Solution

(i) In the given matrix, the number of rows is 3 and the number of columns is 4 .

Therefore, 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 in it.

(iii) $a _{13}=19, a _{21}=35, a _{33}=-5, a _{24}=12, a _{23}=\frac{5}{2}$

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:

$1 \times 24,24 \times 1,2 \times 12,12 \times 2,3 \times 8,8 \times 3,4 \times 6$, 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 $13 \times 1$.

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:

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

$(1,5)$ and $(5,1)$ 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 $5 \times 1$.

Solution

(i) Since it is a $2 \times 2$ matrix it has 2 rows and 2 column. Let matrix be $\mathrm{A}$ Where $A=\left[\begin{array}{ll}a_{11} & a_{12} \ a_{21} & a_{22}\end{array}\right]$ Now it is given that $$ a_{i j}=\frac{(i+j)^2}{2} $$

$$ \begin{array}{|c|c|c|} \hline a_{i j} & i=, j= & a_{i j}=\frac{(i+j)^2}{2} \\ \hline a_{11} & i=1, j=1 & a_{11}=\frac{(1+1)^2}{2}=\frac{(2)^2}{2}=\frac{4}{2}=2 \\ \hline a_{12} & i=1, j=2 & a_{12}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2} \\ \hline a_{21} & i=2, j=1 & a_{21}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2} \\ \hline a_{22} & i=2, j=2 & a_{22}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8\\ \hline \end{array} $$

Hence, the required matrix $\mathrm{A}$ ia $A\left[\begin{array}{ll}a_{11} & a_{12} \ a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{cc}2 & \frac{9}{2} \ \frac{9}{2} & 8\end{array}\right]$ (ii) Since it is a $2 \times 2$ matrix it has 2 rows and 2 column.

Let matrix be A Where $A=\left[\begin{array}{ll}a_{11} & a_{12} \ a_{21} & a_{22}\end{array}\right]$ Now it is given that $\mathrm{a}_{\mathrm{ij}}=\frac{\mathrm{i}}{\mathrm{j}}$

$$ \begin{array}{|c|c|c|} \hline \mathrm{a} _{\mathrm{ij}} & \mathrm{i}=, \mathrm{j}= & \mathrm{a} _{\mathrm{ij}}=\frac{\mathrm{i}}{\mathrm{j}} \\ \hline \mathrm{a} _{11} & \mathrm{i}=1, \mathrm{j}=1 & \mathrm{a} _{11}=\frac{1}{1}=1 \\ \hline \mathrm{a} _{12} & \mathrm{i}=1, \mathrm{j}=2 & \mathrm{a} _{12}=\frac{1}{2} \\ \hline \mathrm{a} _{21} & \mathrm{i}=2, \mathrm{j}=1 & \mathrm{a} _{21}=\frac{2}{1}=2 \\ \hline \mathrm{a} _{22} & \mathrm{i}=2, \mathrm{j}=2 & \mathrm{a} _{22}=\frac{2}{2}=1 \\ \hline \end{array} $$

Hence, the required matrix $\mathrm{A}$ is $$ A\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]=\left[\begin{array}{ll} 1 & \frac{1}{2} \\ 2 & 1 \\ \end{array}\right] $$

(iii) Since it is a $2 \times 2$ matrix it has 2 rows and 2 column. Let matrix be $\mathrm{A}$ Where $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$ Now it is given that $$ a_{i j}=\frac{(i+2 j)^2}{2} $$

$$ \begin{array}{|c|c|c|} \hline a_{i j} & i=, j= & a_{i j}=\frac{(i+2 j)^2}{2} \\ \hline a_{11} & i=1, j=1 & a_{11}=\frac{(1+2(1))^2}{2}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2} \\ \hline a_{12} & i=1, j=2 & a_{12}=\frac{(1+2(2))^2}{2}=\frac{(1+4)^2}{2}=\frac{(5)^2}{2}=\frac{25}{2} \\ \hline a_{21} & i=2, j=1 & a_{21}=\frac{(2+2(1))^2}{2}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8 \\ \hline a_{22} & i=2, j=2 & a_{22}=\frac{(2+2(2))^2}{2}=\frac{(2+4)^2}{2}=\frac{(6)^2}{2}=\frac{36}{2}=18 \\ \hline \end{array} $$

