$X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$       ………… $(1)$

$X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$      ………… $(2)$

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

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

Multiplying equation $(3)$ with $(2)$, we get

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

$\Rightarrow 4 X+6 Y=\left[\begin{array}{ll}4 & 6 \\ 8 & 0\end{array}\right]$     ………… $(5)$

Multiplying equation $(4)$ with $( 3 )$, we get

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

$\Rightarrow 9 X+6 Y=\left[\begin{array}{cc}6 & -6 \\ -3 & 15\end{array}\right]$      ………… $(6)$

From $( 5 )$ and $( 6 )$, we have

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

$ \Rightarrow  – 5X = \left[ {\begin{array}{*{20}{c}}
  {4 – 6}&{6 – ( – 6)} \\ 
  {8 – ( – 3)}&{0 – 15} 
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
  { – 2}&{12} \\ 
  {11}&{ – 15} 
\end{array}} \right]$

$\therefore X =  – \frac{1}{5}\left[ {\begin{array}{*{20}{c}}
  { – 2}&{12} \\ 
  {11}&{ – 15} 
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
  {\frac{2}{5}}&{ – \frac{{12}}{5}} \\ 
  { – \frac{{11}}{5}}&3 
\end{array}} \right]$

$2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$

Now,

$\Rightarrow 2\left[\begin{array}{cc}\frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{array}\right]+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}\frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{array}\right]+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$

$\Rightarrow 3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]-\left[\begin{array}{cc}\frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{array}\right]$

$ \Rightarrow 3Y = $ $\left[ {\begin{array}{*{20}{c}}
  {2 – \frac{4}{5}}&{3 + \frac{{24}}{5}} \\ 
  {4 + \frac{{22}}{5}}&{0 – 6} 
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
  {\frac{6}{5}}&{\frac{{39}}{5}} \\ 
  {\frac{{42}}{5}}&{ – 6} 
\end{array}} \right]$

$\therefore Y=\frac{1}{3}\left[\begin{array}{cc}\frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6\end{array}\right]=\left[\begin{array}{cc}\frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2\end{array}\right]$