Class 12 Maths Exercise 3.1 Solutions

(i) In the given matrix, the number of rows is 3 and the number of columns is 4 .
Order of a matrix \( = \) No of rows \( \times \) No of columns
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) \( \mathrm{a}_{13}=19, \mathrm{a}_{21}=35, \mathrm{a}_{33}=-5, \mathrm{a}_{24}=12, \mathrm{a}_{23}=\frac{5}{2} \).

It is known that if a matrix is of the order \( m \times n \), then it has \( m n \) elements.
Therefore, to find all the possible orders of a matrix having 24 elements, we had 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 \( (6,4) \).
Therefore, the possible orders of a matrix having 24 elements are; \( 1 \times 24,24 \times 1,2 \times 12,12 \times 2,3 \times 8,8 \times 3,4 \times 6,6 \times 4 \)
\( (1,13) \) and \( (13,1) \) are the ordered pairs of natural numbers whose product is 13 .
Therefore, the possible orders of a matrix having 13 elements are \( 1 \times 13 \) and \( 13 \times 1 \).

It is known that if a matrix s of the order \( \mathrm{m} \times \mathrm{n} \), then it has mn elements. Therefore, to find all the possible orders of a matrix having 18 elements, we had 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 \( (6,3) \).
Therefore, the possible orders of a matrix having 18 elements are;
\(
1 \times 18,18 \times 1,2 \times 9,9 \times 2,3 \times 6,6 \times 3\)
\( (1,5) \) and \( (5,1) \) are the ordered pairs of natural numbers whose product is 5 .
Therefore, the possible orders of a matrix having 5 elements are \( 1 \times 5 \) and \( 5 \times 1 \).

In general, \( 2 \times 2 \) matrix is given by \( \mathrm{A}=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right] \)
\( \mathrm{a}_{\mathrm{ij}}=\frac{(i+j)^{2}}{2}, \mathrm{i}, \mathrm{j}=1,2 \)
Therefore,
\( a_{11}=\frac{(1+1)^{2}}{2}=\frac{4}{2} \)
\( a_{12}=\frac{(1+2)^{2}}{2}=\frac{9}{2} \)
\( \mathrm{a}_{21}=\frac{(2+1)^{2}}{2}=\frac{9}{2} \)
\( a_{22}=\frac{(2+2)^{2}}{2}=\frac{16}{2}=8 \)
Therefore, the required matrix is \( \mathrm{A}=\left[\begin{array}{cc}2 & \frac{9}{2} \\ \frac{9}{2} & 8\end{array}\right] \)

In general, a \( 2 \times 2 \) matrix is given by \( \mathrm{A}=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right] \)
\(
\mathrm{a}_{\mathrm{ij}}=\frac{i}{j}, \mathrm{i}, \mathrm{j}=1,2
\)
Therefore,
\( \mathrm{a}_{11}=\frac{1}{1}=1 \)
\( \mathrm{a}_{12}=\frac{1}{2} \)
\( \mathrm{a}_{21}=\frac{2}{1}=2 \)
\( \mathrm{a}_{22}=\frac{2}{2}=1\)
Therefore, the required matrix is \( \mathrm{A}=\left[\begin{array}{ll}1 & \frac{1}{2} \\ 2 & 1\end{array}\right] \).

In general, a \( 2 \times 2 \) matrix is given by \( \mathrm{A}=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right] \)
\(
a_{i j}=\frac{(i+2 j)^{2}}{2}, i, j=1,2
\)
Therefore,
\( a_{11}=\frac{(1+2)^{2}}{2}=\frac{9}{2} \)
\( \mathrm{a}_{12}=\frac{(1+4)^{2}}{2}=\frac{25}{2} \)
\( \mathrm{a}_{21}=\frac{(2+2)^{2}}{2}=\frac{16}{2}=8 \)
\(
a_{22}=\frac{(2+4)^{2}}{2}=\frac{36}{2}=18
\)
Therefore, the required matrix is \( \mathrm{A}=\left[\begin{array}{cc}\frac{9}{2} & \frac{25}{2} \\ 8 & 18\end{array}\right] \)

In general \( 3 \times 4 \) matrix is given by \( \mathrm{A}=\left[\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34}\end{array}\right] \)
\( a_{i j}=\frac{1}{2}|-3 i+j|, i=1,2,3 \), and \( j=1,2,3,4 \)
Therefore,
\(
\begin{array}{l}
\mathrm{a}_{11}=\frac{1}{2}|-3 \times 1+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1 \\
\mathrm{a}_{21}=\frac{1}{3}|-3 \times 2+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|-5|=\frac{5}{2} \\
\mathrm{a}_{31}=\frac{1}{2}|-3 \times 3+1|=\frac{1}{2}|-9+1|=\frac{1}{2}|-8|=\frac{8}{2}=4 \\
\mathrm{a}_{12}=\frac{1}{2}|-3 \times 2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-1|=\frac{1}{2} \\
\mathrm{a}_{22}=\frac{1}{2}|-3 \times 2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2 \\
\mathrm{a}_{32}=\frac{1}{2}|-3 \times 3+2|=\frac{1}{2}|-9+2|=\frac{1}{2}|-7|=\frac{7}{2} \\
\mathrm{a}_{13}=\frac{1}{2}|-3 \times 1+3|=\frac{1}{2}|-3+3|=0 \\
\mathrm{a}_{23}=\frac{1}{2}|-3 \times 2+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2}
\end{array}
\)
\(
\begin{array}{l}
a_{33}=\frac{1}{2}|-3 \times 3+3|=\frac{1}{2}|-9+3|=\frac{1}{2}|-6|=\frac{6}{2}=3 \\
a_{14}=\frac{1}{2}|-3 \times 1+4|=\frac{1}{2}|-3+4|=\frac{1}{2}|1|=\frac{1}{2} \\
a_{24}=\frac{1}{2}|-3 \times 1+4|=\frac{1}{2}|-6+4|=\frac{1}{2}|-2|=\frac{2}{2}=1 \\
a_{34}=\frac{1}{2}|-3 \times 3+4|=\frac{1}{2}|-9+4|=\frac{1}{2}|-5|=\frac{5}{2}
\end{array}
\)
Therefore, required matrix is \( A=\left[\begin{array}{cccc}1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2}\end{array}\right] \)

