Answer
$2A=\begin{bmatrix}
6 & -2
\\
4 & 0
\end{bmatrix}$
Work Step by Step
Multiplying each element of $
A=
\begin{bmatrix}
3 & -1
\\
2 & 0
\end{bmatrix}
$ by $2,$ then
\begin{align*}
2A&=
\begin{bmatrix}
3(2) & -1(2)
\\
2(2) & 0(2)
\end{bmatrix}
\\\\&=
\begin{bmatrix}
6 & -2
\\
4 & 0
\end{bmatrix}
.\end{align*}
Hence, the scalar product $2A$ is equal to $
\begin{bmatrix}
6 & -2
\\
4 & 0
\end{bmatrix}
.$