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

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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 \([latex]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)[/latex]
= – 5I.

Question 7.
If A = [a_ij]_{2 × 2} where a_ij = \(\begin{cases}1, & \text { if } i \neq j \\ 0, & \text { if } i=j\end{cases}\), 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
A^m = \(\)
Now A^m+1 = A^m – 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) 2^n-1A
(d) 2ⁿA
Solution:
(c) 2^n-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 = 2^n-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)]