2 a b ba 2
Solution of Assignment (Set-2)
0
2
ab ab 2
2
ab 2 ab 2
2 ba 2 . a b3 ba 2
1 1 2 (a 3 b 3 ) 3 1 1 a b
3
R1 R1 R2
1 1
3
=4 cos 2 x sin 2 x cos 4 x 42. If the determinant sin 2 x cos 2 x cos 2 x is expanded in powers of sin x then the constant term in the cos 4 x cos 2 x cos 2 x
expansion is (1) 1
(2)
2
(3)
–1
(4)
0
Sol. Answer (3) For constant term, we can substitute x = 0.
1 0 1 0 1 1 The constant term = 1 1 1 = –1
43. If are non-real numbers satisfying
(1) 0
(2)
x3
1 1 is equal to – 1 = 0 then the value of 1
3
(3)
3 + 1
(4)
4
Sol. Answer (2) x3 – 1 = 0 x = 1, , 2 = , = 2 1 + + = 0 and = 1
1 1 1 1 1 1 1 1
C1 C1 C2 C3
1 1 1 1 1
(∵ 1 0)
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Solution of Assignment (Set-2)
Determinants
0 1 0 1 1
229
R2 R2 R1 R3 R3 R1
= [( + – )( + – ) – (1 – )(1 – )] = [2 – ( – )2 – (1 – – + )] = [2 – ( + )2 + 4 – (1 – ( + ) + )] = [2 – (–1)2 + 4 – (1 – (–1) + 1)] = (2 – 1 + 4 – 3) = 3
n 2 U n if U = n 44. The value of n n 1 3 n N
∑
(1) 0
(2)
1 5 2N 1 2N 1 is 3N 2 3N 1
(3)
–1
(4)
2
Sol. Answer (1) N
∑n
1
5
n 1
N
∑U n 1
N
n
=
∑n
2
2N 1 2N 1
n 1 N
∑n
3
3N 2
3N
n 1
N (N 1) 1 5 2 N (N 1)(2N 1) 2N 1 2N 1 6 N 2 (N 1)2 3N 2 3N 4
0 1 5 N (N 1) 0 2N 1 2N 1 12 0 3N 2 3N
6 1 5 N (N 1) 2(2N 1) 2N 1 2N 1 12 3N (N 1) 3N 2 3N
C1 C1 C2 – C3
=0 45. If a > 0, b > 0, c > 0 are respectively the pth, qth, rth terms of a G.P., then the value of the determinant
log a p 1 log b q 1 log c r 1
(1) 1
is
(2)
–1
(3)
abc pqr
(4)
0
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230
Determinants
Solution of Assignment (Set-2)
Sol. Answer (4) Let, a = ARp – 1, b = ARq – 1, c = ARr – 1
log A ( p 1)log R
log a p 1
log A (q 1)log R q 1 log A (r 1)log R r 1
log b q 1 log c r 1
p 1
( p 1)log R
p 1
(q 1)log R q 1 (r 1)log R r 1
C1 C1 (log A)C3
0 p 1 0 q 1 0 r 1
C1 C1 (C2 C3 )log R
=0
a b c 46. If a, b, c are positive and not equal then value of b c a c a b (1) 1
(2)
–3
(3)
may be
2
(4)
4
212 – 211
(4)
5 × 211
Sol. Answer (2)
a b c b c a = – (a3 + b3 + c3 – 3abc) c a b = – (a + b + c) (a2 + b2 + c2 – ab – bc – ca) =
(a b c ) {(a – b)2 + (b – c)2 + (c – a)2} < 0 2
Hence possible value = –3 47. The greatest value of x satisfying the equation 211 – x
– 212
211
– 212
211 – x
211
11
11
2
12
2
–2
(1) 3 × 211
0
is
–x
(2)
211 + 213
(3)
Sol. Answer (1) Let 211 = a, then the given equation reduces to
ax
2a
2a a
ax a 0 a 2a x
a
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Solution of Assignment (Set-2)
Determinants
231
Applying C1 C1 + C2 + C3
x
2a
a
x a x a 0 x a 2a x 1
2a
a
x 1 ax a 0 1 a 2a x
x 0 3a x 0 0 0 3a 3a x
1
2a
a
R2 R2 R1 R3 R3 R1
x[(3a – x) (–3a – x) – 0] = 0 x(x – 3a) (x + 3a) = 0 Greatest value of x satisfying the given equation is x = 3a = 3 × 211 48. The system of equations ax + 4y + z = 0, bx + 3y + z = 0, cx + 2y + z = 0 has non-trival solution if a, b, c are in (1) AP
(2)
GP
(3)
HP
(4)
None of these
(4)
A2 = I
Sol. Answer (1) For non-trivial solutions,
a 4 1 b 3 1 0 ⇒ a 2b c 0 c 2 1 a, b, c are in A.P.
⎡ 0 0 1⎤ ⎥ ⎢ 49. Let A = ⎢ 0 1 0 ⎥ . The only correct statement about the matrix A is ⎢⎣ 1 0 0 ⎥⎦ (1) A is a zero matrix
(2)
A = (–1)I3
(3)
A–1 doesn’t exist
Sol. Answer (4) Clearly, options (1), (2) & (3) are false. Option (4) can be easily verified as
⎡ 0 0 –1⎤ ⎡ 0 0 –1⎤ ⎡ 1 0 0 ⎤ ⎢ 0 –1 0 ⎥ ⎢ 0 –1 0 ⎥ ⎢0 1 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ –1 0 0 ⎥⎦ ⎢⎣ –1 0 0 ⎥⎦ ⎢⎣0 0 1⎥⎦ 50. The largest value of a third order determinant whose elements are equal to 1 or 0 is (1) 0
(2)
2
(3)
4
(4)
6
Sol. Answer (2) 0 1 1 1 0 1 2 1 1 0
Option (2) is correct. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
232
Determinants
xk
x k +2
k 51. If y zk
y k +2
Solution of Assignment (Set-2)
x k +3
⎧ 1 1 1⎫ y k +3 =(x y )(y z )(z x ) ⎨ + + ⎬ , ⎩x y z⎭ z k +3
z k +2
then (1) k = –2
(2)
k = –1
(3)
k = 0
(4)
k = 1
Sol. Answer (2) xK yK zK
xK 2 y K 2 zK 2
x K 3 K K K y K 3 x .y .z ( x y )( y z )( z x ){ xy yz zx } K 3 z ⎧ 1 1 1⎫ ( xyz )K 1( x y )( y z )( z x ) ⎨ ⎬ ⎩x y z⎭
k+ 1 = 0 k = –1 Option (2) is correct. 52. If all elements of a third order determinant are equal to 1 or –1, then deteminant itself is (1) An odd integer
(2)
An even integer
(3) An imaginary number
(4)
A real number
Sol. Answer (2) Let third order determinant be a11 a12 a21 a22 a31 a32
a13 a23 a33
a31 a21 when aij = 1 on –1 then both a and a equals ±1. 11 11 ⎛ a21 ⎞ ⎛ a31 ⎞ ⎟ R1 and R3 R3 ⎜ ⎟ R1 . a ⎝ 11 ⎠ ⎝ a11 ⎠
Apply R2 R2 ⎜
a11 a12 a13 D 0 2 s where , , , are all 1 or –1. 0 2 2
