Students often turn to ML Aggarwal Maths for Class 12 Solutions Chapter 3 Matrices MCQs to clarify doubts and improve problem-solving skills.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices MCQs

Choose the correct answer from the given four options in questions (1 to 31):

Question 1.
The matrix A= $$\left[\begin{array}{lll} 0 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 0 \end{array}\right]$$ is a
(a) scalar matrix
(b) diagonal matrix
(c) square matrix
(d) none of these
Solution:
(c) square matrix

In given matrix,
No. of rows = No. of columns 3
∴ given matrix is a square matrix.

Question 2.
The matrix A = $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right]$$ is a
(a) scalar matrix
(b) symmetric matrix
(c) skew-symmetric matrix
(d) none of these
Solution:
(b) symmetric matrix

Let A = $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right]$$
∴ A’ = $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right]$$ = A

Question 3.
If A = $$\left[\begin{array}{rrr} 2 & -1 & 3 \\ 4 & 5 & -6 \end{array}\right]$$ and B = $$\left[\begin{array}{rr} 1 & 2 \\ 3 & 4 \\ 5 & -6 \end{array}\right]$$, then
(a) only AB is defined
(b) only BA is defined
(c) AB and BA are both defined
(d) AB and BA both are not defined
Solution:
(c) AB and BA are both defined

Given A be a matrix of order 2 × 3
and B be a matrix of order 3 × 2
AB is defined
∴ No. of columns in A = No. of rows in B = 3
and BA is also defined.
∵ No. of columns in B = No. of rows in A = 2.

Question 4.
If A is any m × n matrix and B is a matrix such that AB and BA are both defined, then B is a matrix of order
(a) n × n
(b) m × m
(c) m × n
(d) n × m
Solution:
(d) n × m

Given A be a matrix of order m × n
Let B be a matrix of order p × q.
Since AB is defined
∴ No. of columns in A = No. of rows in B
⇒ n = p
and BA is also defined
∴ No. of columns in B No. of rows in A
⇒ q = m
Thus, B be a matrix of order n × m.

Question 5.
If A = $$\left[\begin{array}{rr} -3 & x \\ y & 5 \end{array}\right]$$ and A = A’, then
(a) x = 5, y = – 3
(b) x = – 3, y = 5
(c) x = y
(d) none of these
Solution:
(c) x = y

Given A = $$\left[\begin{array}{rr} -3 & x \\ y & 5 \end{array}\right]$$
and A’ = $$\left[\begin{array}{rr} -3 & y \\ x & 5 \end{array}\right]$$

Question 6.
If A = $$\left[\begin{array}{ll} 4 & 1 \\ 3 & 2 \end{array}\right]$$ and I = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$, then A2 – 6A is equal to
(a) 3I
(b) – 5I
(c) 5I
(d) none of these
Solution:
(b) – 5I

A2 – 6A = $$\left[\begin{array}{ll} 4 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 3 & 2 \end{array}\right]-6\left[\begin{array}{ll} 4 & 1 \\ 3 & 2 \end{array}\right]$$
= $$\left[\begin{array}{rr} 19 & 6 \\ 18 & 7 \end{array}\right]-\left[\begin{array}{rr} 24 & 6 \\ 18 & 12 \end{array}\right]$$
= $$\left[\begin{array}{rr} -5 & 0 \\ 0 & -5 \end{array}\right]$$
= – 5 $$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$$
= – 5I.

Question 7.
If A = [aij]2 × 2 where aij = $$\begin{cases}1, & \text { if } i \neq j \\ 0, & \text { if } i=j\end{cases}$$, then A2 is equal to
(a) I
(b) A
(c) O
(d) none of these
Solution:
(a) I

A = $$\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]$$
= $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$

∴ A2 = A . A
= $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$
= $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= I.

