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Class 12 Maths Chapter 3 Matrices MCQs

Class 12 Maths Chapter 3 Matrices MCQs are given here to help the students for scoring maximum marks in the board exam 2022-2023. All these multiple-choice questions are prepared based on the NCERT curriculum and as per the latest guidelines released by CBSE for the Term 1 Class 12 Maths exam. Matrix is one of the important concepts of mathematics, so learning matrices will help not only for board exams but also compete with other exams. Thus, objective-type questions on matrices are a useful resource to recall the concepts and practise at any time.

Get Class 12 Maths MCQs for all the chapters here.

MCQs for Chapter 3 Matrices Class 12

Students can access several MCQs on matrices of Class 12 that cover various topics such as fundamentals of matrix and matrix algebra, i.e. mathematical operations on matrices.

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Practising Matrices MCQs based on the Class 12 maths syllabus will help students to score good marks in the board exam 2022-23. Here, you will get multiple-choice questions on matrices and correct options and explanations so that one can practise and verify answers. This kind of preparation will help boost your confidence to attempt any type of question in the examination.

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MCQs for Class 12 Maths Chapter 3 Matrices with Solutions

1. If A is a square matrix such that A2 = A, then (I – A)3 + A is equal to

(a) I

(b) 0

(c) I – A

(d) I + A

Correct option: (a) I

Solution:

Given that, A is a square matrix and A2 = A.

Consider (I + A)3, where I is the identity matrix.

Using the identity of (a + b)3 = a3 + b3 + 3ab (a + b), we get;

(I + A)3 = I3 + A3 + 3A2I + 3AI2

= I + A2(A) + 3AI + 3A

= I + A2 + 3A + 3A

= 7A + I {since it is given that A2 = A}

So, (I + A)3 = 7A + I….(1)

Now,

(I + A)3 – 7A = 7A + I – 7A [From (1)]

= I

2. If A = [aij] is a square matrix of order 2 such that aij = 1, when i ≠ j and aij = 0, when i = j, then A2 is

\(\begin{array}{l}(a) \ \begin{bmatrix} 1 & 0\\ 1& 0 \end{bmatrix}\end{array} \)
\(\begin{array}{l}(b) \ \begin{vmatrix} 1 & 1\\ 0 & 0 \end{vmatrix}\end{array} \)
\(\begin{array}{l}(c) \ \begin{vmatrix} 1 & 1\\ 1 & 0 \end{vmatrix}\end{array} \)
\(\begin{array}{l}(d) \ \begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}\end{array} \)
\(\begin{array}{l}\text{Correct option: (d) }\begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}\end{array} \)

Solution:

Given,

A = [aij] is a square matrix of order 2 such that aij = 1, when i ≠ j and aij = 0, when i = j.

So, a11 = 0, a12 = 1, a21 = 1 and a22 = 0.

Thus,

\(\begin{array}{l}A=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\end{array} \)

A2 = A.A

\(\begin{array}{l}A^2=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}.\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 0+1 & 0+0\\ 0+0 & 1+0 \end{bmatrix}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\end{array} \)

3. Total number of possible matrices of order 3 × 3 with each entry 2 or 0 is

(a) 9

(b) 27

(c) 81

(d) 512

Correct option: (d) 512

Solution:

We know that a matrix 3× 3 contains 9 elements.

Given that each entry of this 3× 3 matrix is either 0 or 2.

Thus, by simple counting principle, we can calculate the total number of possible matrices as:

Total number of possible matrices = Total number of ways in which 9 elements can take possible values

= 29

= 512

4. If A and B are two matrices of the order 3 × m and 3 × n, respectively, and m = n, then the order of matrix (5A – 2B) is

(a) m × 3

(b) 3 × 3

(c) m × n

(d) 3 × n

Correct option: (d) 3 × n

Solution:

Given that, the order of matrix A is 3 × m, and the order of B is 3 × n.

Also, m = n.

So, the order of matrix A and B is the same, i.e. 3 × m.

Thus, subtraction of matrices is possible and (5A – 3B) also has the same order, i.e. 3 × n.