Expand along C1 = 4a11{ – }. which is an even integer hence option (2) is correct. 53. If A is a 3×3 matrix and det (3A) = k{det(A)}, then k is (1) 9
(2)
6
(3)
1
(4)
27
Sol. Answer (4) det (3A) = k{det(A)} 33det(A) = k{det(A)} k = 27, option (4) is correct. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Determinants
233
⎡ x ⎤ ⎡ 1⎤ 54. The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system A ⎢ y ⎥ ⎢ 0 ⎥ has exactly ⎢ ⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ 0 ⎥⎦ two distinct solutions, is (1) 0
[IIT-JEE 2010] (2)
29 – 1
(3)
168
(4)
2
Sol. Answer (1)
⎡x The equation A ⎢ y ⎢ ⎢⎣ z
⎤ ⎡1⎤ ⎥ ⎢ 0 ⎥ has two distinct solutions. It should be noted here that the given equation is linear ⎥ ⎢ ⎥ ⎥⎦ ⎢⎣ 0 ⎥⎦
equation in 3 variables, which may have no solution, or unique solution or infinitely many solutions. Hence there does not exist any matrix A such that the given equation has exactly two solutions and consequently number of 3 × 3 matrices is 0. 55. Let 1 be a cube root of unity and S be the set of all non-singular matrices of the form ⎡1 ⎢ ⎢ ⎢ 2 ⎣
b⎤ ⎥ c⎥ 1⎥⎦
a 1
where each of a, b and c is either or 2. Then the number of distinct matrices in the set S is [IIT-JEE 2011] (1) 2
(2)
6
(3)
4
(4)
8
Sol. Answer (3) We have, ⎡1 ⎢ M= ⎢ ⎢ 2 ⎣
b⎤ ⎥ c⎥ 1⎥⎦
a 1
|M|=
0
a 2 1
b c 1 c
2
1
= –(a – 2 – b – c – ) + 2(ac – 2c – b – c – 1) = –(a + c) + ac2 + 1 a + c 1, ac 1 Since a, b, c are or 2 a=c If a = c = Number of ways of selecting a, b, c = 1 × 1 × 2 = 2 If a = 2, then number of ways = 1 × 1 × 1 = 2 Total number of distinct matrices in the given set S = 4. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
234
Determinants
Solution of Assignment (Set-2)
56. Let P = [aij] be a 3 × 3 matrix and let Q = [bij], where bij = 2i+jaij for 1 i, j 3. If the determinant of P is 2, then the determinant of the matrix Q is [IIT-JEE 2012] (1) 210
(2)
211
⎡ a11 a12 Given, P [aij ]33 ⎢a21 a22 ⎢ ⎢⎣a31 a32
a13 ⎤ a33 ⎥⎥ a33 ⎥⎦
(3)
212
(4)
213
Sol. Answer (4)
i j Now, Q 2 aij [bij ]
⎡ b11 b12 Q ⎢b21 b22 ⎢ ⎢⎣ b31 b32
b13 ⎤ b23 ⎥⎥ b33 ⎥⎦
⎡ 22 a11 23 a12 ⎢ 3 ⎢ 2 a21 24 a22 = ⎢ 4 5 ⎢⎣ 2 a31 2 a32
a11 a12 a21 a22 a31 a32
12
So, Q 2
24 a13 ⎤ ⎥ 25 a23 ⎥ ⎥ 26 a33 ⎥⎦
a13 a23 a33
= 213. (as |P| = 2)
SECTION - B Objective Type Questions (More than one options are correct) 1.
The values of for which the system of equations x + y + z = 1, x + 2y + 4z = , x + 4y + 10z = 2 is consistent, are given by (1) 1, 2
(2)
–1, 2
(3)
1, –2
(4)
–1, –2
Sol. Answer (1) Clearly, the determinant
1 1
1
1 0 0
1 2 4 1 1 3 0 1 4 10 1 3 9
(C2 C2 – C1, C3 C3 – C1)
therefore, the given equations are consistent, if each of the three determinants
1 2
1
1
1
1
1
1 1
1
2 4 , 1 4 , 1 2 are zero. 4 10 1 2 10 1 4 2
Solving, = 1, 2 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
2.
Determinants
235
The system of equations x – y cos + z cos 2 = 0 – x cos + y – z cos = 0 x cos 2 – y cos + z = 0 has non-trivial solution for equals 3
(1)
(2)
6
(3)
2 3
(4)
12
(4)
Sol. Answer (1, 2, 3, 4) For non-trivial solution
1
cos cos 2
cos 1 cos 0 cos 2 cos 1 on expansion, the determinant equals (1 – cos2) + cos(–cos + coscos2) + cos2(cos2 – cos2) = 1 – 2cos2 + 2cos2cos2 – cos22 = 1 – 2cos2 (1 – cos2) – cos22 = 1 – 4cos2sin2 – cos22 = 1 – 4cos2 (1 – cos2) – (2cos2 – 1)2 = 1 – (4cos2 – 4cos4) – (4cos4 – 4cos2 + 1) = 0, for all values of . 3.
If f(x) and g(x) are functions such that f(x + y) = f(x) g(y) + g(x) f(y), then
f ( ) g ( ) f ( ) f () g () f ( ) is independent of f ( ) g ( ) f ( ) (1)
(2)
(3)
Sol. Answer (1, 2, 3, 4)
f ( ) g ( ) f ( ) f () g () f ( ) f ( ) g ( ) f ( ) f ( ) g ( ) f ( )g () g ( )f () g () f ()g () g ()f () g ( ) f ( )g () g ( )f ()
= f () f ()
f ( ) g ( ) 0 = f () g () 0 f () g() 0
C3 C3 – g()C1 – f()C2
=0 Hence, independent of , , and . Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
236 4.
Determinants
Solution of Assignment (Set-2)
The digits A, B, C are such that the three digit numbers A88, 6B8, 86C are divisible by 72, then the determinant
A 6 8 B 8
8 6 is divisible by
8 C
(1) 72
(2)
144
(3)
288
(4)
1216
Sol. Answer (1, 2, 3) 3 digit numbers divisible by 72, include 288, 648 and 864. Hence, the only possibility is A = 2, B = 4, C = 4
2 6 8
8 4 6 288 8 8 4
is divisible by 72, 144 and 288.
5.
a b a b If b c b c 0 then a b b c 0 (1) a, b, c are in A.P.
(2)
a, b, c are in G.P.
(3) is a root of ax2 + 2bx + c = 0
(4)
( x ) is a factor of ax2 + 2bx + c = 0
Sol. Answer (2, 3, 4)
a b a b b c b c 0 a b b c 0 R3 R3 – (R1 + R2)
a b
b c
a b b c
0 0
0
2
(a b b c )
– (a2 + 2b + c) (b2 – ac) = 0 b2 = ac, a2 + 2b + c = 0 Hence option (2, 3, 4) are true
6.
