Question

In this exercise we show that the matrix multiplication is distribute over matrix addition.

a.) Suppose that AandB are m x kmatrices and that C is ak x nmatrix. Show that(A + B)C = AC + BC.

b.) Suppose that C is anm x kmatrix and thatAandBarek x n matrices. Show that C(A + B) = CA + CB .

Short answer

1. $(A+B)C=AC+BC$
   
2. $C\left(A+B\right)=CA+CB$

Step-by-step solution

Step 1

a. Given: A and B are m x k matrices C and is a k x n matrix.

To Prove: $\left(A+B\right)C=AC+BC$

PROOF

$\begin{array}{r}\left[\right(A+B)C{]}_{jj}=\sum _{a=1}^k \left(\left[\right(A+B\left){]}_{ia}\right[C{]}_{qj}\right)\\ =\sum _{a=1}^k \left(\left([A{]}_{ia}+[B{]}_{ij}\right)[C{]}_{qj}\right)\end{array}$

Distributive property real numbers

$\begin{array}{r}=\sum _{a=1}^k \left(\left[A{]}_{ia}\right[C{]}_{qj}+\left[B{]}_{ia}\right[C{]}_{aj}\right)\\ =\sum _{a=1}^k \left(\left[A{]}_{ia}\right[C{]}_{qj}\right)+\sum _{a=1}^k \left(\left[B{]}_{ia}\right[C{]}_{qj}\right)\\ =[AC{]}_{ij}+[BC{]}_{ij}\\ =[AC+BC{]}_{ij}\end{array}$

This then implies that every element of the matrix is identical to the corresponding element of the matrix , which implies that the two matrices are equal.

(A + B) C = AC + BC

Step 2

b. Given: A and B are k x n matrices and C is a m x k matrix.

To prove: C(A + B) = CA + CB

PROOF

$\begin{array}{r}\left[C\right(A+B){]}_{ij}=\sum _{a=1}^k [C{]}_{ia}\left(\left[\right(A+B){]}_{aj}\right)\\ =\sum _{a=1}^k \left([C{]}_{iz}\left([A{]}_{qj}+[B{]}_{qj}\right)\right)\\ =\sum _{a=1}^k \left(\left[C{]}_{iz}\right[A{]}_{qj}+\left[C{]}_{iz}\right[B{]}_{qj}\right)\\ =\sum _{a=1}^k \left(\left[C{]}_{iz}\right[A{]}_{qj}\right)+\sum _{a=1}^k \left(\left[C{]}_{iz}\right[B{]}_{qj}\right)\\ =[CA{]}_{\overrightarrow{j}}+[CB{]}_{ij}\\ =[CA+CB]j\end{array}$

This then implies that every element of the matrix C (A + B) is identical to the corresponding element of the matrix CA + CB , which implies that the two matrices are equal.

C(A + B) = CA + CB