Answer
Solution set: $\{( t-2$ , $t-2$ , $t )$ , $t\in \mathbb{R}\}$
Work Step by Step
Start with the augmented matrix.
Use matrix row operations to get ls down the main diagonal from upper left to lower right, and 0s below the $1\mathrm{s}$ (row-echelon form).
$\left[\begin{array}{lllll}
5 & -11 & 6 & | & 12\\
-1 & 3 & -2 & | & -4\\
3 & -5 & 2 & | & 4
\end{array}\right]$
Swapping R1 and R2 brings a $-1$ into the first diagonal position,
$\left[\begin{array}{lllll}
-1 & 3 & -2 & | & -4\\
5 & -11 & 6 & | & 12\\
3 & -5 & 2 & | & 4
\end{array}\right]\left\{\begin{array}{l}
\times(-1).\\
5R1+R2\rightarrow R2.\\
3R1+R3\rightarrow R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & -3 & 2 & | & 4\\
0 & 4 & -4 & | & -8\\
0 & 4 & -4 & | & -8
\end{array}\right]\left\{\begin{array}{l}
.\\
\div 4.\\
-R2+R3\rightarrow R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & -3 & 2 & | & 4\\
0 & 1 & -1 & | & -2\\
0 & 0 & 0 & | & 0
\end{array}\right]$
Last row: all zeros, so
the system is consistent, but there is no unique solution .
We take z to be any number $t\in \mathbb{R}.$
Substituting into row 2:
$y-t=-2\qquad/+t$
$y=t-2$
Substituting into row $1$:
$x-3(t-2)+2t=4$
$x-t+6=4$
$x=t-2$
Solution set: $\{( t-2$ , $t-2$ , $t )$ , $t\in \mathbb{R}\}$
