Matrices Exercise 3.1 NCERT Solutions Class 12 Math Chapter 3 free PDF Download

Matrices are one of the important chapters of NCERT Solutions for Class 12 Maths. It is also a powerful and essential tool in Mathematics. This Mathematical tool simplifies our work to a great extent when compared with other direct methods. The evolution of the concept of matrices is the result of an attempt to obtain short and simple ways of solving the system of linear equations. Matrices are used not only as a representation of the coefficients in a system of linear equations. They have uses that far exceed that application. Nowadays, this mathematical tool is used in multiple disciplines, including genetics, economics, sociology, modern psychology and industrial management. Coming to education, this topic is extremely important for both the CBSE board exam and competitive exams. So, students can practise these NCERT Solutions for Class 12 and improve their problem-solving skills.

In this article, we will be going to solve the entire Exercise 3.1 of Chapter 3 of the NCERT textbook.

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 a_ij

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.

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 3×4

(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: 1×24, 24×1, 2×12, 12×2, 3×8, 8×3, 4×6, and 6×4

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

Hence possible orders are: 1×13 and 13×1

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: 1×18, 18×1, 2×9, 9×2, 3×6, and 6×3

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

Hence possible orders are: 1×5 and 5×1

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

(i) a_ij = (i + j)²/2

Solution:

Elements in this 2×2 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 2×2 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 2×2 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 3×4 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 Exercise 3.1, helping to solidify your understanding of matrix theory and its applications.

FAQs on Matrices

What is a matrix used for?

Matrices are used in a variety of fields such as engineering, physics, economics, and computer science for solving systems of linear equations, performing linear transformations, and data representation.

What are the basic operations on matrices?

The basic operations include addition, subtraction, and multiplication of matrices, as well as finding their determinants and inverses.

Can all matrices be multiplied?

No, matrix multiplication is only possible when the number of columns in the first matrix is equal to the number of rows in the second matrix.

What is Do NCERT Solutions for Class 12 Maths Chapter 3 help you to score well in the board exam?

NCERT Solutions for Class 12 Maths Chapter 3 Matrices is given by ANAND CLASSES. These solutions help the students to understand the concepts of the NCERT textbook for the Chapter Matrices. ANAND CLASSES provides detailed solutions for all NCERT questions to make students comfortable and confident of getting maximum marks. The NCERT Solutions for Class 12 Maths are solved by our Maths experts using a step-by-step approach, according to the latest update of the latest CBSE syllabus, to assist students in their board examination.

What are the main topics discussed in NCERT Solutions for Class 12 Maths Chapter 3?

In Mathematics, matrices are one of the easiest chapters, which, when understood, would be fun to solve. Matrix, types of matrices, operations on matrices, transpose of a matrix, symmetric and skew-symmetric matrices, elementary operation on matrix and invertible matrices are the main topics discussed in this chapter. These topics are explained in simple language to help students score well in the board exams, irrespective of their intelligence quotient.

Why should we learn about matrices in NCERT Solutions for Class 12 Maths Chapter 3?

Matrices are rectangular arrays of numbers which are represented in rows and columns. Various mathematical operations like multiplication, addition, subtraction and division can be performed using matrices. Representing the data related to infant mortality rate, population etc., are the widely used areas where matrices are used to simplify the calculation of complex data. The other substantial use of matrices are statistics, plotting graphs and various scientific research purposes. The method of solving difficult linear equations is also made simple using the matrices.