Answer
X = $\begin{bmatrix}
\frac{10}{3} & \frac{7}{3} & -13\\
\frac{23}{3} & 0 & -\frac{14}{3}\\
\end{bmatrix}$
Work Step by Step
5A = 6B - 3X
3X = 6B - 5A
X = $\frac{1}{3}$(6B - 5A)
First multiply matrix A by the scalar multiple 5:
$\begin{bmatrix}
-2(5) & 1(5) & 3(5)\\
-1(5) & 0(5) & 4(5)\\
\end{bmatrix}$ = $\begin{bmatrix}
-10 & 5 & 15\\
-5 & 0 & 20\\
\end{bmatrix}$
Then multiply matrix B by the scalar multiple 6:
$\begin{bmatrix}
0(6) & 2(6) & -4(6)\\
3(6) & 0(6) & 1(6)\\
\end{bmatrix}$ = $\begin{bmatrix}
0 & 12 & -24\\
18 & 0 & 6\\
\end{bmatrix}$
Then perform 6B - 5A to get matrix Y:
Y = $\begin{bmatrix}
0+10 & 12-5 & -24-15\\
18+5 & 0-0 & 6-20\\
\end{bmatrix}$ = $\begin{bmatrix}
10 & 7 & -39\\
23 & 0 & -14\\
\end{bmatrix}$
Then multiply Y by the scalar $\frac{1}{3}$ to get X:
X = $\begin{bmatrix}
10(\frac{1}{3}) & 7(\frac{1}{3}) & -39(\frac{1}{3})\\
23(\frac{1}{3}) & 0(\frac{1}{3}) & -14(\frac{1}{3})\\
\end{bmatrix}$ = $\begin{bmatrix}
\frac{10}{3} & \frac{7}{3} & -13\\
\frac{23}{3} & 0 & -\frac{14}{3}\\
\end{bmatrix}$
