Answer
$\begin{bmatrix} 4 & 8 &-10 \end{bmatrix}$
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} 0 & -3 & 4\end{bmatrix} $ and $B=\begin{bmatrix} -2&6&3 \\0&4&2\\-1&1&4\end{bmatrix} $.
Recall that if $A=\begin{bmatrix}a_{11} &a_{12}&a_{13}\end{bmatrix}$ 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
\begin{align*}
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}\\
\end{bmatrix}
\end{align*}
Thus, using the formula above gives:
$AB=\begin{bmatrix} -2 & 4 &1\end{bmatrix} \begin{bmatrix} 3&-2&4 \\2&1&4\\0&-1&4\end{bmatrix} \\=\begin{bmatrix} 0+0+4 &0+12-4 &0+6-6 \end{bmatrix}\\=\begin{bmatrix} 4 & 8 &-10 \end{bmatrix}$
Therefore, $AB$ is equal to $\begin{bmatrix} 4 & 8 &-10 \end{bmatrix}$
