\(\mathbb{S}_1\):
\(\begin{eqnarray*} x + ~~y + z &=& ~~~0\\ - x + 2 y \quad~~~ &=& -1\\ x ~\quad\quad + z &=& ~~~1 \end{eqnarray*}\)
The coefficient matrix is:
\(\left[\begin{array}{rrr|r} 1 & 1 & 1 & 0\\ -1 & 2 & 0 & -1\\ 1 & 0 & 1 & 1 \end{array}\right]\)
Performing the following two row operations
\(E_2 \to E_1 + E_2\)
\(E_3 \to -E_1 + E_3\)
clears out the entries below the main diagonal in column one with the following result:
\(\left[\begin{array}{rrr|r} 1 & 1 & 1 & 0\\ 0 & 3 & 1 & -1\\ 0 & -1 & 0 & 1 \end{array}\right]\)
\(E_2\leftrightarrow E_3\)
\(\left[\begin{array}{rrr|r} 1 & 1 & 1 & 0\\ 0 & -1 & 0 & 1 \\ 0 & 3 & 1 & -1 \end{array}\right]\)
\(E_2 \to -E_2\)
\(\left[\begin{array}{rrr|r} 1 & 1 & 1 & 0\\ 0 & 1 & 0 & -1\\ 0 & 3 & 1 & -1 \end{array}\right]\)
The following operation clears out the entry below the main diagonal in column 2.
\( E_3 \to -3E_2 + E_3\)
This results in the following triangularized matrix.
\(\left[\begin{array}{rrr|r} 1 & 1 & 1 & 0\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 2 \\ \end{array}\right]\)
Now we begin the process of back-substitution, which zeros out the non-zero entries above the main diagonal.
\(E_1 \to -E_3 + E_1\)
\(\left[\begin{array}{rrr|r} 1 & 1 & 0 & -2\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 2 \end{array}\right]\)
\(E_1 \to -E_2 + E_1\)
\(\left[\begin{array}{rrr|r} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 2 \end{array}\right]\)
Which corresponds to a solution
\(\begin{eqnarray*} x \quad\quad &=& -1\\ y \quad &=& -1\\ z &=&~~~2 \end{eqnarray*}\)