Answer
$X
=
\begin{bmatrix}
-10 & 19
\\
-20 & 7
\end{bmatrix}$
Work Step by Step
Using the properties of matrices, the given matrix equation, $
2X+3\begin{bmatrix}
4 & -6
\\
8 & -3
\end{bmatrix}
=
\begin{bmatrix}
-8 & 20
\\
-16 & 5
\end{bmatrix}
,$ is equivalent to
\begin{align*}\require{cancel}
2X+\begin{bmatrix}
4(3) & -6(3)
\\
8(3) & -3(3)
\end{bmatrix}
&=
\begin{bmatrix}
-8 & 20
\\
-16 & 5
\end{bmatrix}
\\\\
2X+\begin{bmatrix}
12 & -18
\\
24 & -9
\end{bmatrix}
&=
\begin{bmatrix}
-8 & 20
\\
-16 & 5
\end{bmatrix}
\\\\
2X
&=
\begin{bmatrix}
-8 & 20
\\
-16 & 5
\end{bmatrix}
-
\begin{bmatrix}
12 & -18
\\
24 & -9
\end{bmatrix}
\\\\
2X
&=
\begin{bmatrix}
-8-12 & 20-(-18)
\\
-16-24 & 5-(-9)
\end{bmatrix}
\\\\
2X
&=
\begin{bmatrix}
-8-12 & 20+18
\\
-16-24 & 5+9
\end{bmatrix}
\\\\
2X
&=
\begin{bmatrix}
-20 & 38
\\
-40 & 14
\end{bmatrix}
\\\\
\left(\dfrac{1}{2}\right)(2X)
&=
\dfrac{1}{2}\begin{bmatrix}
-20 & 38
\\
-40 & 14
\end{bmatrix}
\\\\
X
&=
\begin{bmatrix}
-20\left(\dfrac{1}{2}\right) & 38\left(\dfrac{1}{2}\right)
\\
-40\left(\dfrac{1}{2}\right) & 14\left(\dfrac{1}{2}\right)
\end{bmatrix}
\\\\
X
&=
\begin{bmatrix}
-10 & 19
\\
-20 & 7
\end{bmatrix}
.\end{align*}
Hence, the solution is $
X
=
\begin{bmatrix}
-10 & 19
\\
-20 & 7
\end{bmatrix}
.$
