Answer
$\begin{bmatrix} -20&10&-8\\-15&15&9\end{bmatrix}$
Work Step by Step
Let us consider two matrices $A$ and $C$ and their multiplication can be possible when the number of columns of matrix $A$ is the same as the number of rows of the matrix $C$.
We are given that $A=\begin{bmatrix} 4 & -2\\3&1\end{bmatrix} $ and $C=\begin{bmatrix} -5 &4&1 \\0&3&6\end{bmatrix} $
We can see that the size of matrix $A$ is $2 \times 2$ and that of matrix $B$ is $2 \times 3$. This shows that the number of columns of matrix-$A$ is the same as the number of rows of the matrix-$C$. So, their product $AC$ can be computed.
Recall that if $A=\begin{bmatrix}a_{11} &a_{12}\\a_{21} &a_{22}\end{bmatrix}$ and $C=\begin{bmatrix}c_{11} &c_{12}&c_{13}\\c_{21} &c_{22}&c_{23}\end{bmatrix}$
then \begin{align*} AC=\begin{bmatrix}a_{11}c_{11} +a_{12}c_{21} & a_{11}c_{12}+a_{12}c_{22}& a_{11}c_{13}+a_{12}c_{23}\\a_{21}c_{11} +a_{22}c_{21} & a_{21}c_{12}+a_{22}c_{22}& a_{21}c_{13}+a_{22}c_{23}\\ \end{bmatrix} \end{align*}
Thus, using the formula above gives: $AC=\begin{bmatrix} 4 & -2\\3&1\end{bmatrix} \begin{bmatrix} -5 &4&1 \\0&3&6\end{bmatrix}\\=\begin{bmatrix} 4 & -2\\3&1\end{bmatrix} \begin{bmatrix} -20+0&16-6&4-12\\-15+0&12+3&3+6\end{bmatrix} \\=\begin{bmatrix} -20&10&-8\\-15&15&9\end{bmatrix}$
Therefore, $AC$ is equal to $\begin{bmatrix} -20&10&-8\\-15&15&9\end{bmatrix}$
