Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.1

In this article, we will be going to solve the entire Miscellaneous Exercise 3.1 of Chapter 3 of the NCERT textbook. A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. The numbers, symbols, or expressions in the matrix are called elements or entries. Matrices are used in various fields of mathematics, engineering, physics, computer science, and economics for representing and solving systems of linear equations, performing linear transformations, and more.

What is a Matrix?

A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. Each element in the matrix is denoted by two subscripts, where the first represents the row and the second represents the column. Matrices are typically represented by capital letters, such as A, and their elements by lowercase letters, like 𝑎𝑖𝑗

Types of Matrices

Matrices come in different forms, such as square matrices, diagonal matrices, and identity matrices, depending on the arrangement of their elements. Understanding these types will help in solving a variety of mathematical problems efficiently.

Chapter 3 Matrices - Exercise 3.1 Solution

The solution for the NCERT Exercise 3.1 is added below:

Question 1. In the matrix A = \begin{vmatrix} 2&5&19&-7\\ 35&-2&5/2&12\\ \sqrt{3}&1&-5&17\\ \end{vmatrix}  , write:

(i) The order of the matrix 

Solution:

We can see that matrix contains 3 rows and 4 columns So, the order of this matrix is 3x4

(ii) The number of elements 

Solution:

We know that number of elements in the matrix = product of number of rows and number of columns in matrix So, number of elements = 3 x 4 =12.

(iii) Write the elements a₁₃ , a₂₁, a₃₃ , a₂₄, a₂₃

Solution:

a₁₃ = Element in first row and third column i.e, 19

a₂₁ = Element in second row and first column i.e, 35

a₃₃ = Element in third row and third column i.e, -5

a₂₄ = Element in second row and fourth column i.e, 12

a₂₃ = Element in second row and  third column i.e, 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:

We know that number of elements in the matrix is the product of number 

of rows and number of columns in the matrix .

If matrix has order mxn then number  of elements are mn in that matrix.

So we have to find the ordered pairs of natural number whose product is 24.  

The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4)

Hence possible orders are: 1x24, 24x1, 2x12, 12x2, 3x8, 8x3, 4x6, and 6x4

If matrix has 13 elements then ordered pairs will be (1, 13) and (13, 1)

Hence possible orders are: 1x13 and 13x1

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

Solution:

We know that number of elements in the matrix is the product of number 

of rows and number of columns in the matrix .

If matrix has order mxn then number  of elements are mn in that matrix.

So we have to find the ordered pairs of natural number whose product is 18.  

The ordered pairs are:(1, 18), (18, 1), (2, 9), (9, 2), (3, 6), and (6, 3)

Hence possible orders are: 1x18, 18x1, 2x9, 9x2, 3x6, and 6x3

If matrix has 5 elements then ordered pairs will be (1, 5) and (5, 1)  

Hence possible orders are: 1x5 and 5x1

Question 4. Construct a 2x2 matrix , A = [a_ij]  whose elements are given by :

(i) a_ij = (i + j)²/2

Solution:

Elements in this 2x2 matrix = a₁₁ , a₁₂ ,a₂₁ ,a₂₂

a₁₁ ⇒ i = 1 and j = 1 ⇒ (1 + 1)²/2 = 4/2 = 2

a₁₂ ⇒ i = 1 and j = 2 ⇒ (1 + 2)²/2 = 9/2

a₂₁ ⇒ i = 2 and j = 1 ⇒ (2 + 1)²/2 = 9/2

a₂₂ ⇒ i = 2 and j = 2 ⇒(2 + 2)²/2 = 16/2 = 8

Resultant Matrix is:

\begin{vmatrix} 2&9/2\\ 9/2&8\\ \end{vmatrix}

(ii) a_ij = i/j

Solution:

Elements in this 2x2 matrix = a₁₁ , a₁₂ ,a₂₁ ,a₂₂

a₁₁ ⇒ i = 1 and j = 1 = 1/1 = 1

a₁₂ ⇒ i = 1 and j = 2 = 1/2

a₂₁ ⇒ i = 2 and j = 1 = 2/1 = 2

a₂₂ ⇒ i = 2 and j = 2 = 2/2 = 1

Resultant Matrix is:  

\begin{vmatrix} 1&1/2\\ 2&1\\ \end{vmatrix} 

(iii) a_ij = (i + 2j)²/2

Solution:

