Exercise 3.1 (Revised) - Chapter 3 - Matrices - Ncert Solutions class 12 - Maths

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NCERT Solutions Class 12 Maths: Chapter 3 - Matrices | Comprehensive Guide

Ex 3.1 Question 1:

In the matrix $A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & 5 / 2 & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]$, write:
(i) The order of the matrix.
(ii) The number of elements.
(iii) Write the elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$

Answer.

(i) There are 3 horizontal lines (rows) and 4 vertical lines (columns) in the given matrix A.

Therefore, Order of the matrix is $3 \times 4$.
(ii) The number of elements in the matrix $\mathrm{A}$ is $3 \times 4=12$.
(iii) $a_{13} \rightarrow$ Element in first row and third column $=19$
$a_{21} \rightarrow$ Element in second row and first column $=35$
$a_{33} \rightarrow$ Element in third row and third column $=-5$
$a_{24} \rightarrow$ Element in second row and fourth column $=12$
$a_{23} \rightarrow$ Element in second row and third column $=\frac{5}{2}$

Ex 3.1 Question 2:

If a matrix has 24 elements, what are possible orders it can order? What, if it has 13 elements?

Answer:

Since, a matrix having $m n$ element is of order $m \times n$.
(i) Therefore, there are 8 possible matrices having 24 elements of orders $1 \times 24,2 \times 12,3 \times 8$, $4 \times 6,24 \times 1,12 \times 2,8 \times 3,6 \times 4$.
(ii) Prime number $13=1 \times 13$ and $13 \times 1$

Therefore, there are 2 possible matrices of order $1 \times 13$ (Row matrix) and $13 \times 1$ (Column matrix).

Ex 3.1 Question 3:

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

Answer:

Since, a matrix having $m n$ element is of order $m \times n$.
(i) Therefore, there are 6 possible matrices having 18 elements of orders $1 \times 18,2 \times 9,3 \times 6,18$ x $1,9 \times 2,6 \times 3$.
(ii) Prime number $5=1 \times 5$ and $5 \times 1$

Therefore, there are 2 possible matrices of order $1 \times 5$ (Row matrix) and $5 \times 1$ (Column matrix).

Ex 3.1 Question 4:

Construct a $2 \times 2$ matrix $\mathrm{A}=\left[a_{i j}\right]$ whose elements are given by:
(i) $a_{i j}=\frac{(i+j)^2}{2}$
(ii) $a_{i j}=\frac{i}{j}$
(iii) $a_{i j}=\frac{(i+2 j)^2}{2}$

Answer: 

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

Putting $i=1, j=1$ in eq. (i) $a_{11}=\frac{(1+1)^2}{2}=\frac{(2)^2}{2}=\frac{4}{2}=2$
Putting $i=1, j=2$ in eq. (i) $a_{12}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
Putting $i=2, j=1$ in eq. (i) $a_{11}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
Putting $i=2, j=2$ in eq. (i) $a_{11}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
$
\therefore \mathrm{A}_{2 \times 2}=\left[a_{i j}\right]=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]=\left[\begin{array}{cc}
2 & 9 / 2 \\
9 / 2 & 8
\end{array}\right]
$

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

Putting $i=1, j=1$ in eq. (i) $a_{11}=\frac{1}{1}=1$
Putting $i=1, j=2$ in eq. (i) $a_{12}=\frac{1}{2}$
Putting $i=2, j=1$ in eq. (i) $a_{11}=\frac{2}{1}=2$
Putting $i=2, j=2$ in eq. (i) $a_{11}=\frac{2}{2}=1$
$
\therefore \mathrm{A}_{2 \times 2}=\left[a_{i j}\right]=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 / 2 \\
2 & 1
\end{array}\right]
$

(iii) Given: $a_{i j}=\frac{(i+2 j)^2}{2} \ldots \ldots \ldots .$. (i)

Putting $i=1, j=1$ in eq. (i) $a_{11}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
Putting $i=1, j=2$ in eq. (i) $a_{12}=\frac{(1+4)^2}{2}=\frac{(5)^2}{2}=\frac{25}{2}$
Putting $i=2, j=1$ in eq. (i) $a_{11}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
Putting $i=2, j=2$ in eq. (i) $a_{11}=\frac{(2+4)^2}{2}=\frac{(6)^2}{2}=\frac{36}{2}=18$
$
\therefore \mathrm{A}_{2 \times 2}=\left[a_{i j}\right]=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]=\left[\begin{array}{cc}
9 / 2 & 25 / 2 \\
8 & 18
\end{array}\right]
$

Ex 3.1 Question 5:

Construct a $3 \times 4$ matrix, whose elements are given by:
(i) $a_{i j}=\frac{1}{2}|-3 i+j|$
(ii) $a_{i j}=2 i-j$

Answer.

