Given $A = \left[ \begin{array} { c c }
- 9 & 8 \
- 6 & - 5 \
10 & - 16
\end{array} \right]$ and $B = \left[ \begin{array} { l l }
1 & 0 \
2 & 3 \
0 & 4
\end{array} \right]$

, solve $2 X - A = 3 B$, for $X$

A) $X = \left[ \begin{array} { l l l }
- 3 & 4 & 0 \
5 & 2 & 2
\end{array} \right]$

B) $X = \left[ \begin{array} { l l }
5 & 4 \
0 & 1 \
2 & 2
\end{array} \right]$

C) $X = \left[ \begin{array} { l l }
5 & \frac { 1 } { 2 } \
0 & 2 \
5 & 2
\end{array} \right]$

D) $X = \left[ \begin{array} { c c }
- 3 & 4 \
0 & 2 \
5 & - 2
\end{array} \right]$

E) none

Question

Given $A = \left[ \begin{array} { c c } 
- 9 & 8 \
- 6 & - 5 \
10 & - 16
\end{array} \right]$ and $B = \left[ \begin{array} { l l } 
1 & 0 \
2 & 3 \
0 & 4
\end{array} \right]$

, solve $2 X - A = 3 B$, for $X$

A) $X = \left[ \begin{array} { l l l } 
- 3 & 4 & 0 \
5 & 2 & 2
\end{array} \right]$

B) $X = \left[ \begin{array} { l l } 
5 & 4 \
0 & 1 \
2 & 2
\end{array} \right]$

C) $X = \left[ \begin{array} { l l } 
5 & \frac { 1 } { 2 } \
0 & 2 \
5 & 2
\end{array} \right]$

D) $X = \left[ \begin{array} { c c } 
- 3 & 4 \
0 & 2 \
5 & - 2
\end{array} \right]$

E) none

Quizplus · Accepted Answer

We can first rearrange the given equation: $$2X = 3B + A$$ Then, we can solve for X by multiplying by $\frac{1}{2}$: $$X = \frac{1}{2}(3B + A)$$ We can then substitute in the values of A and B: $$X = \frac{1}{2}\left(3\left[\begin{array}{cc}1&0\2&3\0&4\end{array}ight] + \left[\begin{array}{cc}-9&8\-6&-5\10&-16\end{array}ight]ight) = \frac{1}{2}\left(\left[\begin{array}{cc}0&9\9&5\3&12\end{array}ight]ight)$$ Simplifying further, we get: $$X = \left[\begin{array}{cc}-3&4\0&2\5&-2\end{array}ight]$$ Therefore, the answer is D.

Given A=[−98−6−510−16]A = \left[ \begin{array} { c c } - 9 & 8 \\- 6 & - 5 \\10 & - 16\end{array} \right]A=​−9−610​8−5−16​​

Given $A = \left[ \begin{array} { c c } - 9 & 8 \\- 6 & - 5 \\10 & - 16\end{array} \right]$