If |adj(adj A)| = |A|⁹, then the order of the square matrix A is

(a)

3

(b)

4

(c)

2

(d)

5

If A is a 3 \(\times\) 3 non-singular matrix such that AA^T = A^TA and B = A^-1A^T, then BB^T = 

(a)

A

(b)

B

(c)

I₃

(d)

B^T

If A = \(\left[ \begin{matrix} 3 & 5 \\ 1 & 2 \end{matrix} \right] \), B = adj A and C = 3A, then \(\frac { \left| adjB \right| }{ \left| C \right| } \) = 

(a)

\(\frac { 1 }{ 3 } \)

(b)

\(\frac { 1 }{ 9 } \)

(c)

\(\frac { 1 }{ 4 } \)

(d)

1

If A\(\left[ \begin{matrix} 1 & -2 \\ 1 & 4 \end{matrix} \right] =\left[ \begin{matrix} 6 & 0 \\ 0 & 6 \end{matrix} \right] \), then A = 

(a)

\(\left[ \begin{matrix} 1 & -2 \\ 1 & 4 \end{matrix} \right] \)

(b)

\(\left[ \begin{matrix} 1 & 2 \\ -1 & 4 \end{matrix} \right] \)

(c)

\(\left[ \begin{matrix} 4 & 2 \\ -1 & 1 \end{matrix} \right] \)

(d)

\(\left[ \begin{matrix} 4 & -1 \\ 2 & 1 \end{matrix} \right] \)

If A = \(\left[ \begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix} \right] \), then 9I₂ - A = 

(a)

A^-1

(b)

\(\frac { { A }^{ -1 } }{ 2 } \)

(c)

3A^-1

(d)

2A^-1

If A = \(\left[ \begin{matrix} 2 & 0 \\ 1 & 5 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 1 & 4 \\ 2 & 0 \end{matrix} \right] \) then |adj (AB)| = 

(a)

-40

(b)

-80

(c)

-60

(d)

-20

If P = \(\left[ \begin{matrix} 1 & x & 0 \\ 1 & 3 & 0 \\ 2 & 4 & -2 \end{matrix} \right] \) is the adjoint of 3 × 3 matrix A and |A| = 4, then x is

(a)

15

(b)

12

(c)

14

(d)

11

If A = \(\left[ \begin{matrix} 3 & 1 & -1 \\ 2 & -2 & 0 \\ 1 & 2 & -1 \end{matrix} \right] \) and A^-1 = \(\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right] \) then the value of a₂₃ is

(a)

0

(b)

-2

(c)

-3

(d)

-1

If A, B and C are invertible matrices of some order, then which one of the following is not true?

(a)

adj A = |A|A^-1

(b)

adj(AB) = (adj A)(adj B)

(c)

det A^-1 = (det A)^-1

(d)

(ABC)^-1 = C^-1B^-1A^-1

If (AB)^-1 = \(\left[ \begin{matrix} 12 & -17 \\ -19 & 27 \end{matrix} \right] \) and A^-1 = \(\left[ \begin{matrix} 1 & -1 \\ -2 & 3 \end{matrix} \right] \), then B^-1 = 

(a)

\(\left[ \begin{matrix} 2 & -5 \\ -3 & 8 \end{matrix} \right] \)

(b)

\(\left[ \begin{matrix} 8 & 5 \\ 3 & 2 \end{matrix} \right] \)

(c)

\(\left[ \begin{matrix} 3 & 1 \\ 2 & 1 \end{matrix} \right] \)

(d)

\(\left[ \begin{matrix} 8 & -5 \\ -3 & 2 \end{matrix} \right] \)

If A^TA⁻¹ is symmetric, then A² =

(a)

A^-1

(b)

(A^T)²

(c)

A^T

(d)

(A^-1)²

If A is a non-singular matrix such that A^-1 = \(\left[ \begin{matrix} 5 & 3 \\ -2 & -1 \end{matrix} \right] \), then (A^T)⁻¹ =

(a)

\(\left[ \begin{matrix} -5 & 3 \\ 2 & 1 \end{matrix} \right] \)

(b)

\(\left[ \begin{matrix} 5 & 3 \\ -2 & -1 \end{matrix} \right] \)

(c)

\(\left[ \begin{matrix} -1 & -3 \\ 2 & 5 \end{matrix} \right] \)

(d)

\(\left[ \begin{matrix} 5 & -2 \\ 3 & -1 \end{matrix} \right] \)

If A = \(\left[ \begin{matrix} \frac { 3 }{ 5 } & \frac { 4 }{ 5 } \\ x & \frac { 3 }{ 5 } \end{matrix} \right] \) and A^T = A⁻¹, then the value of x is

(a)

\(\frac { -4 }{ 5 } \)

(b)

\(\frac { -3 }{ 5 } \)

(c)

\(\frac { 3 }{ 5 } \)

(d)

\(\frac { 4 }{ 5 } \)

If A = \(\left[ \begin{matrix} 1 & \tan { \frac { \theta }{ 2 } } \\ -\tan { \frac { \theta }{ 2 } } & 1 \end{matrix} \right] \) and AB = I₂, then B = 

(a)

