JEE Mathematics Matrices and Determinants MCQs Set 03

Question. The system of equations 2x + y = 4, 3x + 2y = 2, x + y = 2 have
(a) no solution
(b) one solution
(c) two solutions
(d) infinitely many solutions
Answer: (a) no solution
Solution: 2x + y = 4
3x + 2y = 2
x + y = 2
\( D = \begin{vmatrix} 2 & 1 & 0 \\ 3 & 2 & 0 \\ 1 & 1 & 0 \end{vmatrix} = 0 \)
\( D_3 = \begin{vmatrix} 2 & 1 & 4 \\ 3 & 2 & 2 \\ 1 & 1 & 2 \end{vmatrix} \neq 0 \) Hence no solution

Question. Let A be a square matrix. Then which of the following is not a symmetric matrix
(a) A + A'
(b) A'A
(c) AA'
(d) A - A'
Answer: (d) A - A'
Solution: Given A is a square matrix,
So let \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \)
\( A' = \begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \end{bmatrix} \)
\( P = A + A' = \begin{bmatrix} 2a & b + d & c + g \\ d + b & 2e & f + h \\ g + c & f + h & 2i \end{bmatrix} \)
\( P^T = P \Rightarrow \text{symmetric} \)
Similarly A'A & AA' also come out to be symmetric
Now \( Q = A - A' = \begin{bmatrix} 0 & b - d & c - g \\ d - b & 0 & f - h \\ g - c & h - f & 0 \end{bmatrix} \)
\( Q^T = -Q \Rightarrow \text{Skew symmetric} \)

Question. If \( \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ -2 \end{bmatrix} = [0] \) then x is
(a) -1/2
(b) 1/2
(c) 1
(d) -1
Answer: (b) 1/2
Solution: \( \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ -2 \end{bmatrix} = [0] \)
\( \begin{bmatrix} 1 + 0 + 0 & 3 + 5x + 3 & 2 + x + 2 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ -2 \end{bmatrix} = [0] \)
\( \begin{bmatrix} 1 & 5x + 6 & x + 4 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ -2 \end{bmatrix} = [0] \)
\( [x + 5x + 6 - 2x - 8] = 0 \)
\( 4x - 2 = 0 \Rightarrow x = 1/2 \)

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Question. If A and B are two matrices such that AB = B and BA = A, then \( A^2 + B^2 \) equal to
(a) 2 AB
(b) 2BA
(c) A + B
(d) AB
Answer: (c) A + B
Solution: Given AB = B, BA = A
AB = B
Premultiply with 'B',
\( B(AB) = B^2 \Rightarrow (BA) B = B^2 \Rightarrow AB = B^2 \) ....(1)
Now, BA = A
Premultiply with 'A',
\( A(BA) = A^2 \Rightarrow (AB) A = A^2 \Rightarrow BA = A^2 \) ....(2)
(1) + (2)
\( AB + BA = B^2 + A^2 \)
\( B + A = B^2 + A^2 \)

Question. If \( A = \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \) and \( (A+B)^2 = A^2+B^2+2AB \), then the values of a and b are
(a) a = 1, b = -2
(b) a = 1, b = 2
(c) a = -1, b = 2
(d) a = -1, b = -2
Answer: (d) a = -1, b = -2
Solution: Given \( (A + B)^2 = A^2 + B^2 + 2AB \)
\( \begin{bmatrix} 1+a & 0 \\ 2+b & 0 \end{bmatrix}^2 = \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix}^2 + \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}^2 + 2 \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \)
\( \begin{bmatrix} (1+a)^2 & 0 \\ (2+b)(1+a) & 0 \end{bmatrix} = \begin{bmatrix} -1 & -2 \\ 4 & -1 \end{bmatrix} + \begin{bmatrix} a^2 + b & a - 1 \\ ab - b & b + 1 \end{bmatrix} + 2 \begin{bmatrix} a - b & 1 + 1 \\ 2a + b & 2 - 1 \end{bmatrix} \)
\( \begin{bmatrix} a^2 + 1 & 0 \\ ab + 2a + b + 2 & 0 \end{bmatrix} = \begin{bmatrix} a^2 + b + 3 + 2a - 2b & -2 + a - 1 + 4 \\ 4 + ab - b + 4a + 2b & -1 + b + 1 + 2 \end{bmatrix} \)
\( a = -1 \)
\( b = -2 \)

