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

[adartsesirhc]

Homework Statement

Prove the following theorem:
A(BC)=(AB)C.

Homework Equations

http://en.wikipedia.org/wiki/Matrix_multiplication#Ordinary_matrix_product"

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!

 

[tiny-tim]

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?

Hi adartsesirhc!

(nice LaTeX, btw!)

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

 

[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?

 

[tiny-tim]

(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?

Perfect!

Woohoo! ​