<s>[INST] Let $S$ be the subset of $M(\mathbb{R})$ consisting of matrices of the form: The proof I read on slader seemed like it only used either $1$ or $2$ matrices to prove this when the question asks for all matrices of a certain form (when $x+y = 1$).
Let $S$ be the subset of $M(\mathbb{R})$ consisting of matrices of the form: $\begin{pmatrix}a & a\\\ b & b\end{pmatrix}$. 
If $x + y = 1$, show that $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ is not a left identity in $S$. 
If $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ is a left identity, then we have 
$\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$$\begin{pmatrix}0 & 0\\\ 1 & 1\end{pmatrix}$ = $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ = $\begin{pmatrix}0 & 0\\\ 1 & 1\end{pmatrix}$.
But $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$$\begin{pmatrix}1 & 1\\\ 0 & 0\end{pmatrix}$ = $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ = $\begin{pmatrix}1 & 1\\\ 0 & 0\end{pmatrix}$
which implies the contradiction that  $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ = $\begin{pmatrix}0 & 0\\\ 1 & 1\end{pmatrix}$ = $\begin{pmatrix}1 & 1\\\ 0 & 0\end{pmatrix}$. 
Hence $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ is not a left identity in $S$.
Is this a correct proof without using the fact that $x+y = 1$?
 [/INST] Your proof is correct and you haven't used the condition that $x+y=1$. In fact, the set of matrices in question has not any left identity. So, whether $x+y=1$ is irrelevant.
I believe that the "left" in the question is a typo. The question makes more sense if it is talking about right identities.
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