Solve the system of linear equations
$\left\{ \begin{array} { l }
6 x _ { 1 } - 12 x _ { 2 } - 6 x _ { 3 } - 12 x _ { 4 } = 0 \
18 x _ { 1 } - 30 x _ { 2 } - 12 x _ { 3 } - 18 x _ { 4 } = - 3 \
12 x _ { 1 } - 30 x _ { 2 } - 12 x _ { 3 } - 30 x _ { 4 } = 2 \
- 6 x _ { 1 } + 24 x _ { 2 } + 24 x _ { 3 } + 66 x _ { 4 } = 0
\end{array} \right.$
using the inverse matrix $\frac { 1 } { 6 } \left[ \begin{array} { c c c c } - 24 & 7 & 1 & - 2 \ - 10 & 3 & 0 & - 1 \ - 29 & 7 & 3 & - 2 \ 12 & - 3 & - 1 & 1 \end{array} \right]$.
A) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 19 } { 6 } \ - \frac { 3 } { 2 } \ - \frac { 5 } { 2 } \ \frac { 7 } { 6 } \end{array} \right]$
B) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 7 } { 6 } \ - \frac { 5 } { 6 } \ \frac { 3 } { 2 } \ - \frac { 3 } { 2 } \end{array} \right]$
C) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 3 } { 2 } \ - \frac { 19 } { 6 } \ 0 \ \frac { 7 } { 6 } \end{array} \right]$
D)
$\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 1 } { 2 } \ \frac { 1 } { 6 } \ 0 \ \frac { 1 } { 3 } \end{array} \right]$
E) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } 0 \ - \frac { 1 } { 6 } \ 0 \ - \frac { 1 } { 3 } \end{array} \right]$

Question

Solve the system of linear equations
$\left\{ \begin{array} { l } 
6 x _ { 1 } - 12 x _ { 2 } - 6 x _ { 3 } - 12 x _ { 4 } = 0 \
18 x _ { 1 } - 30 x _ { 2 } - 12 x _ { 3 } - 18 x _ { 4 } = - 3 \
12 x _ { 1 } - 30 x _ { 2 } - 12 x _ { 3 } - 30 x _ { 4 } = 2 \
- 6 x _ { 1 } + 24 x _ { 2 } + 24 x _ { 3 } + 66 x _ { 4 } = 0
\end{array} \right.$
using the inverse matrix $\frac { 1 } { 6 } \left[ \begin{array} { c c c c } - 24 & 7 & 1 & - 2 \ - 10 & 3 & 0 & - 1 \ - 29 & 7 & 3 & - 2 \ 12 & - 3 & - 1 & 1 \end{array} \right]$.
A) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 19 } { 6 } \ - \frac { 3 } { 2 } \ - \frac { 5 } { 2 } \ \frac { 7 } { 6 } \end{array} \right]$
B) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 7 } { 6 } \ - \frac { 5 } { 6 } \ \frac { 3 } { 2 } \ - \frac { 3 } { 2 } \end{array} \right]$
C) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 3 } { 2 } \ - \frac { 19 } { 6 } \ 0 \ \frac { 7 } { 6 } \end{array} \right]$ 
D)
$\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 1 } { 2 } \ \frac { 1 } { 6 } \ 0 \ \frac { 1 } { 3 } \end{array} \right]$
E) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } 0 \ - \frac { 1 } { 6 } \ 0 \ - \frac { 1 } { 3 } \end{array} \right]$

Quizplus · Accepted Answer

To solve the system of equations using the inverse matrix method, you multiply the inverse matrix by the constants on the right side of the equations. The given inverse matrix is $\frac { 1 } { 6 } \left[ \begin{array} { c c c c } - 24 & 7 & 1 & - 2 \ - 10 & 3 & 0 & - 1 \ - 29 & 7 & 3 & - 2 \ 12 & - 3 & - 1 & 1 \end{array} \right]$ and the constants are $[0, -3, 2, 0]^T$. Multiplying the inverse matrix by the constants vector gives the solution for $x_1, x_2, x_3, x_4$, which matches option A.

Solve the System of Linear Equations Using the Inverse Matrix