# Prove: A(BC)=(AB)C for matrix multiplication

1. Jul 27, 2008

### adartsesirhc

1. The problem statement, all variables and given/known data
Prove the following theorem:
A(BC)=(AB)C.

2. Relevant equations
Definition of matrix multiplication

3. The attempt at a solution
Let A be of order m by n, B be of order n by p, and C be of order p by q. Then
$$(A(BC))_{ij}=\sum^{n}_{k=1} A_{ik}(BC)_{kj}=\sum^{n}_{k=1} (A_{ik}(\sum^{p}_{r=1} B_{kr}C_{rj}))$$
After this, I'm stuck. How do I put the summations together, and how should I change the indices of summation?

Thanks!

2. Jul 27, 2008

### tiny-tim

Hi adartsesirhc!

(nice LaTeX, btw!)

Hint: You can move the ∑s as far to the left as you like.

3. Jul 27, 2008

### adartsesirhc

$$(A(BC))_{ij}=\sum^{n}_{k=1} A_{ik}(BC)_{kj}=\sum^{n}_{k=1} (A_{ik}(\sum^{p}_{r=1} B_{kr}C_{rj}))=\sum^{n}_{k=1} \sum^{p}_{r=1} A_{ik}B_{kr}C_{rj}=\sum^{p}_{r=1} (\sum^{n}_{k=1} A_{ik}B_{kr})C_{rj}=\sum^{p}_{r=1} (AB)_{ir}C_{rj}=((AB)C)_{ij}$$.

How's that?

4. Jul 27, 2008

### tiny-tim

Perfect!
Woohoo!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?