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) $x = \frac { 3} { 2 }$
, $y = - \frac { 3 } { 2 }$
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) $x = \frac { 3} { 2 }$
, $y = - \frac { 3 } { 2 }$
E) none

Quizplus · Accepted Answer

To solve for $x$ and $y$, we multiply the matrices and compare the result with the given matrix. The multiplication gives us:$$\left[ \begin{array} { c c } 2y + (-2)x & 6 + 2x \6y & -12 + 0\end{array} \right] = \left[ \begin{array} { c c } 1 & 3 \-6 & 18\end{array} \right]$$From the first element (top left), we get $2y - 2x = 1$, and from the third element (bottom left), we get $6y = -6$, which simplifies to $y = -1$.Substituting $y = -1$ into $2y - 2x = 1$, we get $2(-1) - 2x = 1$, which simplifies to $-2 - 2x = 1$, and further to $-2x = 3$, giving $x = -\frac{3}{2}$.Therefore, $x = -\frac{3}{2}$ and $y = -1$, which matches option C.

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

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