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!

HINT: use Rouché–Capelli theorem

A system of linear equations with $\,n\,$ variables has a solution if and only if the rank of its coefficient matrix $\,A\,$ is equal to the rank of its augmented matrix $\,\left[\,A \mid b\,\right].\,$


Let us rewrite your system in matrix form $\,\mathbf A\,\vec{\boldsymbol{x}} = \vec{\boldsymbol{b}}$:
\begin{align}
\left\lbrace 
\begin{array}{ccccccc}
x_1 & - & x_2 & +& 2\, x_3 & = & 2 \\
x_1 & + & 2\,x_2 & -&  x_3 & = & 2 \\
x_1 & + & 2\,x_2 & & & = & 2  \\
x_1 & & &+&  x_3 & = & \alpha \\
\end{array}
\right.
\iff
%\underbrace{
\begin{pmatrix}
1 & -1 &  2 \\
1 &  2 & -1 \\
1 &  1 &  0 \\
1 &  0 &  1 \\
\end{pmatrix}
%}_{\;\ \mathbf A}
%\overbrace{
\begin{matrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\\ 
\phantom{x_4}\end{matrix}
%}^{\vec{\boldsymbol{b}}}
=
%\underbrace{
\begin{pmatrix} 2 \\ 2 \\ 2 \\ \alpha \end{pmatrix}
%}_{\vec{\boldsymbol{b}}}
\end{align}
$$
\mathbf{A} = \begin{pmatrix}
1 & -1 &  2 \\
1 &  2 & -1 \\
1 &  1 &  0 \\
1 &  0 &  1 \\
\end{pmatrix}
, \quad b =\begin{pmatrix} 2 \\ 2 \\ 2 \\ \alpha \end{pmatrix}
\implies 
\mathbf{B} := \left[ \, \mathbf A \,\left\lvert \;\,  \vec{\boldsymbol{b}} \right.\, \right] = \left[
\begin{array}{ccc|c}
1 & -1 & 2 & 2 \\
1 & 2 & -1 & 2 \\
1 & 1 & 0 & 2 \\
1 & 0 & 1 & \alpha \\
\end{array}
\right]
$$
It is easy to see that rank  $\,\mathbf A = 2\,$ because its first row is linear combination of the second and the third rows: $\,\mathbf A\left[1\right] = 3\mathbf A\left[2\right] - 2\mathbf A\left[2\right]\,$
$$
\mathbf A\left[1\right] = 3\mathbf A\left[2\right] - 2\mathbf A\left[2\right]
\quad\iff\qquad
\begin{array}{ccc}
  & \begin{pmatrix} 1 & \phantom{-}1 & \phantom{-}0 \end{pmatrix} &  \times \,3 \\
- & \begin{pmatrix} 1 & \phantom{-}2 & -1 \end{pmatrix} &  \times \,2 \\
\hline
  & \begin{pmatrix} 1 & -1 & \phantom{-}2 \end{pmatrix} &  \\
\end{array}
$$
Therefore we need to find values of $\,\alpha\,$ for which rank $\,\mathbf B = 2$.
I hope you can pick it from here.