In general \( 3 \times 4 \) matrix is given by \( A=\left[\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34}\end{array}\right] \)
\( a_{i j}=2 \mathrm{i}-\mathrm{j}, \mathrm{i}=1,2,3 \) and \( \mathrm{j}=1,2,3,4 \)
Therefore,
\(
\begin{array}{l}
a_{11}=2 \times 1-1=2-1=1 \\
a_{21}=2 \times 2-1=4-1=3 \\
a_{31}=2 \times 3-1=6-1=5 \\
a_{12}=2 \times 1-2=2-2=0 \\
a_{22}=2 \times 2-2=4-2=2 \\
a_{32}=2 \times 3-2=6-2=4
\end{array}
\)
\(
\begin{array}{l}
\mathrm{a}_{13}=2 \times 1-3=2-3=-1 \\
\mathrm{a}_{23}=2 \times 2-3=4-3=1 \\
\mathrm{a}_{33}=2 \times 3-3=6-3=3 \\
\mathrm{a}_{14}=2 \times 1-4=2-4=-2 \\
\mathrm{a}_{24}=2 \times 2-4=4-4=0 \\
\mathrm{a}_{34}=2 \times 3-4=6-4=2
\end{array}
\)
Therefore, required matrix is \( \mathrm{A}=\left[\begin{array}{cccc}1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2\end{array}\right] \)

\(
\left[\begin{array}{ll}
4 & 3 \\
x & 5
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 5
\end{array}\right]
\)
Since, the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding element, we have:
\( x=1, y=4 \) and \( z=3 \)

Since, the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding element, we have:
\( \left[\begin{array}{cc}x+y & 2 \\ 5+z & x y\end{array}\right]= {\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]}\)
\(x+y=6, x y=8,5+z=5\)
Now, \( 5+z=5 \)
\(
\Rightarrow z=0\)
\(x+y=6 \Rightarrow x=6-y \text { Also, } x y=8 \Rightarrow(6-y) y=8 \Rightarrow 6 y-y^{2}=8\)
\(\Rightarrow y^{2}-6 y+8=0\)
\(\Rightarrow y^{2}-4 y-2 y+8=0\)
\(\Rightarrow y(y-4)-2(y-4)=0 \Rightarrow(y-2)(y-4)=0\)
Hence, \(y=2 \text { or } y=4 \text { when } y =2 x=6-2=4 \text { when } x=4 x=6-4=2
\)

\(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]
\)
Since, the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding element, we have:
\(
x+y+z=9 \ldots(1)\)
\(
x+z=5 \ldots(2)\)
\(y+z=7 \ldots (3)\)
Putting the value of equation 2 in equation 1,
\(y+5=9\)
\(\Rightarrow y=4
\)
Then, putting the value of \( y \) in equation 3, we get,
\(4+z=7\)
\(\Rightarrow z=3
\)
Therefore, \( x+z=5 \)
\(
\Rightarrow x=2
\)
Therefore, \( x=2, y=4 \) and \( z=3 \).

\(
\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]
\)
Since, the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding element, we have:
\(
\mathrm{a}-\mathrm{b}=-1 \ldots(1)\)
\(2 \mathrm{a}-\mathrm{b}=0 \ldots(2)\)
\(2 \mathrm{a}+\mathrm{c}=5 \ldots(3)\)
\(
3 c+d=13 \ldots (4)\)
From equation (2), we get:
\(
\mathrm{b}=2 \mathrm{a}
\)
Then, from eq. (1), we get,
\(
a-2 a=-1\)
\(\Rightarrow a=1\)
\(\Rightarrow b=2
\)
Now, from eq. (3), we get:
\(
2 \times 1+c=5\)
\(\Rightarrow c=3
\)
From (4), we get,
\(
3 \times 3+d=13\)
\(\Rightarrow 9+\mathrm{d}=13\)
\(\Rightarrow \mathrm{d}=4
\)
Therefore, \( a=1, b=2, \mathrm{c}=3 \) and \( \mathrm{d}=4 \).

We know that if a given matrix is said to be square matrix if the number of rows is equal to the number of columns.
Therefore, \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{m} \times \mathrm{n} \) is a square matrix, if \( \mathrm{m}=\mathrm{n} \).

Now, \( \left[\begin{array}{cc}3 x+7 & 5 \\ y+1 & 2-3 x\end{array}\right]=\left[\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right] \)
Since, the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding element, we have:
\(
3 x+7=0\)
\(x=-\frac{7}{3}\)
\(5=y-2\)
\(\Rightarrow y=7\)
\(y+1=8
\)
\(
\Rightarrow y=7
\)
And \( 2-3 x=4 \)
\(
x=-\frac{2}{3}
\)
Thus, on comparing the corresponding elements of the two matrices, we get different values of \( x \), which is not possible.
Therefore, it is not possible to find the values of \(x\) and \(y\) for which the given matrices are equal.

The given matrix of the order \( 3 \times 3 \) has 9 elements with each entry 0 or 1 .
Now, each of the 9 elements can be filled in two possible ways.
Hence, the required number of possible matrices is \( 2^{9}=512 \).