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

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

Answer:

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

(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

Answer:

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

(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

Answer:

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

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

$a_{13} = 19$, $a_{21} = 35$

$a_{33} = -5$, $a_{24} = 12$

$a_{23} = \frac{5}{2}$

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

Answer:

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?

Answer:

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:

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

Answer:

$A = [a_{ij} ]$

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

Each element of this matrix is calculated as follows

$a_{11} = \frac{(1+1)^2}{2} = \frac{2^2}{2} = \frac{4}{2} = 2$, $a_{22} = \frac{(2+2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8$

$a_{12} = \frac{(1+2)^2}{2} = \frac{3^2}{2} = \frac{9}{2} = 4.5$, $a_{21} = \frac{(2+1)^2}{2} = \frac{3^2}{2} = \frac{9}{2} = 4.5$

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:

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

Answer:

A 2 × 2 matrix, $A = [a_{ij} ]$

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

$a_{11} = \frac{1}{1} = 1$, $a_{22} = \frac{2}{2} = 1$

$a_{12} = \frac{1}{2}$, $a_{21} = \frac{2}{1} = 2$

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:

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

Answer:

(iii)

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

$a_{11} = \frac{(1 + (2 \times 1))^2}{2} = \frac{(1 + 2)^2}{2} = \frac{3^2}{2} = \frac{9}{2}$,

$a_{22} = \frac{(2 + (2 \times 2))^2}{2} = \frac{(2 + 4)^2}{2} = \frac{6^2}{2} = \frac{36}{2} = 18$,

$a_{21} = \frac{(2 + (2 \times 1))^2}{2} = \frac{(2 + 2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8$,

$a_{12} = \frac{(1 + (2 \times 2))^2}{2} = \frac{(1 + 4)^2}{2} = \frac{5^2}{2} = \frac{25}{2}$

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:

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

Answer:

(i)

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

$a_{11} = \frac{\left| -3 + 1 \right|}{2} = \frac{2}{2} = 1$,
$a_{12} = \frac{\left| (-3 \times 1) + 2 \right|}{2} = \frac{1}{2}$,
$a_{13} = \frac{\left| (-3 \times 1) + 3 \right|}{2} = 0$

$a_{21} = \frac{\left| (-3 \times 2) + 1 \right|}{2} = \frac{5}{2}$,
$a_{22} = \frac{\left| (-3 \times 2) + 2 \right|}{2} = \frac{4}{2} = 2$,
$a_{23} = \frac{\left| (-3 \times 2) + 3 \right|}{2} = \frac{\left| -6 + 3 \right|}{2} = \frac{\left| -3 \right|}{2} = \frac{3}{2}$

$a_{31} = \frac{\left| (-3 \times 3) + 1 \right|}{2} = \frac{8}{2} = 4$,
$a_{32} = \frac{\left| (-3 \times 3) + 2 \right|}{2} = \frac{7}{2}$,
$a_{33} = \frac{\left| (-3 \times 3) + 3 \right|}{2} = \frac{\left| -9 + 3 \right|}{2} = \frac{\left| -6 \right|}{2} = \frac{6}{2} = 3$

$a_{14} = \frac{\left| (-3 \times 1) + 4 \right|}{2} = \frac{\left| -3 + 4 \right|}{2} = \frac{\left| 1 \right|}{2} = \frac{1}{2}$,
$a_{24} = \frac{\left| (-3 \times 2) + 4 \right|}{2} = \frac{\left| -6 + 4 \right|}{2} = \frac{\left| -2 \right|}{2} = \frac{2}{2} = 1$,
$a_{34} = \frac{\left| (-3 \times 3) + 4 \right|}{2} = \frac{\left| -9 + 4 \right|}{2} = \frac{\left| -5 \right|}{2} = \frac{5}{2}$

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:

$a_{ij} = 2i - j$

Answer:

A 3 × 4 matrix,

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

$a_{11} = 2 \times 1 - 1 = 2 - 1 = 1$, $a_{12} = 2 \times 1 - 2 = 2 - 2 = 0$, $a_{13} = 2 \times 1 - 3 = 2 - 3 = -1$

$a_{21} = 2 \times 2 - 1 = 4 - 1 = 3$, $a_{22} = 2 \times 2 - 2 = 4 - 2 = 2$, $a_{23} = 2 \times 2 - 3 = 4 - 3 = 1$

$a_{31} = 2 \times 3 - 1 = 6 - 1 = 5$, $a_{32} = 2 \times 3 - 2 = 6 - 2 = 4$, $a_{33} = 2 \times 3 - 3 = 6 - 3 = 3$

$a_{14} = 2 \times 1 - 4 = 2 - 4 = -2$, $a_{24} = 2 \times 2 - 4 = 4 - 4 = 0$, $a_{34} = 2 \times 3 - 4 = 6 - 4 = 2$

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:

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

Answer:

(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:

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

Answer:

(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

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

Answer:

(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:

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

Answer:

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

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) $m <n$

(B) $m >n$

(C) $m =n$

(D) None of these

Answer:

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

Answer:

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

Answer:

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.