# NCERT Solutions for Exercise 3.1 Class 12 Maths Chapter 3 - Matrices

Matrix is an ordered rectangular array of numbers or functions called the elements or the entries of the matrix. It is a very important tool used in genetics, science, sociology, modern psychology, economics, and industrial management. In NCERT solutions for class 12 maths matrices exercise 3.1, you will get questions solved in a step-by-step manner. It covers the matrix, the order of the matrix, and different types of matrix. First, try to solve NCERT textbook problems on your own. If you are finding difficulties in solving them, you can take help from exercise 3.1 chapter 3 maths solutions. The important topics like Square matrix, Row matrix, Column matrix, Diagonal matrix, Scalar matrix, Zero matrix, Identity matrix, and Equality of matrices are also covered in the exercise 3.1 class 12 maths. You can also check for NCERT solutions.

Also, see

Matrices Exercise 3.2

Matrices Exercise 3.3

Matrices Exercise 3.4

Matrices Miscellaneous Exercise

### Matrices Exercise 3.1

Question:1(i).In the matrix $A = \begin{bmatrix} 2& 5 &19 &-7 \\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt3& 1 &-5 &17 \end{bmatrix}$, write:

The order of the matrix

$A = \begin{bmatrix} 2& 5 &19 &-7 \\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt3& 1 &-5 &17 \end{bmatrix}$

(i) The order of the matrix = number of row $\times$ number of columns $= 3\times 4$.

Question 1(ii). In the matrix $A = \begin{bmatrix}2&5&19&-7&\\ 35& -2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$, write:

The number of elements

$A = \begin{bmatrix}2&5&19&-7&\\ 35& -2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$

(ii) The number of elements $3\times 4=12$.

Question 1(iii). In the matrix $A = \begin{bmatrix}2&5&19&-7&\\35&-2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$, write:

Write the elements a13, a21, a33, a24, a23

$A = \begin{bmatrix}2&5&19&-7&\\35&-2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}$

(iii) An element $a_{ij}$ implies the element in raw number i and column number j.

$a_1_3= 19$ $a_2_1= 35$

$a_3_3= -5$ $a_2_4= 12$

$a_2_3= \frac{5}{2}$

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

A matrix has 24 elements.

The possible orders are :

$1\times 24,24\times 1,2\times 12,12\times 2,3\times 8,8\times 3,4\times 6 \, \, and\, \, 6\times 4$.

If it has 13 elements, then possible orders are :

$1\times 13\, \, \, and \, \, \, \, 13\times 1$.

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

A matrix has 18 elements.

The possible orders are as given below

$1\times 18,18\times 1,2\times 9,9\times 2,3\times 6\, \, \, and\, \, \, \, 6\times 3$

If it has 5 elements, then possible orders are :

$1\times 5\, \, \, and \, \, \, \, 5\times 1$.

Question 4(i). Construct a 2 × 2 matrix, $A = [a_{ij} ]$ whose elements are given by:

$a_{ij} = \frac{(i + j)^2}{2}$

$A = [a_{ij} ]$

(i) $a_{ij} = \frac{(i + j)^2}{2}$

Each element of this matrix is calculated as follows

$a_1_1 = \frac{(1+1)^{2}}{2} =\frac{2^{2}}{2}=\frac{4}{2}=2$ $a_2_2 = \frac{(2+2)^{2}}{2} =\frac{4^{2}}{2}=\frac{16}{2}=8$

$a_1_2 = \frac{(1+2)^{2}}{2} =\frac{3^{2}}{2}=\frac{9}{2}=4.5$ $a_2_1 = \frac{(2+1)^{2}}{2} =\frac{3^{2}}{2}=\frac{9}{2}=4.5$

Matrix A is given by

$A = \begin{bmatrix} 2&4.5 \\4.5 & 8 \end{bmatrix}$

Question 4(ii). Construct a 2 × 2 matrix, $A = [a_{ij} ]$, whose elements are given by:

$a_{ij} = \frac{i}{j}$

A 2 × 2 matrix, $A = [a_{ij} ]$

(ii) $a_{ij} = \frac{i}{j}$

$a_1_1 = \frac{1}{1}=1$ $a_2_2 = \frac{2}{2}=1$

$a_1_2 = \frac{1}{2}$ $a_2_1 = \frac{2}{1}=2$

Hence, the matrix is

$A = \begin{bmatrix} 1& \frac{1}{2} \\ 2 & 1 \end{bmatrix}$

Question 4(iii). Construct a 2 × 2 matrix, $A = [a_{ij} ]$, whose elements are given by:

$a_{ij} = \frac{(i+2j)^2}{2}$

(iii)

