Answer
$(1,1+t,t,0)$
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&-1&-1&2\\ 1&-1&1&-1&0\\2&0&0&2&2\\2&4&-4&-2&6 \end{bmatrix}
\begin{array} \\ \\(-R_2+R_1)/2\to R_2\\(-R_3+2R_1)/2\to R_3\\(R_4-2R_1)/2\to R_4 \\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&-1&-1&2\\ 0&1&-1&0&1\\0&1&-1&-2&1\\0&1&-1&0&1 \end{bmatrix}
\begin{array} \\ \\ \\(-R_3+R_2)\to R_3\\(R_4-R_2)\to R_4 \\ \end{array}$
Step 3. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&-1&-1&2\\ 0&1&-1&0&1\\0&0&0&2&0\\0&0&0&0&0 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \\ \end{array}$
Step 4. The third row gives $w=0$. Let $z=t$, the second row gives $y-t=1$ or $y=1+t$, and the first row gives $x+(1+t)-t=2$ or $x=1$
Step 5. The solutions are $(1,1+t,t,0)$
