Answer
$(-1,0,0,1)$
Work Step by Step
Step 1. Establish the augmented matrix of the system and use the Gauss Eliminations method:
$\begin{vmatrix} 1 & 1 & 2 & -1 & -2 \\0 & 3 & 1 & 2 & 2\\1 & 1 & 0 & 3 & 2 \\-3 & 0 & 1 & 2 & 5 \end{vmatrix} \begin{array}(\\. \\(R_1-R_3)/2\to R_3\\3R_1+R_4\to R_4\\ \end{array}$
Step 2. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & 2 & -1 & -2 \\0 & 3 & 1 & 2 & 2\\0 & 0 & 1 & -2 & -2 \\0 & 3 & 7 & -1 & -1 \end{vmatrix} \begin{array}(\\. \\.\\(R_4+R_2)/3\to R_4\\ \end{array}$
Step 3. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & 2 & -1 & -2 \\0 & 3 & 1 & 2 & 2\\0 & 0 & 1 & -2 & -2 \\0 & 0 & 2 & -1 & -1 \end{vmatrix} \begin{array}(\\. \\.\\R_4-2R_3\to R_4\\ \end{array}$
Step 4. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & 2 & -1 & -2 \\0 & 3 & 1 & 2 & 2\\0 & 0 & 1 & -2 & -2 \\0 & 0 & 0 & 3 & 3 \end{vmatrix} \begin{array}(\\. \\.\\.\\ \end{array}$
Step 5. The last row of step 4 gives $w=1$, back substitute the results to row 3 to get $z=0$, to row 2 to get $y=0$, to row 1 to get $x=-1$. Thus the solution is $(-1,0,0,1)$
