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|A. adj. A| = |A||adj A| = |A||A|^n-1 = |A|^1+n-1 = |A|ⁿ = 23 = 8

Students can practice these questions for the board examination to attain well. The important MCQ Questions are taken as per the syllabus of CBSE board. Important Class 12 Maths MCQ Questions of Matrices with Answers are provided here. These Multiple-choice Questions, which are asked within the previous year’s board examination. 

Practice the important MCQ Questions for Class 12 Maths to score good marks in the board examinations. Be thorough with these MCQ Questions of Matrices, as these objective-type questions are taken from the previous year’s Maths examination papers.

Practice MCQ Question for Class 12 Maths chapter-wise

1. If a matrix has 6 elements, then number of possible orders of the matrix can be

(a) 2
(b) 4
(c) 3
(d) 6

2. Total number of possible matrices of order 2 × 3 with each entry 1 or 0 is

(a) 6
(b) 36
(c) 32
(d) 64

3. The restrictions on n, k and p so that PY + WY will be defined are

(a) k = 3, p = n
(b) k is arbitrary, p = 2
(c) p is arbitrary
(d) k = 2,p = 3.

4. If A is a square matrix such that A²=A, then (I + A)² – 3A is

(a) I
(b) 2A
(c) 3I
(d) A

5. If the order of matrix A is m×p. And the order of B is p×n. Then the order of matrix AB is ?

(a) n × p
(b) m × n
(c) n × p
(d) n × m

6. Transpose of a rectangular matrix is a

(a) rectangular matrix
(b) diagonal matrix
(c) square matrix
(d) scaler matrix

7. A square matrix in which all elements except at least one element in diagonal are zeros is said to be a

(a) identical matrix
(b) null/zero matrix
(c) column matrix
(d) diagonal matrix

8. If A = [a_ij]_{m × n} is a square matrix, if:

(a) m < n
(b) m > n
(c) m = n
(d) None of these.

9. If matrices A and B are inverse of each other then

(a) AB = BA
(b) AB = BA = I
(c) AB = BA = 0
(d) AB = 0, BA = I

10. The diagonal elements of a skew symmetric matrix are

(a) all zeroes
(b) are all equal to some scalar k(≠ 0)
(c) can be any number
(d) none of these

11. If a matrix A is both symmetric and skew-symmetric then matrix A is

(a) a scalar matrix
(b) a diagonal matrix
(c) a zero matrix of order n × n
(d) a rectangular matrix.

12. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is

(a) 27
(b) 18
(c) 81
(d) 512.

13. In matrices, columns are denoted by

(a) A
(b) B
(c) R
(d) C

14. In matrices (AB)⁻¹ equals to

(a) A⁻¹
(b) B⁻¹
(c) A⁻¹ B⁻¹
(d) B ^{- 1} A ^{- 1}

15. If A and B are two symmetric matrices of the same order. Then, the matrix AB - BA is equal to

(a) a symmetric matrix
(b) a skew-symmetric matrix
(c) a null matrix
(d) the identity matrix

16. If A is a square matrix, then A – A’ is a

(a) diagonal matrix
(b) skew-symmetric matrix
(c) symmetric matrix
(d) none of these

17. If A is any square matrix, then which of the following is skew-symmetric?

(a) A + A^T
(b) A – A^T
(c) AA^T
(d) A^TA

18. For any square matrix A, AA^T is a

(a) unit matrix
(b) symmetric matrix
(c) skew-symmetric matrix
(d) diagonal matrix

19. If A is a matrix of order 3 × 4 , then each row of A has

(a) 12 elements
(b) 3 elements
(c) 4 elements
(d) 7 elements

20. If n =p, then the order of the matrix 7X – 5Z is:

(a) p × 2
(b) 2 × n
(c) n × 3
(d) p × n.

Answer:

1. Answer: (b) 4

Explanation: As 6 → 1 × 6, 2 × 3, 3 × 2, 6 × 1.

2. Answer: (d) 64

Explanation: The order of the matrix =2×3

The number of elements =2×3=6

Each place can have either 0 or 1.

Each place can be filled in 2 ways.

Thus, the number of possible matrices
=2⁶ =64

3. Answer: (a) k = 3, p = n

Explanation: In this, order of P=p×k
Order of W=n×3
Order of Y=3×k
Thus, order of PY=p×k, when k=3.
And the order of WY=p×k, where p=n

4. Answer: (a) I

Explanation: (a), as (I + A)² -3A 

= I² + IA + AI + A² – 3A 

= I + A + A + A – 3A=I

5. Answer: (b) m × n

Explanation: If A is matrix of order m×n and B is a matrix of order n×p, then the order of AB is m×p.

