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

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: $$ (1 \times 24),(24 \times 1),(2 \times 12),(12 \times 2),(3 \times 8),(8 \times 3),(4 \times 6) \text { and }(6 \times 4) \text {. } $$ $(1,13)$ 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 $(13 \times 1)$.

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

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: $$ (1 \times 18),(18 \times 1),(2 \times 9),(9 \times 2),(3 \times 6) \text { and }(6 \times 3) $$ $(1 \times 5)$ 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 $(5 \times 1)$.

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

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) $\quad a_{i j}=\frac{(i+j)^{2}}{2} ; \quad i, j=1,2$ Therefore, $$ \begin{aligned} & a_{11}=\frac{(1+1)^{2}}{2}=\frac{4}{2}=2 \\ & a_{12}=\frac{(1+2)^{2}}{2}=\frac{9}{2} \\ & a_{21}=\frac{(2+1)^{2}}{2}=\frac{9}{2} \\ & a_{22}=\frac{(2+2)^{2}}{2}=\frac{16}{2}=8 \end{aligned} $$ Thus, the required matrix is $$ A=\left(\begin{array}{cc} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{array}\right) $$ $a_{i j}=\frac{i}{j} ; \quad i, j=1,2$ Therefore, $$ \begin{aligned} & a_{11}=\frac{1}{1}=1 \\ & a_{12}=\frac{1}{2} \\ & a_{21}=\frac{2}{1}=2 \\ & a_{22}=\frac{2}{2}=1 \end{aligned} $$ Thus, the required matrix is $$ A=\left(\begin{array}{ll} 1 & \frac{1}{2} \\ 2 & 1 \end{array}\right) $$ (iii) $a_{i j}=\frac{(i+2 j)^{2}}{2} ; i, j=1,2$ Therefore, $$ \begin{aligned} & a_{11}=\frac{(1+2)^{2}}{2}=\frac{9}{2} \\ & a_{12}=\frac{(1+4)^{2}}{2}=\frac{25}{2} \\ & a_{21}=\frac{(2+2)^{2}}{2}=8 \\ & a_{22}=\frac{(2+4)^{2}}{2}=18 \end{aligned} $$ Thus, the required matrix is $A=\left(\begin{array}{cc}\frac{9}{2} & \frac{25}{2} \\ 8 & 18\end{array}\right)$

Q: Find the value of $x, y$ and $z$ from the following equation: (i) $\quad\left(\begin{array}{ll}4 & 3 \\ x & 5\end{array}\right)=\left(\begin{array}{ll}y & z \\ 1 & 5\end{array}\right)$ (ii) $\quad\left(\begin{array}{cc}x+y & 2 \\ 5+z & x y\end{array}\right)=\left(\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right)$ (iii) $\left(\begin{array}{c}x+y+z \\ x+z \\ y+z\end{array}\right)=\left(\begin{array}{l}9 \\ 5 \\ 7\end{array}\right)$

$\left(\begin{array}{ll}4 & 3 \\ x & 5\end{array}\right)=\left(\begin{array}{ll}y & z \\ 1 & 5\end{array}\right)$ (i) 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) $\quad\left(\begin{array}{cc}x+y & 2 \\ 5+z & x y\end{array}\right)=\left(\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right)$ As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: $$ \begin{aligned} & x+y=6 \\ & x y=8 \\ & 5+z=5 \end{aligned} $$ Hence, $$ \begin{aligned} & \Rightarrow 5+z=5 \\ & \Rightarrow z=0 \end{aligned} $$ We know that $(a-b)^{2}=(a+b)^{2}-4 a b$ $$ \begin{aligned} & \Rightarrow(x-y)^{2}=(6)^{2}-8 \times 4 \\ & \Rightarrow(x-y)^{2}=36-32 \\ & \Rightarrow(x-y)^{2}=4 \\ & \Rightarrow(x-y)= \pm 2 \end{aligned} $$ 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 $x=2, y=4, z=0$ $$ \left(\begin{array}{c} x+y+z \\ x+z \\ y+z \end{array}\right)=\left(\begin{array}{l} 9 \\ 5 \\ 7 \end{array}\right) $$ As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: $$ \begin{align*} & x+y+z=9 \tag{1}\\ & x+z=5 \tag{2}\\ & y+z=7 \tag{3} \end{align*} $$ From (1) and (2), we have $$ \begin{aligned} & \Rightarrow y+5=9 \\ & \Rightarrow y=4 \end{aligned} $$ From (3), we have $$ \begin{aligned} & \Rightarrow 4+z=7 \\ & \Rightarrow z=3 \end{aligned} $$ Therefore, $$ \begin{aligned} & \Rightarrow x+z=5 \\ & \Rightarrow x+3=5 \\ & \Rightarrow x=2 \end{aligned} $$ Thus, $x=2, y=4, z=3$