1
bc
a(b c )
The value of the determinant 1
ca
b(a c ) doesn’t depend on
1
ab
c (a b )
(1) a
(2)
b
(3)
c
(4)
a+b+c
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Solution of Assignment (Set-2)
Determinants
237
Sol. Answer (1, 2, 3)
1 bc a(b c )
1 ca b(a c ) 1 ab c(a b ) 0 c ( b a ) c (a b ) 0 a(c b ) a(b c ) 1 ab c (a b )
(R1 R1 – R2, R2 R1 – R3)
c ( b a ) c (a b ) a(c b ) a(b c )
ac (a b )(b c )
1 1 1 1
=0 does not depend on a, b, c.
7.
⎡ 1 0 0⎤ ⎡1 0 0⎤ ⎢ ⎥ Let A ⎢0 1 1⎥ and I ⎢0 1 0 ⎥ , then ⎢ ⎥ ⎢⎣0 0 1⎥⎦ ⎢⎣0 2 4 ⎥⎦ 1 2 ( A 6 A 11I ) 6
(1) A3 – 6A2 + 11A – 6I = 0
(2)
A 1
(3) A2 is non-singular
(4)
A is singular
Sol. Answer (1, 2, 3)
1 0 0 A I 0 1 1 0 0 2 4 (1 – ) {(1 – ) (4 – ) + 2} = 0 (1 – ) (2 – 5 + 4 + 2) = 0 ( – 1) (2 – 5 + 6) = 0 3 – 52 + 6 + 5 – 2 – 6 = 0 3 – 62 + 11 – 6 = 0 Hence the characteristics equation of the matrix is A3 – 6A2 + 11A – 6 = 0 A–1 =
1 2 (A – 6A + 11I) 6
|A2| 0 |A| 0 Hence options (1, 2, 3) are correct Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
238 8.
Determinants
Solution of Assignment (Set-2)
A is a matrix of order 3 × 3. If A = A and five entries in the matrix are of one kind and remaining four are of another kind, then the maximum number of such matrices is greater than or equal to (1) 9
(2)
10
(3)
11
(4)
8
Sol. Answer (1, 2, 3, 4) Since, the matrix is symmetric, it can be of the following type
⎡a x y ⎤ A ⎢⎢ x b z ⎥⎥ ⎢⎣ y z c ⎥⎦ Case (i) : Let a = b = c = x y = z In this case total matrices = 3C1 = 3 Case (ii) : Let a = x = y b = c = z In this case total matrices = 3C2 × 3C1 = 9 Hence, maximum number of such matrices = 3 + 9 = 12
9.
⎡ 1 4 4⎤ ⎢ ⎥ If the adjoint of a 3 × 3 matrix P is ⎢2 1 7 ⎥ , then the possible value(s) of the determinant of P is (are) ⎢⎣ 1 1 3 ⎥⎦ [IIT-JEE 2012] (1) –2
(2)
–1
(3)
1
(4)
2
Sol. Answer (1, 4)
⎡ 1 4 4⎤ adj P ⎢⎢ 2 1 7 ⎥⎥ ⎢⎣ 1 1 3 ⎥⎦ |adj P| = 1(3 – 7) – 4(6 – 7) + 4(2 – 1) = –4 + 4 + 4 = 4 |adj P| = |P|2 |P|2 = 4 |P| = ±2
SECTION - C Linked Comprehension Type Questions Comprehension-I Consider a matrix A = [aij]n × n. Form the matrix A – I, being a number, real or complex. ⎡a11 – ⎢ ⎢ a21 A – I = ⎢ ... ⎢ ⎢ a ⎣ n1
a12
...
a22 –
...
...
...
an 2
...
a1n
⎤ ⎥ a2 n ⎥ ⎥ ... ⎥ ann – ⎥⎦
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Solution of Assignment (Set-2)
Determinants
239
Then det(A – I) = (–1)n[n + b1n – 1 + b2n – 2 + ... + bn]. An important theorem tells us that the matrix A satisfies the equation xn + b1xn – 1 + b2xn – 2 + ... + bn = 0. This equation
⎡1 is called the characteristic equation of A. For all the questions on this passage, take A = ⎢ ⎢⎣2 1.
4⎤ ⎥ 3 ⎥⎦
The matrix A satisfies the matrix equation (1) A2 – 4A – 5I = 0
(2)
A2 – 4A – 5 = 0
(3) A2 + 4A – 5I = 0
(4)
A2 + 4A – 5 = 0
(3)
1 ( A – 5I ) 4
Sol. Answer (1) | A I | 0 ⇒
1 4 0 2 3
(1 – ) (3 – ) – 8 = 0 2 – 4l – 5 = 0 A2 – 4A – 5I = 0 2.
Which of the following is inverse of A? (1)
1 ( A – 4I ) 5
1 ( A 4I ) 5
(2)
(4)
1 ( A 5I ) 4
Sol. Answer (1) A2 – 4A – 5I = 0 A–1(A2 – 4A – 5I) = 0 A – 4I – 5A–1 = 0 5A–1 = A – 4I 3.
A 1
1 ( A 4I ) 5
The matrix A5 – 4A4 – 7A3 + 11A2 – A – 10I, when expressed as a linear polynomial in A , is (1) A + 5I
(2)
A – 5I
(3)
–A + 5I
(4)
–A – 5I
Sol. Answer (1) A5 – 4A4 – 7A3 + 11A2 – A – 10I = A3(A2 – 4A – 5) – 2A3 + 11A2 – A – 10I = – 2A(A2 – 4A – 5I) + 3A2 – 11A – 10I = 3(4A + 5I) – 11A – 10I = A + 5I Comprehension-II Matrix theory can be applied to investigate the conditions under which a given system of linear equations possesses unique, infinite or no solutions. Consider the system of 3 non-homogeneous linear equations in 3 unknowns x+y+z=6 x + 2y + 3z = 10 x + 2y + z = and answer the questions that follow. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
240 1.
Determinants
Solution of Assignment (Set-2)
The system possesses a unique solution if (1) 3
(2)
3, 10
(3)
10
(4)
= 10
Sol. Answer (1) For unique solution
1 1 1 1 2 3 0 1 2 1 0
0 (C2 C2 – C1, C3 C3 – C2)
1 1 1 0 1 1 2
( – 2) – 1 0 3 2.
The system possesses no solutions if (1) 3, = 10
(2)
= 3, 10
(3)
= 10
(4)
= 3, 10
(3)
= 3, = 10
(4)
3, 10
(3)
8A
(4)
16A
Sol. Answer (2) For no solution
1 1 1 1 2 3 0 and 10 1 2 i.e., = 3 and 10. The last two equations must not be identical. 3.
The system possesses infinite solutions if (1) = 3, 10
(2)
3, = 10
Sol. Answer (3) For infinite solutions
1 1 1 1 2 3 0 and = 10 1 2 i.e., = 3 and = 10 The last two equations must be identical. Comprehension-III A and B are two matrices of same order 3 × 3, where
⎡1 2 3⎤ ⎢ ⎥ A = ⎢2 3 4 ⎥ , B = ⎢5 6 8 ⎥ ⎣ ⎦ 1.
