JEE Mathematics Matrices and Determinants MCQs Set 02

Question. The number of different orders of a matrix having 12 elements is
(a) 3
(b) 1
(c) 6
(d) None of the options
Answer: (c) 6
Solution: The matrix will be any one of the following type
\( 1 \times 12, 12 \times 1, 2 \times 6, 6 \times 2, 3 \times 4, 4 \times 3 \)

Question. \( \begin{bmatrix} x^2 + x & x \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ -x + 1 & x \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 5 & 1 \end{bmatrix} \) then x is equal to
(a) -1
(b) 2
(c) 1
(d) No value of x
Answer: (a) -1
Solution: Given, \( \begin{bmatrix} x^2 + x & x \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ -x + 1 & x \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 5 & 1 \end{bmatrix} \)
\( \therefore x^2 + x = 0, x - 1 = -2, -x + 4 = 5, x + 2 = 1 \)
\( \Rightarrow x = -1 \)

Question. If \( A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \) and \( B = \begin{bmatrix} -5 & 4 & 0 \\ 0 & 2 & -1 \\ 1 & -3 & 2 \end{bmatrix} \), then
(a) \( AB = \begin{bmatrix} -5 & 8 & 0 \\ 0 & 4 & -2 \\ 3 & -9 & 6 \end{bmatrix} \)
(b) \( AB = \begin{bmatrix} -2 & -1 & 4 \end{bmatrix} \)
(c) \( AB = \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} \)
(d) AB does not exist
Answer: (d) AB does not exist
Solution: Given \( A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \) and \( B = \begin{bmatrix} -5 & 4 & 0 \\ 0 & 2 & -1 \\ 1 & -3 & 2 \end{bmatrix} \)
order of A is 3 × 1 and B is 3 × 3 product of AB is defined only when the number of columns in A is equal to the number of rows in B so AB does not exist.

Question. If \( \text{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), \( J = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \), then B equal to
(a) \( \text{I}\cos\theta + J\sin\theta \)
(b) \( \text{I}\cos\theta - J\sin\theta \)
(c) \( \text{I}\sin\theta + J\cos\theta \)
(d) \( -\text{I}\cos\theta + J\sin\theta \)
Answer: (a) \( \text{I}\cos\theta + J\sin\theta \)
Solution: Given \( \text{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), \( J = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \)
then \( B = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \cos\theta + \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \sin\theta = \text{I}\cos\theta + J\sin\theta \)

Question. If A and B are square matrices of order 2, then \( (A + B)^2 \) equal to
(a) \( A^2 + 2AB + B^2 \)
(b) \( A^2 + AB + BA + B^2 \)
(c) \( A^2 + 2BA + B^2 \)
(d) None of the options
Answer: (b) \( A^2 + AB + BA + B^2 \)
Solution: Given A and B are square matrices of order 2
then \( (A + B)^2 = (A + B) (A + B) \)
\( = A.A + A.B + B.A + B.B \)
\( = A^2 + AB + BA + B^2 \)

Question. If A is a skew - symmetric matrix, then trace of A is equal to
(a) 1
(b) -1
(c) 0
(d) None of the options
Answer: (c) 0
Solution: Since in a skew symmetric matrix all elements along the principal diagonal are zero. If A is skew symmetric matrix then trace of A is zero.

Question. If \( A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \), then adj A equal to
(a) \( \begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix} \)
(b) \( \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \)
(c) \( \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix} \)
(d) \( \begin{bmatrix} -1 & 2 \\ 2 & -1 \end{bmatrix} \)
Answer: (a) \( \begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix} \)
Solution: Given \( A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \) then \( \text{adj } A = \begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix} \)

Question. If \( A = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \), then adj A equal to
(a) \( A' \)
(b) I
(c) O
(d) \( A^2 \)
Answer: (a) \( A' \)
Solution: Given \( A = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
then \( \text{adj } A = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} = A' \)

Question. If A is a square matrix such that \( A^2 = \text{I} \), then \( A^{-1} \) equal to
(a) 2A
(b) A
(c) O
(d) A + I
Answer: (b) A
Solution: Given A is a square matrix such that \( A^2 = \text{I} \)
\( \Rightarrow AA = \text{I} \)
By definition of inverse \( A^{-1} A = \text{I} \) then \( A^{-1} = A \)

