Solve the system of linear equations $$\left\{ \begin{array} { l }
2 x + 3 y = - 15 \
4 x - 3 y = 10
\end{array} \right.$$ using the inverse matrix $$\left[ \begin{array} { c c }
\frac { 1 } { 6 } & \frac { 1 } { 6 } \\
\frac { 2 } { 9 } & - \frac { 1 } { 9 }
\end{array} \right]$$
A) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { l }
\frac { 5 } { 6 } \\
\frac { 5 } { 2 }
\end{array} \right]$$
B) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { c }
- \frac { 5 } { 6 } \\
- \frac { 40 } { 9 }
\end{array} \right]$$
C) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { c }
\frac { 25 } { 18 } \\
\frac { 5 } { 2 }
\end{array} \right]$$
D) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { l }
\frac { 5 } { 2 } \
0
\end{array} \right]$$
E) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { c }
- \frac { 5 } { 3 } \\
\frac { 5 } { 6 }
\end{array} \right]$$

Question

​Solve the system of linear equations $$\left\{ \begin{array} { l } 
2 x + 3 y = - 15 \
4 x - 3 y = 10
\end{array} \right.$$ using the inverse matrix $$\left[ \begin{array} { c c } 
\frac { 1 } { 6 } & \frac { 1 } { 6 } \\
\frac { 2 } { 9 } & - \frac { 1 } { 9 }
\end{array} \right]$$
A)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 5 } { 6 } \\
\frac { 5 } { 2 }
\end{array} \right]$$
B)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
- \frac { 5 } { 6 } \\
- \frac { 40 } { 9 }
\end{array} \right]$$
C)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 25 } { 18 } \\
\frac { 5 } { 2 }
\end{array} \right]$$
D)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 5 } { 2 } \
0
\end{array} \right]$$
E)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
- \frac { 5 } { 3 } \\
\frac { 5 } { 6 }
\end{array} \right]$$

Quizplus · Accepted Answer

To solve the system using the inverse matrix, multiply the inverse matrix by the constants matrix $\left[ \begin{array} { c } -15 \ 10 \end{array} \right]$. The calculation is $\left[ \begin{array} { c c } \frac { 1 } { 6 } & \frac { 1 } { 6 } \ \frac { 2 } { 9 } & - \frac { 1 } { 9 } \end{array} \right] \times \left[ \begin{array} { c } -15 \ 10 \end{array} \right] = \left[ \begin{array} { c } - \frac { 5 } { 6 } \ - \frac { 40 } { 9 } \end{array} \right]$.

​Solve the System of Linear Equations {2x+3y=−154x−3y=10\left\{ \begin{array} { l } 2 x + 3 y = - 15 \\4 x - 3 y = 10\end{array} \right.{2x+3y=−154x−3y=10​

Solve the System of Linear Equations $\left\{ \begin{array} { l } 2 x + 3 y = - 15 \\4 x - 3 y = 10\end{array} \right.$