Q: Find the value of $a, b, c$ and $d$ from the equation: $\left(\begin{array}{cc}a-b & 2 a+c \\ 2 a-b & 3 c+d\end{array}\right)=\left(\begin{array}{cc}-1 & 5 \\ 0 & 13\end{array}\right)$

$\left(\begin{array}{cc}a-b & 2 a+c \\ 2 a-b & 3 c+d\end{array}\right)=\left(\begin{array}{cc}-1 & 5 \\ 0 & 13\end{array}\right)$ As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: $$ \begin{align*} & a-b=-1 \tag{1}\\ & 2 a-b=0 \tag{2}\\ & 2 a+c=5 \tag{3}\\ & 3 c+d=13 \tag{4} \end{align*} $$ From (2), $$ b=2 a $$ Putting this value in (1), $$ \begin{aligned} & \Rightarrow a-2 a=-1 \\ & \Rightarrow a=1 \end{aligned} $$ Hence, $$ \Rightarrow b=2 $$ Putting $a=1$ in (3), $$ \begin{aligned} & \Rightarrow 2(1)+c=5 \\ & \Rightarrow c=3 \end{aligned} $$ Putting $c=3$ in (4), $$ \begin{aligned} & \Rightarrow 3(3)+d=13 \\ & \Rightarrow d=4 \end{aligned} $$ Thus, $a=1, b=2, c=3$ and $d=4$.

Q: Which of the given values of $x$ and $y$ make the following pair of matrices equal $\left\lceil\begin{array}{cc}3 x+7 & 5 \\ y+1 & 2-3 x\end{array}\right\rceil,\left\lceil\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right\rceil$ (A) $x=\frac{-1}{3}, y=7$ (B) Not possible to find (C) $y=7, x=\frac{-2}{3}$ (D) $x=\frac{-1}{3}, y=\frac{-2}{3}$

The given matrices are $\left[\begin{array}{cc}3 x+7 & 5 \\ y+1 & 2-3 x\end{array}\right]$ and $\left[\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right]$ Equating the corresponding elements, we get: $$ \begin{aligned} & 3 x+7=0 \Rightarrow x=\frac{-7}{3} \\ & y-2=5 \Rightarrow y=7 \\ & y+1=8 \Rightarrow y=7 \\ & 2-3 x=4 \Rightarrow x=\frac{-2}{3} \end{aligned} $$ We find that on comparing the corresponding elements of the two matrices, we get two different values of $x$, which is not possible. Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal. Thus, the correct option is B.

Q: 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

The given matrix of the order $3 \times 3$ has 9 elements and each of these elements can be either 0 or 1 . Now, each of the 9 elements can be filled in two possible ways. Hence, by the multiplication principle, the required number of possible matrices is $2^{9}=512$. Thus, the correct option is D. \section*{EXERCISE 3.2}

Question 1

$A=\left[a_{i j}ight]_{m 	imes n}$ is a square matrix, if

(A) $m<n$

(B) $m>n$

(C) $m=n$

(D) None of these

Accepted Answer

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

Thus, the correct option is C.

Question 2

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

(i) The order of the matrix

(ii) The number of elements

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

Accepted Answer

(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 $3 	imes 4$.

