Answer
$(1,3,0,-2)$
Work Step by Step
Step 1. Establish the augmented matrix of the system and use the Gauss Eliminations method:
$\begin{vmatrix} 1 & 1 & -1 & -1 & 6 \\ 2 & 0 & 1 & -3 & 8 \\1 & -1 & 0 & 4 & -10 \\3 & 5 & -1 & -1 & 20 \end{vmatrix} \begin{array}(.\\2R_1-R_2\to R_4 \\R_1-R_3\to R_3\\(R_4-3R_1)/2\to R_2\\ \end{array}$
Step 2. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & -1 & -1 & 6 \\0 & 1 & 1 & 1 & 1 \\0 & 2 & -1 & -5 & 16 \\ 0 & 2 & -3 & 1 & 4\end{vmatrix} \begin{array}(.\\. \\5(2R_2-R_3)\to R_3\\3(2R_2-R_4)\to R_4\\ \end{array}$
Step 3. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & -1 & -1 & 6 \\0 & 1 & 1 & 1 & 1 \\0 & 0 & 15 & 35 & -70 \\ 0 & 0 & 15 & 3 & -6\end{vmatrix} \begin{array}(.\\. \\.\\R_3-R_4\to R_4\\ \end{array}$
Step 4. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 1 & -1 & -1 & 6 \\0 & 1 & 1 & 1 & 1 \\0 & 0 & 15 & 35 & -70 \\ 0 & 0 & 0 & 32 & -64\end{vmatrix} \begin{array}(.\\. \\.\\.\\ \end{array}$
Step 5.The last row of step 4 gives $w=-2$, back substitute the known quantities to row 3 gives $z=0$, to row 2 gives $y=3$, and to row 1 gives $x=1$. Thus the solution is $(1,3,0,-2)$
