We have $({\text{X}} + {\text{Y}}) + ({\text{X}} – {\text{Y}}) = $ $\left[ {\begin{array}{*{20}{l}}
  5&2 \\ 
  0&9 
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  3&6 \\ 
  0&{ – 1} 
\end{array}} \right]$

or            $(X + X) + (Y – Y) = $ $\left[ {\begin{array}{*{20}{l}}
  8&8 \\ 
  0&8 
\end{array}} \right] \Rightarrow 2X = $ $\left[ {\begin{array}{*{20}{l}}
  8&8 \\ 
  0&8 
\end{array}} \right]$

or          $X = \frac{1}{2}\left[ {\begin{array}{*{20}{l}}
  8&8 \\ 
  0&8 
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  4&4 \\ 
  0&4 
\end{array}} \right]$

Also            $(X + Y) – (X – Y) = $ $\left[ {\begin{array}{*{20}{c}}
  5&2 \\ 
  0&9 
\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}
  3&6 \\ 
  0&{ – 1} 
\end{array}} \right]$

or          $(X – X) + (Y + Y) = $ $\left[ {\begin{array}{*{20}{c}}
  {5 – 3}&{2 – 6} \\ 
  0&{9 + 1} 
\end{array}} \right] \Rightarrow $ $2{\text{Y}} = \left[ {\begin{array}{*{20}{c}}
  2&{ – 4} \\ 
  0&{10} 
\end{array}} \right]$

or             $Y = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
  2&{ – 4} \\ 
  0&{10} 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&{ – 2} \\ 
  0&5 
\end{array}} \right]$