Question 8.
The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is
(a) 18
(b) 27
(c) 81
(d) 512
Solution:
(d) 512

A matrix of order 3 x 3 containing 9 elements.
Since each entry of given matrix is filled by two numbers either 0 or 1 so there are two ways to fill each entry of given matrix.
Thus, required no. of such possible matrices = 29 = 512

Question 9.
If A and B are square matrices of same order, then (A + B) (A – B) ¡s equal to
(a) A2 – B2
(b) A2 – BA – AB + B2
(c) A2 – AB + BA – B2
(d) A2 + AB – BA – B2
Solution:
(c) A2 – AB + BA – B2

(A + B) (A – B) = A (A – B) + B (A – B)
=A2 – AB + BA – B2
[distributive law holds in matrix multiplication over addition]

Question 10.
If A is a square matrix such that A2 = I, then (A + I)3 + (A – I)3 – 7A is equal to
(a) A
(b) A – I
(c) A + I
(d) 3A
Solution:
(a) A

(A + I)2 = (A + I) (A + I)
= A (A + I) + 1 (A + I)
= A2 + A1+ IA + I2
= I + A + A + I [∵ A2 = I ; I2 = I]
= 2 (A + I)

∴ (A + I)3 = (A + I)2 (A + I)
= 2 (A + I) (A + I)
= 2 (A + I)2
= 4 (A + I)

(A – I)2 = (A – I) (A – 1)
= A2 – AI – IA + I2
= I – A – A+ I
= 2 (I – A)

∴ (A – I)3 = (A – I)2 (A – I)
= 2 (I – A) (A – I)
= – 2(A – I)2
= – 4(I – A)
= 4 (A – I)

Thus,(A + I)3 + (A – I)3 – 7A
= 4 (A + I) + 4 (A – I) – 7A
= 8A – 7A
= A

Question 11.
If A = $$\left[\begin{array}{rr} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right]$$ and A + A’ = I2, then the general value of x is
(a) nπ, n ∈ I
(b) (2n + 1) $$\frac{\pi}{2}$$, n ∈ I
(c) 2nπ ± $$\frac{\pi}{6}$$, n ∈ I
(d) 2nπ ± $$\frac{\pi}{3}$$, n ∈ I
Solution:
(d) 2nπ ± $$\frac{\pi}{3}$$, n ∈ I

Given A = $$\left[\begin{array}{rr} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right]$$
and A’ = $$\left[\begin{array}{rr} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right]$$
and A + A’ = I2
⇒ $$\left[\begin{array}{rr} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right]+\left[\begin{array}{rr} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
$$\left[\begin{array}{cc} 2 \cos x & 0 \\ 0 & 2 \cos x \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
⇒ 2 cos x = 1
⇒ cos x = $$\frac{1}{2}$$ = cos $$\frac{\pi}{3}$$
⇒ x = 2nπ ± $$\frac{\pi}{3}$$, n ∈ I.

Question 12.
If A = $$\left[\begin{array}{cc} \frac{1}{3} & 2 \\ 0 & 2 x-3 \end{array}\right][/late[latex]$$x] and B = $$\left[\begin{array}{rr} 3 & 6 \\ 0 & -1 \end{array}\right]$$ and AB = I2, then the value of x is
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

Given AB = I2
⇒ $$\left[\begin{array}{cc} \frac{1}{3} & 2 \\ 0 & 2 x-3 \end{array}\right]\left[\begin{array}{rr} 3 & 6 \\ 0 & -1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
⇒ $$\left[\begin{array}{cc} 1 & 0 \\ 0 & -2 x+3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Thus, their corresponding elements are equal.
∴ – 2x + 3 = 1
⇒ – 2x = -2
⇒ x = 1.

Question 13.
If A = $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]$$ and A2 – xA = I2, then the value of x is
(A) 4
(B) 2
(C) – 2
(D) – 4
Solution:
(A) 4

Given A2 – xA = I2
⇒ $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]-x\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
⇒ $$\left[\begin{array}{rr} 5 & 8 \\ 8 & 13 \end{array}\right]-\left[\begin{array}{rr} x & 2 x \\ 2 x & 3 x \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
⇒ $$\left[\begin{array}{rr} 5-x & 8-2 x \\ 8-2 x & 13-3 x \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Thus their corresponding elements are equal.
∴ 5 – x = 1
⇒ x = 4 and
8 – 2x = 0
⇒ x = 4
13 – 3x = 1
⇒ x = 4.