6. Answer: (a) rectangular matrix

Explanation: If we exchange the role of the row and column of a matrix then we transpose it. This operation is called transposition and the matrix is called transposition matrix.
If we take a rectangular matrix and transpose it then we get a rectangular matrix.

7. Answer: (d) diagonal matrix

Explanation: A diagonal matrix is a matrix in which all of the elements not on the diagonal of a square matrix are 0.

8. Answer: (c) m = n

Explanation: MatrixA=[a_ij]_m×n square matrix

Number of columns = number of rows

n=m

9. Answer: (b) AB = BA = I

Explanation: We know that if A is a square of order m, and if there exists another square matrix B of the same order m, such that AB=I, then B is said to be the inverse of A. 
In this case, it is clear that A is the inverse of B.
Thus , matrices A and B will be inverses of each other only if AB=BA=I.

10. Answer: (a) all zeroes

Explaination: As in skew symmetric matrix, a_ij = -a_ji

⇒ a_ii = – a_ii

⇒ 2a_ii = 0

⇒ a_ii = 0, i.e. diagonal elements are zeroes.

11. Answer: c

Explanation: If matrix A is symmetric
A^T=A
If matrix A is skew-symmetric
A^T=−A
Also, diagonal elements are zero
Now, it is given that a matrix A is both symmetric as well as skew-symmetric
∴A=A^T=−A
which is only possible if A is zero matrix
A=[(0,0)(0,0)]=A^T=−A

12. Answer: (d) 512

Explanation: The number of possible entries of 3×3 matrix is 9.

Every entry has two choices, 0 or 1

Thus, total number of choices is

2×2×2×2×2×2×2×2×2 = 2⁹ = 512

13. Answer: (d) C

14. Answer: (d) B ^{- 1} A ^{- 1}  

15. Answer: (b) a skew-symmetric matrix

Explanation: A and B are two symmetric matrices.

A = A’ and B = B’

(AB – BA)’ = (AB)’ – (BA)’ = B’A’ – A’B’

= BA – AB = – (AB – BA)

(AB – BA)’ = – (AB – BA)

AB – BA is a skew-symmetric matrix.

16. Answer: (b) skew-symmetric matrix

Explanation: Let's say that B = A - A', where A is a square matrix. We know that if the transpose is equal to the negative of a matrix, then it is a skew-symmetric matrix.

17. Answer: (b) A – A^T  

18. Answer: (b) symmetric matrix

Explanation: Let A be any matrix.

Also let B=AA^T.

Now B^T=(AA^T)^T=(A^T)^TA^T=AA^T=B. [ Since (A^T)^T=A]

So AA^T is a symmetric matrix.

19. Answer: (c) 4 elements

Explanation: Its order is 3*4 which means it has three rows and four columns So every row has four elements.

20. Answer: (b) 2 × n

Explanation: Given, order of X = 2×n
and order of Z = 2×p
Therefore, n=p
Hence the order of 7X−5Z = 2×n.

Click Here to Practice more MCQ Question for Matrices Class 12

Given,

A = B + C, BC = CB and C² = O. 

We need to prove that,

Aⁿ⁺¹ = Bⁿ(B + (n + 1)C). 

We will prove this result using the principle of mathematical induction.

Step 1: 

When n = 1, we have Aⁿ⁺¹ = A¹⁺¹ 

⇒ Aⁿ⁺¹ = B¹(B + (1 + 1)C) 

∴ Aⁿ⁺¹ = B(B + 2C)

For the given equation to be true for n = 1,

Aⁿ⁺¹ must be equal to A². 

It is given that,

A = B + C 

And we know, 

A² = A × A.

⇒ A² = (B + C)(B + C) 

⇒ A² = B(B + C) + C(B + C) 

⇒ A² = B2 + BC + CB + C² 

However, 

BC = CB and C² = O. 

⇒ A² = B² + CB + CB + O 

⇒ A² = B² + 2CB 

∴ A² = B(B + 2C) 

Hence, 

Aⁿ⁺¹ = A² and the equation is true for n = 1.

Step 2 : 

Let us assume the equation true for some n = k, where k is a positive integer. 

⇒ A^k+1 = B^k (B + (k + 1)C) 

To prove the given equation using mathematical induction, we have to show that,

A^k+2 = B^k+1(B + (k + 2)C). 

We know,

A^k+2 = A^k+1 × A. 