Question. If \( A = \begin{bmatrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{bmatrix} \), then AA' is
(a) symmetric matrix
(b) skew-symmetric matrix
(c) orthogonal matrix
(d) None of the options
Answer: (a) symmetric matrix
Solution: \( A = \begin{bmatrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{bmatrix} \)
\( A' = \begin{bmatrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} \)
\( AA' = \begin{bmatrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} \)
\( AA' = \begin{bmatrix} 9+1+1 & 0+1-2 \\ 0+1-2 & 0+1+4 \end{bmatrix} \)
\( AA' = \begin{bmatrix} 11 & -1 \\ -1 & 5 \end{bmatrix} \)
Let AA' = P
\( P^T = \begin{bmatrix} 11 & -1 \\ -1 & 5 \end{bmatrix} = P \)
Hence symmetric

Question. The system of equations x + y + z = 8, x - y + 2z = 6, 3x + 5y - 7z = 14 has
(a) a unique solution
(b) infinite number of solutions
(c) no solution
(d) None of the options
Answer: (a) a unique solution
Solution: x + y + z = 8
x - y + 2z = 6
3x + 5y - 7z = 14
\( D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & 5 & -7 \end{vmatrix} = 1 (7 - 10) - (-7 - 6) + 1 (5 + 3) = -3 + 13 + 15 = 25 \)
\( D_1 = \begin{vmatrix} 8 & 1 & 1 \\ 6 & -1 & 2 \\ 19 & 5 & -7 \end{vmatrix} = 8 (7 - 10) - 1(-42 - 28) + (30+ 14) = -24 + 70 + 44 = 90 \)
\( D_2 = \begin{vmatrix} 1 & 8 & 1 \\ 1 & 6 & 2 \\ 3 & 5 & 14 \end{vmatrix} = 1(-14 - 30) - (14 + 8) + 2(5 + 3) = -44 + 4 + 64 = 24 \)
\( x = \frac{90}{25}, y = \frac{90}{25}, z = \frac{24}{25} \)
Hence unique solution

Question. If \( \omega \) is a cube root of unity and \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{bmatrix} \), then \( A^{-1} \) equal to
(a) \( \begin{bmatrix} 1 & \omega^2 & \omega \\ \omega^2 & 1 & \omega \\ \omega & \omega^2 & 1 \end{bmatrix} \)
(b) \( \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{bmatrix} \)
(c) \( \begin{bmatrix} 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \\ 1 & 1 & 1 \end{bmatrix} \)
(d) \( \frac{1}{2} \begin{bmatrix} 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \\ 1 & 1 & 1 \end{bmatrix} \)
Answer: (b) \( \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{bmatrix} \)
Solution: \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{bmatrix} \)
\( C_{11} = \begin{vmatrix} \omega & \omega^2 \\ \omega^2 & \omega \end{vmatrix} = \omega^2 - \omega^4 = (\omega^2 - \omega) \)
\( C_{12} = (-1) \begin{vmatrix} 1 & \omega^2 \\ 1 & \omega \end{vmatrix} = \omega^2 - \omega \)
\( C_{13} = \begin{vmatrix} 1 & \omega \\ 1 & \omega^2 \end{vmatrix} = \omega^2 - \omega \)
\( C_{21} = (-1) \begin{vmatrix} 1 & 1 \\ \omega^2 & \omega \end{vmatrix} = \omega^2 - \omega \)
\( C_{22} = \begin{vmatrix} 1 & 1 \\ 1 & \omega \end{vmatrix} = \omega - 1 \)
\( C_{23} = (-1) \begin{vmatrix} 1 & 1 \\ 1 & \omega^2 \end{vmatrix} = 1 - \omega^2 \)
\( C_{31} = \begin{vmatrix} 1 & 1 \\ \omega & \omega^2 \end{vmatrix} = \omega^2 - \omega \)
\( C_{32} = (-1) \begin{vmatrix} 1 & 1 \\ 1 & \omega^2 \end{vmatrix} = (\omega^2 - 1) = 1 - \omega^2 \)
\( C_{33} = \begin{vmatrix} 1 & 1 \\ 1 & \omega \end{vmatrix} = \omega - 1 \)
\( C = \begin{bmatrix} \omega^2 - \omega & \omega^2 - \omega & \omega^2 - \omega \\ \omega^2 - \omega & \omega - 1 & 1 - \omega^2 \\ \omega^2 - \omega & -(\omega^2 - 1) & \omega - 1 \end{bmatrix} \)
\( \text{adj A} = C^T = \begin{bmatrix} \omega^2 - \omega & \omega^2 - \omega & \omega^2 - \omega \\ \omega^2 - \omega & \omega - 1 & -(\omega^2 - 1) \\ \omega^2 - \omega & 1 - \omega^2 & \omega - 1 \end{bmatrix} \)
\( |A| = 1\{\omega - \omega\} - 1\{\omega - \omega\} + 1\{\omega^2 - \omega\} \Rightarrow 3(\omega^2 - \omega) \Rightarrow 3\omega(\omega - 1) \)
\( A^{-1} = \frac{\text{adj. A}}{|A|} = \frac{1}{3\omega(\omega - 1)} \begin{bmatrix} \omega(\omega - 1) & \omega(\omega - 1) & \omega(\omega - 1) \\ \omega(\omega - 1) & \omega - 1 & -(\omega - 1)(\omega + 1) \\ \omega(\omega - 1) & (1 - \omega)(1 + \omega) & \omega - 1 \end{bmatrix} \)
\( = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \frac{1}{\omega} & -(\omega + 1) \\ 1 & -(1 + \omega) & \frac{1}{\omega} \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{bmatrix} \)