⎡3 2 5 ⎤ ⎢ ⎥ ⎢2 3 8 ⎥ ⎢7 2 9 ⎥ ⎣ ⎦
The value of adj(adj A) is, (1) –A
(2)
4A
Sol. Answer (1) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
2.
Determinants
241
The value of |adj(AB)| is (1) 24
(2)
242
(3)
243
(4)
65
29
(3)
1
(4)
219
Sol. Answer (2) 3.
Value of |adj(adj(adj(adjA)))| is (1) 24
(2)
Sol. Answer (3) Comprehension-IV Let A be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. [IIT-JEE 2009] 1.
The number of matrices in A is (1) 12
(2)
6
(3)
9
(4)
3
Sol. Answer (1) The matrix is of order 3 × 3 where there are 9 elements. This has to be symmetric matrix with five of the entries as 1 and four of the entries as zero. For the matrix to be symmetric (i) either all the three elements along principal diagonal should be (1) or exactly one element along principal diagonal should be 1. The possible matrixes are as follows
⎡ 1 1 0⎤ ⎡ 1 0 1⎤ ⎡ 1 0 0⎤ ⎢ ⎥ ⎢ ⎥ A1 ⎢ 1 1 0 ⎥ , A2 ⎢0 1 0 ⎥ , A3 ⎢⎢0 1 1⎥⎥ ⎢⎣0 0 1⎥⎦ ⎢⎣ 1 0 1⎥⎦ ⎢⎣0 1 1⎥⎦ ⎡1 1 1⎤ ⎡ 1 1 0⎤ ⎡ 1 0 1⎤ ⎢ ⎥ ⎢ ⎥ B1 ⎢1 0 0 ⎥ , B2 ⎢ 1 0 1⎥ , B3 ⎢⎢0 0 1⎥⎥ ⎢⎣1 0 0 ⎥⎦ ⎢⎣0 1 0 ⎥⎦ ⎢⎣ 1 1 0 ⎥⎦ ⎡ 0 1 1⎤ ⎡0 1 0 ⎤ ⎡0 0 1⎤ ⎢ ⎥ ⎢ ⎥ C1 ⎢ 1 1 0 ⎥ , C2 ⎢ 1 1 1⎥ , C3 ⎢⎢0 1 1⎥⎥ ⎢⎣ 1 0 0 ⎥⎦ ⎢⎣0 1 0 ⎥⎦ ⎢⎣ 1 1 0 ⎥⎦ ⎡ 0 1 1⎤ ⎡0 1 0 ⎤ ⎡0 0 1⎤ ⎢ ⎥ ⎢ ⎥ D1 ⎢ 1 0 0 ⎥ , D2 ⎢ 1 0 1⎥ , D3 ⎢⎢0 0 1⎥⎥ ⎣⎢ 1 0 1⎥⎦ ⎣⎢0 1 1⎦⎥ ⎣⎢ 1 1 1⎦⎥ There are 12 such matrices. 2.
The number of matrices A in A for which the system of linear equations
⎡ x ⎤ ⎡ 1⎤ A ⎢⎢ y ⎥⎥ ⎢⎢0 ⎥⎥ ⎢⎣ z ⎥⎦ ⎢⎣0 ⎥⎦ has a unique solution, is (1) Less than 4
(2)
At least 4 but less than 7
(3) At least 7 but less than 10
(4)
At least 10
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242
Determinants
Solution of Assignment (Set-2)
Sol. Answer (2) The determinant corresponding to matrices B2, B3, C1, C3, D1 and D2 are non-zero. Therefore in these three cases, the given linear equations will have unique solution. Number of required matrices in this case is 6. 3.
The number of matrices A in A for which the system of linear equations
⎡ x ⎤ ⎡ 1⎤ A ⎢⎢ y ⎥⎥ ⎢⎢0 ⎥⎥ ⎢⎣ z ⎥⎦ ⎢⎣0 ⎥⎦ is inconsistent, is (1) 0
(2)
More than 2
(3)
2
(4)
1
Sol. Answer (2)
⎡1 1 0⎤ (i) If we take (i) A1 ⎢ 1 1 0 ⎥ ⎢ ⎥ ⎣⎢0 0 1⎥⎦ then |A1| = 0
1 1 0 x 0 1 0 1 0 , hence in this case system is inconsistent. 0 0 1 1 0 1 (ii) If we take A2 0 1 0 ⇒ | A2 | 0
1 0 1
1 0 1 x 0 1 0 1 0 , hence system is inconsistent. 0 0 1 1 0 0 (iii) If we take A3 0 1 1 ⇒ | A3 | 0 0 1 1 1 0 0 x 0 1 1 0 , Similarly y = A3 = 0. In this case we get infinite solution. 0 1 1
1 1 1 (iv) If we take B1 1 0 0 ⇒ | B1 | 0 . In this case x = y = z = 0 1 0 0 Hence infinite solution consistent system. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Determinants
243
Comprehension-V Let a, b and c be three real numbers satisfying
[a
1.
⎡1 c ] ⎢⎢8 ⎢⎣7
b
9
7⎤ 7 ⎥⎥ [0 7 ⎥⎦
2 3
0
0]
...(E)
[IIT-JEE 2011]
If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is (1) 0
(2)
12
(3)
7
(4)
6
Sol. Answer (4) From the given condition, a + 8b + 7c = 0 9a + 2b + 3c = 0 7a + 7b + 7c = 0 a+b+c=0 Also, 2a + b + c = 1 a=1 7a + b + c = 6a + (a + b + c) = 6
2.
Let be a solution of x3 – 1 = 0 with Im() > 0. If a = 2 with b and c satisfying (E), then the value of
3 a
1 b
3 c
is equal to (1) –2
(2)
2
(3)
3
(4)
–3
Sol. Answer (1) When a = 2 b + c = –2 Also, 8b + 7c = –2 b = 12 and c = –14 Thus,
3 a
1 b
3 c
=
3 2
1 12
3 14
= 1 + 3 + 32 =1–3 = –2 3.
Let b = 6, with a and c satisfying (E). If and are the roots of the quadratic equation ax2 + bx + c = 0, then
⎛ 1 1⎞ ⎜ ⎟ ⎠ n 0 ⎝
∑
(1) 6
n
(2)
7
(3)
6 7
(4)
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244
Determinants
Solution of Assignment (Set-2)
Sol. Answer (2) For b = 6 a + c = –6 a + 7c = –48 a + c = –6 when a = 1, c = –7 and are roots of the given equation ax2 + bx + c = 0, hence + = –6, = –7
1 1 6 7 n
2
⎛ 1 1⎞ 6 ⎛6⎞ ⎜ ⎟ 1 ⎜ ⎟ ...... 7 ⎝7⎠ ⎠ n 0 ⎝
∑
1 1
6 7
7
SECTION - D Assertion-Reason Type Questions
1.
