Solving a linear system by Gauss-Jordan

Augmented matrix.
$$\left(\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 2 & 5 & -1 & 7 \\ 1 & 1 & 3 & 6 \end{array}\right).$$

Elimination. $L_2 \leftarrow L_2 - 2L_1$, $L_3 \leftarrow L_3 - L_1$:
$$\left(\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 0 & 1 & -3 & -1 \\ 0 & -1 & 2 & 2 \end{array}\right).$$
$L_3 \leftarrow L_3 + L_2$:
$$\left(\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 0 & 1 & -3 & -1 \\ 0 & 0 & -1 & 1 \end{array}\right).$$

Back-substitution. From the third row, $z = -1$. From the second, $y - 3(-1) = -1$, so $y = -4$. From the first, $x + 2(-4) + (-1) = 4$, so $x = 13$.

Conclusion. $(x,y,z) = (13, -4, -1)$.