# NCERT 12 Maths Matrices Chapter 3 – Exercise 3.1 Sol. (i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4.

(ii) Since, the order of the matrix is 3 × 4, so there are 3 × 4 = 12 elements in it. 2. If the matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Sol. We know that if a matrix is of the order m × n, it has mn elements. Hence, a matrix containing 24 elements can have any one of the following orders:

1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2 or 24 × 1.

Similarly, a matrix containing 13 elements can have order 1 × 13 or 13 × 1.

3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Sol. Since, a matrix containing 18 elements can have any one of the following orders:

1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, 6 × 3

Similarly, a matrix containing 5 elements can have order 1 × 5 or 5 × 1          as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get

x+ y = 6   ……(i)

5 + z = 5   ……(ii)

And xy = 8    …..(iii)

From eq. (ii), we get z = 0

From eq. (i),  y = 6-x    …..(iv)

Substituting the value of y in eq. (iii), we obtain

x(6-x) = 8 → x2 -6x+8 = 0

→ (x-2)(x-4) = 0 → x=2 or x=4

When x=2, then from eq. (iv), y = 6-2 = 4 and

When x=4, then from eq. (iv), y = 6-4 = 2

So, either x=2, y=4 and z=0

Or x=4, y=2 and z=0 as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get

x+y+z = 9     …..(i)

x+z = 5       ……..(ii)

y+z = 7      ……….(iii)

Subtracting eq. (ii) from eq. (i), we get y=4

Subtracting eq. (iii) from eq. (i), we get x=2

Substituting y=4 in eq. (iii), we get

4+z = 7 → z = 7-4 = 3

Hence, x = 2, y = 4, z = 3  By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get

a-b = -1      …….(i)

2a-b = 0      …….(ii)

2a+c = 5      …….(iii)

and 3c+d = 13   …….(iv)

Subtracting eq. (i) from eq. (ii), we get a = 1

Putting a = 1 in eq. (i) and eq. (iii), we get

1-b = -1  and 2+c = 5

→ b = 2 and c = 3

Substituting c = 3 in eq. (iv), we obtain

33+d = 13 → d = 13-9 = 4

Hence, a=1, b=2, c=3 and d=4 Sol. Since, in a square matrix, the number of rows is equal to the number of columns, therefore, we must have m = n

Therefore,  A=

So, correct option is (c)  Since, x can have only one value at a time. Hence, it is not possible to find the values of x and y for which the given matrices are equal.

So, correct option is (b).

Note: Sometimes on solving an equation, we get more than one values of the variables. This means that such a matrix does not exist.

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

• 27 (b) 18    (c) 81    (d) 512
Sol. As order of 3 × 3 matrix contains 9 elements. Each element can be selected in 2 ways (it can be either 0 or 1).

Hence, all the nine entries can be chosen in 29 = 512 ways.

(By the multiplication principle)

Required number of the matrices in 512. So, the correct option is (d).

Updated: October 16, 2020 — 5:38 pm