About right identity which is not left identity in a ring Let $S$ be the subset of $M_2(\mathbb{R})$ consisting of all matrices of the form 
$\begin{pmatrix}
a & a \\ 
b & b
\end{pmatrix}$
The matrix $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is right identity in $S$ if and only if $x+y=1$. Fine, I can see that.

But I cannot see why "If $x+y=1$ , then $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is not a left identity in $S$".

I have tried that, if $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is a left inverse then : $\begin{pmatrix}
x & x \\ 
y & y
\end{pmatrix}\begin{pmatrix}
a & a \\ 
b & b
\end{pmatrix}=\begin{pmatrix}
x(a+b) & x(a+b) \\ 
y(a+b) & y(a+b)
\end{pmatrix}=\begin{pmatrix}
a & a \\ 
b & b
\end{pmatrix}$ in which case we have $x(a+b)=a$ and $y(a+b)=b$. What can i do with $x+y=1$?

Your result:
$\begin{pmatrix}
x & x \\ 
y & y
\end{pmatrix}\begin{pmatrix}
a & a \\ 
b & b
\end{pmatrix}=\begin{pmatrix}
x(a+b) & x(a+b) \\ 
y(a+b) & y(a+b)
\end{pmatrix}$
shows that in $S$ does not exists a left identity.
