Solve the system of linear equations
$\left\{ \begin{array} { l }
4 x _ { 1 } - 8 x _ { 2 } - 4 x _ { 3 } - 8 x _ { 4 } = 0 \
12 x _ { 1 } - 20 x _ { 2 } - 8 x _ { 3 } - 12 x _ { 4 } = - 9 \
8 x _ { 1 } - 20 x _ { 2 } - 8 x _ { 3 } - 20 x _ { 4 } = 6 \
- 4 x _ { 1 } + 16 x _ { 2 } + 16 x _ { 3 } + 44 x _ { 4 } = 0
\end{array} \right.$
using the inverse matrix $\frac { 1 } { 4 } \left[ \begin{array} { c c c c } - 24 & 7 & 1 & - 2 \ - 10 & 3 & 0 & - 1 \ - 29 & 7 & 3 & - 2 \ 12 & - 3 & - 1 & 1 \end{array} \right]$.

A)
$\left[\begin{array}{l}
x_{1} \
x_{2} \
x_{3} \
x_{4}
\end{array}\right]=\left[\begin{array}{r}
-\frac{57}{4} \
-\frac{27}{4} \
-\frac{45}{4} \
\frac{21}{4}
\end{array}\right]$

B) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 21 } { 4 } \ - \frac { 15 } { 4 } \ \frac { 27 } { 4 } \ - \frac { 27 } { 4 } \end{array} \right]$

C) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 27 } { 4 } \ - \frac { 57 } { 4 } \ 0 \ \frac { 21 } { 4 } \end{array} \right]$
D)
$\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 9 } { 4 } \ \frac { 3 } { 4 } \ 0 \ \frac { 3 } { 2 } \end{array} \right]$

E) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } 0 \ - \frac { 3 } { 4 } \ 0 \ - \frac { 3 } { 2 } \end{array} \right]$

Question

Solve the system of linear equations
$\left\{ \begin{array} { l } 
4 x _ { 1 } - 8 x _ { 2 } - 4 x _ { 3 } - 8 x _ { 4 } = 0 \
12 x _ { 1 } - 20 x _ { 2 } - 8 x _ { 3 } - 12 x _ { 4 } = - 9 \
8 x _ { 1 } - 20 x _ { 2 } - 8 x _ { 3 } - 20 x _ { 4 } = 6 \
- 4 x _ { 1 } + 16 x _ { 2 } + 16 x _ { 3 } + 44 x _ { 4 } = 0
\end{array} \right.$ 
using the inverse matrix $\frac { 1 } { 4 } \left[ \begin{array} { c c c c } - 24 & 7 & 1 & - 2 \ - 10 & 3 & 0 & - 1 \ - 29 & 7 & 3 & - 2 \ 12 & - 3 & - 1 & 1 \end{array} \right]$.

A) 
$\left[\begin{array}{l}
x_{1} \
x_{2} \
x_{3} \
x_{4}
\end{array}\right]=\left[\begin{array}{r}
-\frac{57}{4} \
-\frac{27}{4} \
-\frac{45}{4} \
\frac{21}{4}
\end{array}\right]$

B) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 21 } { 4 } \ - \frac { 15 } { 4 } \ \frac { 27 } { 4 } \ - \frac { 27 } { 4 } \end{array} \right]$

C) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 27 } { 4 } \ - \frac { 57 } { 4 } \ 0 \ \frac { 21 } { 4 } \end{array} \right]$ 
D)
$\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 9 } { 4 } \ \frac { 3 } { 4 } \ 0 \ \frac { 3 } { 2 } \end{array} \right]$

E) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } 0 \ - \frac { 3 } { 4 } \ 0 \ - \frac { 3 } { 2 } \end{array} \right]$

Quizplus · Accepted Answer

We can solve the linear system by multiplying both sides of the equation by the inverse matrix. Let matrix A be the coefficient matrix and x be the vector of variables, then we have:
$$
A x = b
$$
where
$$
A=\left[\begin{array}{cccc}
4 & -8 & -4 & -8 \
12 & -20 & -8 & -12 \
8 & -20 & -8 & -20 \
-4 & 16 & 16 & 44
\end{array}ight], x=\left[\begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
\end{array}ight], b=\left[\begin{array}{c}
0 \
-9 \
6 \
0
\end{array}ight]
$$
The inverse of A is given:
$$
A^{-1}=\frac{1}{4}\left[\begin{array}{cccc}
-24 & 7 & 1 & -2 \
-10 & 3 & 0 & -1 \
-29 & 7 & 3 & -2 \
12 & -3 & -1 & 1
\end{array}ight]
$$
Multiplying both sides by the inverse matrix, we obtain:
$$
x=A^{-1} b
$$
Using the given inverse matrix, we can compute the solution:
$$
x = \frac { 1 } { 4 } \left[ \begin{array} { c c c c } - 24 & 7 & 1 & - 2 \ - 10 & 3 & 0 & - 1 \ - 29 & 7 & 3 & - 2 \ 12 & - 3 & - 1 & 1 \end{array} ight] \left[ \begin{array} { c } 0 \ -9 \ 6 \ 0 \end{array} ight] = \frac { 1 } { 4 } \left[ \begin{array} { c } -57 \ -27 \ -45 \ 21 \end{array} ight] = \left[ \begin{array} { c } -\frac{57}{4} \ -\frac{27}{4} \ -\frac{45}{4} \ \frac{21}{4} \end{array} ight]
$$
Therefore, the answer is choice (A).

Solve the System of Linear Equations Using the Inverse Matrix