Answer
$(0, -3, 0, -3)$
Work Step by Step
Step 1. Establish the augmented matrix of the system and use the Gauss Eliminations method:
$\begin{vmatrix} -1 & 2 & 1 & -3 & 3 \\ 3 & -4 & 1 & 1 & 9 \\-1 & -1 & 1 & 1 & 0 \\2 & 1 & 4 & -2 & 3 \end{vmatrix} \begin{array}(-R_3\leftrightarrow R_1 \\R_4\leftrightarrow R_2\\.\\.\\ \end{array}$
Step 2. exchange -row3 with row 1 and row 4 with row 2:
$\begin{vmatrix} 1 & 1 & -1 & -1 & 0 \\2 & 1 & 4 & -2 & 3 \\-1 & 2 & 1 & -3 & 3 \\ 3 & -4 & 1 & 1 & 9\end{vmatrix} \begin{array} \\.\\2R_1-R2\to R_2\\R_3+R_1\to R_3\\3R_1-R_4\to R_4 \end{array}$
Step 3. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & -1 & -1 & 0 \\ 0 & 1 & -6 & 0 & -3 \\0 & 3 & 0 & -4 & 3 \\0 & 7 & -4 & -4 & -9 \end{vmatrix} \begin{array} . \\ .\\ R_3-3R_2\to R_3\\R_4-7R2\to R_4\end{array}$
Step 4. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & -1 & -1 & 0 \\ 0 & 1 & -6 & 0 & -3 \\0 & 0 & 18 & -4 & 12 \\0 & 0 & 38 & -4 & 12 \end{vmatrix}\begin{array} . \\ .\\ .\\R_4-R_3\to R_4\end{array}$
Step 5. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & -1 & -1 & 0 \\ 0 & 1 & -6 & 0 & -3 \\0 & 0 & 18 & -4 & 12 \\0 & 0 & 20 & 0 & 0 \end{vmatrix}$
Step 6. The last row of step 5 gives $z=0$, back-substitute to row 3 to get $w=-3$, to row 2 to get $y=-3$, and to row 1 to get $x=0$, thus the solutions to the system is $(0, -3, 0, -3)$
