Solve $\left[ \begin{array} { l l }
2 & x \
6 & 0
\end{array} \right] \left[ \begin{array} { l l }
y & 3 \
- 2 & 2
\end{array} \right] = \left[ \begin{array} { l l }
1 & 3 \
- 6 & 18
\end{array} \right]$ for $x$
And $y$
A) $x = - 2$, $y = - 1$

B) $x = - \frac { 3 } { 2 }$, $y = - 2$

C) $x = - \frac { 3 } { 2 }$, $y = - 1$

D) $x = \frac { J } { 2 }$, $y = - \frac { 3 } { 2 }$

E) none

Question

Solve $\left[ \begin{array} { l l } 
2 & x \
6 & 0
\end{array} \right] \left[ \begin{array} { l l } 
y & 3 \
- 2 & 2
\end{array} \right] = \left[ \begin{array} { l l } 
1 & 3 \
- 6 & 18
\end{array} \right]$ for $x$
And $y$
A) $x = - 2$, $y = - 1$

B) $x = - \frac { 3 } { 2 }$, $y = - 2$

C) $x = - \frac { 3 } { 2 }$, $y = - 1$

D) $x = \frac { J } { 2 }$, $y = - \frac { 3 } { 2 }$

E) none

Quizplus · Accepted Answer

By multiplying the matrices and equating them to the given matrix, we solve for $x$ and $y$ to find $x = -\frac{3}{2}$ and $y = -1$.

Solve For xxx And yyy A) x=−2x = - 2x=−2 y=−1y = - 1y=−1

Solve For $x$ And $y$ A) $x = - 2$ $y = - 1$