Q. No. 1 109 102 95

6 13 20

1 -6 -13

is equals to

Option 1 Constant other than zero

Option 2 Zero

Option 3 100

Option 4 -1997

Correct Answer 2

Explanation C + CC C -

2

→→→→

1 32 2

109 0 95

6 0 20 = 0

1 0 -13

Q. No. 2 18 40 89

= 40 89 198

89 198 440

∆∆∆∆ is equals to

Option 1 1

Option 2 -1

Option 3 Zero

Option 4 = 2∆

Correct Answer 2

Explanation RR R - 2R +

2

→→→→

13 3 2

18 40 89

= 40 89 198

0 0 -0.5

(((( ))))1

= - 18 89 - 40 402

× ×× ×× ×× ×

= -1

Q. No. 3

If ( )1ω ω ω ω ≠ is a cube root of unity, then

1 1+i+

1- i -1 -1

-i -i+ -1 -1

ω ωω ωω ωω ω

ωωωω

ωωωω

2 2

2=

Option 1 0

Option 2 1

Option 3 i

Option 4 ωωωω

Correct Answer 1

Explanation 1 1+ i+

1- i -1 -1 =

-i -i+ -1 -1

ω ωω ωω ωω ω

ωωωω

ωωωω

2 2

2

(((( ))))R R - R +R→→→→2 2 1 3

1 1+ i+

0 0 0 = 0

-i -i+ -1 -1

ω ωω ωω ωω ω

ωωωω

2 2

Q. No. 4

The value of

1

1

1

ω ωω ωω ωω ω

ω ωω ωω ωω ω

ω ωω ωω ωω ω

2

2

2

, ωωωω being a cube root of unity, is

Option 1 0

Option 2 1

Option 3 ωωωω2

Option 4 ωωωω

Correct Answer 1

Explanation R R +R +R→→→→1 1 2 3

0 0 0

1 = 0

1

ω ωω ωω ωω ω

ω ωω ωω ωω ω

2

2

since 1 + + = 0 ω ωω ωω ωω ω 2

Q. No. 5 If a + b + c = 0, one root of :

-

-

-

a x c b

c b x a

b a c x

= 0 is

Option 1 x = 1

Option 2 x = 2

Option 3 x = a2 + b

2 + c

2

Option 4 x = 0

Correct Answer 4

Explanation R R + R +R→→→→1 1 2 3

+ + - + + - + + -

- =

-

a b c x a b c x a b c x

c b x a

b a c x

( )1 1 1

= + + - - = 0

-

a b c x c b x a

b a c x

+ + + = 0

= 0

⇒⇒⇒⇒

⇒⇒⇒⇒

a b c x

x

Q. No. 6 0 - -

= - 0 -

- - 0

∆∆∆∆

p q a b

q p x y

b a y x

is equals to

Option 1 0

Option 2 a + b

Option 3 x + y

Option 4 p + q

Correct Answer 1

Explanation 0 - -

= - 0 -

- - 0

∆∆∆∆

p q a b

q p x y

b a y x

skew symmetric matrix of odd order has determinant = 0

Q. No. 7

The determinant

+ + +

+ + +

+ + +

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

b c c a a b

b c c a a b

b c c a a b

=

Option 1 1 1 1

2 2 2

3 3 3

a b c

a b c

a b c

Option 2

2

1 1 1

2 2 2

3 3 3

a b c

a b c

a b c

Option 3

3

1 1 1

2 2 2

3 3 3

a b c

a b c

a b c

Option 4 None of these

Correct Answer 2

Explanation c c + c + c→→→→1 1 2 3

(((( ))))(((( ))))(((( ))))