Hence, the required matrix $\mathrm{A}$ ia $$ A\left[\begin{array}{ll} \mathrm{a} _{11} & \mathrm{a} _{12} \\ \mathrm{a} _{21} & \mathrm{a} _{22} \end{array}\right]=\left[\begin{array}{cc} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \\ \end{array}\right] $$

Solution

In general, a $3 \times 4$ matrix is given by $A= \begin{cases} 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{cases} $

(i) $a _{i j}=\frac{1}{2}|-3 i+j|, i=1,2,3$ and $j=1,2,3,4$

$ \begin{aligned} & \therefore a _{11}=\frac{1}{2}|-3 \times 1+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1 \\ & a _{21}=\frac{1}{2}|-3 \times 2+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|-5|=\frac{5}{2} \\ & a _{31}=\frac{1}{2}|-3 \times 3+1|=\frac{1}{2}|-9+1|=\frac{1}{2}|-8|=\frac{8}{2}=4 \\ & a _{12}=\frac{1}{2}|-3 \times 1+2|=\frac{1}{2}|-3+2|=\frac{1}{2}|-1|=\frac{1}{2} \\ & a _{22}=\frac{1}{2}|-3 \times 2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2 \\ & a _{32}=\frac{1}{2}|-3 \times 3+2|=\frac{1}{2}|-9+2|=\frac{1}{2}|-7|=\frac{7}{2} \\ & a _{13}=\frac{1}{2}|-3 \times 1+3|=\frac{1}{2}|-3+3|=0 \\ & a _{23}=\frac{1}{2}|-3 \times 2+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2} \\ & a _{33}=\frac{1}{2}|-3 \times 3+3|=\frac{1}{2}|-9+3|=\frac{1}{2}|-6|=\frac{6}{2}=3 \\ & a _{14}=\frac{1}{2}|-3 \times 1+4|=\frac{1}{2}|-3+4|=\frac{1}{2}|1|=\frac{1}{2} \\ & a _{24}=\frac{1}{2}|-3 \times 2+4|=\frac{1}{2}|-6+4|=\frac{1}{2}|-2|=\frac{2}{2}=1 \\ & a _{34}=\frac{1}{2}|-3 \times 3+4|=\frac{1}{2}|-9+4|=\frac{1}{2}|-5|=\frac{5}{2} \end{aligned} $

Therefore, the required matrix is $A= \begin{cases} 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{cases} $

(ii) $a _{i j}=2 i-j, i=1,2,3$ and $j=1,2,3,4$

$ \begin{aligned} & \therefore a _{11}=2 \times 1-1=2-1=1 \\ & a _{21}=2 \times 2-1=4-1=3 \\ & a _{31}=2 \times 3-1=6-1=5 \\ & a _{12}=2 \times 1-2=2-2=0 \\ & a _{22}=2 \times 2-2=4-2=2 \\ & a _{32}=2 \times 3-2=6-2=4 \\ & a _{13}=2 \times 1-3=2-3=-1 \\ & a _{23}=2 \times 2-3=4-3=1 \\ & 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 \end{aligned} $

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

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: $x=1, y=4$, 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:

$x+y=6, x y=8,5+z=5$

Now, $5+z=5 \Rightarrow z=0$

We know that:

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

$\Rightarrow(x-y)^{2}=36-32=4$

$\Rightarrow x-y= \pm 2$

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

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

$\therefore x=4, y=2$, and $z=0$ or $x=2, y=4$, and $z=0$

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

As the two 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:

$y+5=9$

$\Rightarrow y=4$

Then, from (3), we have:

$4+z=7$

$\Rightarrow z=3$

$\therefore x+z=5$

$\Rightarrow x=2$

$\therefore x=2, y=4$, and $z=3$

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), we have:

$b=2 a$

Then, from (1), we have:

$a-2 a=-1$

$\Rightarrow a=1$

$\Rightarrow b=2$

Now, from (3), we have:

$2 \times 1+c=5$

$\Rightarrow c=3$

From (4) we have:

$3 \times 3+d=13$

$\Rightarrow 9+d=13 \Rightarrow d=4$

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

Solution

The correct answer is $C$.

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

Solution

The correct answer is B.

It is given that $ \begin{bmatrix} 3 x+7 & 5 \\ y+1 & 2-3 x\end{bmatrix} = \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix} $

Equating the corresponding elements, we get:

$ \begin{aligned} & 3 x+7=0 \Rightarrow x=-\frac{7}{3} \\ & 5=y-2 \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.

Solution

The correct answer is D.

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.

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