# Given A=((-1, 2), (3, 4)) and B=((-4, 3), (5, -2)), how do you find 3A?

$3 A = \left(\begin{matrix}- 3 & 6 \\ 9 & 12\end{matrix}\right)$
If $k$ is a constant and $M = \left(\begin{matrix}{m}_{11} & \ldots & {m}_{c 1} \\ \ldots & \ldots & \ldots \\ {m}_{r 1} & \ldots & {m}_{r c}\end{matrix}\right)$ is a matrix
$\textcolor{w h i t e}{\text{XXX}} k \cdot M = \left(\begin{matrix}k \cdot {m}_{11} & \ldots & k \cdot {m}_{c 1} \\ \ldots & \ldots & \ldots \\ k \cdot {m}_{r 1} & \ldots & k \cdot {m}_{r c}\end{matrix}\right)$