Solve for $X$ in the equation given.
$$16 A + 12 B = - 4 X , A = \left[ \begin{array} { c c c }
5 & 6 & 3 \
- 2 & 5 & - 4
\end{array} \right] \text { and } B = \left[ \begin{array} { c c c }
- 5 & 8 & - 1 \
- 3 & - 9 & - 5
\end{array} \right]$$
A) $\left[ \begin{array} { c c c } 20 & 192 & 36 \ - 68 & - 28 & - 124 \end{array} \right]$
B) $\left[ \begin{array} { c c c } 0 & 14 & 2 \ - 5 & - 4 & - 9 \end{array} \right]$
C) $\left[ \begin{array} { c c c } - 5 & - 48 & - 9 \ 17 & 7 & 31 \end{array} \right]$
D) $\left[ \begin{array} { c c c } - 35 & 0 & - 15 \ - 1 & - 47 & 1 \end{array} \right]$
E) not possible

Question

Solve for $X$ in the equation given.
$$16 A + 12 B = - 4 X , A = \left[ \begin{array} { c c c } 
5 & 6 & 3 \
- 2 & 5 & - 4
\end{array} \right] \text { and } B = \left[ \begin{array} { c c c } 
- 5 & 8 & - 1 \
- 3 & - 9 & - 5
\end{array} \right]$$
A) $\left[ \begin{array} { c c c } 20 & 192 & 36 \ - 68 & - 28 & - 124 \end{array} \right]$
B) $\left[ \begin{array} { c c c } 0 & 14 & 2 \ - 5 & - 4 & - 9 \end{array} \right]$
C) $\left[ \begin{array} { c c c } - 5 & - 48 & - 9 \ 17 & 7 & 31 \end{array} \right]$
D) $\left[ \begin{array} { c c c } - 35 & 0 & - 15 \ - 1 & - 47 & 1 \end{array} \right]$
E) not possible

Quizplus · Accepted Answer

First, we need to multiply matrices $A$ and $B$ by their respective scalar values, $16$ and $12$, and then add the resulting matrices. For $A$, multiplying by $16$ gives:$$16A = \left[ \begin{array} { c c c } 80 & 96 & 48 \ -32 & 80 & -64 \end{array} \right]$$For $B$, multiplying by $12$ gives:$$12B = \left[ \begin{array} { c c c } -60 & 96 & -12 \ -36 & -108 & -60 \end{array} \right]$$Adding $16A$ and $12B$ gives:$$\left[ \begin{array} { c c c } 20 & 192 & 36 \ -68 & -28 & -124 \end{array} \right]$$Since the equation is $16A + 12B = -4X$, we divide the resulting matrix by $-4$ to solve for $X$:$$X = \left[ \begin{array} { c c c } -5 & -48 & -9 \ 17 & 7 & 31 \end{array} \right]$$Therefore, the correct answer is $C$.

Solve For XXX In the Equation Given A) [2019236−68−28−124]\left[ \begin{array} { c c c } 20 & 192 & 36 \\ - 68 & - 28 & - 124 \end{array} \right][20−68​192−28​36−124​]

Solve For $X$ In the Equation Given A) $\left[ \begin{array} { c c c } 20 & 192 & 36 \\ - 68 & - 28 & - 124 \end{array} \right]$