Assuming matrices $A,B,C \;\text{and}\;D\;$ are $ n\times n,$ show two ways that $(A+B)(C+D)=AC+AD+BC+BD$

Question

My way first of showing this is by letting $A,B,C \;\text{and}\;D\;$ equal \begin{bmatrix} a_{ij} \end{bmatrix} \begin{bmatrix} b_{ij} \end{bmatrix} \begin{bmatrix} c_{ij} \end{bmatrix} \begin{bmatrix} d_{ij} \end{bmatrix}

Respectively. I then used the distributive laws to show that both side are equal, but I cannot think of another way.

Answer 1

You can use the distributive property of matrices to show this the second way.

Observe that $(A+B)$ and $(C+D)$ are each themselves $n\times n$ matrices. Using right distribution, $(A+B)(C+D) = (A+B)C + (A+B)D$. Then distributing to the left of each term, we obtain the desired result.