\(\left( \cos ^{ 2 }{ \frac { \theta }{ 2 } } \right) A\)

(b)

\(\left( \cos ^{ 2 }{ \frac { \theta }{ 2 } } \right) { A }^{ T }\)

(c)

\(\left( \cos ^{ 2 }{ \theta } \right) I\)

(d)

(Sin²\(\frac { \theta }{ 2 } \))A

If A = \(\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \) and A(adj A) =  \(\left[ \begin{matrix} k & 0 \\ 0 & k \end{matrix} \right] \), then k =

(a)

0

(b)

sin θ

(c)

cos θ

(d)

1

If A = \(\left[ \begin{matrix} 2 & 3 \\ 5 & -2 \end{matrix} \right] \) be such that λA⁻¹ = A, then λ is

(a)

17

(b)

14

(c)

19

(d)

21

If adj A = \(\left[ \begin{matrix} 2 & 3 \\ 4 & -1 \end{matrix} \right] \) and adj B = \(\left[ \begin{matrix} 1 & -2 \\ -3 & 1 \end{matrix} \right] \) then adj (AB) is

(a)

\(\left[ \begin{matrix} -7 & -1 \\ 7 & -9 \end{matrix} \right] \)

(b)

\(\left[ \begin{matrix} -6 & 5 \\ -2 & -10 \end{matrix} \right] \)

(c)

\(\left[ \begin{matrix} -7 & 7 \\ -1 & -9 \end{matrix} \right] \)

(d)

\(\left[ \begin{matrix} -6 & -2 \\ 5 & -10 \end{matrix} \right] \)

The rank of the matrix \(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ -2 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 6 \\ -3 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -4 \end{matrix} \end{matrix} \right] \) is

(a)

1

(b)

2

(c)

4

(d)

3

If x^a y^b = e^m, x^c y^d = eⁿ, Δ₁ = \(\left| \begin{matrix} m & b \\ n & d \end{matrix} \right| \), Δ₂ = \(\left| \begin{matrix} a & m \\ c & n \end{matrix} \right| \), Δ₃ = \(\left| \begin{matrix} a & b \\ c & d \end{matrix} \right| \), then the values of x and y are respectively,

(a)

e^{(Δ₂  / Δ₁)}, e^{(Δ₃ / Δ₁)}

(b)

log (Δ₁ / Δ₃), log (Δ₂ / Δ₃)

(c)

log (Δ₂ / Δ₁), log(Δ₃ / Δ₁)

(d)

e^{(Δ₁ / Δ₃)},e^{(Δ₂ / Δ₃)}

Which of the following is/are correct?
(i) Adjoint of a symmetric matrix is also a symmetric matrix.
(ii) Adjoint of a diagonal matrix is also a diagonal matrix.
(iii) If A is a square matrix of order n and λ is a scalar, then adj(λA) = λⁿ adj(A).
(iv) A(adjA) = (adjA)A = |A| I

(a)

Only (i)

(b)

(ii) and (iii)

(c)

(iii) and (iv)

(d)

(i), (ii) and (iv)

If \(\rho\) (A) = \(\rho\)([A| B]), then the system AX = B of linear equations is

(a)

consistent and has a unique solution

(b)

consistent

(c)

consistent and has infinitely many solution

(d)

inconsistent

If 0 ≤ θ  ≤ π and the system of equations x + (sinθ)y - (cosθ)z = 0, (cosθ)x - y +z = 0, (sinθ)x + y - z = 0 has a non-trivial solution then θ is

(a)

\(\frac { 2\pi }{ 3 } \)

(b)

\(\frac { 3\pi }{ 4 } \)

(c)

\(\frac { 5\pi }{ 6 } \)

(d)

\(\frac { \pi }{ 4 } \)

The augmented matrix of a system of linear equations is \(\left[\begin{array}{cccc} 1 & 2 & 7 & 3 \\ 0 & 1 & 4 & 6 \\ 0 & 0 & \lambda-7 & \mu+5 \end{array}\right]\). The system has infinitely many solutions if

(a)

\(\lambda=7, \mu \neq-5\)

(b)

\(\lambda=-7, \mu=5\)

(c)

\(\lambda \neq 7, \mu \neq-5\)

(d)

\(\lambda=7, \mu=-5\)

Let A = \(\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \) and 4B = \(\left[ \begin{matrix} 3 & 1 & -1 \\ 1 & 3 & x \\ -1 & 1 & 3 \end{matrix} \right] \). If B is the inverse of A, then the value of x is

(a)

2

(b)

4

(c)

3

(d)

1

If A = \(\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{matrix} \right] \), then adj(adj A) is

(a)

\(\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{matrix} \right] \)

(b)

\(\left[ \begin{matrix} 6 & -6 & 8 \\ 4 & -6 & 8 \\ 0 & -2 & 2 \end{matrix} \right] \)

(c)

\(\left[ \begin{matrix} -3 & 3 & -4 \\ -2 & 3 & -4 \\ 0 & 1 & -1 \end{matrix} \right] \)

(d)

\(\left[ \begin{matrix} 3 & -3 & 4 \\ 0 & -1 & 1 \\ 2 & -3 & 4 \end{matrix} \right] \)