2 + + + +

2 + + + +

2 + + + +

1 1 1 1 1 1 1

2 2 2 2 2 2 2

3 3 3 3 3 3 3

a b c c a a b

a b c c a a b

a b c c a a b

+ + + +

= 2 + + + +

+ + + +

1 1 1 1 1 1 1

2 2 2 2 2 2 2

3 3 3 3 3 3 3

a b c c a a b

a b c a c a b

a b c a c a b

c c - c and c c - c→ →→ →→ →→ →2 2 1 3 3 1

+ + + -

= 2 + + + -

+ + + -

1 1 1 1 1

2 2 2 2 2

3 3 3 3 3

a b c b c

a b c b c

a b c b c

- -

= 2 - - = 2

- -

1 2 3 1 2 3

1 2 3 1 2 3

1 3 3 1 2 3

a b a a b a

b b b b b b

c b c c b c

Q. No. 8 If αααα , ββββ & γγγγ are the roots of the equation x3+ px + q = 0 then the value of the

determinant

α β γα β γα β γα β γ

β γ αβ γ αβ γ αβ γ α

γ α βγ α βγ α βγ α β

=

Option 1 p

Option 2 q

Option 3 p2 - 2q

Option 4 None

Correct Answer 4

Explanation are roots of x3+ + = 0α,β, γα,β, γα,β, γα,β, γ px q

= 0α,β, γα,β, γα,β, γα,β, γ

α β γα β γα β γα β γ

β γ αβ γ αβ γ αβ γ α

γ α βγ α βγ α βγ α β

c c + c + c→→→→1 1 2 3

+ + 0

+ + = 0 = 0

+ + 0

α β γ β γ β γα β γ β γ β γα β γ β γ β γα β γ β γ β γ

α β γ γ α γ αα β γ γ α γ αα β γ γ α γ αα β γ γ α γ α

α β γ α β α βα β γ α β α βα β γ α β α βα β γ α β α β

Q. No. 9

Let f ( )θθθθ = f

cos cos sin -sin

cos sin sin cos then 6

sin -cos 0

θ θ θ θθ θ θ θθ θ θ θθ θ θ θππππ

θ θ θ θθ θ θ θθ θ θ θθ θ θ θ

θ θθ θθ θθ θ

2

2 =

Option 1 0

Option 2 1

Option 3 2

Option 4 None

Correct Answer 2

Explanation

(((( ))))

cos cos sin -sin

= cos sin sin cos

sin -cos

θ θ θ θθ θ θ θθ θ θ θθ θ θ θ

θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θ

θ θ θθ θ θθ θ θθ θ θ

2

2f

3 1 - 3

4 2 2

1 1 1=

6 2 4 2

1 - 30

2 2

ππππ

f

Now solving

4 3 1 1 - 3 - 3 10 + - 0 - - - = 1

3 4 2 4 2 4 8

Q. No. 10

If a b c d+ + + =µ µ µµ µ µµ µ µµ µ µ3 2

3 +1 -1

- 3 -2 + 2

+ 3 - 4 5

µ µ µµ µ µµ µ µµ µ µ

µ µ µµ µ µµ µ µµ µ µ

µ µ µµ µ µµ µ µµ µ µ

be an identity in µµµµ , where a, b, c are

constants, then the value of d is

Option 1 5

Option 2 -6

Option 3 9

Option 4 0

Correct Answer 2

Explanation 3 +1 -1

+ + c + d = - 3 -2 + 2 put = 0

+ 3 - 4 5

µ µ µµ µ µµ µ µµ µ µ

µ µ µ µ µ µ µµ µ µ µ µ µ µµ µ µ µ µ µ µµ µ µ µ µ µ µ

µ µ µµ µ µµ µ µµ µ µ

3 2a b

0 1 -1

d = -3 0 2

3 -4 0

(((( )))) [[[[ ]]]]= -1 -6 -1 12 = -6

Q. No. 11

If P =

2 1 0

3 1 2

5 2 3

, then

2 2 0

9 6 6

5 4 3

is equal to

Option 1 2P

Option 2 3P

Option 3 5P

Option 4 6P

Correct Answer 4

Explanation 2 1 0

P = 3 1 2

5 2 3

Multiplying column II by 2

2 2 0

2P = 3 2 2

5 4 3

Multiplying row II by 3

(((( ))))2 2 0

3 2P = 9 6 6 = 6P

5 4 3

Q. No. 12

Given a, b, c are in A.P then determinant

x x x a

x x x b

x x x c

+1 + 2 +

+ 2 + 3 +

+ 3 + 4 +

in its simplied form is :