\(\begin{array}{l}\text{ 5. If  }\begin{bmatrix} 2p+q & p-2q\\ 5r – s& 4r + 3s \end{bmatrix} = \begin{bmatrix} 4 & -3\\ 11& 24 \end{bmatrix}, \text{ then the value of p + q – r + 2s is }\end{array} \)

(a) 8

(b) 10

(c) 4

(d) -8

Correct option: (a) 8

Solution:

Given,

\(\begin{array}{l}\begin{bmatrix} 2p+q & p-2q\\ 5r – s& 4r + 3s \end{bmatrix} = \begin{bmatrix} 4 & -3\\ 11& 24 \end{bmatrix}\end{array} \)

Now, by equating the corresponding elements of these two matrices, we get;

2p + q = 4….(1)

p – 2q = -3….(2)

5r – s = 11….(3)

4r + 3s = 24….(4)

By equ (1) × 2 + equ (2), we get;

4p + 2q + p – 2q = 8 – 3

5p = 5

p = 1

Substituting p = 1 in (1),

2 + q = 4

q = 4 – 2 = 2

By equ (3) × 3 + equ (4), we get;

15r – 3s + 4r + 3s = 33 + 24

19r = 57

r = 3

Substituting r = 3 in (3),

15 – s = 11

s = 15 – 11 = 4

Now,

p + q – r + 2s = 1 + 2 – 3 + 2(4) = 8

\(\begin{array}{l}\text{6. The matrix  }\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 4 \end{bmatrix} \text{ is a }\end{array} \)

(a) identity matrix

(b) symmetric matrix

(c) skew symmetric matrix

(d) none of these

Correct option: (b) symmetric matrix

Solution:

Let the given matrix be:

\(\begin{array}{l}A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 4 \end{bmatrix}\end{array} \)

Let us find the transpose of A.

\(\begin{array}{l}A^T=\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 4 \end{bmatrix}^T=\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 4 \end{bmatrix}= A\end{array} \)

Therefore, A is a symmetric matrix.

7. For any two matrices A and B, we have

(a) AB = BA

(b) AB ≠ BA

(c) AB = O

(d) None of the above

Correct option: (d) None of the above

Solution:

For any two matrices A and B,

AB = BA and AB ≠ BA are not valid unless they follow the condition of matrix multiplication.

Also, AB = O is not true in all cases.

8. If A and B are symmetric matrices of the same order, then (AB′ –BA′) is a

(a) Skew symmetric matrix

(b) Null matrix

(c) Symmetric matrix

(d) None of these

Correct option: (a) Skew symmetric matrix

Solution:

Given that A and B are symmetric matrices of the same order.

Let’s find the transpose of (AB′ –BA′).

(AB′ –BA′)′ = (AB′)′ – (BA′)′

= (BA′ – AB′)

= – (AB′ –BA′)

As (AB′ –BA′)′ = – (AB′ –BA′), the matrix (AB′ –BA′) is skew symmetric.

9. If A is a skew-symmetric matrix, then A2 is a

(a) Skew symmetric matrix

(b) Symmetric matrix

(c) Null matrix

(d) Cannot be determined

Correct option: (b) Symmetric matrix

Solution:

Given that A is a skew-symmetric matrix, so A′ =-A.

Consider the transpose of A2.

(A2)′ = (AA)′

= A′A′

=(-A)(-A)

= A2

⇒ (A2)′ = A2

Therefore, A2 is a symmetric matrix.

\(\begin{array}{l}\text{10. If  }A=\begin{bmatrix} 0 & 2\\ 3 & -4 \end{bmatrix} \text{ and } kA=\begin{bmatrix} 0 & 3a\\ 2b & 24 \end{bmatrix}, \text{ then the values of k, a and b respectively are}\end{array} \)

(a) -6, -12, -18

(b) -6, -4, -9

(c) -6, 4, 9

(d) -6, 12, 18

Correct option: (b) -6, -4, -9

Solution:

Given,

\(\begin{array}{l}A=\begin{bmatrix} 0 & 2\\ 3 & -4 \end{bmatrix}\end{array} \)

And

\(\begin{array}{l}kA=\begin{bmatrix} 0 & 3a\\ 2b & 24 \end{bmatrix}\end{array} \)

Now,

\(\begin{array}{l}kA=\begin{bmatrix} 0 & 2k\\ 3k & -4k \end{bmatrix}\end{array} \)

Thus,

\(\begin{array}{l}\begin{bmatrix} 0 & 2k\\ 3k & -4k \end{bmatrix}=\begin{bmatrix} 0 & 3a\\ 2b & 24 \end{bmatrix}\end{array} \)

By equating the corresponding elements,

-4k = 24

k = -6

Also, 2k = 3a

2(-6) = 3a

3a = -12

a = -4

And

3k = 2b

3(-6) = 2b

2b = -18

b = -9

Therefore, k = -6, a = -4, and b = -9.

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