I've tried finding $\alpha$ in the following example
$$A=\left[ {\begin{array}{*r} 2 & { - 1} & 2 & 1 \\ 4 & { - 3} & 0 & 7 \\ { - 1} & 0 & { - 3} & 2 \\ \end{array}} \right] ,\quad b=\begin{bmatrix} 0 \\ 6 \\ \alpha\\ \end{bmatrix}.$$ When using the row reduced echleon form I get the following augmented matrix: $$A|b= \left[ {\begin{array}{*r} 1 & 0 & 3 & { - 2} \\ 0 & 1 & 4 & { - 5} \\ 0 & 0 & 0 & 0 \\ \end{array}\left| {\begin{array}{*c} 0 \\ 0 \\ 1 \\ \end{array}} \right.} \right]. $$
But I would like something like the following, I don't understand why $\alpha$ disapears for me?
You system can be written as
$\begin{bmatrix} 2 && -1 && 2 && 1 \\ 4 && -3 && 0 && 7 \\ -1 && 0 && -3 && 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 && 0 \\ 6 && 0 \\ 0 && 1 \end{bmatrix} \begin{bmatrix} 1 \\ \alpha \end{bmatrix}$
The augmented matrix of which is
$\left[ \begin{array}{cccc|cc} 2 & -1 & 2 & 1 & 0 & 0 \\ 4 & -3 & 0 & 7 & 6 & 0 \\ -1 & 0 & -3 & 2 & 0 & 1 \end{array}\right]$
Reducing this using the usual Gauss-Jordan elimination, results in
$\left[ \begin{array}{cccc|cc} 1 & 0 & 3 & -2 & 0 & -1 \\ 0 & 1 & 4 & -5 & 0 & -2 \\ 0 & 0 & 0 & 0 & 1 & -\dfrac{1}{3} \end{array}\right]$
Thus the system can have solutions only if $ 1 - \dfrac{\alpha}{3} = 0 $, i.e. $\alpha = 3 $.