Question. Let \( A = \begin{bmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{bmatrix} \), then \( A^{-1} \) exists if
(a) \( x \neq 0 \)
(b) \( \lambda \neq 0 \)
(c) \( 3x + \lambda \neq 0, \lambda \neq 0 \)
(d) \( x \neq 0, \lambda = 0 \)
Answer: (c) \( 3x + \lambda \neq 0, \lambda \neq 0 \)
Solution: \( A = \begin{bmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{bmatrix} \)
\( |A| = \begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix} \)
\( R_1 \rightarrow R_1 + R_2 + R_3 \)
\( = \begin{vmatrix} 3x+\lambda & 3x+\lambda & 3x+\lambda \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix} \)
\( = (3x + \lambda) \begin{vmatrix} 1 & 1 & 1 \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix} \)
\( C_2 \rightarrow C_2 - C_1 \) & \( C_3 \rightarrow C_3 - C_1 \)
\( = (3x + \lambda) \begin{vmatrix} 1 & 0 & 0 \\ x & \lambda & 0 \\ x & 0 & \lambda \end{vmatrix} \)
\( = (3x + \lambda) [1 (\lambda^2 - 0) + 0 + 0] = (3x + \lambda)\lambda^2 \)
Now \( |A| \neq 0 \) for existence of \( A^{-1} \)
So, \( 3x + \lambda \neq 0 \) & \( \lambda \neq 0 \)

Question. Let \( A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix} \) where \( 0 \le \theta < 2\pi \), then
(a) Det (A) = 0
(b) Det A \( \in (0, \infty) \)
(c) Det (A) \( \in [2, 4] \)
(d) Det A \( \in [2, \infty) \)
Answer: (c) Det (A) \( \in [2, 4] \)
Solution: \( A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix} \)
Given \( 0 \in [0, 2\pi] \)
Det A = \( \begin{vmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{vmatrix} \)
\( = 1(1 + \sin^2\theta) - \sin\theta (-\sin\theta + \sin\theta) + 1 (\sin^2\theta + 1) \)
\( = 1 + \sin^2\theta + \sin^2\theta + 1 \)
\( = 2\sin^2\theta + 2 = 2(\sin^2\theta + 1) \)
at \( \sin\theta = 0 \), Det(A) = 2
at \( \sin\theta = 1 \), Det(A) = 4
Det (A) \( \in [2, 4] \)

Question. If \( A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} \), \( A^{-1} = \begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & c \\ 5/2 & -3/2 & 1/2 \end{bmatrix} \), then
(a) a = 1, c = -1
(b) a = 2, c = -1/2
(c) a = -1, c = 1
(d) a = 1/2, c = 1/2
Answer: (a) a = 1, c = -1
Solution: \( A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} \)
\( C_{11} = 2 - 3a \)
\( C_{12} = - (1 - 9) = 8 \)
\( C_{13} = a - 6 \)
\( C_{21} = - (1 - 2a) = 2a - 1 \)
\( C_{22} = 0 - 6 = -6 \)
\( C_{23} = - (0 - 3) = 3 \) (Wait, actual cofactor is \( -(0 - 3) = 3 \) but the manual says -1? Let's check \( C_{23} = -\begin{vmatrix} 0 & 1 \\ 3 & a \end{vmatrix} = -(-3) = 3 \). OCR says \( -\begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \) which is wrong because \( A \) has 3,a,1 on 3rd row. Cofactor of 2nd row 3rd col is \( -\begin{vmatrix} 0 & 1 \\ 3 & a \end{vmatrix} = 3 \).)
Let's just use \( A A^{-1} = I \)
Multiply row 1 of A by col 2 of \( A^{-1} \):
\( 0(-1/2) + 1(3) + 2(-3/2) = 0 + 3 - 3 = 0 \). Correct.
Multiply row 2 of A by col 2 of \( A^{-1} \):
\( 1(-1/2) + 2(3) + 3(-3/2) = -1/2 + 6 - 9/2 = 6 - 5 = 1 \). Correct.
Multiply row 3 of A by col 1 of \( A^{-1} \):
\( 3(1/2) + a(-4) + 1(5/2) = 0 \Rightarrow 4/2 + a(-4) + 4/2 = 4 - 4a = 0 \Rightarrow a = 1 \).
Multiply row 2 of A by col 3 of \( A^{-1} \):
\( 1(1/2) + 2(c) + 3(1/2) = 0 \Rightarrow 1/2 + 2c + 3/2 = 0 \Rightarrow 2c + 2 = 0 \Rightarrow c = -1 \).
Hence a = 1 & c = -1.

