If $A=\left[\begin{array}{rr}2 & -2 \\ 4 & 2 \\ -5 & 1\end{array}\right], B=\left[\begin{array}{rr}8 & 0 \\ 4 & -2 \\ 3 & 6\end{array}\right]$, find matrix $X$ such that $2 A+3 X=5 B$.
Given : $2 A+3 X=5 B$
$\Rightarrow 2\left[\begin{array}{cc}2 & -2 \\ 4 & 2 \\ -5 & 1\end{array}\right]+3 X=5\left[\begin{array}{cc}8 & 0 \\ 4 & -2 \\ 3 & 6\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}4 & -4 \\ 8 & 4 \\ -10 & 2\end{array}\right]+3 X=\left[\begin{array}{cc}40 & 0 \\ 20 & -10 \\ 15 & 30\end{array}\right]$
$\Rightarrow 3 X=\left[\begin{array}{cc}40 & 0 \\ 20 & -10 \\ 15 & 30\end{array}\right]-\left[\begin{array}{cc}4 & -4 \\ 8 & 4 \\ -10 & 2\end{array}\right]$
$\Rightarrow 3 X=\left[\begin{array}{cc}40-4 & 0+4 \\ 20-8 & -10-4 \\ 15+10 & 30-2\end{array}\right]$
$\Rightarrow 3 X=\left[\begin{array}{cc}36 & 4 \\ 12 & -14 \\ 25 & 28\end{array}\right]$
$\Rightarrow X=\frac{1}{3}\left[\begin{array}{cc}36 & 4 \\ 12 & -14 \\ 25 & 28\end{array}\right]$
$\Rightarrow X=\left[\begin{array}{cc}12 & \frac{4}{3} \\ 4 & \frac{-14}{3} \\ \frac{25}{3} & \frac{28}{3}\end{array}\right]$