Prove the associative la of multiplication for 2x2 matrices, show that (AB)C=A(BC)? Let
$$A= \left[
\begin{array}{cc|c}
  a_{11}&a_{12}\\
  a_{21}&a_{22}
\end{array}
\right] , B= \left[
\begin{array}{cc|c}
  b_{11}&b_{12}\\
  b_{21}&b_{22}
\end{array}
\right],C= \left[
\begin{array}{cc|c}
  c_{11}&c_{12}\\
  c_{21}&c_{22}
\end{array}
\right] $$
I did the exercise, but I have doubts if I have the procedure correctly, if you can tell me if I have an error it would be very helpful. This is my answer:
$$ AB= \left[
\begin{array}{cc|c}
  a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\
  a_{21}b_{21}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}
\end{array}
\right] $$
$$ (AB)C= \left[
\begin{array}{cc|c}
  a_{11}b_{11}C_{11}+a_{12}b_{21}C_{21}&a_{11}b_{12}C_{12}+a_{12}b_{22}C_{22}\\
  a_{21}b_{21}C_{11}+a_{22}b_{21}C_{21}&a_{21}b_{12}C_{12}+a_{22}b_{22}C_{22}
\end{array}
\right] $$
$$ BC= \left[
\begin{array}{cc|c}
  b_{11}C_{11}+b_{21}C_{21}&b_{12}C_{12}+b_{22}C_{22}\\
  b_{21}C_{11}+b_{21}C_{21}&b_{12}C_{12}+b_{22}C_{22}
\end{array}
\right] $$
$$ A(BC)= \left[
\begin{array}{cc|c}
  a_{11}b_{11}C_{11}+a_{12}b_{21}C_{21}&a_{11}b_{12}C_{12}+a_{12}b_{22}C_{22}\\
  a_{21}b_{21}C_{11}+a_{22}b_{21}C_{21}&a_{21}b_{12}C_{12}+a_{22}b_{22}C_{22}
\end{array}
\right] $$
Therefore (AB)C= A(BC)

It's wrong, this is the correct solution for $(AB)C$:
$$ (AB)C=\left[
\begin{array}{cc|c}
  (a_{11}b_{11}+a_{12}b_{21})C_{11}+(a_{11}b_{12}+a_{12}b_{22})C_{21}&(a_{11}b_{11}+a_{12}b_{21})C_{12}+(a_{11}b_{12}+a_{12}b_{22})C_{22}\\
  (a_{21}b_{21}+a_{22}b_{21})C_{11}+(a_{21}b_{12}+a_{22}b_{22})C_{21}&
(a_{21}b_{21}+a_{22}b_{21})C_{12}+(a_{21}b_{12}+a_{22}b_{22})C_{22}
\end{array}
\right] $$
For the matrix $A(BC)$ is the same.