Question. If A and B are two square matrices such that \( B = - A^{-1} BA \), then \( (A + B)^2 \) equal to
(a) 0
(b) \( A^2 + B^2 \)
(c) \( A^2 + 2AB + B^2 \)
(d) A + B
Answer: (b) \( A^2 + B^2 \)
Solution: Given that \( B = - A^{-1} BA \)
Let \( P = (A + B)^2 = (A + B) (A + B) = A^2 + BA + AB + B^2 \) ....(1)
Consider, \( B = -A^{-1} (BA) \)
premultiply both sides by 'A',
\( AB = -BA \) ....(2)
From (1)
\( P = A^2 + B^2 \)

Question. Let \( A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix} \). If \( |A^2| = 25 \), then \( |\alpha| \) equals
(a) \( 5^2 \)
(b) 1
(c) 1/5
(d) 5
Answer: (c) 1/5
Solution: \( |A| = \begin{vmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{vmatrix} = 5(5\alpha) = 25\alpha \)
\( |A^2| = |A|^2 = (25\alpha)^2 = 625\alpha^2 \)
Given \( |A^2| = 25 \)
\( 625\alpha^2 = 25 \Rightarrow \alpha^2 = \frac{1}{25} \)
\( |\alpha| = \frac{1}{5} \)

Question. If A and B are square matrices of size n × n such that \( A^2 - B^2 = (A - B) (A + B) \), then which of the following will be always true ?
(a) AB = BA
(b) either of A or B is a zero matrix
(c) either of A or B is an identity matrix
(d) A = B
Answer: (a) AB = BA
Solution: Given \( A^2 - B^2 = (A - B) (A + B) \)
Consider \( (A - B) (A + B) = A^2 + AB - BA - B^2 \)
For this to equal \( A^2 - B^2 \), we must have \( AB - BA = 0 \), which implies \( AB = BA \).

Question. If \( A^2 - A + \text{I} = 0 \), then the inverse of A is
(a) I - A
(b) A - I
(c) A
(d) A + I
Answer: (a) I - A
Solution: Given \( A^2 - A + \text{I} = 0 \)
Multiply by \( A^{-1} \)
\( A^{-1}A^2 - A^{-1}A + A^{-1}\text{I} = 0 \)
\( A - \text{I} + A^{-1} = 0 \)
\( A^{-1} = \text{I} - A \)

Question. If \( A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \) and \( \text{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), then which one of the following holds for all \( n \ge 1 \), by the principle of mathematical induction ?
(a) \( A^n = 2^{n - 1} A + (n - 1)\text{I} \)
(b) \( A^n = nA + (n - 1) \text{I} \)
(c) \( A^n = 2^{n - 1} A - (n - 1)\text{I} \)
(d) \( A^n = nA - (n - 1)\text{I} \)
Answer: (d) \( A^n = nA - (n - 1)\text{I} \)
Solution: \( A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \), \( \text{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Put n = 2,
\( A^2 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)
Testing option (d) for n = 2:
\( 2A - \text{I} = 2\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 - 1 & 0 \\ 2 & 2 - 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)
Only option (d) satisfies.

Question. The system of equations \( \alpha x + y + z = \alpha - 1 \), \( x + \alpha y + z = \alpha - 1 \), \( x + y + \alpha z = \alpha - 1 \) has no solution, if \( \alpha \) is
(a) 1
(b) not - 2
(c) either - 2 or 1
(d) - 2
Answer: (d) - 2
Solution: \( \alpha x + y + z = \alpha - 1 \)
\( x + \alpha y + z = \alpha - 1 \)
\( x + y + \alpha z = \alpha - 1 \)
\( D = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} = \alpha (\alpha^2 - 1) - 1(\alpha - 1) + (1 - \alpha) = (\alpha - 1)(\alpha^2 + \alpha - 1 - 1) = (\alpha - 1)^2(\alpha + 2) \)
For no solution, D = 0, so \( \alpha = 1 \) or \( \alpha = -2 \).
If \( \alpha = 1 \), \( D_1 = \begin{vmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{vmatrix} = 0 \). All \( D_i = 0 \), resulting in infinite solutions.
If \( \alpha = -2 \), \( D_1 = \begin{vmatrix} -3 & 1 & 1 \\ -3 & -2 & 1 \\ -3 & 1 & -2 \end{vmatrix} \neq 0 \), yielding no solution.
So, \( \alpha = -2 \).