Option 1 x3 + 3ax + 7c

Option 2 0

Option 3 15

Option 4 10x2 + 5x + 2c

Correct Answer 2

Explanation If a, b, c, are in A.P then

2b = a + c

a+ cb - = 0

2

⇒⇒⇒⇒

R +RR R -

2

→→→→

1 32 2

+1 + 2 ++1 + 2 +

2 + +0 0 + = 0 0 0 = 0

2+ 3 + 4 +

+ 3 + 4 +

x x x ax x x a

x a cx b -

x x x cx x x c

Q. No. 13 11 12 13

12 13 14

13 14 15

is equals to

Option 1 1

Option 2 0

Option 3 -1

Option 4 67

Correct Answer 2

Explanation R +RR R -

2

→→→→

1 32 2

11 12 13

0 0 0 - = 0

13 14 15

Q. No. 14 If every element of a third order determinant of value ∆∆∆∆ is multiplied by 5, then the

value of new determinant is

Option 1 ∆∆∆∆

Option 2 5 ∆∆∆∆

Option 3 25 ∆∆∆∆

Option 4 125 ∆∆∆∆

Correct Answer 4

Explanation Let matrix be

and let its determinant A

a b c

d e f

g h i

Multiplying every element by 5 gives

5 5 5

5 5 5

5 5 5

a b c

d e f

g h i

Taking out 5 cannon row 1, 2 and 3

5 = 125 A =125

∆∆∆∆

3

a b c

d e f

g h i

Q. No. 15

If

y z x x

y z x y

z z x y

+

+

+

= k (xyz), then k is equal to

Option 1 4

Option 2 -4

Option 3 Zero

Option 4 None of these

Correct Answer 1

Explanation

(((( ))))+

+ =

+

y z x x

y x z y k xyz

z z x y

(((( ))))R R - R +R→→→→1 1 2 3

(((( ))))0 -2 -2

+ =

+

y y

y z x y k xyz

z z y x

Calculating determinant

(((( )))) (((( ))))(((( )))) (((( ))))0 + + - 3 + 23 + -2

z x y x y y xy yz (((( ))))-2 - 3 - 32

y yz x

= 4 xyz k = 4.

Q. No. 16 If X and Y two matrices are such that X - Y =

3 2

-1 0

and X + Y = 1 -2

3 4

then Y is

given by

Option 1 2 0

1 2

Option 2 -1 -2

3 4

Option 3 -1 -2

2 2

Option 4 None of these

Correct Answer 3

Explanation 32X - Y = (i)

-10

1 -2X + Y = (ii)

3 4

Equation (2) Equation (1)

-2 -42Y =

4 4

-1 -2Y =

2 2

⇒⇒⇒⇒

Q. No. 17

If A = k

1 -3 2

2 5

4 2 1

is a singular matrix, then k is equal to

Option 1 -1

Option 2 8

Option 3 4

Option 4 -8

Correct Answer 4

Explanation 1 -3 2

A = 2 k 5

4 2 1

Singular means (A) = 0

(A) = 1 (k -10) + 3 (2 - 20) + 2 (4 - 4k)

0 = k - 10 + 6 - 60 + 8 - 8k

= -8k

Q. No. 18 A matrix A = a i j of order 2 ×××× 3 whose elements are such that ai j = i + j is -

Option 1 2 3 4

3 4 5

Option 2 2 3 4

5 4 3

Option 3 2 3 4

5 5 3

Option 4 None of these

Correct Answer 1

Explanation [[[[ ]]]]A = j1 2×3a

2 3 4=

3 4 5

1+1 1+2 1+3

2+1 2+2 2+3

a a a

a a a

Q. No. 19 In the following , singular matrix is -

Option 1 2 3

1 3

Option 2 3 2

2 3

Option 3 1 2

1 0

Option 4 2 6

4 12

Correct Answer 4

Explanation 2 3= A A 0

1 3

⇒⇒⇒⇒ ≠≠≠≠

3 2= B B 0

2 3

⇒⇒⇒⇒ ≠≠≠≠

1 2= C C 0

1 0

⇒⇒⇒⇒ ≠≠≠≠

2 6= D D = 0

4 12

⇒⇒⇒⇒

Singular

Q. No. 20 If A + B =

7 4

8 9

and A – B = 1 2

0 3

then the value of A is -

Option 1 3 1

4 3

Option 2 4 3

4 6

Option 3 6 2

8 6

Option 4 7 6

8 6

Correct Answer 2

Explanation (((( ))))

7 4A +B = i

8 9

(((( ))))1 2

A -B = ii0 3

Adding equation (i) and equation (ii)

8 6A =

8 12

4 3A =

4 6

Q. No. 21

If [[[[ ]]]]x2 3 1

1 2 0 4 2 1

0 3 2 -1

××××

= o, then the value of x is

Option 1 -1

Option 2 0

Option 3 1

Option 4 2

Correct Answer 1

Explanation

[[[[ ]]]]2 3 1

1 2 0 4 1 1 = 0

0 3 2 -1××××

× ×× ×× ×× ×

x

x1 3

3 3 3 1

[[[[ ]]]]2 9 + 4 5+ 2 1 = 0

-1××××

x

x x1 3

2 +9 + 4 - 5-2 = 0⇒⇒⇒⇒ x x x

4 = -4x

= -1x

Q. No. 22 If

1 2

2 1

x

y

= 5

4

then

Option 1 x = 2, y = 1

Option 2 x = 1, y = 2

Option 3 x = 3, y = 2

Option 4 x = 2, y = 3

Correct Answer 2

Explanation 1 2 5=

2 1 4

x

y

+ 2 5=

2 + 4

x y

x y

Solving we get x = 1, y = 2.