(i) Given: $a_{i j}=\frac{1}{2}|-3 i+j|$ $\qquad$
Putting $i=1, j=1$ in eq. (i) $a_{11}=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{1}{2}(2)=1$
Putting $i=1, j=2$ in eq. (i) $a_{12}=\frac{1}{2}|-3+2|=\frac{1}{2}|-1|=\frac{1}{2}(1)=\frac{1}{2}$

Putting $i=1, j=3$ in eq. (i) $a_{13}=\frac{1}{2}|-3+3|=\frac{1}{2}|0|=\frac{1}{2}(0)=0$
Putting $i=1, j=4$ in eq. (i) $a_{14}=\frac{1}{2}|-3+4|=\frac{1}{2}|1|=\frac{1}{2}(1)=\frac{1}{2}$
Putting $i=2, j=1$ in eq. (i) $a_{21}=\frac{1}{2}|-6+1|=\frac{1}{2}|-5|=\frac{1}{2}(5)=\frac{5}{2}$
Putting $i=2, j=2$ in eq. (i) $a_{22}=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{1}{2}(4)=2$
Putting $i=2, j=3$ in eq. (i) $a_{23}=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{1}{2}(3)=\frac{3}{2}$
Putting $i=2, j=4$ in eq. (i) $a_{24}=\frac{1}{2}|-6+4|=\frac{1}{2}|-2|=\frac{1}{2}(2)=1$
Putting $i=3, j=1$ in eq. (i) $a_{31}=\frac{1}{2}|-9+1|=\frac{1}{2}|-8|=\frac{1}{2}(8)=4$
Putting $i=3, j=2$ in eq. (i) $a_{32}=\frac{1}{2}|-9+2|=\frac{1}{2}|-7|=\frac{1}{2}(7)=\frac{7}{2}$
Putting $i=3, j=3$ in eq. (i) $a_{33}=\frac{1}{2}|-9+3|=\frac{1}{2}|-6|=\frac{1}{2}(6)=3$
Putting $i=3, j=4$ in eq. (i) $a_{34}=\frac{1}{2}|-9+4|=\frac{1}{2}|-5|=\frac{1}{2}(5)=\frac{5}{2}$
$
\therefore \mathrm{A}_{3 \times 4}=\left[a_{34}\right]=\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]=\left[\begin{array}{cccc}
1 & 1 / 2 & 0 & 1 / 2 \\
5 / 2 & 2 & 3 / 2 & 1 \\
4 & 7 / 2 & 3 & 5 / 2
\end{array}\right]
$

(ii) Given: $a_{i j}=2 i-j$ $\qquad$
Putting $i=1, j=1$ in eq. (i) $a_{11}=2-1=1$

Putting $i=1, j=2$ in eq. (i) $a_{12}=2-2=0$
Putting $i=1, j=3$ in eq. (i) $a_{13}=2-3=-1$
Putting $i=1, j=4$ in eq. (i) $a_{14}=2-4=-2$
Putting $i=2, j=1$ in eq. (i) $a_{21}=4-3=3$
Putting $i=2, j=2$ in eq. (i) $a_{22}=4-2=2$
Putting $i=2, j=3$ in eq. (i) $a_{23}=4-3=1$
Putting $i=2, j=4$ in eq. (i) $a_{24}=4-4=0$
Putting $i=3, j=1$ in eq. (i) $a_{31}=6-1=5$
Putting $i=3, j=2$ in eq. (i) $a_{32}=6-2=4$
Putting $i=3, j=3$ in eq. (i) $a_{33}=6-3=3$
Putting $i=3, j=4$ in eq. (i) $a_{34}=6-4=2$
$
\therefore \mathrm{A}_{3 \times 4}=\left[a_{34}\right]=\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]=\left[\begin{array}{cccc}
1 & 0 & -1 & -2 \\
3 & 2 & 1 & 0 \\
5 & 4 & 3 & 2
\end{array}\right]
$

Ex 3.1 Question 6:

