Answer
$(20-12t,31-19t,2t-2,t)$
Work Step by Step
Step 1. Establish the augmented matrix of the system and use the Gauss Eliminations method:
$\begin{vmatrix} 2 & -1 & 2 & 1 & 5 \\-1 & 1 & 4 & -1 & 3\\3 & -2 & -1 & 0 & 0 \end{vmatrix} \begin{array}(-R_2\leftrightarrow R_1\\R_1+2R_2\to R_2\\R_3+3R_3\to R_3\\ \end{array}$
Step 2. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & -1 & -4 & 1 & -3 \\0 & 1 & 10 & -1 & 11\\0 & 1 & 11 & -3 & 9 \end{vmatrix} \begin{array}(.\\.\\R_3-R_2\to R_3\\ \end{array}$
Step 3. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & -1 & -4 & 1 & -3 \\0 & 1 & 10 & -1 & 11\\0 & 0 & 1 & -2 & -2 \end{vmatrix} \begin{array}(.\\.\\.\\ \end{array}$
Step 4. Let $w=t$, the third row becomes $z-2t=-2$ and $z=2t-2$. Back substitute to the second row, $y+10z-t=11$, we have $y=31-19t$. The first row becomes $x-y-4z+t=-3$ and $x=20-12t$. Thus the solutions are $(20-12t,31-19t,2t-2,t)$
