If $A = \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -2 & 4 \\ 3 & 1 \end{bmatrix}$, find a matrix $X$ such that $3 A + 4 X = 5 B$
Given: $\;$ $A = \begin{bmatrix}
2 & 0 \\
-3 & 1
\end{bmatrix}$, $\;$ $B = \begin{bmatrix}
-2 & 4 \\
3 & 1
\end{bmatrix}$
The matrix equation is $\;$ $3 A + 4 X = 5 B$
i.e. $\;$ $4 X = 5 B - 3 A$
$\implies$ $X = \dfrac{5}{4} B - \dfrac{3}{4}A$ $\;\;\; \cdots \; (1)$
Now,
$\begin{aligned}
\dfrac{5}{4} B & = \dfrac{5}{4} \begin{bmatrix}
-2 & 4 \\
3 & 1
\end{bmatrix} \\\\
& = \begin{bmatrix}
-2 \times \dfrac{5}{4} & 4 \times \dfrac{5}{4} \\
3 \times \dfrac{5}{4} & 1 \times \dfrac{5}{4}
\end{bmatrix} \\\\
& = \begin{bmatrix}
-10/4 & 20 / 4 \\
15 / 4 & 5 / 4
\end{bmatrix} \;\;\; \cdots \; (2a)
\end{aligned}$
$\begin{aligned}
\dfrac{3}{4} A & = \dfrac{3}{4} \begin{bmatrix}
2 & 0 \\
-3 & 1
\end{bmatrix} \\\\
& = \begin{bmatrix}
2 \times \dfrac{3}{4} & 0 \times \dfrac{3}{4} \\
-3 \times \dfrac{3}{4} & 1 \times \dfrac{3}{4}
\end{bmatrix} \\\\
& = \begin{bmatrix}
6/4 & 0 \\
-9 / 4 & 3 / 4
\end{bmatrix} \;\;\; \cdots \; (2b)
\end{aligned}$
$\therefore \;$ In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes,
$\begin{aligned}
X & = \begin{bmatrix}
-10/4 & 20/4 \\
15/4 & 5/4
\end{bmatrix} - \begin{bmatrix}
6/4 & 0 \\
-9/4 & 3/4
\end{bmatrix} \\\\
& = \begin{bmatrix}
-16/4 & 20/4 \\
24/4 & 2/4
\end{bmatrix} \\\\
& = \begin{bmatrix}
-4 & 5 \\
6 & 1/2
\end{bmatrix}
\end{aligned}$