Solve the system of linear equations $$\left\{ \begin{array} { l }
4 x + 4 y = 3 \
8 x - 20 y = - 9
\end{array} \right.$$ using an inverse matrix.
A) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { l }
\frac { 3 } { 14 } \\
\frac { 15 } { 28 }
\end{array} \right]$$
B) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { l }
\frac { 9 } { 28 } \\
\frac { 12 } { 7 }
\end{array} \right]$$
C) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { c }
\frac { 9 } { 28 } \\
\frac { - 3 } { 14 }
\end{array} \right]$$
D) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { c }
\frac { 9 } { 7 } \\
\frac { 15 } { 28 }
\end{array} \right]$$
E) $$\left[ \begin{array} { l }
x \
y
\end{array} \right] = \left[ \begin{array} { c }
\frac { - 9 } { 7 } \\
\frac { - 3 } { 14 }
\end{array} \right]$$

Question

​Solve the system of linear equations $$\left\{ \begin{array} { l } 
4 x + 4 y = 3 \
8 x - 20 y = - 9
\end{array} \right.$$ using an inverse matrix.
A)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 3 } { 14 } \\
\frac { 15 } { 28 }
\end{array} \right]$$
B)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 9 } { 28 } \\
\frac { 12 } { 7 }
\end{array} \right]$$
C)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 9 } { 28 } \\
\frac { - 3 } { 14 }
\end{array} \right]$$
D)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 9 } { 7 } \\
\frac { 15 } { 28 }
\end{array} \right]$$
E)​ $$\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 9 } { 7 } \\
\frac { - 3 } { 14 }
\end{array} \right]$$

Quizplus · Accepted Answer

The correct answer is A because the inverse matrix of the coefficient matrix is $$\left[ \begin{array} { c c } \frac { 1 } { 14 } & \frac { 1 } { 7 } \\frac { 1 } { 28 } & \frac { 3 } { 14 }\end{array} \right]$$and the solution to the system of linear equations is $$\left[ \begin{array} { l } x \y\end{array} \right] = \left[ \begin{array} { c c } \frac { 1 } { 14 } & \frac { 1 } { 7 } \\frac { 1 } { 28 } & \frac { 3 } { 14 }\end{array} \right] \left[ \begin{array} { l } 3 \- 9\end{array} \right] = \left[ \begin{array} { l } \frac { 3 } { 14 } \\\frac { 15 } { 28 }\end{array} \right]$$

​Solve the System of Linear Equations {4x+4y=38x−20y=−9\left\{ \begin{array} { l } 4 x + 4 y = 3 \\8 x - 20 y = - 9\end{array} \right.{4x+4y=38x−20y=−9​

Solve the System of Linear Equations $\left\{ \begin{array} { l } 4 x + 4 y = 3 \\8 x - 20 y = - 9\end{array} \right.$