Answer
$\begin{bmatrix} -15 & -16 & 3 \\-1 & 0 & 9 \\7 & 6 & 12 \end{bmatrix} $
Work Step by Step
Let us consider two matrices $A$ and $B$ 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-$B$.
We are given that $A=\begin{bmatrix} -2 & -3 & -4\\ 2&-1&0\\4 &-2&3 \end{bmatrix} $ and $B=\begin{bmatrix} 0 & 1 & 4 \\ 1 & 2 & -1\\ 3& 2& -2\end{bmatrix} $
We can see that the size of matrix $A$ is $3 \times 3$ and that of matrix $B$ is $3 \times 3$. This shows that the number of columns of matrix-$A$ is the same as the number of rows of the matrix-$B$. So, their product $AB$ can be computed.
Recall that if $A=\begin{bmatrix}a_{11} &a_{12}&a_{13}\\a_{21} &a_{22}&a_{23}\\a_{31} &a_{32}&a_{33}\end{bmatrix}$ and $B=\begin{bmatrix}b_{11} &b_{12}&b_{13}\\b_{21} &b_{22}&b_{23}\\b_{31} &b_{32}&b_{33}\end{bmatrix}$, then
$$AB=\begin{bmatrix}a_{11}b_{11} +a_{12}b_{21}+a_{13}b_{31}&a_{11}b_{12} +a_{12}b_{22}+a_{13}b_{32}&a_{11}b_{13} +a_{12}b_{23}+a_{13}b_{33}\\
a_{21}b_{11} +a_{22}b_{21}+a_{23}b_{31}&a_{21}b_{12} +a_{22}b_{22}+a_{23}b_{32}&a_{21}b_{13} +a_{22}b_{23}+a_{23}b_{33}\\a_{31}b_{11} +a_{32}b_{21}+a_{33}b_{31}&a_{31}b_{12} +a_{32}b_{22}+a_{33}b_{32}&a_{31}b_{13} +a_{32}b_{23}+a_{33}b_{33}
\end{bmatrix}$$
Thus, using the formula above gives:
$AB=\begin{bmatrix} -2 & -3 & -4\\ 2&-1&0\\4 &-2&3 \end{bmatrix} \begin{bmatrix} 0 & 1 & 4 \\ 1 & 2 & -1\\ 3& 2& -2\end{bmatrix}$
$AB=\begin{bmatrix} 0-3-12 & -2-6-8& -8+3+8 \\0-1+0 & 2-2+0 &8+1+0\\0-2+9&4-4+6&16+2-6 \end{bmatrix} $
$AB=\begin{bmatrix} -15 & -16 & 3 \\-1 & 0 & 9 \\7 & 6 & 12 \end{bmatrix} $
Therefore, $AB$ is equal to $\begin{bmatrix} -15 & -16 & 3 \\-1 & 0 & 9 \\7 & 6 & 12 \end{bmatrix} $.