Question. If A and B are square matrices of order 3 such that \( |A| = -1 \), \( |B| = 3 \), then \( |3AB| \) is equal to
(a) -9
(b) -81
(c) -27
(d) 81
Answer: (b) -81
Solution: Given \( |A| = -1 \) & \( |B| = 3 \)
\( |3 AB| = 3^3 (-1) 3 = -81 \)

Question. If \( A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \), then value of \( A^{-1} \) is equal to
(a) A
(b) \( A^2 \)
(c) \( A^3 \)
(d) \( A^4 \)
Answer: (c) \( A^3 \)
Solution: Given \( A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \)
Now \( |A| = \begin{vmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{vmatrix} = 3(-3+4) - 2(-3+4) + 0(-12+12) = 1 \)
\( \text{adj A} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}^T = \begin{bmatrix} 1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3 \end{bmatrix} \)
then \( A^{-1} = \frac{\text{adjA}}{|A|} = \begin{bmatrix} 1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3 \end{bmatrix} \)
and \( A^3 = A.A.A. \)
\( = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3 \end{bmatrix} \)
Hence \( A^{-1} = A^3 \)

Question. Let \( A = \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix} \) and \( \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \) and X be a matrix such that \( A = BX \) is equal to
(a) \( \frac{1}{2} \begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix} \)
(b) \( \frac{1}{2} \begin{bmatrix} -2 & 4 \\ 3 & 5 \end{bmatrix} \)
(c) \( \begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix} \)
(d) None of the options
Answer: (a) \( \frac{1}{2} \begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix} \)
Solution: Given \( A = \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \) and \( A = BX \)
Let \( x = \begin{bmatrix} a & c \\ b & d \end{bmatrix} \)
then \( A = Bx \Rightarrow \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} a & c \\ b & d \end{bmatrix} = \begin{bmatrix} a & c \\ 2b & 2d \end{bmatrix} \)
\( \therefore a = 1, b = \frac{3}{2}, c = 2, d = -\frac{5}{2} \)
Hence \( x = \begin{bmatrix} 1 & 2 \\ 3/2 & -5/2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix} \)

Question. If B is a non-singular matrix and A is a square matrix, then det \( (B^{-1} AB) \) is equal to
(a) det \( (A^{-1}) \)
(b) det \( (B^{-1}) \)
(c) det (A)
(d) det (B)
Answer: (c) det (A)
Solution: det \( (B^{-1} AB) = \text{det } (B^{-1}) \text{ det } (A) \text{ det } (B) \)
\( = (\text{det}(B))^{-1} \text{det}(A) \text{ det } (B) \)
\( = \text{det } (A) \)

Question. The system of equation -2x + y + z = 1, x - 2y + z = -2, x + y + \( \lambda z \) = 4 will have no solution if
(a) \( \lambda = -2 \)
(b) \( \lambda = -1 \)
(c) \( \lambda = 3 \)
(d) None of the options
Answer: (a) \( \lambda = -2 \)
Solution: Given equations are -2x + y + z = 1
x - 2y + z = -2
x + y + \( \lambda z \) = 4
Here \( D = \begin{vmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{vmatrix} \)
\( D = -2(-2\lambda -1) - 1(\lambda - 1) + 1(1 + 2) \)
\( D = 3\lambda + 6 \)
The system of equation will have no solution if D = 0
so \( 3\lambda + 6 = 0 \Rightarrow \lambda = -2 \)

Question. The system of the linear equations x + y - z = 6, x + 2y - 3z = 14 and 2x + 5y - \( \lambda z \) = 9 (\( \lambda \in \mathbb{R} \)) has a unique solution if
(a) \( \lambda = 8 \)
(b) \( \lambda \neq 8 \)
(c) \( \lambda = 7 \)
(d) \( \lambda \neq 7 \)
Answer: (b) \( \lambda \neq 8 \)
Solution: Given equations are x + y - z = 6
x + 2y - 3z = 14
2x + 5y - \( \lambda z \) = 9
Here \( D = \begin{vmatrix} 1 & 1 & -1 \\ 1 & 2 & -3 \\ 2 & 5 & -\lambda \end{vmatrix} \)
\( D = 1(-2\lambda + 15) - 1(-\lambda + 6) -1(5 - 4) \)
\( D = -\lambda + 8 \)
The system of equation has a unique solution if
\( D \neq 0 \) so \( -\lambda + 8 \neq 0 \Rightarrow \lambda \neq 8 \)

