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 $c=0$.