Solve the linear system

\begin{aligned*} x + 2y + z & = 1 \\ 3x + y + 4x & = 0 \\ 2x + 2y + 3z & = 2. \end{aligned*}

Phase 1. Towards the row echelon form.

1. Write an augmented matrix.

   \begin{equation*} M = \begin{pmatrix} 1 & 2 & 1 & 1 \\ 3 & 1 & 4 & 0 \\ 2 & 2 & 3 & 2 \end{pmatrix}. \end{equation*}

2. Move all zero rows.

   Nothing to do.

3. Enumerating rows, top to bottom:

   For \(M_1\):

   1. Scale the selected row.

      Nothing to do.

   2. Replace the rows below the selected row.

      \(M_2 \gets M_2 + kM_1\) with \(k = -3\), so

      \begin{align*} M & \gets \begin{pmatrix} 1 & 2 & 1 & 1 \\ 3 + (-3) \cdot 1 & 1 + (-3) \cdot 2 & 4 + (-3) \cdot 1 & 0 + (-3) \cdot 1 \\ 2 & 2 & 3 & 2 \end{pmatrix} \\[2ex] & = \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & -5 & 1 & -3 \\ 2 & 2 & 3 & 2 \end{pmatrix}. \end{align*}

      \(M_3 \gets M_3 + kM_1\) with \(k = -2\), so

      \begin{align*} M & \gets \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & -5 & 1 & -3 \\ 2 + (-2) \cdot 1 & 2 + (-2) \cdot 2 & 3 + (-2) \cdot 1 & 2 + (-2) \cdot 1 \end{pmatrix} \\[2ex] & = \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & -5 & 1 & -3 \\ 0 & -2 & 1 & 0 \end{pmatrix}. \end{align*}

   For \(M_2\):

   1. Scale the selected row.

      \(M_2 \gets kM_2\) with \(k = -\frac{1}{5}\), so

      \begin{align*} M & \gets \begin{pmatrix} 1 & 2 & 1 & 1 \\ -\frac{1}{5} \cdot 0 & -\frac{1}{5} \cdot (-5) & -\frac{1}{5} \cdot 1 & -\frac{1}{5} \cdot (-3) \\ 0 & -2 & 1 & 0 \end{pmatrix} \\[2ex] & = \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & 1 & -\frac{1}{5} & \frac{3}{5} \\ 0 & -2 & 1 & 0 \end{pmatrix}. \end{align*}

   2. Replace the rows below the selected row.

      \(M_3 \gets M_3 + kM_2\) with \(k = 2\), so

      \begin{align*} M & \gets \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & 1 & -\frac{1}{5} & \frac{3}{5} \\ 0 + 2 \cdot 0 & -2 + 2 \cdot 1 & 1 + 2 \cdot (-\frac{1}{5}) & 0 + 2 \cdot \frac{3}{5} \end{pmatrix} \\[2ex] & = \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & 1 & -\frac{1}{5} & \frac{3}{5} \\ 0 & 0 & \frac{3}{5} & \frac{6}{5} \end{pmatrix}. \end{align*}

   For \(M_3\):

   1. Scale the selected row.

      \(M_3 \gets kM_3\) with \(k = \frac{5}{3}\), so

      \begin{align*} M & \gets \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & 1 & -\frac{1}{5} & \frac{3}{5} \\ \frac{5}{3} \cdot 0 & \frac{5}{3} \cdot 0 & \frac{5}{3} \cdot \frac{3}{5} & \frac{5}{3} \cdot \frac{6}{5} \end{pmatrix} \\[2ex] & = \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & 1 & -\frac{1}{5} & \frac{3}{5} \\ 0 & 0 & 1 & 2 \end{pmatrix}. \end{align*}

   2. Replace the rows below the selected row.

      Nothing to do.

Phase 2.

4. Is the last column is a pivot column?

   The system is consistent.

5. Enumerating rows, bottom to top:

   For \(M_3\):

   1. Replace the rows above.

      \(M_2 \gets M_2 + kM_3\) using \(k = \frac{1}{5}\), so

      \begin{align*} M & \gets \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 + \frac{1}{5} \cdot 0 & 1 + \frac{1}{5} \cdot 0 & -\frac{1}{5} + \frac{1}{5} \cdot 1 & \frac{3}{5} + \frac{1}{5} \cdot 2 \\ 0 & 0 & 1 & 2 \end{pmatrix} \\[2ex] & = \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{pmatrix}. \end{align*}

      \(M_1 \gets M_1 + kM_3\) using \(k = -1\), so

      \begin{align*} M & \gets \begin{pmatrix} 1 + (-1) \cdot 0 & 2 + (-1) \cdot 0 & 1 + (-1) \cdot 1 & 1 + (-1) \cdot 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{pmatrix} \\ & = \begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{pmatrix}. \end{align*}

   For \(M_2\):

   1. Replace the rows above.

   \(M_1 \gets M_1 + kM_2\) using \(k = -2\), so

   \begin{align*} M & \gets \begin{pmatrix} 1 + (-2) \cdot 0 & 2 + (-2) \cdot 1 & 0 + (-2) \cdot 0 & -1 + (-2) \cdot 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{pmatrix} \\ & = \begin{pmatrix} 1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{pmatrix}. \end{align*}

   For \(M_1\):

   1. Replace the rows above.

      Nothing to do.

The solution is \((x, y, z) = (-3, 1, 2)\).