Question 14.
If A = $$\frac{1}{\pi}\left[\begin{array}{ll} \sin ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x) \end{array}\right]$$ and B = $$\frac{1}{\pi}\left[\begin{array}{cc} -\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x) \end{array}\right]$$ then A – B is equal to
(a) I
(b) $$\frac{1}{2}$$ I
(c) O
(d) 2I
Solution:
(b) $$\frac{1}{2}$$ I

A – B = $$\frac{1}{\pi}\left[\begin{array}{cc} \sin ^{-1} \pi x & \tan ^{-1} \frac{x}{\pi} \\ \sin ^{-1} \frac{x}{\pi} & \cot ^{-1} \pi x \end{array}\right]-\frac{1}{\pi}\left[\begin{array}{cc} -\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1} \frac{x}{\pi} & -\tan ^{-1}(\pi x) \end{array}\right]$$
= $$\frac{1}{\pi}\left[\begin{array}{cc} \sin ^{-1} \pi x+\cos ^{-1} \pi x & 0 \\ 0 & \cot ^{-1}(\pi x)+\tan ^{-1}(\pi x) \end{array}\right]$$
= $$\frac{1}{\pi}\left[\begin{array}{ll} \frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2} \end{array}\right]$$
= $$\frac{1}{\pi} \times \frac{\pi}{2}\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= $$\frac{1}{2}$$ I

Question 15.
If A = $$\left[\begin{array}{ll} a & b \\ b & a \end{array}\right]$$ and A2 = $$\left[\begin{array}{ll} \alpha & \beta \\ \beta & \alpha \end{array}\right]$$, then
(a) α = a2 + b2, β = ab
(b) α = a2 + b2, β = 2ab
(c) α = a2 + b2, β = a2 – b2
(d) α = 2ab, β = a2 + b2
Solution:
(b) α = a2 + b2, β = 2ab

⇒ $$\left[\begin{array}{ll} a & b \\ b & a \end{array}\right]\left[\begin{array}{ll} a & b \\ b & a \end{array}\right]=\left[\begin{array}{ll} \alpha & \beta \\ \beta & \alpha \end{array}\right]$$
⇒ $$\left[\begin{array}{cc} a^2+b^2 & 2 a b \\ 2 a b & b^2+a^2 \end{array}\right]=\left[\begin{array}{cc} \alpha & \beta \\ \beta & \alpha \end{array}\right]$$
Thus their corresponding entries are equal.
∴ α = a2 + b2
and β = 2ab

Question 16.
If A = , then An (where n ∈ N) is equal to
(a) $$\left[\begin{array}{ll} 1 & n a \\ 0 & 1 \end{array}\right]$$
(b) $$\left[\begin{array}{cc} 1 & n a \\ 1 & 0 \end{array}\right]$$
(c) $$\left[\begin{array}{cc} n & n a \\ 0 & n \end{array}\right]$$
(d) $$\left[\begin{array}{cc} 1 & n^2 a \\ 0 & 1 \end{array}\right]$$
Solution:
(a) $$\left[\begin{array}{ll} 1 & n a \\ 0 & 1 \end{array}\right]$$

A’ = $$\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]$$

∴ A2 = A . A
= $$\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{cc} 1 & 2 a \\ 0 & 1 \end{array}\right]$$

A3 = A2 . A
= $$\left[\begin{array}{cc} 1 & 2 a \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{cc} 1 & 3 a \\ 0 & 1 \end{array}\right]$$

Thus An = $$\left[\begin{array}{cc} 1 & n a \\ 0 & 1 \end{array}\right]$$

Aliter:
We shall use mathematical induction on n ∈ N
For n = 1 ;
A’ = $$\left[\begin{array}{cc} 1 & 1 \times a \\ 0 & 1 \end{array}\right]$$
result is true for n = 1
Let us assume that result is true for n = m
Am = 
Now Am+1 = Am – A
= $$\left[\begin{array}{cc} 1 & m a \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{cc} 1 & (m+1) a \\ 0 & 1 \end{array}\right]$$

∴ result is true for n = m + 1
Hence by mathematical induction result is true for all n ∈ N.

