EXERCISE 3.1 (MATRIX)

Class 12 ncert solution math exercise 3.1 matrix

Question 1: (i) The order of the matrix

$A=\begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix} \text {, write: }$

(ii) The number of elements

(iii) Write the elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$

Solution: (i) Since, in the given matrix, the number of rows is 3 and the number of columns is 4 , the order of the matrix is $3 \times 4$ .

(ii) Since the order of the matrix is $3 \times 4$ , there are $3 \times 4=12$ elements.

(iii) Here,

$a_{13}=19$

$a_{21}=35$

$a_{33}=-5$

$a_{24}=12$

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

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

Solution: We know that if a matrix is of the order $m \times n$ , it has $m n$ elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24 .

The ordered pairs are: $(1,24),(24,1),(2,12),(12,2),(3,8),(8,3),(4,6)$ and $(6,4)$ .

Hence, the possible orders of a matrix having 24 elements are:

$(1 \times 24),(24 \times 1),(2 \times 12),(12 \times 2),(3 \times 8),(8 \times 3),(4 \times 6) \text { and }(6 \times 4).$

$(1,13)$ and $(13,1)$ are the ordered pairs of natural numbers whose product is 13 . Hence, the possible orders of a matrix having 13 elements are $(1 \times 13)$ and $(13 \times 1)$ .

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

Solution: We know that if a matrix is of the order $m \times n$ , it has $m n$ elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18 .

The ordered pairs are: $(1,18),(18,1),(2,9),(9,2),(3,6)$ and $(6,3)$ .

Hence, the possible orders of a matrix having 18 elements are:

$(1 \times 18),(18 \times 1),(2 \times 9),(9 \times 2),(3 \times 6) \text { and }(6 \times 3) \text {. }$

$(1 \times 5)$ and $(5 \times 1)$ are the ordered pairs of natural numbers whose product is 5 .

Hence, the possible orders of a matrix having 5 elements are $(1 \times 5)$ and $(5 \times 1)$ .

Question 4: Construct a $2 \times 2$ matrix, $A=\left[a_{i j}\right]$ , whose elements are given by:

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

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

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

Solution: In general, a $2 \times 2$ matrix is given by $A=\left(\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right)$

(i) $a_{i j}=\frac{(i+j)^{2}}{2} ; i, j=1,2$

Therefore,

$a_{11}=\frac{(1+1)^{2}}{2}=\frac{4}{2}=2$

$a_{12}=\frac{(1+2)^{2}}{2}=\frac{9}{2}$

$a_{21}=\frac{(2+1)^{2}}{2}=\frac{9}{2}$

$a_{22}=\frac{(2+2)^{2}}{2}=\frac{16}{2}=8$

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

(ii) $a_{i j}=\frac{i}{j} ; \quad i, j=1,2$

Therefore,

$a_{11}=\frac{1}{1}=1$

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

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

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

Thus, the required matrix is $A=\begin{bmatrix}1 & \frac{1}{2} \\ 2 & 1\end{bmatrix}$

(iii) $\quad a_{i j}=\frac{(i+2 j)^{2}}{2} ; i, j=1,2$

Therefore,

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

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

$a_{21}=\frac{(2+2)^{2}}{2}=8$

$a_{22}=\frac{(2+4)^{2}}{2}=18$

Thus, the required matrix is $A=\begin{bmatrix}\frac{9}{2} & \frac{25}{2} \\ 8 & 18\end{bmatrix}$

Question 5: In general, a $3 \times 4$ matrix whose elements are given by

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

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

Solution: $A=\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$

(i) Given $a_{i j}=\frac{1}{2}|-3 i+j| ; i=1,2,3 \quad j=1,2,3,4$

$a_{11}=\frac{1}{2}|-3(1)+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1$

$a_{21}=\frac{1}{2}|-3(2)+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|-5|=\frac{5}{2}$

$a_{31}=\frac{1}{2}|-3(3)+1|=\frac{1}{2}|-9+1|=\frac{1}{2}|-8|=\frac{8}{2}=4$

$a_{12}=\frac{1}{2}|-3(1)+2|=\frac{1}{2}|-3+2|=\frac{1}{2}|-1|=\frac{1}{2}$

$a_{22}=\frac{1}{2}|-3(2)+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2$

$a_{32}=\frac{1}{2}|-3(3)+2|=\frac{1}{2}|-9+2|=\frac{1}{2}|-7|=\frac{7}{2}$

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

$a_{23}=\frac{1}{2}|-3(2)+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2}$

