Simplifying calculation for searching for a matrix The exercise it the following: I should find all $$A \in M^{}_{22}(\mathbb{R}) $$ with 
$$
A
    \begin{pmatrix}
    1 & 1 \\
    0 & 1 \\
    \end{pmatrix}
=
    \begin{pmatrix}
    1 & 1 \\
    0 & 1 \\
    \end{pmatrix}
A
$$
I solved it like the following but I wonder whether there is a more elegant way:
$$
    \begin{pmatrix}
    a & b \\
    c & d \\
    \end{pmatrix}
    \begin{pmatrix}
    1 & 1 \\
    0 & 1 \\
    \end{pmatrix}
=
    \begin{pmatrix}
    1 & 1 \\
    0 & 1 \\
    \end{pmatrix}
    \begin{pmatrix}
    a & b \\
    c & d \\
    \end{pmatrix}
$$
$$
=>  \begin{pmatrix}
    a & a + b \\
    c & c + d \\
    \end{pmatrix}
=
    \begin{pmatrix}
    a + c & b + d \\
    c & d \\
    \end{pmatrix}
$$
=>
I) 
$$
a = a + c => c = 0
$$
II) 
$$
a + b = b + d => a = d
$$
III) 
$$
c = c
$$
IV) 
$$
d = c + d => d = d = a
$$
$$
A
=
    \begin{pmatrix}
    a & b \\
    0 & a \\
    \end{pmatrix}
$$
with 
$$
a, b \in (\mathbb{R})
$$

Perhaps not more elegant, but you can simplify the problem a little further. Since $$\begin{pmatrix}1&1\\0&1\end{pmatrix}=\mathbf{I}_2+\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ you have, using that $\mathbf{A}\mathbf{I}_2=\mathbf{I}_2\mathbf{A}$,
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ which simplifies to $$\begin{pmatrix}0&a\\0&c\end{pmatrix}=\begin{pmatrix}c&d\\0&0\end{pmatrix}$$
This means that $a=d$ and $c=0$.
