Solve the system of linear equations

\left\{ \begin{array} { l 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 } & = - 15 \\8 x _ { 1 } - 20 x _ { 2 } - 8 x _ { 3 } - 20 x _ { 4 } & = 10 \\- 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]

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A)

\left[ \begin{array} { l } x _ { 1 } \\x _ { 2 } \\x _ { 3 } \\x _ { 4 }\end{array} \right] = \left[ \begin{array} { c } \frac { - 95 } { 4 } \\\\\frac { - 45 } { 4 } \\\\\frac { - 75 } { 4 } \\\\\frac { 35 } { 4 }\end{array} \right]

B)

\left[ \begin{array} { l } x _ { 1 } \\x _ { 2 } \\x _ { 3 } \\x _ { 4 }\end{array} \right] = \left[ \begin{array} { c } 0 \\\frac { - 5 } { 4 } \\0 \\\frac { - 5 } { 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 { - 45 } { 4 } \\\\\frac { - 95 } { 4 } \\\\0 \\\\\frac { 35 } { 4 }\end{array} \right]

D)

\left[ \begin{array} { l } x _ { 1 } \\x _ { 2 } \\x _ { 3 } \\x _ { 4 }\end{array} \right] = \left[ \begin{array} { r } \frac { - 15 } { 4 } \\\frac { 5 } { 4 } \\0 \\\frac { 5 } { 2 }\end{array} \right]

E)

\left[ \begin{array} { l } x _ { 1 } \\x _ { 2 } \\x _ { 3 } \\x _ { 4 }\end{array} \right] = \left[ \begin{array} { c } \frac { - 35 } { 4 } \\\\\frac { - 25 } { 4 } \\\\\frac { 45 } { 4 } \\\\\frac { - 45 } { 4 }\end{array} \right]

​Use an Inverse Matrix to Solve (If Possible)the System of Linear

Multiple Choice