Chapter 3: Matrices

For a given matrix of order m x n, it has mn elements, where m and n are natural numbers.

Here we have, m x n = 28

(m, n) = {(1, 28), (2, 14), (4, 7), (7, 4), (14, 2), (28, 1)}

So, the possible orders are 1 x 28, 2 x 14, 4 x 7, 7 x 4, 14 x 2, 28 x 1.

Also, if it has 13 elements, then m x n = 13

(m, n) = {(1, 13), (13, 1)}

Thus, the possible orders are 1 x 13, 13 x 1.

For the given matrix,

(i) The order of the matrix A is 3 x 3.

(ii) The number of elements of the matrix = 3 x 3 = 9

(iii) Elements: a₂₃ = x² – y, a₃₁ = 0, a₁₂ = 1

We have,

A = [a_ij]_2×2

(i) Such that, a_ij = (i – 2j)²/ 2; where 1 ≤ i ≤ 2; 1 ≤ j ≤ 2

So, the terms of the matrix are

(ii) Here, a_ij = |-2i + 3j|

So, the terms of the matrix are

Let A be a 3 x 2 matrix

Such that, a_ij = e^i.xsin jx; where where 1 ≤ i ≤ 3; 1 ≤ j ≤ 2

So, the terms are given as

Given, matrix A = matrix B

Then their corresponding elements are equal.

So, we have

a₁₁ = b₁₁; a + 4 = 2a + 2 ⇒ a = 2

a₁₂ = b₁₂; 3b = b² + 2 ⇒ b² – 3b + 2 = 0 ⇒ b = 1, 2

a₂₂ = b₂₂; -6 = b² – 5b ⇒ b² – 5b + 6 = 0 ⇒ b = 2, 3

Hence, a = 2 and b = 2 (common value)

The given two matrices, A and B, are of different orders. Two matrices can be added only if the order of both matrices is the same. Thus, the sum of matrices A and B is not possible.

Given,

On comparing the corresponding elements, we get

2x + 10x = 48

12x = 48

Thus, x = 4

It’s also seen that this value od x also satisfies the equation 3x + 8 = 20 and x² + 8x = 12x.

Therefore, x = 4 (common) is the solution of the given matrix equation.

Hence, from (i) and (ii),

(A + B) (A – B) ≠ A² – B²

x² + 16x + 28 = 0

(x + 2) (x + 14) = 0

Therefore, x = -2, -14