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

(iii) Here,

$$

\begin{aligned}

& a_{13}=19 \

& a_{21}=35 \

& a_{33}=-5 \

& a_{24}=12 \

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

\end{aligned}

$$

Question 3

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

Accepted Answer

We know that if a matrix is of the order $m 	imes n$, 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 	imes 24),(24 	imes 1),(2 	imes 12),(12 	imes 2),(3 	imes 8),(8 	imes 3),(4 	imes 6) 	ext { and }(6 	imes 4) 	ext {. }

$$

$(1,13)$ 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 $(1 	imes 13)$ and $(13 	imes 1)$.

Question 4

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

Accepted Answer

We know that if a matrix is of the order $m 	imes n$, 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 	imes 18),(18 	imes 1),(2 	imes 9),(9 	imes 2),(3 	imes 6) 	ext { and }(6 	imes 3)

$$

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

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

Question 5

Construct a $2 	imes 2$ matrix, $A=\left[a_{i j}ight]$, whose elements are given by:

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

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

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

Accepted Answer

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

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

Therefore,

$$

\begin{aligned}

& 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

\end{aligned}

$$

Thus, the required matrix is

$$

A=\left(\begin{array}{cc}

2 & \frac{9}{2} \

\frac{9}{2} & 8

\end{array}ight)

$$

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

Therefore,

$$

\begin{aligned}

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

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

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

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

\end{aligned}

$$

Thus, the required matrix is

$$

A=\left(\begin{array}{ll}

1 & \frac{1}{2} \

2 & 1

\end{array}ight)

$$

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

Therefore,

$$

\begin{aligned}

& 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

\end{aligned}

$$

Thus, the required matrix is $A=\left(\begin{array}{cc}\frac{9}{2} & \frac{25}{2} \ 8 & 18\end{array}ight)$

Question 6

In general, a $3 	imes 4$ matrix whose elements are given by

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

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

Accepted Answer

In general, a $3 	imes 4$ matrix is given by $A=\left(\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \ a_{21} & a_{22} & a_{23} & a_{24} \ a_{31} & a_{32} & a_{33} & a_{34}\end{array}ight)$

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

$$

\begin{aligned}

& 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}

\end{aligned}

$$

Thus, the required matrix is $A=\left(\begin{array}{cccc}1 & \frac{1}{2} & 0 & \frac{1}{2} \ \frac{5}{2} & 2 & \frac{3}{2} & 1 \ 4 & \frac{7}{2} & 3 & \frac{5}{2}\end{array}ight)$

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

$$

\begin{aligned}

& a_{11}=2(1)-1=2-1=1 \

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

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

\end{aligned}

$$

\begin{aligned}

& a_{12}=2(1)-2=2-2=0 \

& a_{22}=2(2)-2=4-2=2 \

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

\end{aligned}

$$

\begin{aligned}

& a_{13}=2(1)-3=2-3=-1 \

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

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

\end{aligned}

$$

\begin{aligned}

& a_{14}=2(1)-4=2-4=-2 \

& a_{24}=2(2)-4=4-4=0 \

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

\end{aligned}

$$

Thus, the required matrix is

$$

A=\left(\begin{array}{cccc}

1 & 0 & -1 & -2 \

3 & 2 & 1 & 0 \

5 & 4 & 3 & 2

\end{array}ight)

$$

Question 7

Find the value of $x, y$ and $z$ from the following equation:

(i) $\quad\left(\begin{array}{ll}4 & 3 \ x & 5\end{array}ight)=\left(\begin{array}{ll}y & z \ 1 & 5\end{array}ight)$

(ii) $\quad\left(\begin{array}{cc}x+y & 2 \ 5+z & x y\end{array}ight)=\left(\begin{array}{ll}6 & 2 \ 5 & 8\end{array}ight)$

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

Accepted Answer

$\left(\begin{array}{ll}4 & 3 \ x & 5\end{array}ight)=\left(\begin{array}{ll}y & z \ 1 & 5\end{array}ight)$

