Answer
$(2-t, 1-t, t)$
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&0&1\\ 1&1&2&3\\ 1&-3&-2&-1 \end{bmatrix}
\begin{array} \\ \\R_2-R_1\to R_2\\-R_3+R_1\to R_3\\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&0&1\\ 0&2&2&2\\ 0&2&2&2 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \end{array}$
Step 3. The last two rows are the same. Let $z=t$, we get $y+t=1$ or $y=1-t$. Use row-1 to get $x-(1-t)=1$ or $x=2-t$
Step 4. Conclusion: the solutions are $(2-t, 1-t, t)$
