Class 12 NCERT Maths – Chapter 3 (Matrices) – Exercise 3.1 Solutions

In this post we have provided solution of Class 12 NCERT Maths – Exercise 3.1. This exercise is based on the basic concepts like equality of matrix, order of matrix and formation of matrices.
These are step by step solutions that will help you apply the concepts and score good marks in your board exams

Question 1. In the matrix

write:
(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements a_{13},\ a_{21},\ a_{33},\ a_{24},\ a_{23}

Solution:
(i) The order of the matrix is 3 \times 4 
(ii) The number of elements is 3 \times 4 = 12 
(iii) Elements: 
a_{13} = -19,\quad a_{21} = 35,\quad a_{33} = -5,\quad a_{24} = 12,\quad a_{23} = \dfrac{5}{2}

Question 2. If a matrix has 24 elements, what are the possible orders it can have? 
What, if it has 13 elements?

Solution:
If a matrix has 24 elements, we list all possible ordered pairs (m, n) such that m \times n = 24: 
(1,24),\ (2,12),\ (3,8),\ (4,6),\ (6,4),\ (8,3),\ (12,2),\ (24,1) 

If the matrix has 13 elements: 
Since 13 is a prime number, only possible orders are 1 \times 13\ \text{and}\ 13 \times 1

Question 3. If a matrix has 18 elements, what are the possible orders it can have? 
What, if it has 5 elements?

Solution:
If the matrix has 18 elements, we find all pairs (m, n) such that 
m \times n = 18

Possible orders: 
(1,18),\ (2,9),\ (3,6),\ (6,3),\ (9,2),\ (18,1)

If the matrix has 5 elements (and 5 is a prime number), possible orders are: 
(1,5)\ \text{and}\ (5,1)

Question 4. Construct a 2 \times 2 matrix A = [a_{ij}] whose elements are given by: 

(i) a_{ij} = \dfrac{(i + j)^2}{2} 
(ii) a_{ij} = \dfrac{i}{j} 
(iii) a_{ij} = \dfrac{(i + 2j)^2}{2}

Solution
(i)   A = \begin{bmatrix}  \dfrac{(1+1)^2}{2} & \dfrac{(1+2)^2}{2} \\\\  \dfrac{(2+1)^2}{2} & \dfrac{(2+2)^2}{2}  \end{bmatrix}  =  \begin{bmatrix}  2 & \dfrac{9}{2} \\\\  \dfrac{9}{2} & 8  \end{bmatrix} 

(ii)   A = \begin{bmatrix}  \dfrac{1}{1} & \dfrac{1}{2} \\\\  \dfrac{2}{1} & \dfrac{2}{2}  \end{bmatrix}  =  \begin{bmatrix}  1 & \dfrac{1}{2} \\\\  2 & 1  \end{bmatrix} 

(iii)   A = \begin{bmatrix}  \dfrac{(1 + 2)^2}{2} & \dfrac{(1 + 4)^2}{2} \\\\  \dfrac{(2 + 2)^2}{2} & \dfrac{(2 + 4)^2}{2}  \end{bmatrix}  =  \begin{bmatrix}  \dfrac{9}{2} & \dfrac{25}{2} \\\\  8 & 18  \end{bmatrix} 

Question 5. Construct a 3 \times 4 matrix, whose elements are given by: 

(i) a_{ij} = \dfrac{1}{2} \left| -3i + j \right| 
(ii) a_{ij} = 2i - j

Solution:
(i)   A = \begin{bmatrix}  \dfrac{1}{2}|-3(1)+1| & \dfrac{1}{2}|-3(1)+2| & \dfrac{1}{2}|-3(1)+3| & \dfrac{1}{2}|-3(1)+4| \\\\  \dfrac{1}{2}|-3(2)+1| & \dfrac{1}{2}|-3(2)+2| & \dfrac{1}{2}|-3(2)+3| & \dfrac{1}{2}|-3(2)+4| \\\\  \dfrac{1}{2}|-3(3)+1| & \dfrac{1}{2}|-3(3)+2| & \dfrac{1}{2}|-3(3)+3| & \dfrac{1}{2}|-3(3)+4|  \end{bmatrix}  =  \\\\\begin{bmatrix}  1 & \dfrac{1}{2} & 0 & \dfrac{1}{2} \\\\  \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1 \\\\  4 & \dfrac{7}{2} & 3 & \dfrac{5}{2}  \end{bmatrix} 

