\(\mathbb{S}_1\):
\(\begin{eqnarray*} x + 2y - z &=& 1 &\\ x ~\quad\quad + z &=& 3 &\quad E_2 \to -E_1 + E_2\\ y + z &=& 1 & \end{eqnarray*}\)
\(\mathbb{S}_2\):
\(\begin{eqnarray*} x + 2y - ~~z &=& 1 &\\ - 2y + 2z &=& 2 &\quad E_2 \leftrightarrow E_3\\ y + ~~z &=& 1 \end{eqnarray*}\)
\(\mathbb{S}_3\):
\(\begin{eqnarray*} x + 2y - ~~z &=& 1 &\\ y + ~~z &=& 1 &\\ - 2y + 2z &=& 2 &\quad E_3 \to 2E_2 + E_3 \end{eqnarray*}\)
\(\mathbb{S}_4\):
\(\begin{eqnarray*} x + 2y - z &=& ~~~1 &\\ y - z &=& -1 &\\ 4z &=& ~~~4 &\quad E_3 \to \left(\frac{1}{4}\right)\cdot E_3 \end{eqnarray*}\)
\(\mathbb{S}_5\):
\(\begin{eqnarray*} x + 2y - z &=& ~~~1\\ y - z &=& -1\\ z &=& ~~~1 \end{eqnarray*}\)
Solving the last system for \(y\) by back-substitution, \(y = 0\).
Using the values of \(z\) and \(y\), we solve the first equation for \(x\), getting \(x = 2\).
Therefore, the solution of the system is \((2,0,1)\).