Answer
X = $\begin{bmatrix}
-\frac{8}{3} & \frac{8}{3} & \frac{4}{3}\\
\frac{2}{3} & 0 & 6\\
\end{bmatrix}$
Work Step by Step
3X - 4A = 2B
3X = 2B + 4A
X = $\frac{1}{3}$(2B + 4A)
First multiply matrix A by the scalar multiple 4:
$\begin{bmatrix}
-2(4) & 1(4) & 3(4)\\
-1(4) & 0(4) & 4(4)\\
\end{bmatrix}$ = $\begin{bmatrix}
-8 & 4 & 12\\
-4 & 0 & 16\\
\end{bmatrix}$
Then multiply matrix B by the scalar multiple 2:
$\begin{bmatrix}
0(2) & 2(2) & -4(2)\\
3(2) & 0(2) & 1(2)\\
\end{bmatrix}$ = $\begin{bmatrix}
0 & 4 & -8\\
6 & 0 & 2\\
\end{bmatrix}$
Then perform 2B + 4A to get matrix Y:
Y = $\begin{bmatrix}
0-8 & 4+4 & -8+12\\
6-4 & 0+0 & 2+16\\
\end{bmatrix}$ = $\begin{bmatrix}
-8 & 8 & 4\\
2 & 0 & 18\\
\end{bmatrix}$
Then multiply Y by the scalar $\frac{1}{3}$ to get X:
X = $\begin{bmatrix}
-8(\frac{1}{3}) & 8(\frac{1}{3}) & 4(\frac{1}{3})\\
2(\frac{1}{3}) & 0(\frac{1}{3}) & 18(\frac{1}{3})\\
\end{bmatrix}$ = $\begin{bmatrix}
-\frac{8}{3} & \frac{8}{3} & \frac{4}{3}\\
\frac{2}{3} & 0 & 6\\
\end{bmatrix}$
