Important Questions for Class 12 Maths Chapter 3 Matrices

Very Short Answer Questions [1 Mark Questions]

Ques. Define “Square Matrix.”

Ans. A square matrix is one in which the number of rows equals the number of columns, i.e. m = n.

Ques. Define the Diagonal matrix and the principal diagonal of a matrix.

Ans. A diagonal matrix is a square matrix in which every non-diagonal entry is zero. The principal diagonal of a matrix is defined as the first diagonal element of the first row to the final diagonal element of the last row in a square matrix.

Ques. What are the possible orders of a matrix that has 8 elements?

Ans. Number of rows x number of columns = 8.
The possible number of rows and columns are the possible factors of 4.
Therefore, 1 x 8, 8 x 1, 4 x 2, 2 x 4 are the possible orders of a matrix that has 8 elements.

Ques. Provide an example of a matrix A and B where, AB = 0 but A ≠ 0, B ≠ 0

Ans. \(\alpha - \beta\)

A = [ 0 -1 0 2 ], B = [ 3 5 0 0 ]

AB = [ 0 0 0 0 ]

Ques. A = [ay] m × n is a square matrix, if:

Ans. A matrix is said to be a square matrix if the number of rows equals the number of columns. Therefore, A = [ay] m × n is a square matrix, conditioned that, m=, where m is the number of rows and n denotes the number of columns.

Ques. What are all the possible orders a matrix can have if it has 28 elements?

Ans. To write all possible orders,
So, number of rows x number of columns = 28
Where the possible number of rows and columns are the possible factors of 28.
Therefore, 1 x 28, 2 x 14, 4 x 7, 7 x 4, 14 x 2, 28 x 1 are the possible ordered pairs of 28 elements.

Also read:

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Short Answer Questions [2 Mark Questions]

Ques. If one given matrix A is both symmetric and also skew-symmetric, then find the value of A.

Ans. For any square matrix taken, we know that A+A^T is symmetric and A-A^T is skew-symmetric.

A₁ = A 

A₁ = -A

A = -A 

2A = 0 

Thus, A = 0

Ques. Give the elements a₂₃ of a 3 x 3 matrix A = [aij] whose elements atj are given by: |i − j|2

Ans. I and j are used to denote the row and column number
We have [aij] = |i−j|2

∴ a₂₃ = |2−3|2=|−1|2 = 12

Ques. Prove that AA 1 is symmetric if A is any square matrix.

Ans.
Let P = AA’
P’ = (AA)’
= [ (A’)’ A’]
= AA’
= P, Hence, Proved.

Ques. If A and B are symmetric matrices of the same order, prove that AB + BA is symmetric

Ans. Let P= AB + BA
P’ = ( AB + BA )’
= ( AB)’ + ( BA)’
= A’B’ + B’A’
= BA + AB ( A= A’, B’= B)

= AB + BA

= P

Hence, proved that AB + BA is symmetric.

Ques. What possible orders can a matrix have if it has 24 elements? What if it has 13 elements?

Ans. The order of a matrix is in the form of m x n where m is the number of rows and n is the number of columns. 

To find the possible orders of a matrix, we will have to find all the ordered pairs of natural numbers that are factors of 24.

1x 24).(24×1). (2×12). (12×2). (3×8).(8×3). (4×6). (6×4) are all the possible ordered pairs.

Also, for 13 elements, since the product of factors can be only 1 and 13 so the possible ordered pairs are ( 1 × 13 ) and ( 13 × 1 ).

Ques. A = [ 2 5 4 6 ], Prove that A + A’ is a symmetric matrix.

Ans. P = A + A’

= [ 2 5 4 6 ] + [ 2 4 5 6 ]

P = [ 4 9 9 12 ]

P’ = [ 4 9 9 12 ]

P = P’ Hence, proved.

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Long Answer Type Questions [3 Marks Questions]

Ques. Solve for x [ 2 -3 1 1 ][ x 4 ]  = [ 1 3 ]

Ans. [ 2x -3y x+ y ] = [ 1 3 ]

2x – 3y = 1 

x + y = 3 

x = 3 – y 

2 (3 – y) – 3y = 1

-5y = -5 

y = 1 

x = 3 – 1 

Thus, x = 2

Ques. Solve for x and y, given that [ x y 3y x ][ 1 2 ]  = [ 3 5 ]
[ x y 3y x ] = [ 3 5 ]

Ans.

x + 2y = 3

3y + 2x = 5

2x + 4y = 6

2x + 3y = 5

y =1 

x + 2 (1) = 3 

Therefore, x = 1

Ques. What values of x and y will make the following 

[ 3x + 75y + 12 – 3x ] = [ 0 y-2 8 4 ]

Ans. We will equate these equations as they are equal, 

So, 

3x + 7=0

x = -7⁄3

y - 2 = 5

y = 7 

y + 1 = 8

y = 7

2 - 3x = 4

\(X = \frac{-2}{3}\)

Since, there cannot be two values of x and y, therefore, they are impossible to find.

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Very Long Answer Type Questions [5 Marks Questions]

Ques. If A = [ cos cos a a sin sin a cos cos a ] and A + A’ = I, then find the value of α

Ans. It is given that A + A’ = I, therefore, we shall first equate this in the matrix form by substituting A. 

[cos cos a a sin sin a cos cos a] + [cos cos a sin sin a a cos cos a] = [1 0 0 1]

On Adding the two matrices together we get,

On equating the terms to find the value of α

cos a =\(\frac{1}{2}\)

We know that, \(cos\frac{\pi}{3}=\frac{1}{2}\)

Therefore \(\text{cos a} = \text{cos cos}\frac{\pi}{3}\)

\(\alpha = \frac{\pi}{3}\)

Answer, Hence, the value of \(\alpha = \frac{\pi}{3}\)

Ques. Find a matrix A in which, 2A – 3B + 5C = 0,

And Where B = [ -2 2 0 3 1 4 ]  and C = [ 2 0 -2 7 1 6 ]

Ans. First we shall take the equation 2A – 3B + 5C = 0
2A = 3B – 5C
2A = 3 [-2 2 0 3 1 4] – 5 [2 0 -2 7 1 6]

= [-6 6 0 9 3 12] + [-10 0 10 35 -5 -30] = [-6-10 6+0 0+10 9-35 3-5 12-30]

Therefore, we get 2A = [-16 6 10 -26 -2 -18]

Hence, A = [-8 3 5 -13 -1 -9]

Ques. Using the elementary transformations, find the inverse of each of the matrices given, conditioned that it exists.

[2 0 -1 5 1 0 0 1 3]

Ans. Using the row operation,

Consider, A = [2 0 -1 5 1 0 0 1 3]

We know, A = IA

[2 0 -1 5 1 0 0 1 3] = [1 0 0 0 1 0 0 0 1]
So, using equation, R₂ → R₂ – 2R₁
R₁ ⇔ R₂

[1 1 2 2 0 -1 0 1 3] = [-2 1 0 1 0 0 0 0 1] A

So, using equation, R₂ → R₂ – 2R₁

R₂ ⇔ R₃

[1 1 2 0 1 3 0 -2-5] = [-2 1 0 5 -2 0 0 0 1] A

So, using equation, R₁ → R₁ – R₂   R₃ → R₃ + 2R₃

[1 0 -1 0 1 3 0 0 1] = [-2 1-1 0 0 1 5 -22] A

So, using equation, R₁ → R₁ + R₃   R₂ → R₂ – 3R₃

[1 0 0 0 1 0 0 0 1] = [3 -1 1 -15 6 -5 5 -2 2] A

A^-1 = [3 -1 1 -15 6 -5 5 -2 2]