Answer
The solution of the system is $S=\{2t-2s+3,s-t+1,s,2+2t,t\}$
Work Step by Step
Given: $A=\begin{bmatrix}
1 & 1 & 1& 1 &-3\\
1 & 1 & 1& 2&-5\\
2 & 3 & 1 & 4 & -9 \\
2 & 2 &2 &3 & -8
\end{bmatrix}$
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
1 & 1 & 1& 1 &-3|6\\
1 & 1 & 1& 2&-5|8\\
2 & 3 & 1 & 4 & -9|17 \\
2 & 2 &2 &3 & -8|14
\end{bmatrix}\approx^1\begin{bmatrix}
1 & 1 & 1& 1 &-3|6\\
0 & 0 & 0& 1&-2|2\\
0 & 1 & -1 & 2 & -3|5 \\
0 & 0 &0 &-1 & 2|-2
\end{bmatrix} \approx^2\begin{bmatrix}
1 & 1 & 1& 1 &-3|6\\
0 & 1 & -1 & 2 & -3|5\\
0 & 0 & 0& 1&-2|2 \\
0 & 0 &0 &-1 & 2|-2
\end{bmatrix} \approx^3 \begin{bmatrix}
1 & 1 & 1& 1 &-3|6\\
0 & 1 & -1 & 2 & -3|5\\
0 & 0 & 0& 1&-2|2 \\
0 & 0 &0 &0& 0|0
\end{bmatrix} $
$1.A_{12}(-1),A_{13}(-2),A_{14}(-2)$
$2. P_{23}$
$3.A_{34}(1)$
The equivalent system for this matrix is now:
$x_1+x_2+x_3+x_4-3x_5=6$
$x_2-x_3+2x_4-3x_5=5$
$x_3=s$
$x_4-2x_5=2$
$x_5=t$
There is one free variable, which we take to be $x_3 = s, x_5=t$, where s and t can assume any complex value. Applying back substitution yields:
$x_3=s$
$x_4-2t=2 \rightarrow x_4=2+2t$
$x_5=t$
$x_2-s+2(2+2t)-3t=5 \rightarrow x_2=s-t+1$
$x_1+s-t+1+s+2+2t-3t=6 \rightarrow x_1=2t-2s+3$
The solution of the system is $S=\{2t-2s+3,s-t+1,s,2+2t,t\}$
