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 $x+y=1$?
 A: To prove that $S$ contains no left identity, let an arbitrary 
element 
$
A
=
\begin{bmatrix}
   x & x \\
   y & y
\end{bmatrix}
\in S $ be given.  Now, either $x = 0$ or $x \ne 0$.  If $x = 0$,
then note that
$
\begin{bmatrix}
   1 & 1 \\
   0 & 0
\end{bmatrix}
\in S
$, 
but
\begin{equation*}
   A
   \begin{bmatrix}
      1 & 1 \\
      0 & 0
   \end{bmatrix}
   =
   \begin{bmatrix}
      0 & 0 \\
      y & y
   \end{bmatrix}
   \begin{bmatrix}
      1 & 1 \\
      0 & 0
   \end{bmatrix}
   =
   \begin{bmatrix}
      0 & 0 \\
      y & y
   \end{bmatrix}
   \ne 
   \begin{bmatrix}
      1 & 1 \\
      0 & 0
   \end{bmatrix},
\end{equation*}
so $A$ is not a left identity in $S$.  If, on the other hand, $x
\ne 0$, then note that
$
\begin{bmatrix}
   0 & 0 \\
   1 & 1
\end{bmatrix}
\in S
$,
but
\begin{equation*}
   A
   \begin{bmatrix}
      0 & 0 \\
      1 & 1
   \end{bmatrix}
   =
   \begin{bmatrix}
      x & x \\
      y & y
   \end{bmatrix}
   \begin{bmatrix}
      0 & 0 \\
      1 & 1
   \end{bmatrix}
   =
   \begin{bmatrix}
      x & x \\
      y & y
   \end{bmatrix}
   \ne 
   \begin{bmatrix}
      0 & 0 \\
      1 & 1
   \end{bmatrix},
\end{equation*}
so $A$ is not a left identity in $S$. Therefore, no element of 
$S$ can be a left identity in $S$. That is, $S$ does not contain 
a left identity.
A: 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.
