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 $x+y = 1$?
 A: Let $S = \left\{ A \in \mathcal{M}_{2 \times 2}(\mathbb{R}) :\ A= \begin{bmatrix}a &a\\\ b& b \end{bmatrix}\right\}$.
Then, $\begin{bmatrix}x &x\\\ y& y \end{bmatrix}$ is a left identity if $\begin{bmatrix}x &x\\\ y& y \end{bmatrix} \cdot\begin{bmatrix}a &a\\\ b& b \end{bmatrix} = \begin{bmatrix}ax+bx & ax+bx\\\ ay+by& ay+by \end{bmatrix} = \begin{bmatrix}a &a\\\ b& b \end{bmatrix}$ for all $\begin{bmatrix}a &a\\\ b& b \end{bmatrix} \in S, \begin{bmatrix}a &a\\\ b& b \end{bmatrix}\ne 0_{S}.$ 
$\Rightarrow ax+bx = a$ and $ay+by = b \Rightarrow x= \frac{a}{a+b}, y = \frac{b}{a+b} \Rightarrow x+y = 1$.
So one may conclude that $\begin{bmatrix}x &x\\\ y& y \end{bmatrix}$ is an identity of $S$ if $x+y =1$. But then even if $x+y =1$, by taking 
$\begin{bmatrix}a &a\\\ b& b \end{bmatrix} = \begin{bmatrix}0 &0\\\ 1& 1 \end{bmatrix}$ once and $\begin{bmatrix}1 &1\\\ 0& 0 \end{bmatrix}$ second, we obtain $\begin{bmatrix}x &x\\\ y& y \end{bmatrix} = \begin{bmatrix}0 &0\\\ 1& 1 \end{bmatrix}$ and
$\begin{bmatrix}x &x\\\ y& y \end{bmatrix} = \begin{bmatrix}1 &1\\\ 0& 0 \end{bmatrix}$ respectively, which is not possible.
A: 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.