Q. No. 23 If A =

2 -1

-1 2

and A2 - 4A - nI = 0, then n is equal to -

Option 1 3

Option 2 -3

Option 3 1/3

Option 4 -1/3

Correct Answer 2

Explanation 2 -1A =

-1 2

2 -1 2 -1 5 -4A = =

-1 2 -1 2 -4 5

2

A = 4A - I= 0x2

5 -4 8 -4 0 0 0- - =

-4 5 -4 8 0 0 0

⇒⇒⇒⇒

x

x

Comparing we get x = -3

Q. No. 24 If A =

-1 2

3 -4

then element a21 of A2 is -

Option 1 22

Option 2 -15

Option 3 -10

Option 4 7

Correct Answer 2

Explanation -1 2 -1 2A =

3 -4 3 -4

2

7 -10

=-15 22

q = -1521

Q. No. 25 If E (((( ))))θθθθ =

cos sin

-sin cos

θ θθ θθ θθ θ θ θθ θθ θθ θ

, then value of E (((( ))))αααα . E (((( ))))ββββ is -

Option 1 E (0O)

Option 2 E ( 90O)

Option 3 E (((( ))))+θ βθ βθ βθ β

Option 4 E (((( ))))-α βα βα βα β

Correct Answer 3

Explanation (((( ))))

cos sinE =

-sin cos

θ θθ θθ θθ θ θθθθ θ θθ θθ θθ θ

(((( )))) (((( ))))cos sin cos sin

E E =-sin cos -sin cos

α α β βα α β βα α β βα α β β α βα βα βα β α α β βα α β βα α β βα α β β

cos cos - sin sin cos sin + cos sin

=-sin cos - cos sin -sin sin + cos cos

α β α β α β β αα β α β α β β αα β α β α β β αα β α β α β β α α β α β α β α βα β α β α β α βα β α β α β α βα β α β α β α β

(((( )))) (((( ))))(((( )))) (((( ))))

