Answer
$\begin{bmatrix} 2 & 7 &-4 \end{bmatrix}$
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 & 4 &1\end{bmatrix} $ and $B=\begin{bmatrix} 3&-2&4 \\2&1&4\\0&-1&4\end{bmatrix} $
We can see that the size of matrix $A$ is $1 \times 3$ 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 $A=\begin{bmatrix}a_{11} &a_{12}&a_{13}\end{bmatrix}$ 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:
$AB=\begin{bmatrix} -2 & 4 &1\end{bmatrix} \begin{bmatrix} 3&-2&4 \\2&1&4\\0&-1&4\end{bmatrix} \\=\begin{bmatrix} -6+8+0 &4+4-1 &-8+0+4 \end{bmatrix}\\=\begin{bmatrix} 2 & 7 &-4 \end{bmatrix}$
Therefore, $AB$ is equal to $\begin{bmatrix} 2 & 7 &-4 \end{bmatrix}$
