Answer
$\begin{bmatrix}12 & -6 &3 \\-4 & 12 & 14\end{bmatrix}$
Work Step by Step
We have $AB=\begin{bmatrix}3 & 0\\-1 & 5\end{bmatrix} \cdot \begin{bmatrix}4 & -2 & 1\\0 & 2 & 3\end{bmatrix} $
or, $=\begin{bmatrix}12+0 & -6+0 & 3+0 \\-4+0 & 2+10 & -1+15\end{bmatrix}$
After simplification, we obtain:
$=\begin{bmatrix}12 & -6 &3 \\-4 & 12 & 14\end{bmatrix}$
