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!
 A: 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.
A: I'd argue like the following.
From the first two rows it follows, that whatever value $x_1$ has, $x_2$ must equal $x_3$.
If we insert this knowledge into the last two rows, (replacing $x_3$ by $x_2$) then we see, that $\alpha$ must equal $2$. And if we insert also that, 
$$\left\{\begin{array}{c}
x_1 &−  x_2 &+ 2x_3 &= 2 \\
x_1 &+ 2x_2 &− x_3  &= 2 \\
x_1 &+  x_2 &      &= 2 \\
x_1 & +x_2 &&= 2&=α
\end{array}\right.$$
then it follows $x_1 = 2 - x_2$ and (unfortunately) nothing more - and thus we have an infinitude of solutions.
A: 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.
A: 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.