(ii)   A = \begin{bmatrix}  2(1) - 1 & 2(1) - 2 & 2(1) - 3 & 2(1) - 4 \\\\  2(2) - 1 & 2(2) - 2 & 2(2) - 3 & 2(2) - 4 \\\\  2(3) - 1 & 2(3) - 2 & 2(3) - 3 & 2(3) - 4  \end{bmatrix}  =  \begin{bmatrix}  1 & 0 & -1 & -2 \\  3 & 2 & 1 & 0 \\  5 & 4 & 3 & 2  \end{bmatrix} 

Question 6. Find the values of x, y, z from the following equations:

(i) \begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}
(ii) \begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}
(iii) \begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}

Solution: (i) Comparing corresponding elements of the matrices:

4 = y \Rightarrow y = 4

3 = z \Rightarrow z = 3

x = 1 (from the bottom-left elements)

(ii) Comparing matrices:

x + y = 6 \quad \text{(1)}

z + 5 = 5 \Rightarrow z = 0 \quad \text{(2)}

xy = 8 \quad \text{(3)}

From (1):
y = 6 - x

Now, substitute into (3):
x(6 - x) = 8 \\\\ x^2 - 6x + 8 = 0

Solving the quadratic:
x = 2 \quad and \quad x = 4

y = 6 - 2 = 4 \quad and \quad y = 6 - 4 =2

Therefore, possible solutions:
(x, y, z) = (2, 4, 0) \text{ or } (4, 2, 0)

(iii) Given:
x + y + z = 9 \quad \text{(i)}
x + z = 5 \quad \text{(ii)}
y + z = 7 \quad \text{(iii)}

From (i) and (ii):

Now using y in (iii): 4 + z = 7 \Rightarrow z = 3
Then from (ii): x = 5 - z = 2

Final solution: x = 2, y = 4, z = 3

Question 7. Find the value of a, b, c, and d from the equation:

Solution: Compare elements:

a - b = -1 \quad (1)
2a + c = 5 \quad (2)
2a - b = 0 \quad (3)
3c + d = 13 \quad (4)

From (1) and (3):

Subtract (1) from (3):

(2a - b) - (a - b) = 0 - (-1) \Rightarrow a = 1
Then
From (1): 1 - b = -1 \Rightarrow b = 2
From (2): 2(1) + c = 5 \Rightarrow c = 3
From (4): 3(3) + d = 13 \Rightarrow d = 4

Final values: a = 1, b = 2, c = 3, d = 4

Question 8. A = [a_{ij}]_{m \times n} is a square matrix, if:

(A) m < n  
(B) m > n  
(C) m = n  
(D) None of these

Solution: A square matrix is defined as a matrix with equal number of rows and columns.
So, the correct condition is: m = n
✅ Correct Option: (C)

Question 9. Which of the given values of x and y make the following pair of matrices equal?

Solution: Compare elements:

3x + 7 = 0 \quad (1)
\Rightarrow x = -\dfrac{7}{3}

y + 1 = 8 \quad (2)
\Rightarrow y = 7

y - 2 = 5 \quad (3)
\Rightarrow y = 7 \quad (OK)

2 - 3x = 4 \quad (4)
\Rightarrow x = -\dfrac{2}{3} \quad (Contradiction)

Equations (1) and (4) are inconsistent. Hence,
❌ No consistent values of x and y satisfy all conditions.

✅ Correct Option: (B) Not possible to find

Question 10. The number of all possible matrices of order 3 \times 3 with each entry 0 or 1 is:

(A) 27   (B) 18   (C) 81   (D) 512

Solution: Each element has 2 possibilities: 0 or 1

A 3 \times 3 matrix has 9 elements.
So total number of such matrices = 2^9 = 512

✅ Correct Option: (D) 512