The following deals with finding all the solutions to a linear systems of equations using reduced row echelon form.
I have an augmented matrix, $[A| \vec b]$, defined as $[A| \vec b]:=\left[\begin{array}{rrr|r} 0 & 2 & 2 & 4 \\ 1 & 2 & 3 & 3 \\ 2 & 0 & 2 & -2 \\ \end{array}\right]$, where the first row is $x_1$, second one is $x_2$ and third one is $x_3$.
I used the reduced row echelon form to make it easier to solve the equation and I get $[A| \vec b]:=\left[\begin{array}{rrr|r} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right]$.
Now I have $x_1+x_2=0 => x_1=-x_2 => s$, $x_2+x_3=0 => x_2=-x_3 => t$ and $x_3=-x_2=>u$ and the solution set is therefore $\begin{pmatrix}-s \\ -t \\ -u \end{pmatrix}$ but I'm not sure if the last part of the working out is correct.