$\underline{Thm}$
Let $A \in M_{m \times n},\, B,C \in M_{n \times p},\, D,E \in M_{q \times m}$.
Then
$(a)$ $A(B+C)=AB+AC,\, (D+E)A=DA+EA$
$(b)$ $a(AB)=(aA)B=A(aB)$ $(a \in \mathbb{F})$
$(c)$ $I_{m}A=A=AI_{n}$
$(d)$ If $V$ is an n-dimensional vector space with an ordered basis, then $[I_{V}]_{\beta}=I_{n}$.
$\underline{Proof \, of \, (c)}$
$$\displaystyle \begin{equation} \begin{split} (I_{m}A)_{ij} & = \sum_{k=1}^{m} (I_{m})_{ik} A_{kj} \\ & = \sum_{k=1}^{m} \delta_{ik} A_{kj} \\ & = A_{ij} \end{split} \end{equation}$$
$\underline{Coro}$
Let $A \in M_{m \times n},\, B_{1},\cdots,B_{k} \in M_{n \times p},\, C_{1},\cdots,C_{k} \in M_{q \times m},\, a_{1},\cdots,a_{k} \in \mathbb{F}.$
Then
$$\displaystyle \begin{equation} \begin{split} A(\sum_{i=1}^{k} a_{i}B_{i}) & = \sum_{i=1}^{k} a_{i} AB_{i} \\ & = (\sum_{i=1}^{k} a_{i}C_{i})A \\ & = \sum_{i=1}^{k} a_{i}C_{i}A \end{split} \end{equation}$$