Question

Find matrix $A$ such that$\left(\begin{array}{cc}2& -1\\ 1& 0\\ -3& 4\end{array}\right)A=\left(\begin{array}{cc}-1& -8\\ 1& -2\\ 9& 22\end{array}\right)$

Answer 1

Here, multiplier has $3\times 2$ order matrix and product matrix has $3\times 2$ order matrix. So, A will be $2\times 2$ order matrix.

Let $X$ = $\left(\begin{array}{cc}x& y\\ z& t\end{array}\right)$

We have,

$\left(\begin{array}{cc}2& -1\\ 1& 0\\ -3& 4\end{array}\right)\left(\begin{array}{cc}x& y\\ z& t\end{array}\right)=\left(\begin{array}{cc}-1& -8\\ 1& -2\\ 9& 22\end{array}\right)$

$\Rightarrow \left(\begin{array}{cc}2x-z& 2y-t\\ x& y\\ -3x+4z& -3y+4t\end{array}\right)=\left(\begin{array}{cc}-1& -8\\ 1& -2\\ 9& 22\end{array}\right)$

Equating correncponding elements, we have

$\begin{array}{ccc}& & \\ 2x-z=-1& & 2y-t=-8\\ x=1& \text{and}& y=-2\\ -3x+4z=9& & -3y+4t=22\end{array}$

On solving these equations, we have

$x=1,y=-2z=3,t=4$

So, $A=\left(\begin{array}{cc}1& -2\\ 3& 4\end{array}\right)$