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

Question:1(i).In the matrix $A=\left[\begin{array}{cccc}2& 5& 19& -7\\ 35& -2& \frac{5}{2}& 12\\ \sqrt{3}& 1& -5& 17\end{array}\right]$, write:

The order of the matrix

Answer:

$A=\left[\begin{array}{cccc}2& 5& 19& -7\\ 35& -2& \frac{5}{2}& 12\\ \sqrt{3}& 1& -5& 17\end{array}\right]$

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

Question 1(ii). In the matrix $A=\left[\begin{array}{ccccc}2& 5& 19& -7& \\ 35& -2& \frac{5}{2}& 12\\ \sqrt{3}& 1& -5& 17\end{array}\right]$, write:

The number of elements

Answer:

$A=\left[\begin{array}{ccccc}2& 5& 19& -7& \\ 35& -2& \frac{5}{2}& 12\\ \sqrt{3}& 1& -5& 17\end{array}\right]$

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

Question 1(iii). In the matrix $A=\left[\begin{array}{ccccc}2& 5& 19& -7& \\ 35& -2& \frac{5}{2}& 12\\ \sqrt{3}& 1& -5& 17\end{array}\right]$, write:

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

Answer:

$A=\left[\begin{array}{ccccc}2& 5& 19& -7& \\ 35& -2& \frac{5}{2}& 12\\ \sqrt{3}& 1& -5& 17\end{array}\right]$

(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 :

$1\times 24,24\times 1,2\times 12,12\times 2,3\times 8,8\times 3,4\times 6and6\times 4$.

If it has 13 elements, then possible orders are :

$1\times 13and13\times 1$.

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

$1\times 18,18\times 1,2\times 9,9\times 2,3\times 6and6\times 3$

If it has 5 elements, then possible orders are :

$1\times 5and5\times 1$.

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

$A=\left[\begin{array}{cc}2& 4.5\\ 4.5& 8\end{array}\right]$

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

$A=\left[\begin{array}{cc}1& \frac{1}{2}\\ 2& 1\end{array}\right]$

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

$A=\left[\begin{array}{cc}\frac{9}{2}& \frac{25}{2}\\ 8& 18\end{array}\right]$

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

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

Answer:

(i)

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

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

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

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

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

Hence, the required matrix of the given order is

$A=\left[\begin{array}{cccc}1& \frac{1}{2}& 0& \frac{1}{2}\\ \frac{5}{2}& 2& \frac{3}{2}& 1\\ 4& \frac{7}{2}& 3& \frac{5}{2}\end{array}\right]$

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

$A=\left[\begin{array}{cccc}1& 0& -1& -2\\ 3& 2& 1& 0\\ 5& 4& 3& 2\end{array}\right]$

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

$\left[\begin{array}{cc}4& 3\\ x& 5\end{array}\right]=\left[\begin{array}{cc}y& z\\ 1& 5\end{array}\right]$

Answer:

(i) $\left[\begin{array}{cc}4& 3\\ x& 5\end{array}\right]=\left[\begin{array}{cc}y& z\\ 1& 5\end{array}\right]$

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

$\therefore$ $x=1,y=4andz=3$

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

$\left[\begin{array}{cc}x+y& 2\\ 5+z& xy\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 5& 8\end{array}\right]$

Answer:

(ii)

$\left[\begin{array}{cc}x+y& 2\\ 5+z& xy\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 5& 8\end{array}\right]$

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,

$y=4andy=2$

Putting the values of y, we get

$x=2andx=4$

And also equating the first element of the second raw

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

Hence,

$x=2,y=4,z=0andx=4,y=2,z=0$

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

$\left[\begin{array}{c}x+y+z\\ x+z\\ y+z\end{array}\right]=\left[\begin{array}{c}9\\ 5\\ 7\end{array}\right]$

Answer:

(iii)

$\left[\begin{array}{c}x+y+z\\ x+z\\ y+z\end{array}\right]=\left[\begin{array}{c}9\\ 5\\ 7\end{array}\right]$

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:

$\left[\begin{array}{cc}a-b& 2a+c\\ 2a-b& 3c+d\end{array}\right]=\left[\begin{array}{cc}-1& 5\\ 0& 13\end{array}\right]$

Answer:

$\left[\begin{array}{cc}a-b& 2a+c\\ 2a-b& 3c+d\end{array}\right]=\left[\begin{array}{cc}-1& 5\\ 0& 13\end{array}\right]$

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

$a=1andb=2$

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

$\left[\begin{array}{cc}3x+7& 5\\ y+1& 2-3x\end{array}\right]$, $\left[\begin{array}{cc}0& y-2\\ 8& 4\end{array}\right]$

(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, $\left[\begin{array}{cc}3x+7& 5\\ y+1& 2-3x\end{array}\right]$ $=\left[\begin{array}{cc}0& y-2\\ 8& 4\end{array}\right]$

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.