Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 1 & -4 & 2 & -3 & 0 & -8 \\ -1 & -3 & -2 & 0 & 0 & 1 \\ 2 & 2 & 2 & -5 & 0 & 0 \\ 0 & -2 & 2 & 2 & 2 & -8 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 0 & 0 & -\frac{64}{3} & 2 \\ 0 & 1 & 0 & 0 & 2 & 1 \\ 0 & 0 & 1 & 0 & \frac{23}{3} & -3 \\ 0 & 0 & 0 & 1 & -\frac{14}{3} & 0 \end{array}\right] \]

1. The vector equation \( x_{1} \left[\begin{array}{c} 1 \\ -1 \\ 2 \\ 0 \end{array}\right] + x_{2} \left[\begin{array}{c} -4 \\ -3 \\ 2 \\ -2 \end{array}\right] + x_{3} \left[\begin{array}{c} 2 \\ -2 \\ 2 \\ 2 \end{array}\right] + x_{4} \left[\begin{array}{c} -3 \\ 0 \\ -5 \\ 2 \end{array}\right] + x_{5} \left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 2 \end{array}\right] = \left[\begin{array}{c} -8 \\ 1 \\ 0 \\ -8 \end{array}\right] \)has a solution.

2. \( \left[\begin{array}{c} -8 \\ 1 \\ 0 \\ -8 \end{array}\right] \) is a linear combination of the vectors \( \left[\begin{array}{c} 1 \\ -1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -4 \\ -3 \\ 2 \\ -2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -2 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} -3 \\ 0 \\ -5 \\ 2 \end{array}\right] , \text{ and } \left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 2 \end{array}\right] \).