Answer
\[\left[ {\begin{array}{*{20}{c}}
{14}&{ - 2} \\
{ - 12}&6
\end{array}} \right]\]
Work Step by Step
\[\begin{gathered}
{\text{Let }}A = \left[ {\begin{array}{*{20}{c}}
{ - 2}&4 \\
0&3
\end{array}} \right]{\text{ and }}B = \left[ {\begin{array}{*{20}{c}}
{ - 6}&2 \\
4&0
\end{array}} \right] \hfill \\
{\text{Find }}2A - 3B \hfill \\
2A - 3B = 2\left[ {\begin{array}{*{20}{c}}
{ - 2}&4 \\
0&3
\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}
{ - 6}&2 \\
4&0
\end{array}} \right] \hfill \\
{\text{Use the scalar multiplication}} \hfill \\
2A - 3B = \left[ {\begin{array}{*{20}{c}}
{ - 4}&8 \\
0&6
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{18}&{ - 6} \\
{ - 12}&0
\end{array}} \right] \hfill \\
{\text{Add the corresponding elements}} \hfill \\
2A - 3B = \left[ {\begin{array}{*{20}{c}}
{ - 4 + 18}&{8 - 6} \\
{0 - 12}&{6 - 0}
\end{array}} \right] \hfill \\
{\text{Simplify}} \hfill \\
2A - 3B = \left[ {\begin{array}{*{20}{c}}
{14}&{ - 2} \\
{ - 12}&6
\end{array}} \right] \hfill \\
\end{gathered} \]