Question. If the system of equations x + 2y + 3z = 4, x + \( \lambda y \) + 2z = 3, x + 4y + \( \mu z \) = 3 has an infinite a number of solutions then
(a) \( \lambda = 2, \mu = 3 \)
(b) \( \lambda = 2, \mu = 4 \)
(c) \( 3\lambda = 2\mu \)
(d) None of the options
Answer: (d) None of the options
Solution: Given equations are x + 2y + 3z = 4
x + \( \lambda y \) + 2z = 3
x + 4y + \( \mu z \) = 3
Here \( D = \begin{vmatrix} 1 & 2 & 3 \\ 1 & \lambda & 2 \\ 1 & 4 & \mu \end{vmatrix} \)
\( D = (\lambda\mu - 8) - 2(\mu - 2) + 3(4 - \lambda) \)
\( = \lambda\mu - 2\mu - 3\lambda + 8 \)
\( = (\lambda - 2)\mu - 3\lambda + 8 \)
The system of equation has an infinite a number of solution if D = 0 which not gives value of \( \lambda \) and \( \mu \).

Question. The matrix \( A = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix} \) is
(a) idempotent matrix
(b) involutory matrix
(c) nilpotent matrix
(d) None of the options
Answer: (c) nilpotent matrix
Solution: Given \( A = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ -1 & 3 & -3 \end{bmatrix} \)
\( A^3 = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ -1 & 3 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 20 & 8 & 24 \end{bmatrix} \)
\( A^4 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 20 & 8 & 24 \end{bmatrix} \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 12 & 12 & 36 \end{bmatrix} \)
\( A^5 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 12 & 12 & 36 \end{bmatrix} \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0 \)
Hence, A is nilpotent matrix of order 5.

Question. If A = diag (2, -1, 3), B = diag (-1, 3, 2), then \( A^2 B \) equal to
(a) diag (5, 4, 11)
(b) diag (-4, 3, 18)
(c) diag (3, 1, 8)
(d) B
Answer: (b) diag (-4, 3, 18)
Solution: Given A = diag(2, -1, 3), B = diag(-1, 3, 2)
then \( A^2 B = A.A.B. \)
\( = \text{diag } (2, -1, 3) \text{ diag } (2, -1, 3) \text{ diag}(-1, 3, 2) \)
\( = \text{diag } (4, 1, 9) \text{ diag}(-1, 3, 2) \)
\( = \text{diag } (-4, 3, 18) \)

Question. \( A = \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix} \) & \( B = \begin{bmatrix} 4 & 1 \\ 7 & 2 \end{bmatrix} \) then \( B^T A^T \) is
(a) a null matrix
(b) an identity matrix
(c) scalar, but not an identity matrix
(d) such that Tr \( (B^T A^T) \) = 4
Answer: (b) an identity matrix
Solution: Given \( A = \begin{bmatrix} 2 & -1 \\ -7 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & 1 \\ 7 & 2 \end{bmatrix} \)
then \( A^T = \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix} \) and \( B^T = \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix} \)
Now, \( B^T A^T = \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
which an identity matrix.

Question. If the matrix AB is a zero matrix, then
(a) A = O or B = O
(b) A = O and B = O
(c) It is not necessary that either A = O or B = O
(d) All the above statements are wrong
Answer: (c) It is not necessary that either A = O or B = O
Solution: If the matrix AB is a zero matrix then
It is not necessary that either A = 0 or B = 0

Question. Which relation true for \( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 4 \\ -1 & 1 \end{bmatrix} \)
(a) \( (A + B)^2 = A^2 + 2AB + B^2 \)
(b) \( (A - B)^2 = A^2 - 2AB + B^2 \)
(c) AB = BA
(d) None of the options
Answer: (d) None of the options
Solution: Given \( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 4 \\ -1 & 1 \end{bmatrix} \)
then \( AB = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ -3 & -2 \end{bmatrix} \)
and \( BA = \begin{bmatrix} 1 & 4 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 7 \\ -3 & 3 \end{bmatrix} \)
so AB \(\neq\) BA
Hence \( (A + B)^2 \neq A^2 + 2AB + B^2 \)
\( (A - B)^2 \neq A^2 - 2AB + B^2 \)

