If possible, find $4 A + 2 B$.
$$A = \left[ \begin{array} { c c c }
0 & - 4 & 4 \
1 & 0 & - 3
\end{array} \right] , B = \left[ \begin{array} { c c c }
- 3 & 2 & 5 \
- 5 & 8 & - 2
\end{array} \right]$$
A) $\left[ \begin{array} { c c c } 6 & - 20 & 6 \ 14 & - 16 & - 8 \end{array} \right]$
B) $\left[ \begin{array} { c c c } - 3 & - 2 & 9 \ - 4 & 8 & - 5 \end{array} \right]$
C) $\left[ \begin{array} { l l l } 1 & 0 & 0 \ 0 & 1 & 0 \end{array} \right]$
D) $\left[ \begin{array} { c c c } - 6 & - 12 & 26 \ - 6 & 16 & - 16 \end{array} \right]$
E) not possible

Question

If possible, find $4 A + 2 B$.
$$A = \left[ \begin{array} { c c c } 
0 & - 4 & 4 \
1 & 0 & - 3
\end{array} \right] , B = \left[ \begin{array} { c c c } 
- 3 & 2 & 5 \
- 5 & 8 & - 2
\end{array} \right]$$
A) $\left[ \begin{array} { c c c } 6 & - 20 & 6 \ 14 & - 16 & - 8 \end{array} \right]$
B) $\left[ \begin{array} { c c c } - 3 & - 2 & 9 \ - 4 & 8 & - 5 \end{array} \right]$
C) $\left[ \begin{array} { l l l } 1 & 0 & 0 \ 0 & 1 & 0 \end{array} \right]$
D) $\left[ \begin{array} { c c c } - 6 & - 12 & 26 \ - 6 & 16 & - 16 \end{array} \right]$
E) not possible

Quizplus · Accepted Answer

We can simply use the formula for matrix addition to find $4A+2B$: $$4A+2B=4\left[ \begin{array} { c c c } 
0 & - 4 & 4 \
1 & 0 & - 3
\end{array} ight]+2\left[ \begin{array} { c c c } 
- 3 & 2 & 5 \
- 5 & 8 & - 2
\end{array} ight].$$ Distributing the scalar values, we get: $$4A+2B=\left[ \begin{array} { c c c } 
0 & -16 & 16 \
4 & 0 & -12
\end{array} ight]+\left[ \begin{array} { c c c } 
- 6 & 4 & 10 \
- 10 & 16 & -4
\end{array} ight].$$ Adding the two matrices, we get: $$4A+2B=\boxed{	extbf{(D) } \left[ \begin{array} { c c c } 
-6 & -12 & 26 \
-6 & 16 & -16
\end{array} ight]}.$$

If Possible, Find 4A+2B4 A + 2 B4A+2B A) [6−20614−16−8]\left[ \begin{array} { c c c } 6 & - 20 & 6 \\ 14 & - 16 & - 8 \end{array} \right][614​−20−16​6−8​]

If Possible, Find $4 A + 2 B$ A) $\left[ \begin{array} { c c c } 6 & - 20 & 6 \\ 14 & - 16 & - 8 \end{array} \right]$