cos + sin +=

-sin + cos +

α β α βα β α βα β α βα β α β

α β α βα β α βα β α βα β α β

(((( ))))= E +α βα βα βα β

Q. No. 26

The root of the equation [[[[ ]]]]x 1 2

0 1 1

1 0 1

1 1 0

x

-1

1

= o is

Option 1 1/3

Option 2 -1/3

Option 3 0

Option 4 1

Correct Answer 1

Explanation 0 1 1

12 1 0 1 -1 = 0

1 1 0 1

x

xB×3

3×3 3×1

[[[[ ]]]]3 + 2 +1 -1 = 0

1

x

x x

(((( ))))3 + + 2 + +1 = 0x x x

1=

3x

Q. No. 27 If A, B, C are matrices of order 1 ×××× 3, 3 ×××× 3 and 3 ×××× 1 respectively, the order of ABC will

be -

Option 1 3 ×××× 3

Option 2 1 ×××× 3

Option 3 1 ×××× 1

Option 4 3 ×××× 1

Correct Answer 3

Explanation A B C1×3 3×3 3×1

(((( )))) (((( ))))AB C = ABC3×11×3 1×1

Q. No. 28 If A =

1 2

3 0

and B = 3 4

1 6

then (AB)T equals -

Option 1 5 16

9 16

Option 2 5 9

16 12

Option 3 5 9

4 3

Option 4 None of these

Correct Answer 2

Explanation 1 2 3 4A = and B,

3 0 1 6

(((( )))) (((( ))))AB = first we find ABT

5 16AB =

9 12

(((( ))))5 9

AB =16 12

T

Q. No. 29 If A =

2 -1

-7 4

and B = 4 1

7 2

then BTA

T is equals to -

Option 1 1 0

0 1

Option 2 1 1

1 1

Option 3 0 1

1 0

Option 4 1 0

0 0

Correct Answer 1

Explanation 2 -1 4 1A = and B

-7 4 7 2

(((( ))))B A = ABTT T

2 -1 4 1 1 0AB = =

-7 4 7 2 0 1

(((( ))))1 0

AB = = B A0 1

T T T

Q. No. 30 If A =

-1 7

2 3

, then skew-symmetric part of A is -

Option 1 9-1

2

-93

2

Option 2 -5-0

2

50

2

Option 3 -9-1

2

93

2

Option 4 50

2

-50

2

Correct Answer 4

Explanation -1 7A =

2 3

A - Askew symmetric part

2

T

-1 2A =

7 3

50

A - A 2

52- 0

2

⇒⇒⇒⇒

T

Q. No. 31

Matrix

0 5 -7

-5 0 11

7 -11 0

is a -

Option 1 Diagonal matrix

Option 2 Upper triangular matrix

Option 3 Skew-symmetric matrix

Option 4 symmetric matrix

Correct Answer 3

Explanation 0 -5 7

-5 0 11 Clearly this is skew symmetric matrix

7 -11 0

→→→→

Q. No. 32 If A and B are square matrices of same order, then which of the following is skew-

symmetric -

Option 1 A + A

2

T

Option 2 A +B

2

T T

Option 3 A -B

2

T T

Option 4 B -B

2

T

Correct Answer 4

Explanation B -BClearly is skew symmetric

2

T

Q. No. 33 If A is symmetric as well as skew symmetric matrix, then -

Option 1 A is a diagonal matrix

Option 2 A is null matrix

Option 3 A is a unit matrix

Option 4 A is a triangular matrix

Correct Answer 2

Explanation Only possible if A is full matrix

Q. No. 34 If A and B are 3 3×××× matrices, then AB = o implies :

Option 1 A = O and B = O

Option 2 A = 0 and B = 0

Option 3 Either A = 0 or B = 0

Option 4 A = O or B = O

Correct Answer 3

Explanation If A and B are matrix then AB = 0

22 1 2If A = ,B =

33 -1-2

A 0, B 0, but AB = 0≠ ≠≠ ≠≠ ≠≠ ≠

So A 0 or B 0 is not necessary≠ ≠≠ ≠≠ ≠≠ ≠

either A = 0 or B = 0 fullows.

Because if AB = 0 AB = 0⇒⇒⇒⇒

A B = 0⇒⇒⇒⇒

either one of them has to be zero⇒⇒⇒⇒

Q. No. 35 IF A =

0 1

1 0

, then A4 =

Option 1 1 0

0 1

Option 2 1 1

0 0

Option 3 0 0

1 1

Option 4 0 1

1 0

Correct Answer 1

Explanation 0 1A =

1 0

0 1 0 1 1 0A = =

1 0 1 0 0 1

2

(((( ))))1 0

A = A =0 1

24 2

Q. No. 36

The order of [[[[ ]]]]x y z

a h g

h b f

g f c

x

y

z

is

Option 1 3 1××××

Option 2 1 1××××

Option 3 1 3××××

Option 4 3 3××××

Correct Answer 2

Explanation

[[[[ ]]]], ,

a h g x

x y z h b f y

g f c z1×3

3×3 3×1

A B C1×3 3×3 3×1

(((( )))) [[[[ ]]]]AB C = D3×11×3 1×1

Q. No. 37 If A = a i j is a skew-symmetric matrix of order n, then aii =

Option 1 0 for some i

Option 2 0 for all i = 1,2,……n

Option 3 1 for some i

Option 4 1 for all i = 1,2,……,n.

Correct Answer 2

Explanation Diagonal elements of skew symmetric matrix are always zero.