Question. If AB = A and BA = B, then \( B^2 \) is equal to
(a) B
(b) A
(c) I
(d) 0
Answer: (a) B
Solution: Given AB = A & BA = B
AB = A ....(1)
BA = B ....(2)
(1) × (2)
\( AB^2 A = AB \)
\( B^2 A = B \)
\( B^2 = BA^{-1} \) ....(3)
Consider (2)
BA = B
\( B = BA^{-1} \) ....(4)
So, from (3)
\( B^2 = B \)

Question. If A and B are symmetric matrices, then ABA is
(a) symmetric matrix
(b) skew-symmetric
(c) a diagonal matrix
(d) scalar matrix
Answer: (a) symmetric matrix
Solution: \( (ABA)^T = A^T B^T A^T \)
= ABA (\(\because\) A & B are symmetric matrices)
So, ABA is symmetric matrix.

Question. If A is a skew - symmetric matrix and n is an even positive integer, then \( A^n \) is
(a) a symmetric matrix
(b) a skew-symmetric matrix
(c) a diagonal matrix
(d) None of the options
Answer: (a) a symmetric matrix
Solution: Let \( A^n = P \)
Taking transpose both sides.
\( (A^n)^T = P^T \)
\( (A^T)^n = P^T \Rightarrow P^T = (-A)^n = A^n \)
\( \Rightarrow P^T = P \) Hence symmetric matrix.

Question. If A is a non-singular matrix and \( A^T \) denotes the transpose of A, then
(a) \( |A| \neq |A^T| \)
(b) \( |A . A^T| \neq |A|^2 \)
(c) \( |A^T . A| \neq |A^T|^2 \)
(d) \( |A| + |A^T| \neq 0 \)
Answer: (d) \( |A| + |A^T| \neq 0 \)
Solution: Given A is a non-singular matrix, i.e. \( |A| \neq 0 \)
So, \( |A^T| \neq 0 \) So, \( |A| + |A^T| \neq 0 \)

Question. Which of the following is incorrect
(a) \( A^2 - B^2 = (A + B) (A - B) \)
(b) \( (A^T)^T = A \)
(c) \( (AB)^n = A^nB^n \), where A, B commute
(d) \( (A - \text{I}) (\text{I} + A) = O \Leftrightarrow A^2 = \text{I} \)
Answer: (c) \( (AB)^n = A^nB^n \), where A, B commute
Solution: \( (AB)^n = A^nB^n \)
Given that AB = BA
For n = 1 :
\( (AB)^1 = A^1B^1 \)
For n = m
\( (AB)^m = A^m B^m \)
For n = m + 1 :
\( (AB)^{m+1} = A^{m+1} B^{m+1} \)
LHS
\( (AB)^{m+1} = (AB)^m (AB) \)
\( = (A^mB^m) (AB) \)
\( = A^m(B^m(AB)) \)
\( = A^m (B^m(BA)) \) (\(\because\) AB = BA)
\( = A^m (B^{m+1}A) \)
\( = A^m(AB^{m+1}) = A^{m+1} B^{m+1} \)

Question. If A is square matrix of order 3, then the true statement is (where I is unit matrix).
(a) det (-A) = -det A
(b) det A = 0
(c) det (A + I) = 1 + det A
(d) det 2A = 2 det A
Answer: (a) det (-A) = -det A
Solution: Let \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \)
\( -A = \begin{bmatrix} -a & -b & -c \\ -d & -e & -f \\ -g & -h & -i \end{bmatrix} \)
det (-A) = - det A

