Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 0 & 3 & -1 & 1 & 4 & 8 \\ 1 & 1 & -3 & -2 & -1 & 5 \\ 3 & -2 & -1 & -2 & 2 & -11 \\ -1 & 1 & -3 & 0 & 1 & 11 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 0 & 0 & \frac{12}{5} & -3 \\ 0 & 1 & 0 & 0 & -\frac{1}{5} & 2 \\ 0 & 0 & 1 & 0 & -\frac{6}{5} & -2 \\ 0 & 0 & 0 & 1 & \frac{17}{5} & 0 \end{array}\right] \]

1. The vector equation \( x_{1} \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -1 \end{array}\right] + x_{2} \left[\begin{array}{c} 3 \\ 1 \\ -2 \\ 1 \end{array}\right] + x_{3} \left[\begin{array}{c} -1 \\ -3 \\ -1 \\ -3 \end{array}\right] + x_{4} \left[\begin{array}{c} 1 \\ -2 \\ -2 \\ 0 \end{array}\right] + x_{5} \left[\begin{array}{c} 4 \\ -1 \\ 2 \\ 1 \end{array}\right] = \left[\begin{array}{c} 8 \\ 5 \\ -11 \\ 11 \end{array}\right] \)has a solution.

2. \( \left[\begin{array}{c} 8 \\ 5 \\ -11 \\ 11 \end{array}\right] \) is a linear combination of the vectors \( \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -1 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ -2 \\ -2 \\ 0 \end{array}\right] , \text{ and } \left[\begin{array}{c} 4 \\ -1 \\ 2 \\ 1 \end{array}\right] \).