Answer
Solution set: $\{ ( \displaystyle \frac{1}{3}t+1$ , $\displaystyle \frac{1}{3}t$ , $t )$ , $t\in \mathbb{R}\}$
Work Step by Step
Augmented matrix:
$\left[\begin{array}{lllll}
2 & 1 & -1 & | & 2\\
3 & 3 & -2 & | & 3
\end{array}\right]\left\{\begin{array}{l}
\leftarrow-R_{1}+R_{3}\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & 1\\
3 & 3 & -2 & | & 3
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow-3R_{1}+R_{3}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & 1\\
0 & -3 & 1 & | & 0
\end{array}\right]\left\{\begin{array}{l}
.\\
\div(-3).
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & 1\\
0 & 1 & -1/3 & | & 0
\end{array}\right]$
The system is consistent and does not have a unique solution.
Let $z=t, t\in \mathbb{R}$.
Substituting into row 2,
$y-\displaystyle \frac{1}{3}t=0$
$y=\displaystyle \frac{1}{3}t$
Substituting into row 1,
$x+2(\displaystyle \frac{1}{3}t)-t=1$
$x-\displaystyle \frac{1}{3}t=1$
$x=\displaystyle \frac{1}{3}t+1$
Solution set: $\{ ( \displaystyle \frac{1}{3}t+1$ , $\displaystyle \frac{1}{3}t$ , $t )$ , $t\in \mathbb{R}\}$