Question. If a, b, c are non zeros, then the system of equations \( (\alpha + a) x + \alpha y + \alpha z = 0 \) , \( \alpha x + (\alpha + b)y + \alpha z = 0 \) , \( \alpha x + \alpha y + (\alpha + c) z = 0 \) has a non-trivial solution if
(a) \( \alpha^{-1} = - (a^{-1} + b^{-1} + c^{-1}) \)
(b) \( a^{-1} = a + b + c \)
(c) \( \alpha + a + b + c = 1 \)
(d) None of the options
Answer: (a) \( \alpha^{-1} = - (a^{-1} + b^{-1} + c^{-1}) \)
Solution: \( (\alpha + a) x + \alpha y + \alpha z = 0 \)
\( \alpha x + (\alpha + b)y + \alpha z = 0 \)
\( \alpha x + \alpha y + (\alpha + c) z = 0 \)
From non-trivial solution, D = 0
\( \begin{vmatrix} \alpha + a & \alpha & \alpha \\ \alpha & \alpha + b & \alpha \\ \alpha & \alpha & \alpha + c \end{vmatrix} = 0 \)
\( R_1 \rightarrow R_1 + R_2 + R_3 \)
\( \Rightarrow \begin{vmatrix} 3\alpha + a & 3\alpha + b & 3\alpha + c \\ \alpha & \alpha + b & \alpha \\ \alpha & \alpha & \alpha + c \end{vmatrix} = 0 \)
\( \Rightarrow \begin{vmatrix} 3\alpha & 3\alpha & 3\alpha \\ \alpha & \alpha + b & \alpha \\ \alpha & \alpha & \alpha + c \end{vmatrix} + \begin{vmatrix} a & b & c \\ \alpha & \alpha + b & \alpha \\ \alpha & \alpha & \alpha + c \end{vmatrix} = 0 \)
\( \Rightarrow 3\alpha \begin{vmatrix} 1 & 1 & 1 \\ \alpha & \alpha + b & \alpha \\ \alpha & \alpha & \alpha + c \end{vmatrix} + \begin{vmatrix} a & b & c \\ \alpha & \alpha + b & \alpha \\ \alpha & \alpha & \alpha + c \end{vmatrix} = 0 \)
Applying \( C_2 \rightarrow C_2 - C_1 \) & \( R_2 \rightarrow R_2 - R_3 \)
Both parts respectively
\( \Rightarrow 3\alpha \begin{vmatrix} 1 & 0 & 0 \\ \alpha & b & 0 \\ \alpha & 0 & c \end{vmatrix} + \begin{vmatrix} a & b & c \\ 0 & b & c \\ \alpha & \alpha & \alpha + c \end{vmatrix} = 0 \)
\( \Rightarrow \alpha(ab + bc + ca) + ab c = 0 \)
\( \Rightarrow \alpha(-c^{-1} - a^{-1} - b^{-1}) = + 1 \)
\( \Rightarrow \alpha^{-1} = -(a^{-1} + b^{-1} + c^{-1}) \)

Question. Given \( A = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix} \); \( \text{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). If A - \( \lambda \text{I} \) is a singular matrix then
(a) \( \lambda \in \phi \)
(b) \( \lambda^2 - 3\lambda - 4 = 0 \)
(c) \( \lambda^2 + 3\lambda + 4 = 0 \)
(d) \( \lambda^2 - 3\lambda - 6 = 0 \)
Answer: (b) \( \lambda^2 - 3\lambda - 4 = 0 \)
Solution: Given \( A = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix} \), \( \text{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( P = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( P = \begin{bmatrix} 1 - \lambda & 3 \\ 2 & 2 - \lambda \end{bmatrix} \)
\( |P| = \begin{vmatrix} 1 - \lambda & 3 \\ 2 & 2 - \lambda \end{vmatrix} = 0 = (1 - \lambda) (2 - \lambda) - 6 = 0 \)
\( = 2 - \lambda - 2\lambda + \lambda^2 - 6 = 0 \)
\( = \lambda^2 - 3\lambda - 4 = 0 \)

Question. From the matrix equation AB = AC, we conclude B = C provided
(a) A is singular
(b) A is non-singular
(c) A is symmetric
(d) A is a square
Answer: (b) A is non-singular
Solution: AB = AC
pre-multiply by \( A^{-1} \Rightarrow B = C \)
for existance of \( A^{-1} \Rightarrow |A| \neq 0 \)
\( \Rightarrow \) A is non-singular matrix

Question. Let \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \) and \( 10B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix} \). If B is the inverse of matrix A, then \( \alpha \) is
(a) -2
(b) -1
(c) 2
(d) 5
Answer: (d) 5
Solution: \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \), \( 10B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix} \)
\( B = A^{-1} \Rightarrow AB = \text{I}_3 \)
\( 10BA = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \)
\( 10\text{I}_3 = \begin{bmatrix} 4 + 4 + 2 & -4 + 2 + 2 & 4 - 6 + 2 \\ -5 + \alpha & 5 + \alpha & -5 + \alpha \\ 1 - 4 + 3 & -1 - 2 + 3 & 1 + 6 + 3 \end{bmatrix} \)
\( 19\text{I}_3 = \begin{bmatrix} 10 & 0 & 0 \\ \alpha - 5 & \alpha + 5 & \alpha - 5 \\ 0 & 0 & 10 \end{bmatrix} \)
\( 10 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 10 & 0 & 0 \\ \alpha - 5 & \alpha + 5 & \alpha - 5 \\ 0 & 0 & 10 \end{bmatrix} \)
\( \Rightarrow \alpha = 5 \)