Find the values of $x, y$ and $z$ from the following equations:
(i) $\left[\begin{array}{ll}4 & 3 \\ x & 5\end{array}\right]=\left[\begin{array}{ll}y & z \\ 1 & 5\end{array}\right]$
(ii) $\left[\begin{array}{cc}x+y & 2 \\ 5+z & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]$

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

Answer:

(i)Given: $\left[\begin{array}{ll}4 & 3 \\ x & 5\end{array}\right]=\left[\begin{array}{ll}y & z \\ 1 & 5\end{array}\right]$
By definition of Equal matrices, $x=1, y=4, z=3$
(ii) $\left[\begin{array}{cc}x+y & 2 \\ 5+z & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]$

Equating corresponding entries, $x+y=6$ $\qquad$
$
5+z=5 \Rightarrow z=5-5 \Rightarrow z=0
$

And $x y=8 \Rightarrow x(6-x)=8$ [From eq. (i), $y=6-x$ ]
$
\begin{aligned}
& \Rightarrow 6 x-x^2=8 \\
& \Rightarrow x^2-6 x+8=0 \\
& \Rightarrow(x-4)(x-2)=0 \\
& \Rightarrow x=4 \text { or } x=2
\end{aligned}
$

Putting these values of $x$ in eq. (i), we have $y=2$ and $y=4$
$
\therefore x=2, y=4, z=0 \text { or } x=4, y=2, z=0
$
(iii) Given: $\left[\begin{array}{c}x+y+z \\ x+z \\ y+z\end{array}\right]=\left[\begin{array}{l}9 \\ 5 \\ 7\end{array}\right]$

Equating corresponding entries, $x+y+z=9$ $\qquad$
$
x+z=5
$

And $y+z=7$ $\qquad$ (iii)

Eq. (i) - Eq. (ii) $=y=9-5=4$
Eq. (i) - Eq. (iii) $=x=9-7=2$
Putting values of $x$ and $y$ in eq. (i),
$
\begin{aligned}
& 2+4+z=9 \Rightarrow z=3 \\
& \therefore x=2, y=4, z=3
\end{aligned}
$

Ex 3.1 Question 7:

7. Find the values of $a, b, c$ and $d$ from the equation $\left[\begin{array}{cc}a-b & 2 a+c \\ 2 a-b & 3 c+d\end{array}\right]=\left[\begin{array}{cr}-1 & 5 \\ 0 & 13\end{array}\right]$.

Answer.

Equating corresponding entries,
$
a-b=-1
$
$
2 a-b=0
$
$
2 a+c=5
$
$
3 c+d=13
$

Eq. (i) - Eq. (ii) $=-a=-1$

$
\Rightarrow a=1
$

Putting $a=1$ in eq. (i), $1-b=-1$
$
\Rightarrow-b=-2 \Rightarrow b=2
$

Putting $a=1$ in eq. (iii), $2+c=5$
$
\Rightarrow c=5-2 \Rightarrow c=3
$

Putting $c=3$ in eq. (iv), $9+d=13$
$
\Rightarrow d=13-9 \Rightarrow d=4
$

$\therefore a=1, b=2, c=3, d=4$

Ex 3.1 Question 8:

$\mathbf{A}=\left[a_{i j}\right]_{m_{\times n}}$ is a square matrix if:
(A) $m<n$
(B) $m>n$
(C) $m=n$
(D) None of these

Answer.

By definition of square matrix $m=n$, option (C) is correct.

Ex 3.1 Question 9:

Which of the given values of $x$ and $y$ make the following pairs of matrices equal:
$
\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\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.

Equating corresponding sides,
$
3 x+7=0 \Rightarrow x=\frac{-7}{3}
$

And $5=y-2 \Rightarrow y=7$
Also $y+1=8 \Rightarrow y=7$
And $2-3 x=4 \Rightarrow x=\frac{-2}{3}$
Since, values of $x$ are not equal, therefore, no values of $x$ and $y$ exist to make the two matrices equal.

Therefore, option (B) is correct.

Ex 3.1 Question 10:
The number of all possible matrices of order $3 \times 3$ with each entry 0 or 1 is:
(A) 27
(B) 18
(C) 81
(D) 512

Answer.

Since, general matrix of order $3 \times 3$ is $\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$.
This matrix has 9 elements.
The number of choices for $a_{11}$ is 2 (as 0 or 1 can be used)
Similarly, the number of choices for each other element is 2 .
Therefore, total possible arrangements (matrices) $=2 \times 2 \times 2 \times \ldots 9$ times $=2^9=512$
Therefore, option (D) is correct.