(i)

As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

$$

x=1, y=4 	ext { and } z=3

$$

(ii) $\quad\left(\begin{array}{cc}x+y & 2 \ 5+z & x y\end{array}ight)=\left(\begin{array}{ll}6 & 2 \ 5 & 8\end{array}ight)$

As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

$$

\begin{aligned}

& x+y=6 \

& x y=8 \

& 5+z=5

\end{aligned}

$$

Hence,

$$

\begin{aligned}

& \Rightarrow 5+z=5 \

& \Rightarrow z=0

\end{aligned}

$$

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

$$

\begin{aligned}

& \Rightarrow(x-y)^{2}=(6)^{2}-8 	imes 4 \

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

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

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

\end{aligned}

$$

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

$$

\left(\begin{array}{c}

x+y+z \

x+z \

y+z

\end{array}ight)=\left(\begin{array}{l}

9 \

5 \

7

\end{array}ight)

$$

As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

$$

\begin{align*}

& x+y+z=9  	ag{1}\

& x+z=5  	ag{2}\

& y+z=7 	ag{3}

\end{align*}

$$

From (1) and (2), we have

$$

\begin{aligned}

& \Rightarrow y+5=9 \

& \Rightarrow y=4

\end{aligned}

$$

From (3), we have

$$

\begin{aligned}

& \Rightarrow 4+z=7 \

& \Rightarrow z=3

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

& \Rightarrow x+z=5 \

& \Rightarrow x+3=5 \

& \Rightarrow x=2

\end{aligned}

$$

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

Question 8

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

$\left(\begin{array}{cc}a-b & 2 a+c \ 2 a-b & 3 c+d\end{array}ight)=\left(\begin{array}{cc}-1 & 5 \ 0 & 13\end{array}ight)$

Accepted Answer

$\left(\begin{array}{cc}a-b & 2 a+c \ 2 a-b & 3 c+d\end{array}ight)=\left(\begin{array}{cc}-1 & 5 \ 0 & 13\end{array}ight)$

As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

$$

\begin{align*}

& a-b=-1  	ag{1}\

& 2 a-b=0  	ag{2}\

& 2 a+c=5  	ag{3}\

& 3 c+d=13 	ag{4}

\end{align*}

$$

From (2),

$$

b=2 a

$$

Putting this value in (1),

$$

\begin{aligned}

& \Rightarrow a-2 a=-1 \

& \Rightarrow a=1

\end{aligned}

$$

Hence,

$$

\Rightarrow b=2

$$

Putting $a=1$ in (3),

$$

\begin{aligned}

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

& \Rightarrow c=3

\end{aligned}

$$

Putting $c=3$ in (4),

$$

\begin{aligned}

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

& \Rightarrow d=4

\end{aligned}

$$

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

Question 9

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

$\left\lceil\begin{array}{cc}3 x+7 & 5 \ y+1 & 2-3 x\end{array}ightceil,\left\lceil\begin{array}{cc}0 & y-2 \ 8 & 4\end{array}ightceil$

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

(B) Not possible to find

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

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

Accepted Answer

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

Equating the corresponding elements, we get:

$$

\begin{aligned}

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

& y-2=5 \Rightarrow y=7 \

& y+1=8 \Rightarrow y=7 \

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

\end{aligned}

$$

We find that on comparing the corresponding elements of the two matrices, we get two different values of $x$, which is not possible.

Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal.

Thus, the correct option is B.

Question 10

The number of all possible matrices of order $3 	imes 3$ with each entry 0 or 1 is:

(A) 27

(B) 18

(C) 81

(D) 512

Accepted Answer

The given matrix of the order $3 	imes 3$ has 9 elements and each of these elements can be either 0 or 1 .

Now, each of the 9 elements can be filled in two possible ways.

Hence, by the multiplication principle, the required number of possible matrices is $2^{9}=512$.

Thus, the correct option is D.

\section*{EXERCISE 3.2}