Q. No. 38 If A = a i j is a square matrix of order n n×××× and k is a scalar, then kA =

Option 1 kn

A

Option 2 k A

Option 3 kn-1

A

Option 4 None of these

Correct Answer 1

Explanation [[[[ ]]]]A = ji is square matrix of order KA means××××a x x

Multiply all rows by k

so KA becomes k An

Q. No. 39 If A =

p q

q p-

, B = r s

s r-

then

Option 1 AB = BA

Option 2 AB ≠≠≠≠ BA

Option 3 AB = -BA

Option 4 None of these

Correct Answer 1

Explanation A = B,

-p q r s

-q p -s r

AB =

-p q r s

-q p -s r

+

=+

pr - qs ps qr

-qr - ps -qs pr

+BA = ,

+

r s p q pr - qs ps qr

-s r -q p -qr - ps -qs pr

= AB

Q. No. 40 If A =

0 1

0 0

and a and b are arbitrary constants then (aI + bA)2=

Option 1 a2I + abA

Option 2 a2I + 2abA

Option 3 a2I + b

2A

Option 4 None of these

Correct Answer 2

Explanation 0 1= A

0 0

(((( ))))0 0 b

I+ bA =0 0 0

++++

aa

a

2

2

2= ,

0 0

a b a b a ab

a a a a

2

2

1 0 0 12

0 1 0I + 2 A

0

+ →+ →+ →+ →

a bab a a

2 2

Q. No. 41 If A =

1 2

3 0

; B = 3 4

1 6

then which of the following statements is true -

Option 1 AB = BA

Option 2 A2 = B

Option 3 (AB)

T =

5 9

16 12

Option 4 None of these

Correct Answer 3

Explanation 1 2 3 4A = and B =

3 0 1 6

1 2 3 4 5 16AB = =

3 0 1 6 9 12

(((( ))))5 9

AB =16 12

T

Q. No. 42

If A =

1 3 5

3 5 1

5 1 3

, then adj. A is equal to -

Option 1 14 -4 -22

-4 -22 14

-22 14 -4

Option 2 -14 4 22

4 22 -14

22 -14 4

Option 3 14 4 -22

4 -22 -14

-22 -14 -4

Option 4 None of these

Correct Answer 1

Explanation 1 3 5

A = 3 5 1

5 1 3

For ad joint we in st calculate cofactors and them tale its Transpose

14 4 -22

Cofactor of A = -4 -22 14

-22 14 -4

Transpose of cofactors is called ad joint.

Q. No. 43

If A =

1 2 3

0 3 1

2 1 2

, then A (adj A) equals -

Option 1 9 0 0

0 9 0

0 0 9

Option 2 9 0 0

- 0 9 0

0 0 9

Option 3 0 0 9

0 9 0

9 0 0

Option 4 None of these

Correct Answer 3

Explanation (((( )))) (((( ))))A adj A = A I

(((( ))))A = -g

(((( ))))1 0 0 9 0 0

A adj A = -9 0 1 0 = - 0 9 0

0 0 1 0 0 9

Q. No. 44 If A and B are invertible matrices of the order n, then (AB)-1

is equals to

Option 1 AB-1

Option 2 A-1

B

Option 3 B-1

A-1

Option 4 A-1

B-1

Correct Answer 3

Explanation (((( ))))AB = B A for every invatible matrix-1 -1 -1

(((( ))))AB = AB = I-1

(((( ))))AB AAB = B-1 -1 -1

(((( ))))AB A = B-1 -1

(((( ))))AB = B A-1 -1 -1

Q. No. 45 If A =

2 3

5 -2

, then 19 A-1

is equals to

Option 1 A’

