let $A$, $B$ and $C$ be three matrices, such that $A$ and $B$ can be multiplied, $A$ and $C$ can also be multiplied, and we can add $B$ to $C$.
Prove that $$A(B+C)=AB+AC$$
This is my proof (it's probably wrong.)
since we can add $B$ to $C$ this implies that if $B:n\times s$ then $C:n\times s$ and $A:m\times n$
let $$\cases{A=(a_{ij}) \\ B=(b_{jk}) \\ C=(c_{jk})} \implies B+C=(b_{jk}+c_{jk}), \text{ } i=1,..,m\text{ and } j=1,..,n$$ and using the definition of matrix multiplication $$ A(B+C)= \sum_{j=1}^n a_{ij}(b_{jk}+c_{jk})=\sum_{j=1}^n a_{ij}b_{jk}+\sum_{j=1}^n a_{ij}c_{jk}$$ $$=AB+AC$$
I've interpreted $a_{ij}$ and $b_{jk}$ as numbers so that I can apply the distributive rule between them, but i suspect that $a_{ij}$ and $b_{jk}$ are matrices, so I can't do that, is there any method to make this proof rigorous enough?