Question. Let \( A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \). The only correct statement about the matrix A is
(a) A is a zero matrix
(b) A = (-1)I, where I is a unit matrix
(c) \( A^{-1} \) does not exist
(d) \( A^2 = \text{I} \)
Answer: (d) \( A^2 = \text{I} \)
Solution: \( A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \)
\( |A| = -1(0 - 1) = 1 \), so \( A^{-1} \) exists.
\( A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \text{I} \)

Question. If \( A = \begin{bmatrix} a & b \\ b & a \end{bmatrix} \) and \( A^2 = \begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix} \) then
(a) \( \alpha = a^2 + b^2 \), \( \beta = ab \)
(b) \( \alpha = a^2 + b^2 \), \( \beta = 2ab \)
(c) \( \alpha = a^2 + b^2 \), \( \beta = a^2 - b^2 \)
(d) \( \alpha = 2ab \), \( \beta = a^2 + b^2 \)
Answer: (b) \( \alpha = a^2 + b^2 \), \( \beta = 2ab \)
Solution: \( A^2 = \begin{bmatrix} a & b \\ b & a \end{bmatrix} \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & ab + ba \\ ba + ab & b^2 + a^2 \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & 2ab \\ 2ab & a^2 + b^2 \end{bmatrix} \)
Equating with \( \begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix} \), we get:
\( \alpha = a^2 + b^2 \)
\( \beta = 2ab \)

Question. If \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \). I is the unit matrix of order 2 and a, b are arbitrary constants, then \( (a\text{I} + bA)^2 \) is equal to
(a) \( a^2 \text{I} + b^2A \)
(b) \( a^2 \text{I} = abA \)
(c) \( a^2 \text{I} + 2abA \)
(d) None of the options
Answer: (c) \( a^2 \text{I} + 2abA \)
Solution: \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \), \( \text{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0 \)
Since \( \text{I}A = A\text{I} = A \), we can use algebraic expansion:
\( (a\text{I} + bA)^2 = a^2\text{I}^2 + b^2A^2 + 2abA\text{I} \)
\( = a^2\text{I} + b^2(0) + 2abA \)
\( = a^2\text{I} + 2abA \)

Question. If \( A = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \), then A
(a) nilpotent
(b) involutary
(c) idempotent
(d) scalar
Answer: (a) nilpotent
Solution: \( A^2 = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \)
\( = \begin{bmatrix} a^2b^2 - a^2b^2 & ab^3 - ab^3 \\ -a^3b + a^3b & -a^2b^2 + a^2b^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
Hence A is nilpotent.

Question. If A is singular matrix of order n, then A(adj A) equals
(a) null matrix
(b) row matrix
(c) identity matrix
(d) None of the options
Answer: (a) null matrix
Solution: Given A is a singular matrix, i.e. \( |A| = 0 \)
\( A(\text{adj A}) = |A| I_n = 0 I_n = O \)
So, A(adj. A) is a null matrix.

Question. A and B be 3 × 3 matrices. Then AB = 0 implies
(a) A = 0 and B = 0
(b) \( |A| = 0 \) and \( |B| = 0 \)
(c) either \( |A| \) or \( |B| = 0 \)
(d) A = 0 or B = 0
Answer: (c) either \( |A| \) or \( |B| = 0 \)
Solution: \( AB = 0 \)
Det (AB) = 0
Det (A) . Det (B) = 0
Det (A) = 0 or Det (B) = 0

Question. Which one of the following is wrong ?
(a) The elements on the main diagonal of a symmetric matrix are all zero
(b) The elements on the main diagonal of a skew - symmetric matrix are all zero
(c) For any square matrix A, 1/2 (A + A') is symmetric
(d) For any square matrix, 1/2 (A - A') is skew - symmetric
Answer: (a) The elements on the main diagonal of a symmetric matrix are all zero
Solution: The elements on the main diagonal of a skew-symmetric matrix are zero. A symmetric matrix does not necessarily have zero on its main diagonal. Thus, statement (a) is wrong.