Option 2 2A

Option 3 1

2A

Option 4 A

Correct Answer 2

Explanation 2 3A =

5 -2

A = -4 -15 = -19

A = adj A T-1

[[[[ ]]]]-2 -3-1

= = A-5 219

Q. No. 46 If A is an invertible matrix of order n, then the determinant of Adj. A =

Option 1 An

Option 2 A+1n

Option 3 A-1n

Option 4 A +2n

Correct Answer 3

Explanation Adj A = An-1

Q. No. 47 -6 5

-7 6

-1

=

Option 1 -6 5

-7 6

Option 2 6 -5

-7 6

Option 3 6 5

7 6

Option 4 6 -5

7 -6

Correct Answer 1

Explanation -6 5 6 -5-1=

-7 6 7 -6-1

-1

-6 5

=-7 6

Q. No. 48 Inverse matrix of

2 -3

-4 2

is -

Option 1 2 31-

4 28

Option 2 2 41-

3 28

Option 3 2 31

4 28

Option 4 2 3

4 2

Correct Answer 1

Explanation 2 -3inverse of = A

-4 2

(((( ))))2 31 1

adj A = -4 2det A 8

Q. No. 49

If A = 0 -1 2

2 -2 0

, B =

0 1

1 0

1 1

and M = AB, then M-1

is equal to -

Option 1 2 -2

2 1

Option 2 1 1

3 3

-1 1

3 6

Option 3 1 -1

3 3

1 1

3 6

Option 4 1 -1

3 3

-1 1

3 6

Correct Answer 3

Explanation 0 10 -1 2

A = and -B = 1 02 -2 0

1 1

0 10 -1 2

M= AB 1 02 -2 0

1 1

2×3

1 2=

-2 2

2×2

(((( ))))2 21

M = AB =+2 16

-1-1

1 1-

3 3=

1 1

3 6

Q. No. 50 -tan

12

tan1

2

θθθθ

θθθθ

tan1

2

-tan1

2

θθθθ

θθθθ

-1

is equals to

Option 1 sin -cos

cos sin

θ θθ θθ θθ θ θ θθ θθ θθ θ

Option 2 cos sin

-sin cos

θ θθ θθ θθ θ θ θθ θθ θθ θ

Option 3 cos -sin

sin cos

θ θθ θθ θθ θ θ θθ θθ θθ θ

Option 4 None of these

Correct Answer 3

Explanation -tan tan1 1

2 2

tan -tan1 1

2 2

θ θθ θθ θθ θ

θ θθ θθ θθ θ

-1

-tan -tan1 1

12 2

tan +tan1+ tan1 1

22 2

θ θθ θθ θθ θ

θθθθθ θθ θθ θθ θ

2

1 - tan -2tancos -sin2 2

= cos =sin cos2

2tan 1- tan2 2

θ θθ θθ θθ θ θ θθ θθ θθ θ θθθθ θ θ θ θθ θ θ θθ θ θ θθ θ θ θ

2

2

2

Q. No. 51 If A =

1 2

3 -5

,B =1 0

0 2

and X is a matrix such that A = BX, then X equal -

Option 1 -2 41

3 52

Option 2 2 41

3 -52

Option 3 2 4

3 -5

Option 4 None of these

Correct Answer 2

Explanation 1 2 1 0A = and B =

3 -5 0 2

A =B××××

B A = ××××-1

2 01B =

0 12

-1

2 0 1 2 2 41 1B A = =

0 1 3 -5 3 -52 2

-1

Q. No. 52

Matrix

-1 4

-3 0 1

-1 1 2

λλλλ

is not invertible if -

Option 1 λλλλ = -15

Option 2 λλλλ = -17

Option 3 λλλλ = -16

Option 4 λλλλ = -18

Correct Answer 2

Explanation

(((( ))))-1 4

1A = -3 0 1 is invertible if A = 0

2-1 1 2

λλλλ

(((( )))) (((( )))) [[[[ ]]]] [[[[ ]]]]A = -1 +1 -6 +1 + 4 -3 = 0λλλλ

= - - 5-12 = 0λλλλ

= -17λλλλ

Q. No. 53

If A =

1 2 3

5 0 4

2 6 7

then adj A is equal to -

Option 1 -24 4 8

4 1 2

8 11 -11

Option 2 -24 4 8

4 1 11

30 -2 -10

Option 3 -24 4 8

-27 1 11

30 -2 -10

Option 4 None of these

Correct Answer 3

Explanation 1 2 3

A = 5 0 4

2 6 7

-24 4 8

adj A = -27 1 11

30 -2 -10

Q. No. 54 If A =

1 0

1 1

then A-n

is equal to -

Option 1

n

1 0

1

Option 2

n

1 0

- -1

Option 3

n

1 0

- 1

Option 4 None of these

Correct Answer 3

Explanation 1 0A =

1 1

1 0 1 0 1 01

A = A =-1 1 -1 1 -1 11

-1 -2

1 0

=-2 1

1 0A =

-3 1

-3

1 0A =

-4 1

-4

1 0A -

- 1

x x

-

Q. No. 55 The system of equations kx + y+z =1, x+ ky +z = k and x+ y +kz = k2 have no solution if k

equals

Option 1 0

Option 2 1

Option 3 -1

Option 4 -2

Correct Answer 4

Explanation + + = 1kx y z

+ + =x ky z k

+ + =x y k k2 2

1 1 1

1 1 =

1 1

k x

k y k

k z k2

A = B××××

For no soluti A = 0 i.on e. = 0∆∆∆∆

At least one of , , is non zero∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆x y z

