Consider the following linear program:
$$ \begin{array}{ll} \text{maximize} & x_1 + x_2 + x_3 \\ \text{subject to} & 60x_1 + 20x_2 +50x_3 \leq 1 \\ & 20x_1 + 50x_2 + 10x_3 \leq 1 \\ \text{and} & x_1, x_2, x_3 \geq 0 \end{array} $$
It has the following augmented matrix form:
$$ \left[\begin{array}{cccccc|c} x_1 & x_2 & x_3 & s_1 & s_2 & z \\ \hline 60 & 20 & 50 & 1 & 0 & 0 & 1 \\ 20 & 50 & 10 & 0 & 1 & 0 & 1 \\ \hline -1 & -1 & -1 & 0 & 0 & 1 & 0 \end{array}\right] $$
When solved, the matrix becomes:
$$ \left[\begin{array}{cccccc|c} x_1 & x_2 & x_3 & s_1 & s_2 & z \\ \hline \frac{26}{23} & 0 & 1 & \frac{1}{46} & \frac{-1}{115} & 0 & \frac{3}{230} \\ \frac{4}{23} & 1 & 0 & \frac{-1}{230} & \frac{1}{46} & 0 & \frac{2}{115} \\ \hline \frac{7}{23} & 0 & 0 & \frac{2}{115} & \frac{3}{230} & 1 & \frac{7}{230} \end{array}\right] $$
As you can see, the $z$ column doesn't change. In practice I've seen that this is always the case. So, is there any need to keep the column $z$ around beside pedantic reasons?
I know that the solution is $\{ x_1 = 0, x_2 = \frac{2}{115}, x_3 = \frac{3}{230}, s_1 = 0, s_2 = 0, z = \frac{7}{230} \}$ which is why the equation $\frac{7}{23}x_1 + 0x_2 + 0x_3 + \frac{2}{115}s_1 + \frac{3}{230}s_2 + z = \frac{7}{230}$ holds true.