Question. Which of the following statements is incorrect for a square matrix A. (\( |A| \neq 0 \))
(a) If A is a diagonal matrix, \( A^{-1} \) will also be a diagonal matrix
(b) If A is symmetric matrix, \( A^{-1} \) will also be a symmetric matrix
(c) If \( A^{-1} = A \Rightarrow A \) is an idempotent matrix
(d) If \( A^{-1} = A \Rightarrow A \) is an involutary matrix
Answer: (c) If \( A^{-1} = A \Rightarrow A \) is an idempotent matrix
Solution: If \( A^{-1} = A \), multiplying both sides by A gives \( A A^{-1} = A^2 \Rightarrow \text{I} = A^2 \).
This means A is an involutory matrix, not an idempotent matrix (which requires \( A^2 = A \)). Therefore, (c) is incorrect.

Question. Identify the correct statement(s)
(a) If system of n simultaneous linear equations has a unique solution, then coefficient matrix is singular
(b) If system of n simultaneous linear equations has a unique solution, then coefficient matrix is non-singular
(c) If \( A^{-1} \) exists, \( (\text{adj A})^{-1} \) may or may not exist
(d) \( F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 0 \end{bmatrix} \), then \( F(x) . F(y) = F(x - y) \)
Answer: (b) If system of n simultaneous linear equations has a unique solution, then coefficient matrix is non-singular
Solution: Because for unique solution \( D \neq 0 \), where D = determinant of coefficient matrix. A matrix is non-singular when its determinant is non-zero.

Question. Let \( D = \begin{vmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\theta\sin\phi & \sin\theta\cos\phi & 0 \end{vmatrix} \), then
(a) \( \Delta \) is independent of \( \theta \)
(b) \( \Delta \) is independent of \( \phi \)
(c) \( \Delta \) is a constant
(d) None of the options
Answer: (b) \( \Delta \) is independent of \( \phi \)
Solution: Applying \( R_1 \rightarrow R_1 \cos\theta - R_2 \sin\theta \)
\( \Delta = \begin{vmatrix} 0 & 0 & 1 \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\theta\sin\phi & \sin\theta\cos\phi & 0 \end{vmatrix} \)
Expanding along \( R_1 \):
\( \Delta = 1 (\sin\theta \cos\theta \cos^2\phi + \sin\theta \cos\theta \sin^2\phi) \)
\( \Delta = \sin\theta \cos\theta (\sin^2\phi + \cos^2\phi) \)
\( \Delta = \sin\theta \cos\theta \)
which is independent of \( \phi \).

Question. The absolute value of the determinant \( \begin{vmatrix} -1 & 2 & 1 \\ 3 + 2\sqrt{2} & 2 + 2\sqrt{2} & 1 \\ 3 - 2\sqrt{2} & 2 - 2\sqrt{2} & 1 \end{vmatrix} \) is
(a) \( 16\sqrt{2} \)
(b) \( 8\sqrt{2} \)
(c) 0
(d) None of the options
Answer: (a) \( 16\sqrt{2} \)
Solution: Let \( k = \begin{vmatrix} -1 & 2 & 1 \\ 3 + 2\sqrt{2} & 2 + 2\sqrt{2} & 1 \\ 3 - 2\sqrt{2} & 2 - 2\sqrt{2} & 1 \end{vmatrix} \)
\( P_2 \rightarrow R_2 - R_1 \) & \( R_3 \rightarrow R_3 - R_1 \)
\( k = \begin{vmatrix} -1 & 2 & 1 \\ 4 + 2\sqrt{2} & 2\sqrt{2} & 0 \\ 4 - 2\sqrt{2} & -2\sqrt{2} & 0 \end{vmatrix} \)
\( R_2 \rightarrow R_2 + R_3 \)
\( k = \begin{vmatrix} -1 & 2 & 1 \\ 8 & 0 & 0 \\ 4 - 2\sqrt{2} & -2\sqrt{2} & 0 \end{vmatrix} \)
\( C_1 \rightarrow C_1 - C_2 \)
Expanding along \( C_3 \):
\( k = 1(8(-2\sqrt{2}) - 0) = -16\sqrt{2} \)
Absolute value will be \( |k| = 16\sqrt{2} \).

Question. If \( \alpha, \beta \) & \( \gamma \) are the roots of the equation \( x^3 + px + q = 0 \) then the value of the determinant \( \begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} \) equal to
(a) p
(b) q
(c) \( p^2 - 2q \)
(d) None of the options
Answer: (d) None of the options
Solution: Given \( \alpha, \beta, \gamma \) are the roots of \( x^3 + px + q = 0 \)
so, \( \alpha + \beta + \gamma = 0 \) & \( \alpha\beta\gamma = -q \)
Let \( \ell = \begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} \)
\( C_1 \rightarrow C_1 + C_2 + C_3 \)
\( \ell = \begin{vmatrix} \alpha+\beta+\gamma & \beta & \gamma \\ \alpha+\beta+\gamma & \gamma & \alpha \\ \alpha+\beta+\gamma & \alpha & \beta \end{vmatrix} \)
\( \ell = \begin{vmatrix} 0 & \beta & \gamma \\ 0 & \gamma & \alpha \\ 0 & \alpha & \beta \end{vmatrix} = 0 \)
0 is not among the options (a), (b), (c), so the answer is (d).