1 1

1 1 = 0

1 1

k

k

k

[[[[ ]]]] (((( ))))-1 -1 -1 + 1 - = 0 ⇒⇒⇒⇒ k k k k2

(((( )))) (((( ))))-1 +1 -1-1 = 0 k k k

(((( ))))-1 + k -2 = 0 k k

2

(((( )))) (((( )))) (((( ))))-1 k + 2 k1 = 0k

= 1,-2k

For = 1 = = = 0∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆∆ ∆ ∆k x y 2

so = -2 is solution.k

Q. No. 56 With the help of matrices, the solution of the equations 3x + y +2z = 3 , 2x -3y -z=-3,

x +2y +z = 4 is given by

Option 1 x = 1, y = 2, z = -1

Option 2 x = -1, y = 2, z = 1

Option 3 x = 1, y = -2, z = -1

Option 4 x = -1, y = -2, z = 1

Correct Answer 1

Explanation 3 + +2 = 3x y z

2 + 3 + = -3x y z

+ 2 + = 4x y z

3 1 2 3

2 -3 -1 = -3

1 2 1

x

y

z y

A =B××××

= A B×××× -1

3 1 2 3

= 2 -3 -1 -3

1 2 1

x

y

z y

-1

1

= 2

-1

Q. No. 57 The system of linear equation x + y + z =2, 2x + y - z = 3, 3x + 2y + kz = 4 has a unique

solution if

Option 1 k 0≠≠≠≠

Option 2 k-1 1< << << << <

Option 3 k-2 2< << << << <

Option 4 k = 0

Correct Answer 1

Explanation + + = 2x y z

2 + + = 3x y z

3 + 2 +k = 4x y z

1 1 1 2

2 1 -1 = 3

3 2 k 4

x

y

z

A = B××××

(((( ))))Unique solution A = 0∆ ≠∆ ≠∆ ≠∆ ≠

(((( )))) [[[[ ]]]] [[[[ ]]]]= 0 + 2 - 2 + 3 +1 1 0∆∆∆∆ ⇒⇒⇒⇒ ≠≠≠≠k k

0≠≠≠≠k

Q. No. 58 The value of λλλλ and µµµµ for which the system of equation x + y + z = 6 , x + 2y + 3z = 10

and x + 2y + λλλλ z = µµµµ have no solution are

Option 1 = 3, = 10λ µλ µλ µλ µ

Option 2 = 3, 10λ µ ≠λ µ ≠λ µ ≠λ µ ≠

Option 3 3, = 10λ ≠ µλ ≠ µλ ≠ µλ ≠ µ

Option 4 3, 10λ ≠ µ ≠λ ≠ µ ≠λ ≠ µ ≠λ ≠ µ ≠

Correct Answer 2

Explanation + + = 6x y z

+ 2 + 3 = 10x y z

+2 + 3 =λ µλ µλ µλ µx y

1 1 1 6

1 2 3 = 10

1 2

λ µλ µλ µλ µ

x

y

z

A =B××××

A = 0 and at lestFor no solut one ofi ,on , , 0∆ ∆ ∆ ∆ ≠∆ ∆ ∆ ∆ ≠∆ ∆ ∆ ∆ ≠∆ ∆ ∆ ∆ ≠x y z

(((( )))) [[[[ ]]]] [[[[ ]]]]1 2 -6 -1 - 3 +1 0 = 0λ λλ λλ λλ λ

= 3λλλλ

1 1 6

1 2 10 0

1 2

≠≠≠≠ µµµµ

R R -R→→→→3 3 2

1 1 6

1 2 10 0

0 0 -10

≠≠≠≠ µµµµ

(((( ))))-10 0 10µ ≠µ ≠µ ≠µ ≠ ⇒⇒⇒⇒ µ ≠µ ≠µ ≠µ ≠ =

Q. No. 59 Consider the system of equations a1x + b1y +c1z = 0, a2x + b2y +c2z = 0, a3x + b3y +c3z = 0

If

a b c

a b c

a b c

1 1 1

2 2 2

3 3 3

= 0 then the system has

Option 1 More than two solution

Option 2 Only non trivial solution

Option 3 No solution

Option 4 Only trivial solution (0, 0, 0)

Correct Answer 1

Explanation Since = 0, = 0, = 0, = 0∆ ∆ ∆ ∆∆ ∆ ∆ ∆∆ ∆ ∆ ∆∆ ∆ ∆ ∆x y z

So we have solutions [infinite solutions]

Q. No. 60 Solution of

x y z+ 3 - 2 = 0

x y z2 - + 4 = 0

x y z-11 + 14 = 0 is

Option 1 x y z= = =

8 -10 7λλλλ

Option 2 x y z= = =

-10 8 7λλλλ

Option 3 x y z= = =

7 8 -10λλλλ

Option 4 None of these

Correct Answer 2

Explanation5 + 3 -22 = 0x y

2 - + 4 = 0x y

- +14 = 0x y

From equation (2) and (3)

= = =30 24 -21

λλλλx -y z

= = =-10 8 7

λλλλx y z