Answer
$(-2,1,0,3)$
Work Step by Step
Step 1. Establish the augmented matrix of the system and use the Gauss Eliminations method:
$\begin{vmatrix} 1 & -3 & 2 & 1 & -2 \\1 & -2 & 0 & -2 & -10\\0 & 0 & 1 & 5 & 15 \\3 & 0 & 2 & 1 & -3 \end{vmatrix} \begin{array}(\\R_2-R_1\to R_2 \\.\\R_4-3R_1\to R_4\\ \end{array}$
Step 2. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & -3 & 2 & 1 & -2 \\0 & 1 & -2 & -3 & -8\\0 & 0 & 1 & 5 & 15 \\0 & 9 & -4 & -2 & 3 \end{vmatrix} \begin{array}(\\. \\.\\R_4-9R_2\to R_4\\ \end{array}$
Step 3. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & -3 & 2 & 1 & -2 \\0 & 1 & -2 & -3 & -8\\0 & 0 & 1 & 5 & 15 \\0 & 0 & 14 & 25 & 75 \end{vmatrix} \begin{array}(\\. \\.\\14R_3-R_4\to R_4\\ \end{array}$
Step 4. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & -3 & 2 & 1 & -2 \\0 & 1 & -2 & -3 & -8\\0 & 0 & 1 & 5 & 15 \\0 & 0 & 0 & 45 & 135 \end{vmatrix} \begin{array}(\\. \\.\\.\\ \end{array}$
Step 5. The last row of step 4 gives $w=3$, back substitute the results to row 3 to get $z=0$, to row 2 to get $y=1$, to row 1 to get $x=-2$. Thus the solution is $(-2,1,0,3)$