Elements in this 2x2 matrix = a₁₁ , a₁₂ , a₂₁ , a₂₂

a₁₁ ⇒ i = 1 and j = 1 ⇒ (1 + 2 x 1)²/2 = 9/2

a₁₂ ⇒ i = 1 and j = 2 ⇒ (1 + 2 x 2)²/2 = 25/2

a₂₁ ⇒ i = 2 and j = 1 ⇒(2 + 2 x 1)²/2  = 16/2 = 8

a₂₂ ⇒ i = 2 and j = 2 ⇒(2 + 2 x 2)²/2 = 36/2 = 18

Resultant Matrix is:

\begin{vmatrix} 9/2&25/2\\ 8&18\\ \end{vmatrix} 

Question 5. Construct a 3x4 matrix, whose elements are given by :

(i) a_ij = 1/2 {|-3i + j|}

Solution:

Elements in this 3 x 4 matrix are  a₁₁ , a₁₂ , a₁₃ , a₁₄ , a₂₁ , a₂₂, a₂₃ , a₂₄  , a₃₁ , a₃₂ , a₃₃ , a₃₄

a₁₁ ⇒ i = 1 and j = 1 ⇒ 1/2 (|-3 x 1 + 1|) = 1

a₁₂ ⇒  i = 1 and j = 2 ⇒ 1/2 (|-3 x 1 + 2|) = 1/2

a₁₃ ⇒ i = 1 and j = 3 ⇒ 1/2 (|-3 x 1 + 3) = 0

a₁₄ ⇒ i = 1 and j = 4 ⇒ 1/2 (|-3 x 1 + 4|) = 1/2

a₂₁ ⇒ i = 2 and j = 1 ⇒ 1/2 (|-3 x 2 + 1|) = 5/2

a₂₂ ⇒ i = 2 and j = 2 ⇒ 1/2 (|-3 x 2 + 2|) = 2

a₂₃ ⇒ i = 2 and j = 3 ⇒ 1/2 (|-3 x 2 + 3|) = 3/2

a₂₄ ⇒ i = 2 and j = 4 ⇒ 1/2 (|-3 x 2 + 4|) = 1

a₃₁ ⇒ i = 3 and j = 1 ⇒ 1/2 (|-3 x 3 + 1|) = 4

a₃₂ ⇒ i = 3 and j = 2 ⇒ 1/2 (|-3 x 3 + 2|) = 7/2

a₃₃ ⇒ i = 3 and j = 3 ⇒ 1/2 (|-3 x 3 + 3|) = 3

a₃₄ ⇒ i = 3 and j = 4 ⇒ 1/2 (|-3 x 3 + 4|) = 5/2

Resultant matrix is:  

\begin{vmatrix} 1&1/2&0&1/2\\ 5/2&2&3/2&1\\ 4&7/2&3&5/2\\ \end{vmatrix} 

(ii) a_ij = 2i - j

Solution:

Elements in this 3 x 4 matrix are  a₁₁ , a₁₂ , a₁₃ , a₁₄ , a₂₁ , a₂₂ , a₂₃ , a₂₄  , a₃₁ , a₃₂ , a₃₃ , a₃₄

So,

a₁₁ ⇒ i = 1 and j = 1 ⇒ 2 x 1 - 1 = 1

a₁₂ ⇒ i = 1 and j = 2 ⇒ 2 x 1 - 2 = 0

a₁₃ ⇒ i = 1 and j = 3 ⇒ 2 x 1 - 3 = -1

a₁₄ ⇒ i = 1 and j = 4 ⇒ 2 x 1 - 4 = -2

a₂₁ ⇒ i = 2 and j = 1 ⇒ 2 x 2 - 1 = 3

a₂₂ ⇒ i = 2 and j = 2 ⇒ 2 x 2 - 2 = 2

a₂₃ ⇒ i = 2 and j = 3 ⇒ 2 x 2 - 3 = 1

a₂₄ ⇒ i = 2 and j = 4 ⇒ 2 x 2 - 4 = 0

a₃₁ ⇒ i = 3 and j = 1 ⇒ 2 x 3 - 1 = 5

a₃₂ ⇒ i = 3 and j = 2 ⇒ 2 x 3 - 2 = 4

a₃₃ ⇒ i = 3 and j = 3 ⇒ 2 x 3 - 3 = 3

a₃₄ ⇒ i = 3 and j = 4 ⇒ 2 x 3 - 4 = 2

Resultant matrix is: \begin{vmatrix} 1&0&-1&-2\\ 3&2&1&0\\ 5&4&3&2\\ \end{vmatrix}  

