If X and Y are 2 x 2 matrices,

Given, $\quad 2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$ $\ldots(\mathrm{i})$

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

On subractiong equations (i) and (ii), we get

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

Thus, $\quad X-Y=\left[\begin{array}{rr}-4 & -1 \\ -3 & -5\end{array}\right]$ ...(iii)

On adding equations (i) and (ii), we get

$5 X+5 Y=\left[\begin{array}{cc}0 & 5 \\ 5 & -5\end{array}\right]$

$X+Y=\frac{1}{5}\left[\begin{array}{cc}0 & 5 \\ 5 & -5\end{array}\right]=\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]$$\ldots$ (iv)

On adding equations (iii) and (iv), we get

$(X-Y)+(X+Y)=\left[\begin{array}{cc}-4 & 0 \\ -2 & -6\end{array}\right]$

$2 X=\left[\begin{array}{cc}-4 & 0 \\ -2 & -6\end{array}\right]$

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

From equation (iv), we get

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

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

Thus, $Y=\left[\begin{array}{ll}2 & 1 \\ 2 & 2\end{array}\right]$