We will run the proof entry-by-entry.
\begin{align*}
\matrixentry{\alpha(AB)}{ij}=&
\alpha\matrixentry{AB}{ij}&&\knowl{./knowls/definition.MSM.knowl}{\text{Definition MSM}}\\
=&\alpha\sum_{k=1}^{n}\matrixentry{A}{ik}\matrixentry{B}{kj}&&\knowl{./knowls/theorem.EMP.knowl}{\text{Theorem EMP}}\\
=&\sum_{k=1}^{n}\alpha\matrixentry{A}{ik}\matrixentry{B}{kj}&&\text{Distributivity in $\complex{\null}$}\\
=&\sum_{k=1}^{n}\matrixentry{A}{ik}\alpha\matrixentry{B}{kj}&&\text{Commutativity in $\complex{\null}$}\\
=&\sum_{k=1}^{n}\matrixentry{A}{ik}\matrixentry{\alpha B}{kj}&&\knowl{./knowls/definition.MSM.knowl}{\text{Definition MSM}}\\
=&\matrixentry{A(\alpha B)}{ij}&&\knowl{./knowls/theorem.EMP.knowl}{\text{Theorem EMP}}
\end{align*}
So the matrices $\alpha(AB)$ and $A(\alpha B)$ are equal, entry-by-entry, and by the definition of matrix equality (Definition ME) we can say they are equal matrices.