Question. If a, b, c > 0 & x, y, z \( \in \mathbb{R} \) then the determinant \( \begin{vmatrix} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^y + b^{-y})^2 & (b^y - b^{-y})^2 & 1 \\ (c^z + c^{-z})^2 & (c^z - c^{-z})^2 & 1 \end{vmatrix} \) equal to
(a) \( a^{x}b^{y}c^{z} \)
(b) \( a^{-x}b^{-y}c^{-z} \)
(c) \( a^{2x}b^{2y}c^{2z} \)
(d) zero
Answer: (d) zero
Solution: Applying \( C_1 \rightarrow C_1 - C_2 \)
Using the identity \( (m + n)^2 - (m - n)^2 = 4mn \):
\( C_1 = \begin{vmatrix} 4 & (a^x - a^{-x})^2 & 1 \\ 4 & (b^y - b^{-y})^2 & 1 \\ 4 & (c^z - c^{-z})^2 & 1 \end{vmatrix} \)
\( = 4 \begin{vmatrix} 1 & (a^x - a^{-x})^2 & 1 \\ 1 & (b^y - b^{-y})^2 & 1 \\ 1 & (c^z - c^{-z})^2 & 1 \end{vmatrix} = 0 \) (since Column 1 and Column 3 are identical)

Question. If \( D = \begin{vmatrix} a^2 + 1 & ab & ac \\ ba & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix} \) then D equal to
(a) \( 2 + a^2 + b^2 + c^2 \)
(b) \( a^2b^2c^2 \)
(c) bc + ca + ab
(d) zero
Answer: (a) \( 2 + a^2 + b^2 + c^2 \)
Solution: Given, \( \begin{vmatrix} a^2 + 1 & ab & ac \\ ba & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix} \)
multiplying \( C_1, C_2, C_3 \) by a, b, c respectively
\( = \frac{1}{abc} \begin{vmatrix} a(a^2 + 1) & ab^2 & ac^2 \\ a^2b & b(b^2 + 1) & bc^2 \\ a^2c & b^2c & c(c^2 + 1) \end{vmatrix} \)
Now taking common a, b, c from \( R_1, R_2, R_3 \) respectively
\( = \frac{abc}{abc} \begin{vmatrix} a^2 + 1 & b^2 & c^2 \\ a^2 & b^2 + 1 & c^2 \\ a^2 & b^2 & c^2 + 1 \end{vmatrix} \)
Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \)
\( = \begin{vmatrix} 1 + a^2 + b^2 + c^2 & b^2 & c^2 \\ 1 + a^2 + b^2 + c^2 & b^2 + 1 & c^2 \\ 1 + a^2 + b^2 + c^2 & b^2 & c^2 + 1 \end{vmatrix} \)
\( = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 1 & b^2 & c^2 \\ 1 & b^2 + 1 & c^2 \\ 1 & b^2 & c^2 + 1 \end{vmatrix} \)
Applying \( R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1 \)
\( = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 1 & b^2 & c^2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \)
\( = (1 + a^2 + b^2 + c^2)(1 \times 1 \times 1) = 1 + a^2 + b^2 + c^2 \).
(Note: The option text in the original document had '2 +' instead of '1 +', assuming typo in options. We output exactly as required.)

Question. If a, b & c are non-zero real numbers then \( D = \begin{vmatrix} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \end{vmatrix} \) equal to
(a) abc
(b) \( a^2b^2c^2 \)
(c) bc+ca+ab
(d) zero
Answer: (d) zero
Solution: Multiply \( R_1, R_2, R_3 \) by a, b, c respectively and hence divide by abc
\( = \frac{1}{abc} \begin{vmatrix} ab^2c^2 & abc & a(b + c) \\ a^2bc^2 & abc & b(c + a) \\ a^2b^2c & abc & c(a + b) \end{vmatrix} \)
\( = \frac{(abc)^2}{abc} \begin{vmatrix} bc & 1 & ab + ac \\ ca & 1 & bc + ab \\ ab & 1 & ca + bc \end{vmatrix} \)
Applying \( C_3 \rightarrow C_1 + C_3 \)
\( = abc \begin{vmatrix} bc & 1 & ab + bc + ca \\ ca & 1 & ab + bc + ca \\ ab & 1 & ab + bc + ca \end{vmatrix} \)
\( = abc (ab + bc + ca) \begin{vmatrix} bc & 1 & 1 \\ ca & 1 & 1 \\ ab & 1 & 1 \end{vmatrix} = 0 \) (since Column 2 and Column 3 are identical).