⎡0 h g ⎤ Consider the matrix A ⎢⎢ h 0 f ⎥⎥ ⎢⎣g f 0 ⎥⎦ STATEMENT-1 : det A = 0. and STATEMENT-2 : The value of the determinant of a skew symmetric matrix of odd order is always zero.
Sol. Answer (1)
⎡ 0 h g ⎤ | A | ⎢⎢ h 0 f ⎥⎥ ⎢⎣g f 0 ⎥⎦ 0 h
h g
f h g 0 g
0 f
= 0 + h (0 + gf) – g(hf – 0) =0
2.
2 1 3 2 0 2 Consider the determinants 1 1 1 , 2 1 1 1 1 1
4
1
3
STATEMENT-1 : = 2. and STATEMENT-2 : The determinant formed by the cofactors of the elements of a determinant of order 3 is equal in value to the square of the original determinant. Sol. Answer (1) Clearly, ' is formed by the cofactor of ' = 2 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
3.
Determinants
245
⎡a 0 0 0 ⎤ ⎢ ⎥ STATEMENT-1 : Matrix ⎢0 b 0 0 ⎥ is a diagonal matrix. ⎢0 0 c 0 ⎥ ⎣ ⎦
and STATEMENT-2 : A = [aij] is a square matrix such that aij = 0 i j, then A is called diagonal matrix. Sol. Answer (4) A is not diagonal matrix ∵ A is not square matrix Option (4) is correct.
4.
⎡1 1 1⎤ ⎢ ⎥ STATEMENT-1 : Inverse of A= ⎢1 2 3 ⎥ does not exist. ⎢1 4 7 ⎥ ⎣ ⎦
and STATEMENT-2 : If any matrix is singular, then its inverse does not exist. Sol. Answer (1) ∵ |A| = 0, A–1 does not exists. Option (1) correct. 5.
STATEMENT-1 : The system of equations x + ky + 3z = 0, 3x + ky – 2z = 0, 2x + 3y – 4z = 0, possesses a non-trival solution, then value of k is
31 . 2
and STATEMENT-2 : Three linear equations in x, y, z can never, hence exactly two solutions. Sol. Answer (4)
∵
1 k 3 3 k 2 0 2 3 4
20k + 33 – 22k = 0
k
33 2
Option (4) is correct.
(1+x )21 (1+x )22 6.
STATEMENT-1 : f (x )= (1+x )31 (1+x )32
(1+x )41 (1+x )42
(1+x )23 (1+x )33 , then coefficient of x in f(x) is zero. (1+x )43
and STATEMENT-2 : If F(x) = A0 + A1x + ...... + Anxn, then A1 = F(0), dash denotes differential coefficient. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
246
Determinants
Solution of Assignment (Set-2)
Sol. Answer (1) f'(x) = a1 + 2a2x + or f'(0) = a1
21 22 23 1 1 1 1 1 1 a1 1 1 1 31 32 33 1 1 1 0 1 1 1 1 1 1 41 42 43
Option (1) is correct. 7.
Consider the system of equations x – 2y + 3z = –1 –x + y – 2z = k x – 3y + 4z = 1 STATEMENT-1 : The system of equations has no solution for k 3. and
1
3
STATEMENT-2 : The determinant 1 2
1
4
1 k 0, for k 3. 1
[IIT-JEE 2008]
Sol. Answer (1) The given system of equations can be expressed as
⎡ 1 2 3 ⎤ ⎡ x ⎤ ⎡ 1⎤ ⎢ 1 3 4 ⎥ ⎢ y ⎥ ⎢ 1 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 1 1 2⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ k ⎥⎦ Applying R2 R2 R1, R3 R1 R3
⎡ 1 2 3 ⎤ ~ ⎢⎢0 1 1⎥⎥ ⎣⎢0 1 1⎥⎦
⎡ x ⎤ ⎡ 1 ⎤ ⎢y ⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ z ⎦⎥ ⎢⎣k 1⎥⎦
⎡ 1 2 3 ⎤ ~ ⎢⎢0 1 1⎥⎥ ⎢⎣0 0 0 ⎥⎦
⎡ x ⎤ ⎡ 1 ⎤ ⎢y ⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ by R3 R3 – R2 ⎢⎣ z ⎥⎦ ⎢⎣k 3 ⎥⎦
When k 3 , the given system of equations has no solution. Statement 1 is true. Clearly statement–2 is also true as it is rearrangement of rows and columns of
⎡ 1 2 3 ⎤ ⎢ 1 3 4 ⎥ ⎢ ⎥ ⎢⎣ 1 1 2 ⎥⎦ Hence option (1) is correct. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Determinants
247
SECTION - E Matrix-Match Type Questions 1.
Let f(x) denote the determinant x2 f (x) x 2 1
2x
1 x2
x 1
1
–1
x –1
x
on expansion f(x) is seen to be a 4th degree polynomial given by f(x) = a0x4 + a1x3 + a2x2 + a3x + a4. Using differentiation of determinant or otherwise, match the values of the quantities on the left to those on the right. Column-I
Column-II
(A) a4
(p)
–1
(B) a3
(q)
1
(C) a0
(r)
–3
(D) a12 + a1 + 1
(s)
3
Sol. Answer A(p), B(q), C(r), D(s)
0 0 1 If x = 0, a4 1 1 1 1 0 1 1 f(x) = x2[(x2 – 1) – 1] – 2x [(x2 + 1)(x – 1) – x] + (1 + x2) [(1 + x2)(–1) – x(x + 1)] = x2(x2 – 2) – 2x(x3 – x2 – 1) – (1 + x2)2 – x(1 + x + x2 + x3) = (x4 – 2x2) – (2x4 – 2x3 – 2x) – (1 + 2x2 + x4) – (x + x2 + x3 + x4) = – 3x4 + x3 – 5x2 + x – 1 a4 = –1, a3 = 1, a0 = –3, a1 = 1
a12 + a1 + 1 = 3 2.
If A is a non-singular matrix of order n × n, then match the following Column-I
Column-II
(A) (adj A)–1
(p)
(B) adj (KA)
(q)
(C) adj (adj A)
(r)
(D) adj (A–1)
(s) (t)
kn (adj A) A A
|A|n–2 A adj adj A A
n –1
kn–1 (adj A)
Sol. Answer A(q, s), B(t), C(r), D(q, s) We know that (adj A)(A) = A(adj A) = |A|In and adj(adj A) = |A|n – 2A; |A| 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
248 3.
Determinants
Solution of Assignment (Set-2)
The entries in a 3 × 3 determinant are either 1 or –1, then match the following Column-I
Column-II
(A) Total number of such determinants
(p)
4
(B) The maximum value of such a determinant
(q)
3
(C) The maximum value of trace of such determinant
(r)
512
(D) The minimum value of such determinant
(s)
Zero
(t)
–4
(u)
–3
Sol. Answer A(r), B(p), C(q), D(t) (A) 29 = 512 (B) For maximum value the determinant will be
1
1 1
1 1 1 = 1(1 + 1) +1(1 + 1) + 0 = 4 1 1 1 (C) Maximum trace = 1 + 1 + 1 = 3 (D) Minimum value = –4
SECTION - F Integer Answer Type Questions 1.