Question. The value of 'k' for which the set of equations 3x + ky - 2z = 0, x + ky + 3z = 0, 2x + 3y - 4z = 0 has a non - trivial solution over the set of rational is
(a) 33/2
(b) 31/2
(c) 16
(d) 15
Answer: (a) 33/2
Solution: 3x + ky - 2z = 0
x + ky + 3z = 0
2x + 3y - 4z = 0
D = 0
\( \begin{vmatrix} 3 & k & -2 \\ 1 & k & 3 \\ 2 & 3 & -4 \end{vmatrix} = 0 \)
\( 3(-4k - 9) - k(-4 - 6) - 2(3 - 2k) = 0 \)
\( -12k - 27 + 4k + 6k - 6 + 43k = 0 \)
\( 2k = 33 \)
\( k = \frac{33}{2} \)

Question. The value of a for which system of equations, \( a^3x + (a + 1)^3y + (a + 2)^3z = 0 \), \( ax + (a + 1) y + (a + 2) z = 0 \), x + y + z = 0, has a non-zero solution is
(a) -1
(b) 0
(c) 1
(d) None of the options
Answer: (a) -1
Solution: \( a^3x + (a + 1)^3y + (a + 2)^3z = 0 \)
\( ax + (a + 1)y + (a + 2) z = 0 \)
\( x + y + z = 0 \)
D = 0
\( \begin{vmatrix} a^3 & (a + 1)^3 & (a + 2)^3 \\ a & (a + 1) & (a + 2) \\ 1 & 1 & 1 \end{vmatrix} = 0 \)
\( C_2 \rightarrow C_2 - C_1 \) & \( C_3 \rightarrow C_3 - C_1 \)
\( \begin{vmatrix} a^3 & (a + 1)^3 - a^3 & (a + 2)^3 - a^3 \\ a & 1 & 2 \\ 1 & 0 & 0 \end{vmatrix} = 0 \)
\( \{2(a + 1)^3 - 2a^3 - (a + 2)^3 + a^3\} = 0 \)
\( \{(a + 1)^3 - (a + 2)^3 - a^3\} = 0 \)
\( a^3 + 1 + 3a + 3a^2 - a^3 - 8 - 8 - 6a^2 - 12a - a^3 = 0 \)
\( -a^3 - 3a - 9a - 7 = 0 \)
\( a^3 + 3a^2 + aa + 7 = 0 \)
\( a = -1 \)

Question. If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) satisfies the equation \( x^2 - (a + d) x + k = 0 \), then
(a) k = bc
(b) k = ad
(c) \( k = a^2 + b^2 + c^2 + d^2 \)
(d) ad - bc
Answer: (d) ad - bc
Solution: Given \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
Let, \( f(x) = x^2 - (a + d) x + k = 0 \)
\( f(A) = A^2 - (a + d) A + k = 0 \) ....(1)
Because 'A' satisfies f(x)
\( A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & cb + d^2 \end{bmatrix} \)
\( (a + d) A = (a + d) \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + ad & ab + bd \\ ac + cd & ad + d^2 \end{bmatrix} \)
\( k = k\text{I}_2 = k \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} \)
Putting in (1)
\( \begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & cb + d^2 \end{bmatrix} - \begin{bmatrix} a^2 + ad & ab + bd \\ ac + cd & ab + d^2 \end{bmatrix} + \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} = 0 \)
\( k = a^2 + ad - a^2 - bc \)
\( k = ad - bc \)

Question. Which of the following is a nilpotent matrix
(a) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
(b) \( \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \)
(c) \( \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \)
(d) \( \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \)
Answer: (c) \( \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \)
Solution: Let \( M = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \)
\( M^2 = M.M. = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \)
\( M^2 = \begin{bmatrix} 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 \end{bmatrix} = 0 \)