#### Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question 48 sub question (ii)

$A=\left[\begin{array}{cc}1 & -2 \\ 2 & 0\end{array}\right]$

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

$A\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]_{2 \times 3}=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]_{2 \times 3}$

The matrix given on the RHS of the equation is a ${2 \times 3}$ matrix and the matrix given on the LHS of the equation is ${2 \times 3}$ So, matrix A has to be ${2 \times 2}$ matrix.

Now, let

$A=\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]$

we have,

$\\\\ \Rightarrow\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right] \\\\ \Rightarrow \left[\begin{array}{ccc}a+4 c & 2 a+5 c & 3 a+6 c \\ b+4 d & 2 b+5 d & 3 b+6 d\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$

Equating the corresponding elements of the two matrices, we have

$a+4 c=-7,2 a+5 c=-8,3 a+6 c=-9, b+4 d=2,2 b+5 d=4,3 b+6 d=6$

Now,

$\\\\ \Rightarrow a+4 c=-7 \quad \ldots (i) \\\\ \Rightarrow a=-7-4 c\\\\ 2 a+5 c=-8 \quad \ldots (ii)$

Put the value of a in equation ii

$\begin{array}{l} \Rightarrow 2(-7-4 c)+5 c=-8 \\\\ \Rightarrow-14-8 c+5 c=-8 \\\\ \Rightarrow-3 c=6 \\\\ \Rightarrow c=-2 \end{array}$

Now, put value of a in a=-7-4c

$\\\\ \Rightarrow a=-7-4(-2)=-7+8=1 \\\\ \Rightarrow a=1$

Using (iii) in (iv)

$\\\\ \Rightarrow 4-8 d+5 d=4\\\\ \Rightarrow-3 d=0\\\\ \Rightarrow d=0$

Now use the value of d=0 in b+4d=2

$\therefore b=2-4(0)=2$

Thus, a=1, b=2, c=-2, d=0

Hence, the required matrix A is

$\left[\begin{array}{cc}1 & -2 \\ 2 & 0\end{array}\right]$