Let |A| = |aij|3 × 3 0. Each element aij is multiplied by ki–j. Let |B| be the resulting determinant, where k1|A| + k2|B| = 0, then k1 + k2 is
Sol. Answer (0)
a11 a12 a13 A a21 a22 a23 a31 a32 a33 a11 B k a21 k 2 a31
k 1a12 k 2a13 a22 k a32
k 1 a23 a33
k 2a11 ka12 a13
1 2 k a21 ka22 a23 A k3 2 k a31 ka32 a33
|A| = |B| |A| – |B| = 0 Comparing it with k1|A| + k2|B| = 0 We get k1 + k2 = 0 x 2 6x 5
2.
If
2x 6 2 2 x 2 5 x 9 4 x 5 4 Ax 3 Bx 2 Cx D , then A + B + 2C is equal to 6 x 2 4 x 6 12 x 4 12
Sol. Answer (0) As the value of the determinant is zero A=B=C=D=0 A + B + 2C + 3D = 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
3.
Determinants
249
Let A = [aij]n × n, n is odd. Then det((A – AT)2009) is equal to
Sol. Answer (0) ∵ A – AT is skew symmetric. (A – AT)2009 is also skew symmetric det (A – AT)2009 = 0, as determinant value of every skew symmetric matrix of odd order is 0.
1 4.
3cosθ
If = sinθ
1
1
1 3cosθ , then the
sinθ
1
1 [|maximum value of – minimum value of |3] is equal 1000
to_________. Sol. Answer (1) ∵
= 1{1 – 3sincos} – 3cos{sin – 3cos} + 1{sin2 – 1} = 1 – 6sincos + 9cos2 + sin2 – 1
= (sin – 3sin)2
10 sin 3 cos 10
0 (sin – 3cos)2 10 0 10
1 [|maximum value of – minimum value of |]3. 100
= 5.
1 1000 [| (10 0) |3 ] 1 1000 1000
Given 2x + 4y + z = 1, x + 2y + z = 2, x + y – z = 3, then one of the value of such that the given system of equations has no solution, is_______.
Sol. Answer (0) Since, given system has no solution, = 0 and any one amongst x, y, z is non-zero. 2 4 1 Let 2 1 1 1
=0
42 – 3 = 0 (4 – 3) = 0 0,
3 4
1 4 1 and x 2 2 1 0 3 1
6.
Let (x, y, z) be points with integer coordinates satisfying the system of homogeneous equations: 3x – y – z = 0 –3x + z = 0 –3x + 2y + z = 0 Then the number of such points for which x2 + y2 + z2 100 is
[IIT-JEE 2009]
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250
Determinants
Solution of Assignment (Set-2)
Sol. Answer (7) We have 3x – y – z = 0
...(i)
–3x + z = 0
...(ii)
–3x + 2y + z = 0
...(iii)
Applying (i) and (ii) we get y=0
...(iv)
Also 3x = z Points satisfying x2 + y2 + z2 100 with integral coordinates can be (0, 0, 0), (1, 0, 3), (2, 0, 6), (3, 0, 9), (–1, 0, –3), (–2, 0, –6), (–3, 0, –9) Hence 7 such points exist.
7.
Let be the complex number cos z 1
2
z 2
1
1
z
2
2 2 i sin . Then the number of distinct complex numbers z satisfying 3 3
0 is equal to
[IIT-JEE 2010]
Sol. Answer (1) ⎡1 ⎢ Let A ⎢ ⎢ 2 ⎣⎢
2 1
2 ⎤ ⎥ 1⎥ ⎥ ⎦⎥
⎡0 0 0 ⎤ A ⎢⎢0 0 0 ⎥⎥ , Tr ( A) 0, | A | 0 ⎢⎣0 0 0 ⎥⎦ 2
A3 = 0
z 1
2
z 2
1
2
1
z
| A zI | 0
z3 = 0 z = 0, the number of z satisfying the given equation is 1. 8.
The number of all possible values of , where 0 < < , for which the system of equations (y + z)cos3 = (xyz)sin3
x sin3
2cos3 2sin3 y z
(xyz)sin3 = (y + 2z)cos3 + ysin3 have a solution (x0, y0, z0) with y0z0 0, is
[IIT-JEE 2010]
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Solution of Assignment (Set-2)
Determinants
251
Sol. Answer (3) Given equation can be written as
x sin3
cos3 cos3 0 y z
...(i)
x sin3
2cos3 2sin3 0 y z
...(ii)
x sin3
2 1 cos3 (cos3 sin3) 0 y z
...(iii)
Equations (ii) and (iii) imply 2sin3 = cos3 + sin3 sin3 = cos3
9.
tan3 1 ⇒ 3
or
5 9 , , 4 4 4
5 9 , , 12 12 12
Let k be a positive real number and let ⎡ 2k 1 2 k ⎢ A⎢2 k 1 ⎢ ⎣⎢ 2 k 2k
⎡ 0 2 k⎤ ⎥ ⎢ 2k ⎥ and B ⎢1 2k ⎥ ⎢ 1 ⎦⎥ ⎢⎣ k
2k 1 0 2 k
k ⎤ ⎥ 2 k ⎥. ⎥ 0 ⎦⎥
If det (adj A) + det(adj B) = 106, then [k] is equal to
[IIT-JEE 2010]
[Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k] Sol. Answer (4) |A| = (2k + 1)3, |B| = 0 det (adj A) det (adj B) = (2k + 1)6 = 106 k
9 . 2
[k] = 4 10. Let M be a 3 × 3 matrix satisfying
⎡0 ⎤ ⎡ 1⎤ ⎡1⎤ ⎡1⎤ ⎡1⎤ ⎡ 0 ⎤ M ⎢⎢ 1⎥⎥ ⎢⎢ 2 ⎥⎥ , M ⎢⎢ 1⎥⎥ ⎢⎢ 1 ⎥⎥ , and M ⎢⎢1⎥⎥ ⎢⎢ 0 ⎥⎥ ⎢⎣0 ⎥⎦ ⎢⎣ 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1⎥⎦ ⎢⎣1⎥⎦ ⎢⎣12⎥⎦ Then the sum of the diagonal entries of M is
[IIT-JEE 2011]
Sol. Answer (9)
⎡ a11 Let M ⎢a21 ⎢ ⎢⎣a31
a12 a22 a32
a13 ⎤ a23 ⎥⎥ be the given matrix. a33 ⎥⎦
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252
Determinants
Solution of Assignment (Set-2)
Using the given conditions, we have
⎡ a11 ⎢a ⎢ 21 ⎢⎣a31
a12 a22 a32
a13 ⎤ ⎡0 ⎤ ⎡ 1⎤ a23 ⎥⎥ ⎢⎢ 1⎥⎥ ⎢⎢ 2 ⎥⎥ a33 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 3 ⎥⎦
a12 = –1 a22 = 2 a32 = 3
⎡1⎤ ⎡1⎤ Also, M ⎢ 1⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1⎥⎦ a11 – a12 = 1 a21 – a22 = 1 a31 – a32 = –1 Using above equations, we shall get a11 = 0
⎡1⎤ ⎡ 0 ⎤ Moreover, M ⎢1⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣1⎥⎦ ⎢⎣12⎥⎦ a11 + a12 + a13 = 0 a21 + a22 + a23 = 0 a31 + a32 + a33 = 12 Using above results, we get a33 = 7 Finally, the sum of elements of leading diagonals = a11 + a22 + a33 =0+2+7 =9
SECTION - G Multiple True-False Type Question 1.
