Answer
The solution set is $(t,1.3)$
Work Step by Step
$Ax=b$
$\begin{bmatrix}
0 & 1 & -1\\
0& 5 & 1\\
0 & 2 &1
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}=\begin{bmatrix}
-2\\
8\\
5
\end{bmatrix}$
Converting the given system of equations to an augmented matrix and then using Gauss-Jordan elimination to determine the solution set to the given system.
$\begin{bmatrix}
0 & 1 & -1|-2\\
0& 5 & 1|8\\
0 & 2 &1|5
\end{bmatrix} \approx^1\begin{bmatrix}
0 & 1 & -1|-2\\
0& 0 & 6|18\\
0 & 0 &3|9
\end{bmatrix} \approx^2 \begin{bmatrix}
0 & 1 & -1|-2\\
0& 0 & 1|3\\
0 & 0 &3|9
\end{bmatrix} \approx^3 \begin{bmatrix}
0 & 1 & -1|-2\\
0& 0 & 1|3\\
0 & 0 &0|0
\end{bmatrix}$
1. $A_{12}(-5),A_{13}(-2)$
2. $M_2(\frac{1}{6})$
5. $A_{21}(1),A_{23}(-3)$
The equivalent system is
$x_2-x_3=-2$
$x_3=3$
$x_1 = t$
There is one free variable, which we take to be $x_1 = t$, where t can assume any complex value. Applying back substitution yields:
$x_3=3$
$x_1 = t$
$x_2-3=-2 \rightarrow x_2=1$
The solution set is $(t,1.3)$
