Question

Find the matrix $A$ satisfying the matrix equation $\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]A\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Answer 1

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

so, $\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]A\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}2a+b& 2c+d\\ 3a+2b& 3c+2d\end{array}\right]\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}-6a-3b+10c+5d& 4a+3b-6c-3d\\ -9a-6b+15c+10d& 6a+4b-9c-6d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

so, $-6a-3b+10c+5d=1$

$4a+3b-6c-3d=0$

$-9a-6b+15c+10d=0$

$6a+4b-9c-6d=1$

on squaring, we get

$\Rightarrow a=\frac{1}{3},b=1,c=\frac{1}{3},d=\frac{8}{15}$

so, $A=\left[\begin{array}{cc}\frac{1}{3}& \frac{1}{3}\\ 0& \frac{8}{15}\end{array}\right]=\frac{1}{15}\left[\begin{array}{cc}5& 5\\ 15& 8\end{array}\right].$

Hence, the answer is $\frac{1}{15}\left[\begin{array}{cc}5& 5\\ 15& 8\end{array}\right].$