Answer
The solution set is $(3+r-t,-1-r,r,t)$
Work Step by Step
$Ax=b$
$\begin{bmatrix}
1&1&0 & 1\\
3 & 1 & -2& 3\\
2 & 3 &1 & 1\\
-2 & 3 & 5&-2
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}=\begin{bmatrix}
2\\
8\\
3\\
-9
\end{bmatrix}$
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.
$\begin{bmatrix}
1&1&0 & -1|2\\
3 & 1 & -2& 3|8\\
2 & 3 &1 & 1|3\\
-2 & 3 & 5&-2|-9
\end{bmatrix}\approx^1\begin{bmatrix}
1&1&0 & 1|2\\
0 & -2 & -2& 0|2\\
0 & 1 &1 & 0|-1\\
0 & 5 & 5&-0|-5
\end{bmatrix} \approx^2 \begin{bmatrix}
1&1&0 & 1|2\\
0 & 1 & 1& 0|-1\\
0 & -2 &-2 & 0|2\\
0 & 5 & 5&-0|-5
\end{bmatrix}\approx^3 \begin{bmatrix}
1&0&-1 & 1|3\\
0 & 1 & 1& 0|-1\\
0 &0 &0 & 0|0\\
0 & 0 & 0&0|0
\end{bmatrix}$
1. $A_{12}(-3),A_{13}(-2),A_{14}(2)$
2. $P_{23}$
5. $A_{21}(-1),A_{23}(2),A_{24}(-5)$
The equivalent system is
$x_1-x_3+x_4=3$
$x_2+x_3=-1$
$x_3=r$
$x_4 = t$
There is one free variable, which we take to be $x_3 =r, x_4= t$, where r and t can assume any complex values. Applying back substitution yields:
$x_1-r+t=3 \rightarrow x_1=3+r-t$
$x_2+r=-1 \rightarrow x_2=-1-r$
$x_3=r$
$x_4 = t$
The solution set is $(3+r-t,-1-r,r,t)$