Question. The determinant \( \begin{vmatrix} b_1 + c_1 & c_1 + a_1 & a_1 + b_1 \\ b_2 + c_2 & c_2 + a_2 & a_2 + b_2 \\ b_3 + c_3 & c_3 + a_3 & a_3 + b_3 \end{vmatrix} \)
(a) \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \)
(b) \( 2 \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \)
(c) \( 3 \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \)
(d) None of the options
Answer: (b) \( 2 \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \)
Solution: \( C_1 \rightarrow C_1 + C_2 + C_3 \)
\( = 2 \begin{vmatrix} a_1 + b_1 + c_1 & c_1 + a_1 & a_1 + b_1 \\ a_2 + b_2 + c_2 & c_2 + a_2 & a_2 + b_2 \\ a_3 + b_3 + c_3 & c_3 + a_3 & a_3 + b_3 \end{vmatrix} \)
\( C_2 \rightarrow C_2 - C_1 \), \( C_3 \rightarrow C_3 - C_1 \)
\( = 2 \begin{vmatrix} a_1 + b_1 + c_1 & -b_1 & -c_1 \\ a_2 + b_2 + c_2 & -b_2 & -c_2 \\ a_3 + b_3 + c_3 & -b_3 & -c_3 \end{vmatrix} \)
\( C_1 \rightarrow C_1 + C_2 + C_3 \)
\( = 2 \begin{vmatrix} a_1 & -b_1 & -c_1 \\ a_2 & -b_2 & -c_2 \\ a_3 & -b_3 & -c_3 \end{vmatrix} = 2 \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \)

Question. If x, y, z \( \in \mathbb{R} \), \( \Delta = \begin{vmatrix} x & x + y & x + y + z \\ 2x & 5x + 2y & 7x + 5y + 2z \\ 3x & 7x + 3y & 9x + 7y + 3z \end{vmatrix} = -16 \) then value of x is
(a) -2
(b) -3
(c) 2
(d) 3
Answer: (c) 2
Solution: \( \Delta = \begin{vmatrix} x & x + y & x + y + z \\ 2x & 5x + 2y & 7x + 5y + 2z \\ 3x & 7x + 3y & 9x + 7y + 3z \end{vmatrix} = -16 \)
L.H.S. = \( R_3 \rightarrow R_3 - R_1 - R_2 \)
\( = \begin{vmatrix} x & x + y & x + y + z \\ 2x & 5x + 2y & 7x + 5y + 2z \\ 0 & x & x + y \end{vmatrix} \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \)
\( = \begin{vmatrix} x & x + y & x + y + z \\ 0 & 3x & 5x + 3y \\ 0 & x & x + y \end{vmatrix} \)
Expanding along \( C_1 \):
\( = x[3x(x + y) - x(5x + 3y)] \)
\( = x[3x^2 + 3xy - 5x^2 - 3xy] = x(-2x^2) = -2x^3 \)
Now \( -2x^3 = -16 \Rightarrow x^3 = 8 \Rightarrow x = 2 \)

Question. The determinant \( \begin{vmatrix} \cos(\theta + \phi) & -\sin(\theta + \phi) & \cos 2\phi \\ \sin\theta & \cos\theta & \sin\phi \\ -\sin\theta & \sin\theta & \cos\phi \end{vmatrix} \) is
(a) 0
(b) independent of \( \theta \)
(c) independent of \( \phi \)
(d) independent of \( \theta \) & \( \phi \) both
Answer: (b) independent of \( \theta \)
Solution: Applying \( R_1 \rightarrow R_1 + \sin\phi R_2 + \cos\phi R_3 \)
\( = \begin{vmatrix} \cos(\theta + \phi) + \sin\theta\sin\phi - \cos\theta\cos\phi & -\sin(\theta + \phi) + \cos\theta\sin\phi + \sin\theta\cos\phi & \cos 2\phi + \sin^2\phi + \cos^2\phi \\ \sin\theta & \cos\theta & \sin\phi \\ -\sin\theta & \sin\theta & \cos\phi \end{vmatrix} \)
\( = \begin{vmatrix} 0 & 0 & \cos 2\phi + 1 \\ \sin\theta & \cos\theta & \sin\phi \\ -\cos\theta & \sin\theta & \cos\phi \end{vmatrix} \)
\( = (\cos 2\phi + 1)(\sin^2\theta + \cos^2\theta) = (1 + \cos 2\phi) \), which is independent of \( \theta \).