proof that if $AB=BA$ matrix $A$ must be $\lambda E$ Let $A \in Mat(2\times 2, \mathbb{Q})$ be a matrix with $AB = BA$ for all matrices $B \in Mat(2\times 2, \mathbb{Q})$.
Show that there exists a $\lambda \in \mathbb{Q}$ so that $A = \lambda E_2$.

Let $E_{ij}$ be the matrix with all entries $0$ except $e_{ij} = 1$.
$$
\begin{align}
AE_{11} &= E_{11}A \\
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array} \right)
\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array} \right)
&=
\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array} \right)
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array} \right)
\\
\left( \begin{array}{cc}
a_{11} & 0 \\
a_{21} & 0 \\
\end{array} \right)
&=
\left( \begin{array}{cc}
a_{11} & a_{12} \\
0 & 0 \\
\end{array} \right) \\
\end{align}
$$
$\implies a_{12} = a_{21} = 0$
And then the same for the other three matrices $E_{12}, E_{21}, E_{22}$ …
I guess it's not the most efficient way of argueing or writing it down … ? Where's the trick to simplify this thingy?

You wish to show that the center of the ring of matrices is the set of scalar matrices. Observe that the matrices commuting with diagonal matrices must be diagonal matrices. Hence the center must be contained in the set of diagonal matrices. Now a diagonal matrix that commutes with an arbitrary matrix must be a scalar matrix.