⇒ A^k+2 = [B^k(B + (k + 1)C)](B + C) 

⇒ A^k+2 = [B^k+1 + (k + 1)B^kC)](B + C) 

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^kC(B + C) 

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^kCB + (k + 1)B^kC² 

However, 

BC = CB and C² = O. 

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^kBC + (k + 1)B^kO 

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^k+1C + O 

⇒ A^k+2 = B^k+1(B + C) + B^k+1[(k + 1)C] 

⇒ A^k+2 = B^k+1[(B + C) + (k + 1)C] 

⇒ A^k+2 = B^k+1[B + (1 + k + 1)C] 

∴ A^k+2 = B^k+1[B + (k + 2)C] 

Hence, 

The equation is true for n = k + 1 under the assumption that it is true for n = k. 

Therefore, 

By the principle of mathematical induction, the equation is true for all positive integer values of n. 

Thus, 

Aⁿ⁺¹ = Bⁿ(B + (n + 1)C) for every n ϵ N.

The given matrix is of order 3 x 3, therefore |3I₃| = 3³|I₃| = 27 |I₃| = 27 x 1 = 27 

We know that |adj A| = |A|^n–1, 

n is order of the matrix

∴ 64 = |A|²

⇒ |A| = ± 8.

(A²)^–1 = (A.A)^–1  [ ∵ (AB)^–1 = B^–1 A^–1 ]

= A^–1.A^–1 

= (A^–1)²

We have,

A² – A + I = 0

Pre-multiply both sides, by A^–1, we get

A^–1AA – A^–1A + AI = 0

⇒ A – I + A^–1 = 0

⇒ A^–1 = I – A

Since

|adj A| = |A|^{n – 1}, where n is order of matrix A

⇒ 81 =  |A|^{3 – 1}

⇒ 81 = |A|²

∴ |A| = ± 9

We have,

(A^T)-1 = (A^-1)^T

= \(\begin{bmatrix} 5 &3 \\[0.3em] -2 & -1 \\[0.3em] \end{bmatrix}^T\)

= \(\begin{bmatrix} 5 &-2 \\[0.3em] 3 & -1 \\[0.3em] \end{bmatrix}.\)

|adj A| = |A|²

[ ∵ |adj A| = |A|n-1, where n is the order of matirx.]

\(=\begin{bmatrix} a & 0 & 0 \\[0.3em] 0 & a & 0 \\[0.3em] 0 & 0 & a \end{bmatrix}^2\)

= (a³)²

= a⁶

[∵ Determinant of a triangular matrix is equal to the product of its diagonal elements].

The relation is adj AB = (adj B) (adj A)

Given : 

A = [a_ij] is a skew – symmetric matrix

⇒ a_ij = - a_ji …(i)

[for all values of i, j]

For diagonal elements,

⇒ aii = - a_ii 

[for all values of i] 

⇒ a_ii + a_ii = 0 

⇒ 2a_ii = 0 

⇒ a_ii = 0 …(ii)

Now,

\(\displaystyle\sum_{i}\)\(\displaystyle\sum_{j} a_{ij}=\) a₁₁ + a₁₂ + a₁₃...+ a₂₁ + a₂₂ + a₂₃ ... + a₃₁ + a₃₂ + a₃₃ ...

= 0 + a₁₂ + a₁₃ +...+ (- a₁₂) +0 + a₂₃ +... + (-a₁₃) + (-a₂₃) + 0...

[from (i) and (ii)]

= 0 

Thus,

 \(\displaystyle\sum_{i}\)\(\displaystyle\sum_{j} a_{ij}=0\) 

Hence Proved.

Here, 

We have any square matrix 

To Find : AA^T is symmetric or skew – symmetric

Proof : Firstly, we take the transpose of AA^T, so we get 

(AA^T)^T = (A^T)^T A^T

[∵ (AB)^T = B^TA^T] 

⇒ (AA^T)^T = AA^T 

[∵ (A^T)^T = A] 

∴ AA^T is a symmetric matrix

True

(AA’)’=(A’)’A’

As we know (A^’)^’ = A

(AA’)’=AA’ (Condition of symmetric matrix)

True

For skew symmetric matrix A’=-A

⇒ (A²)’=(AA)’=A’A’

⇒ A’A’=(-A)(-A)=A²

⇒ (A²)’=A² (symmetric matrix)

False

∵ If A and B are invertible matrices then,

⇒ (AB)^-1=B^-1A^-1

Correct option is : (c) x = 1, y = 2

Correct option is : (b) \(\begin{vmatrix}4&3\\4&6\end{vmatrix} \)