We have $2A+3X=5 B$

or         $2A+3X-2A=5B-2A$

or         $2 A-2A+3X=5B-2A$          $($ Matrix addition is commutative $)$

or         $O+3 X=5B-2 A$        $(-2A$ is the additive inverse of $2A)$

or         $3 \mathrm{X}=5 \mathrm{B}-2 \mathrm{A}$                  ( $O$ is the additive identity)

or         $X=\frac{1}{3}(5 B-2 A)$

or         ${\text{X}} = $ $\frac{1}{3}\left( {5\left[ {\begin{array}{*{20}{c}}
  2&{ – 2} \\ 
  4&2 \\ 
  { – 5}&1 
\end{array}} \right] – 2\left[ {\begin{array}{*{20}{l}}
  8&0 \\ 
  4&{ – 2} \\ 
  3&6 
\end{array}} \right]} \right)$ $ = \frac{1}{3}\left( {\left[ {\begin{array}{*{20}{c}}
  {10}&{ – 10} \\ 
  {20}&{10} \\ 
  { – 25}&5 
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  { – 16}&0 \\ 
  { – 8}&4 \\ 
  { – 6}&{ – 12} 
\end{array}} \right]} \right)$

$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
  {10 – 16}&{ – 10 + 0} \\ 
  {20 – 8}&{10 + 4} \\ 
  { – 25 – 6}&{5 – 12} 
\end{array}} \right]$ 

$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
  { – 6}&{ – 10} \\ 
  {12}&{14} \\ 
  { – 31}&{ – 7} 
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
  { – 2}&{\frac{{ – 10}}{3}} \\ 
  4&{\frac{{14}}{3}} \\ 
  {\frac{{ – 31}}{3}}&{\frac{{ – 7}}{3}} 
\end{array}} \right]$