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!

Yes you are correct that if $\alpha=2$, then there would be two zero rows. Therefore, (in this 4x4 augmented matrix case) one of the variables, say $x_3$ can be chosen as a "free" variable.
This can be seen by letting $t:=x_3$. Now you can see that $x_2$ and $x_1$ are fully defined by the equations of your row-reduced matrix and $t$. Thus, you only have one free variable, and your degrees of freedom is 1.
Alternatively, you can reason that because you have 3 variables, and the rank of your matrix is 2 (a 4x3 matrix with two non-zero rows), then you have one free variable or a degree of freedom of 1.
