Answer
$\begin{bmatrix} -20 &10&-8 \\-15 &15&9 \end{bmatrix}$
Work Step by Step
Since, the dimensions of matrix $A$ is $2 \times 2$ and that of matrix $C$ is $2 \times 3$. Thus, the number of columns of matrix-$A$ will be same as the number of rows of the matrix-$C$. and so we can take their product as $AC$.
$AC=\begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -5 & 4& 1 \\ 0 & 3 & 6 \end{bmatrix} \\=\begin{bmatrix} -20 +0 &16-6& 4-12 \\ -15+0 & 12+3 & 3+6 \end{bmatrix}\\=\begin{bmatrix} -20 &10&-8 \\-15 &15&9 \end{bmatrix}$
