If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y.

We have,

$3(2 X+3 Y)-2(3 X+2 Y)=3\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]-2\left[\begin{array}{cc}-2 & 2 \\ 1 & -5\end{array}\right]$

$\Rightarrow 6 X+9 Y-6 X-4 Y=\left[\begin{array}{cc}6 & 9 \\ 12 & 0\end{array}\right]+\left[\begin{array}{cc}4 & -4 \\ -2 & 10\end{array}\right]$

$\Rightarrow 5 Y=\left[\begin{array}{cc}6+4 & 9-4 \\ 12-2 & 0+10\end{array}\right]$

$\Rightarrow Y=\frac{1}{5}\left[\begin{array}{cc}10 & 5 \\ 10 & 10\end{array}\right]$

$\Rightarrow Y=\left[\begin{array}{ll}2 & 1 \\ 2 & 2\end{array}\right]$                       ...(1)

Also,

$2(2 X+3 Y)-3(3 X+2 Y)=2\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]-3\left[\begin{array}{cc}-2 & 2 \\ 1 & -5\end{array}\right]$

$\Rightarrow 4 X+6 Y-9 X-6 Y=\left[\begin{array}{ll}4 & 6 \\ 8 & 0\end{array}\right]+\left[\begin{array}{cc}6 & -6 \\ -3 & 15\end{array}\right]$

$\Rightarrow-5 X=\left[\begin{array}{cc}6+4 & 6-6 \\ 8-3 & 0+15\end{array}\right]$

$\Rightarrow X=\frac{1}{-5}\left[\begin{array}{cc}10 & 0 \\ 5 & 15\end{array}\right]$

$\Rightarrow X=\left[\begin{array}{cc}-2 & 0 \\ -1 & -3\end{array}\right]$                    ...(2)

From (1) and (2), we get

$X=\left[\begin{array}{cc}-2 & 0 \\ -1 & -3\end{array}\right]$ and $Y=\left[\begin{array}{ll}2 & 1 \\ 2 & 2\end{array}\right]$