⎡ 1 1⎤ The matrices which commute with A = ⎢ ⎥ in case of multiplication ⎣0 1⎦
STATEMENT-1 : Are always singular. STATEMENT-2 : Are always non-singular. STATEMENT-3 : Are always symmetric. (1) F F F
(2) T T F
(3)
TTT
(4)
TFT
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Solution of Assignment (Set-2)
Determinants
253
Sol. Answer (1) Let B ⎡ x y ⎤ commute with A ⎢a b ⎥ ⎣ ⎦ AB = BA ⎡x a y b⎤ ⎡x x y ⎤ i.e., ⎢ a b ⎥⎦ ⎢⎣ a a b ⎥⎦ ⎣
x + a = x, y + b = x + y, b = a + b a = 0 and b = x
⎡x y ⎤ B⎢ ⎥ ; option (1) is correct. ⎣0 x ⎦
SECTION - H Aakash Challengers Questions 1.
If the system of equation ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 has non-trivial solution then
find the value of
bc a 2 ca b 2 ab c 2
ca b 2 ab c 2 bc a 2
ab c 2 bc a 2 ca b 2
.
Sol. Answer (0) For non-trivial solution
a b c A b c a 0 c a b And the determinant bc a 2
ca b 2
ab c 2
ca b 2
ab c 2
bc a 2
2
2
2
ab c
bc a
ca b
is determinant of cofactor matrix of A.
Hence its value = 0
2.
If a, b, c are real such that a2 + b2 + c2 = 1 then show that
ax by c bx ay cx a bx ay ax by c cy b 0 cx a cy b ax by c
will represent a straight line. Sol. Applying C1 aC1 + bC2 + cC3 (a 2 b 2 c 2 ) x
bx ay
(a 2 b 2 c 2 )y ax by c (a 2 b 2 c 2 )
cy b
cx a cy b ax by c
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254
Determinants
Solution of Assignment (Set-2)
bx ay
x
cx a
y ax by c
cy b
cy b
1
∵ a2 b2 c 2 1
ax by c
C2 C2 – bC1, C3 C3 – cC1
x
ay
a
y ax c 1
b ax by
cy
R3 xR1 + yR2 + R3 x
ay
a
ax c b
y x2 y 2 1
0
0
Expand w.r.t. R3
= (x2 + y2 + 1) [aby + a2x + ac] = a(x2 + y2 + 1) (ax + by + c)
3.
= 0 ax + by + c = 0 represents a straight line.
Prove that
(i)
(ii)
1+ a
b
c
a
1+ b
c
a
b
1+ c
1+ a
1
1
1
1+ b
1
1
1
1+ c
Sol. (i) Let =
=1+a+b+c
1 1 1⎞ ⎛ = abc ⎜ 1 + + + ⎟ , where abc 0 a b c⎠ ⎝
b c b c 1 a 1 a b c a c c (Performing C1 C1 + C2 + C3) 1 a b c 1 b 1 b a b b 1 c 1 a b c 1 c
1 b c c = (1 a b c ) 1 1 b 1 b 1 c
⎛ Taking 1 a b c common ⎞ ⎜ ⎟ ⎝ from the 1st column ⎠
1 b c = (1 a b c ) 0 1 0 (Performing R2 R2 – R1, R3 R3 – R1) 0 0 1
= (1 + a + b + c) (1.1 – 0) = 1 + a + b + c, expanding along first row. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Determinants
255
(ii) It follows as a corollary of (i)
1 a 1 1 abc 1 1 b 1 = 1 1 1 c
1 1 b 1 c
1 a
1 a 1 1 c
1 b
1 a 1 b 1
⎛ Taking a, b, c common from ⎞ ⎜ ⎟ ⎝ R1, R2 , R3 respectively ⎠
1 c
1 1 1⎞ ⎛ = abc ⎜ 1 ⎟ , from (i) part. a b c⎠ ⎝
If some of a, b, c, say a, is zero, we rewrite the left hand side as abc + ab + bc + ca, and the determinant then evaluates to bc. 4.
Show that
2bc – a 2
c2
b2
bc – a 2
ca – b 2
ab – c 2
c2
2ac – b 2
a2
= ca – b 2
ab – c 2
bc – a 2
b2
a2
2ab – c 2
ab – c 2
bc – a 2
ca – b 2
Sol. A direct evaluation of the determinants involved is not at all attractive. One is required to think out of the box to discover what may be the idea behind the solution. Write LHS as product of two determinants 2bc – a 2 c2 b2
c2 2ac – b 2 a2
b2 – a 2 bc bc – ab ab c 2 a2 – ab c 2 ba – b 2 ac ca 2 2ab – c – ac ca b 2 – bc a 2 bc
– ac b 2 ca – bc bc a 2 – c 2 ab ab
a b c –a c b = b c a – b a c (Row-by-row multiplication) c a b –c b a a b c a b c a b c = b c a b c a b c a c a b c a b c a b
2bc – a 2 c2 i.e. b2
c2 2ca – b 2 a2
b2 a b c a2 b c a c a b 2ab – c 2
bc – a 2 2 Now how to relate the determinant ca – b ab – c 2
2
2
…(A)
ca – b 2 ab – c 2 bc – a 2
2
ab – c 2 a b c 2 bc – a to the determinant b c a ? c a b ca – b 2
a b c Consider = b c a c a b
The determinant formed by cofactors of elements of is Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
256
Determinants
Solution of Assignment (Set-2)
bc – a 2 2 1 = ca – b ab – c 2
ca – b 2 ab – c 2 bc – a 2
ab – c 2 bc – a 2 ca – b 2
and we know that 1 = 2 bc – a 2 2 Thus ca – b ab – c 2
ca – b 2 ab – c 2 bc – a 2
ab – c 2 a b c bc – a 2 b c a c a b ca – b 2
2
…(B)
From (A) and (B) we have 2bc – a 2 c2 b2
5.
