Exercise-3.1, Class 12th, Maths, Chapter 3- Matrices, NCERT

Exercise 3.1


Question 1:

In the matrix

A=[251971517312125]

write:
(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.

Answer:
(i) Order = 3 × 4
(ii) Number of elements = 3 × 4 = 12
(iii) a₁₃ = 19, a₂₁ = 1, a₃₃ = 12, a₂₄ = 3, a₂₃ = 17.


Question 2:

If a matrix has 24 elements, what are the possible orders it can have? What if it has 13 elements?

Answer:
We know that if a matrix has m × n elements, then total elements = m × n.

(i) For 24 elements:
Possible orders = (1,24), (2,12), (3,8), (4,6), (6,4), (8,3), (12,2), (24,1)

(ii) For 13 elements (prime number):
Possible orders = (1,13), (13,1)


Question 3:

If a matrix has 18 elements, what are its possible orders? What if it has 5 elements?

Answer:
(i) For 18 elements: m×n=18
Possible orders = (1,18), (2,9), (3,6), (6,3), (9,2), (18,1)

(ii) For 5 elements (prime): (1,5), (5,1)


Question 4:

Construct a 2 × 2 matrix A = [aᵢⱼ] where:

(i) aᵢⱼ = 2i − j
(ii) aᵢⱼ = i² − 3j
(iii) aᵢⱼ = i² / 2j

Answer:
For i, j = 1, 2

(i)

A=[2(1)12(1)22(2)12(2)2]=[1032]

(ii)

A=[123(1)123(2)223(1)223(2)]=[2512]

(iii)

A=[122×1122×2222×1222×2]=[121421]


Question 5:

Construct a 3 × 4 matrix where:
(i) aᵢⱼ = ½ (i − 3j)
(ii) aᵢⱼ = 2i − j

Answer:

(i)

A=[12.545.50.523.5501.534.5]

(ii)

A=[101232105432]


Question 6:

Find the values of x, y, and z from the following equations:

(i)

[x5zy15]=[262158]

(ii)

[5zx+y58x+y]=[x+zyzxzyz]

Answer:

(i) Comparing elements:
x = 2, y = 1, z = 2.

(ii) On comparing:
x = 3, y = 1, z = 2.


Question 7:

Find a, b, c, d from the equation:

[ab2ab2a+c3cd]=[015013]

Answer:
Equating elements:
a − b = 0
2a − b = −15
2a + c = 0
3c − d = 13

Solving, we get
a = −15, b = −15, c = 30, d = 77.


Question 8:

A = [aᵢⱼ] is a square matrix if:
(A) m < n  (B) m > n  (C) m = n  (D) None of these

Answer:
(C) m = n


Question 9:

Which values of x and y make the following matrices equal?

[3x+75y+12]=[3x028]

Answer:
3x + 7 = −3x ⇒ 6x = −7 ⇒ x = −7/6
y + 1 = −2 ⇒ y = −3

x = −7/6, y = −3


Question 10:

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

Answer:
Each entry can be 0 or 1, i.e., 2 possibilities per element.
Total = 23×3=29=512

Answer: (D) 512

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