Consider the following system and find the values of b for which the system has a solution So I have this system:
$$\left\{\begin{array}{c}
x_1 &−  x_2 &+ 2x_3 &= 2 \\
x_1 &+ 2x_2 &− x_3  &= 2 \\
x_1 &+  x_2 &      &= 2 \\
x_1 &       & +x_3 &= α
\end{array}\right.$$
And we are asked to find the values of $\alpha$ for which the system has a solution. When I do the coefficient matrix and a Gauss-Jordan elimination i get two entirely zero rows with one of the augmented matrix solutions being $\alpha-2$ so am I correct in saying that the system will have an infinite number of solutions if $\alpha=2$?
The thing that concerns me is that we are then asked to find the value of $\alpha$ if the degree of freedom is $2$?
Any help would be greatly appreciated!
PS. Sorry for any formatting errors!

Gaussian elimination.                 
Downwards:
$$\tag 1 \begin{array} {l}
 x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\
 x_1 & +2x_2 & -  x_3 & 0 & = 2 \\
 x_1 &  +1x_2 & 0 & 0 & = 2 \\
 x_1 &   0 &+1 x_3 & -\alpha & = 0 \\
\end{array} $$
Subtract row 1 from rows 2 to 4:
$$\tag 2 \begin{array} {l}
 x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\ \hline
 0  & 3x_2 & - 3 x_3 & 0 & = 0 \\
 0  &   2x_2 & -2x_3 & 0 & = 0 \\
 0  &  1x_2 & -1x_3 & -1\alpha & = -2 \\
\end{array} $$
Cancel the 3 and the 2 because on the rhs are zeros in row 2 and 3:
$$\tag 3 \begin{array} {l}
 x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\ \hline
 0  &  x_2 & -  x_3 & 0 & = 0 \\
 0  &   x_2 & -x_3 & 0 & = 0 \\
 0  &   x_2 & -x_3 & -1\alpha & = -2 \\
\end{array} $$
Subtract row 2 from rows 3 and 4:
$$ \begin{array} {l}
 x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\ 
 0  &  x_2 & -  x_3 & 0 & = 0 \\ \hline
 0  &   0 & 0 & 0 & = 0 \\
 0  &   0 & 0 & -1\alpha & = -2 \\
\end{array} $$
It follows $\alpha = 2$. Also, from row 2 that $x_2=x_3$
From the remaining gaussian-eiminate upwards: add row 2 to row 1
$$ \begin{array} {l}
 x_1 & 0 & +  x_3 & 0 & = 2 \\ \hline
 0  &  x_2 & -  x_3 & 0 & = 0 \\ \hline
 0  &   0 & 0 & 0 & = 0 \\
 0  &   0 & 0 & -1\alpha & = -2 \\
\end{array} $$
We get $x_1 =2-x_3$ with arbitrary many solutions.