$a_{33}=\frac{1}{2}|-3(3)+3|=\frac{1}{2}|-9+3|=\frac{1}{2}|-6|=\frac{6}{2}=3$

$a_{14}=\frac{1}{2}|-3(1)+4|=\frac{1}{2}|-3+4|=\frac{1}{2}|1|=\frac{1}{2}$

$a_{24}=\frac{1}{2}|-3(2)+4|=\frac{1}{2}|-6+4|=\frac{1}{2}|-2|=\frac{2}{2}=1$

$a_{34}=\frac{1}{2}|-3(3)+4|=\frac{1}{2}|-9+4|=\frac{1}{2}|-5|=\frac{5}{2}$

Thus, the required matrix is

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

(ii) $\quad a_{i j}=2 i-j ; i=1,2,3 \quad j=1,2,3,4$

$a_{11}=2(1)-1=2-1=1$

$a_{21}=2(2)-1=4-1=3$

$a_{31}=2(3)-1=6-1=5$

$a_{12}=2(1)-2=2-2=0$

$a_{22}=2(2)-2=4-2=2$

$a_{32}=2(3)-2=6-2=4$

$a_{13}=2(1)-3=2-3=-1$

$a_{23}=2(2)-3=4-3=1$

$a_{33}=2(3)-3=6-3=3$

$a_{14}=2(1)-4=2-4=-2$

$a_{24}=2(2)-4=4-4=0$

$a_{34}=2(3)-4=6-4=2$

Thus, the required matrix is $A=\begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}$

Question 6: Find the value of $x, y$ and $z$ from the following equation:
(i) $\begin{bmatrix}4 & 3 \\ x & 5\end{bmatrix}=\begin{bmatrix}y & z \\ 1 & 5\end{bmatrix}$

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

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

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

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

$x=1, y=4 \text { and } z=3$

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

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

$x+y=6$

$x y=8$

$5+z=5$

Hence,

$\Rightarrow 5+z=5$

$\Rightarrow z=0$

We know that $(a-b)^{2}=(a+b)^{2}-4 a b$

$\Rightarrow(x-y)^{2}=(6)^{2}-8 \times 4$

$\Rightarrow(x-y)^{2}=36-32$

$\Rightarrow(x-y)^{2}=4$

$\Rightarrow(x-y)=\pm 2$

Equating $x-y=2$ and $x+y=6$ , we get $x=4, y=2$

Similarly, Equating $x-y=-2$ and $x+y=6$ , we get $x=2, y=4$

Thus, $x=4, y=2, z=0$ or $x=2, y=4, z=0$

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

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

$x+y+z=9$

$x+z=5$

$y+z=7$

From (1) and (2), we have

$\Rightarrow y+5=9$

$\Rightarrow y=4$

From (3), we have

$\Rightarrow 4+z=7$

$\Rightarrow z=3$

Therefore,

$\Rightarrow x+z=5$

$\Rightarrow x+3=5$

$\Rightarrow x=2$

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

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

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

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

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

$a-b=-1$

$2 a-b=0$

$2 a+c=5$

$3 c+d=13$

From (2),

$b=2 a$

Putting this value in (1),

Hence,

$\Rightarrow a-2 a=-1$
$\Rightarrow a=1$

$\Rightarrow b=2$

Putting
$a=1$ in (3),

$\Rightarrow 2(1)+c=5$

$\Rightarrow c=3$

Putting $c=3$ in $(4)$ ,

$\Rightarrow 3(3)+d=13$

$\Rightarrow d=4$

Thus, $a=1, b=2, c=3$ and $d=4$ .

Question 8: $A=\left[a_{i j}\right]_{m \times n}$ is a square matrix, if
(A) $m<n$
(B) $m>n$
(C) $m=n$
(D) None of these

Solution: It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

Therefore, $A=\left[a_{i j}\right]_{m \times n}$ is a square matrix, if $m=n$ .

Thus, the correct option is {C}.

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

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

(A) $x=\frac{-1}{3}, y=7$
(B) Not possible to find
(C) $y=7, x=\frac{-2}{3}$
(D) $x=\frac{-1}{3}, y=\frac{-2}{3}$

Solution: The given matrices are $\left[\begin{array}{cc}3 x+7 & 5 \\ y+1 & 2-3 x\end{array}\right]$ and $\left[\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right]$