Answer
Solution set: $\{ ( 1$ , $-t-1$ , $2$ , $t )$ , $t\in \mathbb{R}\}$
Work Step by Step
Augmented matrix:
$\left[\begin{array}{llllll}
1 & 1 & -1 & 1 & | & -2\\
2 & -1 & 2 & -1 & | & 7\\
-1 & 2 & 1 & 2 & | & -1
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow-2R_{1}+R_{2}\\
\leftarrow R_{1}+R_{3}
\end{array}\right.$
$\left[\begin{array}{llllll}
1 & 1 & -1 & 1 & | & -2\\
0 & 3 & 0 & 3 & | & -3\\
0 & -3 & 4 & -3 & | & 11
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow\frac{1}{3}R_{2}\\
\leftarrow R_{2}+R_{3}
\end{array}\right.$
$\left[\begin{array}{llllll}
1 & 1 & -1 & 1 & | & -2\\
0 & 1 & 0 & 1 & | & -1\\
0 & 0 & 4 & 0 & | & 8
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\leftarrow\frac{1}{4}R_{3}
\end{array}\right.$
$\left[\begin{array}{llllll}
1 & 1 & -1 & 1 & | & -2\\
0 & 1 & 0 & 1 & | & -1\\
0 & 0 & 1 & 0 & | & 2
\end{array}\right]$
The system is consistent, has more unknowns than equations, does not have a unique solution.
Let $z=t, t\in \mathbb{R}$.
Substituting into row 3,
$y+0=2$
$y=2$
Substituting into row $2$,
$x+0+t=-1$
$x=-t-1$
Substituting into row $1$,
$w+(-t-1)-(2)+(t)=-2$
$w-3=-2$
$w=-2+3=1$
Solution set: $\{ ( 1$ , $-t-1$ , $2$ , $t )$ , $t\in \mathbb{R}\}$