Let
c2 2ca – b 2 a2
b2 bc – a 2 2 a ca – b 2 2 2ab – c ab – c 2
ca – b 2 ab – c 2 bc – a 2
ab – c 2 bc – a 2 ca – b 2
x 2 + 3x
x –1
x +3
x +1
–2x
x – 4 = ax 4 + bx 3 + cx 2 + dx + k
x–3
x+4
3x
where a, b, c, d, k are independent of x. Find the value of d. Sol. Rather than expand the determinant, we differentiate both sides w.r.t. x and then put x = 0 to obtain the value of d. x 2 3x (x) = x 1 x3
x –1 x 3 –2 x x – 4 ax 4 bx 3 cx 2 dx k x 4 3x
Differentiating w.r.t. x 2x 3 1 1 x 2 3x 1 ´(x) = x 1 –2 x x – 4 x – 3 x 4 3x x–3
x – 1 x 3 x 2 3x –2 1 x 1 1 x 4 3x
x –1 x 3 –2x x – 4 1 3
= 4ax3 + 3bx2 + 2cx + d Set x = 0 on both sides to obtain
d=
3 1 1 0 –1 3 0 –1 3 1 0 –4 1 –2 1 1 0 –4 –3 4 0 –3 4 0 1 1 3
= {3(0 + 16) – 1(0 – 12) + 1(4 – 0)} + {0 + 1(0 + 3) + 3(4 – 6)} + {0 + 1(3 + 4) + 3(1 – 0)} = (48 + 12 + 4) + (3 – 6) + (7 + 3) = 64 – 3 + 10 = 71 Hence the value of d is 71. Remarks : (1) If the value of k(the constant term) is to be found out, just put x = 0 in the determinant. (2) If the value of b is to found out, then divide both sides by x4 and write the polynomial in t, where t=
1 . Now differentiate, as in the example above, w.r.t. t to find the desired coefficient. x
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Solution of Assignment (Set-2)
1
2
0
3 , value of co-factor to its minor of element –3 is
3
0
0
6.
In the determinant 1 2
Determinants
(1) –1
(2)
0
(3)
1
(4)
257
2
Sol. Answer (1) Ratio of cofactor to its minor of element –3, which is in 3rd row and 2nd column = (–1)3+2 = –1
7.
1
5
loge e
5
5 is equal to
log10 10 5
e
(1)
(2)
e
(3)
1
(4)
0
(3)
a+b+c
(4)
0
(4)
–1
Sol. Answer (4) 1 5
1 5
5 51 1
5 0
1 5
e
e
1 1 1 1
x + 2 x +3 x + a
8.
If a, b, c are in A.P., then x + 4 x +5 x + b is x + 6 x +7 x + c
(1) x – (a + b + c)
9x2 + a + b + c
(2)
Sol. Answer (4) Apply C2 C2 – C1 x 2 1 x a
x 2 1 x a
R R R
2 2 1 x 4 1 x b R R R
2
0 ba
x6 1 x c
4
0 c a
3
3
1
x 2 1 x a
2
0 b a 1[2(c a ) 4(b a )]
4
0 c a
= 2{2b – c – a} = 0
9.
{∵ 2b = a + c}
1+ax
1+bx
1+cx
If 1+a1x
1+b1x
1+c1x = A0 + A1x + A2x2 + A 3x3, then A0 is
1+a2 x 1+b2 x 1+c2 x
(1) abc
(2)
0
(3)
1
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258
Determinants
Solution of Assignment (Set-2)
Sol. Answer (2) 1 1 1
Put x = 0 on both sides A0 1 1 1 0 1 1 1
x 3 7
10. If –9 is a root of equation 2 x
2 = 0, then other two roots are
7 6 x
(1) 2, 7
(2)
–2, 7
(3)
2, –7
(4)
–2, –7
–1
(4)
0
Sol. Answer (1) x 3 7 2 x
1 1 1
2 0 ⇒ ( x 9) 2 x
7 6 x
2 0
7 6 x
[R1 R1 + R2 + R3] (x + 9){x2 – 12 – (2x – 14) + (12 – 7x)} = 0 (x + 9)[(x – 2)(x – 7)] = 0 x = 2, 7 x
11. The coefficient of x in f (x )= 1 x
1+ sin x
cos x
log(1+x )
2
2
1+x
2
where
0
–1 < x 1, is (1) 1
(2)
–2
(3)
Sol. Answer (2) f(x) = x{–2(1 + x2)} – (1 + sinx)(–2x2) + cosx{1 + x2 – x2log(1 + x)} = –2x – 2x3 + 2x2 + 2x2sinx + cosx{1 + x2 – x2log(1 + x)} Hence, –2 12. If adjB = A, |P| = |Q| = 1, then adj(Q–1BP–1) equals (1) PQ
(2)
QAP
(3)
PAQ
(4)
PA–1Q
(3)
Nilpotent
(4)
Involutary
Sol. Answer (3) adj (Q 1BP 1 ) adjQ 1.adjB.adjP 1
⎡ ⎢ 13. The matrix A = ⎢ ⎢ ⎢ ⎣
(1) Unitary
1 2 1 2
P Q A. PAQ | P | |Q |
1 ⎤ ⎥ 2⎥ is 1 ⎥ ⎥ 2⎦
(2)
Orthogonal
Sol. Answer (2) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Determinants
259
⎡a 2 ⎤ 14. The matrix A = ⎢ ⎥ is singular if ⎣2 4 ⎦
(1) a 1
(2)
a=1
(3)
a=0
(4)
a = –1
k = –1
(3)
k±1
(4)
k = 0
Any odd integer
(4)
Any integer
Sol. Answer (2) A is singular |A| = 0 4a – 4 = 0 a = 1
15.
⎡1 0 k ⎤ ⎢ ⎥ A = ⎢ 2 1 3 ⎥ is invertible for ⎢k 0 1⎥ ⎣ ⎦
(1) k = 1
(2)
Sol. Answer (3) A is invertible |A| 0 K ±1 16. Let a, b, c be such that (b + c) 0 if a
a+1 a 1
a+1
b+1
c 1
b
b+1 b 1 +
a 1
b 1
c +1
c
c 1 c +1
( 1)
n +2
=0
a ( 1) b ( 1) c n +1
n
then value of n is (1) 0
(2)
Any even integer
(3)
Sol. Answer (3) a 1 a 1
a
a 1 b 1 c 1
b b 1 b 1 ( 1) a 1 b 1 c 1 n
c 1 c 1
c a
b
a
a 1 a 1
c
a 1 a 1
a
b b 1 b 1 ( 1) b 1 b 1 b n
c
c 1 c 1
a
a 1 a 1
c 1 c 1 a 1
a
c a 1
b b 1 b 1 ( 1)n 1 b 1 b b 1 {C2 C3} c
c 1 c 1
a n 2
[1 ( 1)
c 1
c
c 1
a 1 a 1
] b b 1 b 1 any odd integer c
c 1 c 1
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260
Determinants
Solution of Assignment (Set-2)
x2 1 + x3
x 17. If y
y2
1 + y 3 = 0 and x, y, z are all distinct, then xyz equals
z
z2
1 + z3
(1) –1
(2)
1
(3)
0
(4)
3
Sol. Answer (1) x
x2 1 x3
y
y 2 1 y 3 0
z
z2
1 z3
x
x2 1
y
y 2 1 xyz 1 y
y2 0
z
z2 1
z2
1 x
(1 xyz ) 1 y 1 z
1 x 1 z
x2
x2 y 2 0 xyz = –1 z2
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