Question 17.
If A = $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$, n ∈ N then An is equal to
(a) nA
(b) 2nA
(c) 2n-1A
(d) 2nA
Solution:
(c) 2n-1A

A’ = $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$
∴ A2 = A . A
= $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right]$$
= 2 $$\left[\begin{array}{ll} 1 & 1 \\ 1& 1 \end{array}\right]$$
= 2A
and A3 = A2 . A
= 2A . A
= 2A2
= 2 (2A)
= 22 A
Thus, An = 2n-1A

Question 18.
If A = $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$ satisfies A5 = kA, then the value of k is
(a) 5
(b) 8
(c) 16
(d) 32
Solution:
(c) 16

Given A = $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$
∴ A2 = A . A
= $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right]$$

⇒ A2 = 2 $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$
= 2A

∴ A4 = A2 . A2
= (2A) (2A)
= 4A2
= 4 (2A)
= 8A
⇒ A5 = A4 . A
= (8A) A
= 8A2
= 8 (2A)
= 16A
Given A5 = KA
⇒ 16A = KA
⇒ K = 16.

Question 19.
If A and B are square matrices of same order, then AB’ – BA’ is a
(a) skew-symmetric matrix
(b) symmetric matrix
(c) null matrix
(d) unit matrix
Solution:
(a) skew-symmetric matrix

Let P = AB’ – BA’
∴ P’ = (AB’ – BA’)’
= (AB’)’ – (BA’)’
= BA’ – AB’
[∵ (A’)’ = A]
= – (AB’ – BA’)
= – P
Thus, P is skew symmetric.

Question 20.
If A and B are symmetric matrices of same order, then AB – BA is a
(a) symmetric matrix
(b) skew-symmetric matrix
(c) zero matrix
(d) identity matrix
Solution:
(b) skew-symmetric matrix

Given A and B are symmetric matrices of order
∴ A’ = A
and B’ = B
Let P = AB – BA
P’ = (AB – BA)’
= (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB
= – (AB – BA)
= – P
Thus, AB – BA is skew-symmetric matrix.

Question 21.
If A is a symmetric matrix and n ∈ N, then An is
(a) symmetric matrix
(b) a skew-symmetric matrix
(c) a diagonal matrix
(d) none of these
Solution:
(a) symmetric matrix

Given A is symmetric matrix
∴ A’ = A
∴ (An) = (A’)n
= An
∴ An is symmetric matrix.

Question 22.
If A is a skew-symmetric matrix and n is a positive even integer, then An is
(a) symmetric matrix
(b) a skew-symmetric matrix
(c) a diagonal matrix
(d) none of these
Solution:
(a) symmetric matrix

∴ A’ = – A …………….(1)
Let P = An
∴ P’ = (An)’
= (A’)n
= (- A)n
= (- 1)n An
= – P
[∵ n is positive even integer
∴ (- 1)n = 1]
Thus P i.e., An is symmetric matrix.

Question 23.
If A is skew-symmetric matrix and n is a odd positive integer, then An is
(a) symmetric matrix
(b) a skew-symmetric matrix
(c) a diagonal matrix
(d) none of these
Solution:
(b) a skew-symmetric matrix

Given A is skew-symmetric matrix
∴ A’ = – A ……………..(1)
Let P = An
P’ = (An)’
= (A’)n
= (- A)n
= (- 1)n An
= – P
[∵ n is odd positive integer
∴ (- 1)n = – 1]
Thus P is skew-symmetric matrix.

Question 24.
If A is a skew-symmetric matrix such that (An)’ = k An, n ∈ N, then the value of k is
(a) 1
(b) – 1
(c) (- 1)n
(d) n
Solution:
(c) (- 1)n

Given A is skew-symmetric matrix
∴ A’ = – A
⇒ (A’)n = kAn
⇒ (- A)n = kA
⇒ (- 1)n An = kAn
⇒ k = (- 1)n.

Question 25.
If A and B are square matrices of same order such that AB = A and BA = B, then A2 + B2 =
(a) AB
(b) A + B
(c) 2AB
(d) 2BA
Solution:
(b) A + B

Given AB = A and BA = B ………..(1)
∴ A2 + B2 = A . A + B . B
= (AB) A + (BA) B
= A (BA) + B (AB) [using (1)]
= A + B [using (1)]