$a_{ij} = \frac{(i+2j)^2}{2}$

$a_1_1 = \frac{(1+(2\times 1))^{2}}{2}= \frac{(1+2)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2}$ $a_2_2 = \frac{(2+(2\times 2))^{2}}{2}= \frac{(2+4)^{2}}{2}=\frac{6^{2}}{2}=\frac{36}{2}=18$

$a_2_1 = \frac{(2+(2\times 1))^{2}}{2}= \frac{(2+2)^{2}}{2}=\frac{4^{2}}{2}=\frac{16}{2}=8$ $a_1_2 = \frac{(1+(2\times 2))^{2}}{2}= \frac{(1+4)^{2}}{2}=\frac{5^{2}}{2}=\frac{25}{2}$

Hence, the matrix is given by

$A = \begin{bmatrix} \frac{9}{2}& \frac{25}{2} \\ 8 & 18 \end{bmatrix}$

Question 5(i). Construct a 3 × 4 matrix, whose elements are given by:

$a_{ij} = \frac{1}{2}|-3i + j|$

(i)

$a_{ij} = \frac{1}{2}|-3i + j|$

$a_1_1 = \frac{\left | -3+1 \right |}{2}=\frac{2}{2}=1$ $a_1_2 = \frac{\left | (-3\times 1)+2 \right |}{2}=\frac{1}{2}$ $a_1_3 = \frac{\left | (-3\times 1)+3 \right |}{2}=0$

$a_2_1 = \frac{\left | (-3\times 2)+1 \right |}{2}=\frac{5}{2}$ $a_2_2 = \frac{\left | (-3\times 2)+2 \right |}{2}=\frac{4}{2}=2$ $a_2_3 = \frac{\left | (-3\times 2)+3 \right |}{2}=\frac{\left | -6+3 \right |}{2}=\frac{\left | -3 \right |}{2} =\frac{3}{2}$

$a_3_1 = \frac{\left | (-3\times 3)+1 \right |}{2}=\frac{8}{2}=4$ $a_3_2 = \frac{\left | (-3\times 3)+2 \right |}{2}=\frac{7}{2}$ $a_3_3 = \frac{\left | (-3\times 3)+3 \right |}{2}=\frac{\left | -9+3 \right |}{2}=\frac{\left | -6 \right |}{2} =\frac{6}{2}=3$

$a_1_4 = \frac{\left | (-3\times 1)+4 \right |}{2}=\frac{\left | -3+4 \right |}{2}=\frac{\left | 1 \right |}{2} =\frac{1}{2}$ $a_2_4 = \frac{\left | (-3\times 2)+4 \right |}{2}=\frac{\left | -6+4 \right |}{2}=\frac{\left | -2 \right |}{2} =\frac{2}{2}=1$

$a_3_4 = \frac{\left | (-3\times 3)+4 \right |}{2}=\frac{\left | -9+4 \right |}{2}=\frac{\left | -5 \right |}{2} =\frac{5}{2}$

Hence, the required matrix of the given order is

$A = \begin{bmatrix} 1& \frac{1}{2} & 0&\frac{1}{2} \\ \frac{5}{2} & 2&\frac{3}{2}&1 \\4&\frac{7}{2}&3&\frac{5}{2}\end{bmatrix}$

Question 5(ii) Construct a 3 × 4 matrix, whose elements are given by:

$a_{ij} = 2i - j$

A 3 × 4 matrix,

(ii) $a_{ij} = 2i - j$

$a_1_1 = 2\times 1-1 =2-1=1$ $a_1_2 = 2\times 1-2 =2-2=0$ $a_1_3 = 2\times 1-3 =2-3=-1$

$a_2_1 = 2\times 2-1 =4-1=3$ $a_2_2= 2\times 2-2 =4-2=2$ $a_2_3 = 2\times 2-3 =4-3=1$ $a_3_1 = 2\times 3-1 =6-1=5$ $a_3_2 = 2\times 3-2 =6-2=4$ $a_3_3 = 2\times 3-3 =6-3=3$

$a_1_4 = 2\times 1-4 =2-4=-2$ $a_2_4= 2\times 2-4 =4-4=0$ $a_3_4= 2\times 3-4 =6-4=2$

Hence, the matrix is

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

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

$\begin{bmatrix}4&3\\x&5 \end{bmatrix} = \begin{bmatrix}y&z\\1&5 \end{bmatrix}$

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

If two matrices are equal, then their corresponding elements are also equal.