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

(i) \begin{vmatrix} 4&3\\ x&5\\ \end{vmatrix} = \begin{vmatrix} y&z\\ 1&5\\ \end{vmatrix} 

Solution:

We can compare or equate both the matrices because both are equal 

So on equating both the matrices we get 

x = 1; y = 4; z = 3

 (ii) \begin{vmatrix} x+y&2\\ 5+z&xy\\ \end{vmatrix} = \begin{vmatrix} 6&2\\ 5&8\\ \end{vmatrix}

Solution:

We can compare or equate both the matrices because both are equal

So, on equating both the matrices. we get

x + y = 6       -(1)

5 + z = 5      -(2)

xy = 8         -(3)

Now, we can solve these equations   

z = 0 from eq(2)

x = 6 - y         -(4)

Now putting value of x from eq(4) in eq(3)

 (6 - y)(y) = 8

6y - y² = 8

y² - 6y + 8 = 0         -(5)

Now we have to factorize this equation

(y - 4)(y - 2) = 0

either y - 4 = 0 or y - 2 = 0

so, y = 2 or y = 4

Put these values in eq(4) we get 

x = 4 and x = 2

Therefore, the value of  x = 2 , y = 4 , z = 0

(iii) \begin{vmatrix} x+y+z\\ x+z\\ y+z\\ \end{vmatrix} = \begin{vmatrix} 9\\ 5\\ 7\\ \end{vmatrix} 

Solution:

We can compare or equate both the matrices because both are equal

So, on equating both the matrices, we get

x + y + z = 9      -(1) 

x + z = 5          -(2) 

y + z = 7          -(3) 

If we put the value of eq(2) in eq(1)

 we get, 5 + y = 9

y = 4 

On putting value of y in eq(3)

4 + z = 7

z = 3

On putting value of z in eq(2)

x + 3 = 5

x = 2

So, the value of  x = 2; y = 4; z = 3

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

\begin{vmatrix} a-b&2a+c\\ 2a-b&3c+d\\ \end{vmatrix} = \begin{vmatrix} -1&5\\ 0&13\\ \end{vmatrix}

Solution:

We can compare or equate both the matrices because both are equal

So, on equating both the matrices, we get

a - b = -1          -(1)   

2a - b = 0          -(2) 

2a + c= 5          -(3) 

3c + d = 13          -(4) 

On solving eq(1) and eq(2) we get 

a = 1

On putting a = 1 in eq(3) we get 

c = 3

On putting a = 1 in eq(2) we get 

b = 2

On putting c = 3 in eq(4) we get 

d = 4

So, the value of  a = 1; b = 2; c = 3; d = 4

Question 8. A = [a_ij]_mxn is a square matrix, if 

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

Solution:

This will be square matrix if number of rows = number of columns

So , m = n is correct option.

Hence, the option answer is C.

Question 9. Which of the given values of x and y make the following pair of matrices equal\begin{vmatrix} 3x+7&5\\ y+1&2-3x\\ \end{vmatrix}, \begin{vmatrix} 0&y-2\\ 8&4\\ \end{vmatrix} 

(A) x = -1/3, y = 7    (B) Not possible to find    (C) y = 7, x = -2/3    (D) x = -1/3, y = -2/3   

Solution:

We can compare or equate both the matrices because both are equal

So on equating both the matrices we get;

3x + 7 = 0           -(1)

y + 1 = 8           -(2)

2 - 3x = 4           -(3)

y - 2 = 5           -(4)

From eq(2) and eq(4) we get same value of y i.e, y=7

but on solving eq(1) we get value of x = -7/3 and on solving eq(3) we get value of x = -2/3

Both the values of x are different for the value of y. So, it is not possible to find.

Hence, the correct option is B

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

Solution:

We know that number of elements in a matrix of order mxn is mn. 

So number of elements in matrix of 3 x 3 is 9.

For each element we have two choices either 0 or 1

So, total number of possible matrices of order 3 x 3 with each entry 0 or 1 = 2⁹ = 512

Correct option is D

Summary

In this article, we have covered the concept of matrices, their types, and basic operations. We also provided step-by-step solutions to Miscellaneous Exercise 3.1, helping to solidify your understanding of matrix theory and its applications.