Question: Given the set of vectors $(1,1), (-1,1), (2,2) \subset \mathbb{R}^2$ demonstrate that this system of vectors forms an Indeterminate Compatible System (Rouché-Fröbenius theorem).
I have tried to demostrate it by using matrices:
$ M^+=\begin{pmatrix} 1 & 1 & 0 \\ -1 & 1 & 0\\ 2 & 2 & 0 \end{pmatrix} $
$ M=\begin{pmatrix} 1 & 1 \\ -1 & 1\\ 2 & 2 \end{pmatrix} $
$ X=\begin{pmatrix} x \\ y \end{pmatrix} $
So, $MX=0$ is:
$ \begin{pmatrix} x & y \\ -x & y\\ 2x & 2y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $
But $rank(M^+)=rank(M)=$ number of variables $=2$ which would mean it's a Determinate Compatible System by the R-F theorem.
Your use of Rouché-Fontené theorem is wrong, (and it is not a very efficient way to do it, cardinal of the family is the best answer, or $v_3 = 2 v_1$). If you wish to use your theorem no matter what, here is how you should do it:
Always use column vectors to avoid confusion $$v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad v_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}, \quad v_3 = \begin{pmatrix} 2 \\ 2 \end{pmatrix}$$ Thus $$M = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$$ $$M_+ = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 1 & 2\end{pmatrix}$$
and $\text{rk} (M_+) = 2 = \text{rk} (M)$ which means your family is related.