$\therefore$ $x=1\, \, \, ,\, \, \, y=4\, \, \, \, and\, \, \, \, z=3$

Question 6(ii) Find the values of x, y and z from the following equations:

$\begin{bmatrix} x +y & 2\\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 &2 \\ 5 & 8 \end{bmatrix}$

(ii)

$\begin{bmatrix} x +y & 2\\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 &2 \\ 5 & 8 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal.

$\therefore$ $x+y=6$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$

$x=6-y$

$xy=8$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$

Solving equation (i) and (ii) ,

$(6-y)y =8$

$6y-y^{2}=8$

$y^{2}-6y+8=0$

solving this equation we get,

$y=4 \, \, and\, \, y=2$

Putting the values of y, we get

$x=2 \, \, and\, \, x=4$

And also equating the first element of the second raw

$5+z = 5$, $z=0$

Hence,

$x=2,y=4,z=0\, \, \, \, \, and\, \, \, \, \, \, x=4,y=2,z=0$

Question 6(iii) Find the values of x, y and z from the following equations

$\begin{bmatrix} x + y + z\\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9\\5 \\7 \end{bmatrix}$

(iii)

$\begin{bmatrix} x + y + z\\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9\\5 \\7 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal

$x+y+z=9........(1)$

$x+z=5..............(2)$

$y+z=7..............(3)$

subtracting (2) from (1) we will get y=4

substituting the value of y in equation (3) we will get z=3

now substituting the value of z in equation (2) we will get x=2

therefore,

$x=2$, $y=4$ and $z=3$

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

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

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

If two matrices are equal, then their corresponding elements are also equal

$a-b=-1$ $.............................1$

$2a+c=5$ $.............................2$

$2a-b=0$ $.............................3$

$3c+d=13$ $.............................4$

Solving equation 1 and 3 , we get

$a=1 \, \, \, \, and \, \, \, \, b=2$

Putting the value of a in equation 2, we get

$c=3$

Putting the value of c in equation 4 , we get

$d=4$

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

(A) $m

(B) $m >n$

(C) $m =n$

(D) None of these

A square matrix has the number of rows and columns equal.

Thus, for $A = [a_{ij}]_{m\times n}$ to be a square matrix m and n should be equal.

$\therefore m=n$

Option (c) is correct.

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

$\begin{bmatrix} 3x + 7 &5 \\ y + 1 & 2 -3x \end{bmatrix}$, $\begin{bmatrix} 0 & y - 2 \\ 8 & 4 \end{bmatrix}$

(A) $x = \frac{-1}{3}, y = 7$

(B) Not possible to find

(C) $y =7, x = \frac{-2}{3}$

(D) $x = \frac{-1}{3}, y = \frac{-2}{3}$

Given, $\begin{bmatrix} 3x + 7 &5 \\ y + 1 & 2 -3x \end{bmatrix}$ $=\begin{bmatrix} 0 & y - 2 \\ 8 & 4 \end{bmatrix}$

If two matrices are equal, then their corresponding elements are also equal

$3x+7=0\Rightarrow x=\frac{-7}{3}$

$y-2=5 \Rightarrow y=5+2=7$

$y+1=8\Rightarrow y=8-1=7$

$2-3x=4\Rightarrow 3x=2-4\Rightarrow 3x=-2\Rightarrow x=\frac{-2}{3}$

Here, the value of x is not unique, so option B is correct.

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

Total number of elements in a 3 × 3 matrix

$=3\times 3=9$

If each entry is 0 or 1 then for every entry there are 2 permutations.

The total permutations for 9 elements

$=2^{9}=512$

Thus, option (D) is correct.

More About NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.1:-

There are 5 solved examples given before this exercise that you can solve. Solving these examples will help you to get conceptual clarity. There are 7 long answer types questions and 3 multiple choice types questions given in the NCERT Solutions for Class 12 Maths Chapter 3 exercise 3.1. These questions are very basic questions based on the matrix and order of the matrix. There are some questions in Class 12th Maths chapter 3 exercise 3.1 based on the equality of the matrices of order 2x2 and 3x3.

Also Read| Matrices Class 12 Maths Chapter Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.1:-

• As most of the questions in the board exams are directly asked from the NCERT textbook, NCERT solutions for Class 12 Maths chapter 3 exercise 3.1 becomes very important for the students to solve the NCERT problems.

• You are advised to solve the CBSE board's previous year's paper to get familiar with the exam pattern.

• Solving more problems will help students to get conceptual clarity.

• Class 12 Maths chapter 3 exercise 3.1 solutions are created based on the CBSE guideline which you can rely upon.

Also see-

• NCERT solutions for class 12 maths chapter 3

• NCERT exemplar solutions class 12 maths chapter 3

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