# Thread: Matrix Notation and Matrix Multiplication

1. ## Matrix Notation and Matrix Multiplication

If A and B are n by n matrices with all entries equal to 1, find $\ (ab)_{ij}$. Summation notation turns the product AB, and the law (AB)C=A(BC), into
$\\ (ab)_{ij}= \displaystyle \sum_{k}\ a_{ik}b_{kj}\ \displaystyle \sum_{j}\left(\sum_{k}\ a_{ik}b_{kj}\right)\ c_{jl}\ =\displaystyle \sum_{k}\ a_{ik}\left(\sum_j\ b_{kj}c_{il}\right)$
Compute both sides if C is also n by n with every $\ c_{jl}=2$
I haven't even done any work on this yet, because first I don't even understand what the question wants, and second I don't even know where to begin.

2. Could you please insert some punctuation into your big equation there? It's impossible to parse as is.

3. That's how it's written in my book. I am not really after a solution, more like am more interested in how I could start working this out. What I think that this question is basically asking, is for a proof of the associative property using the above formulas, but I don't know where to start.

4. It looks like you're asked to assume that $A_{ij}=B_{ij}=1$ for all $i,j,$ and $C_{ij}=2$ for all $i,j$. Then you're asked to compute $(AB)_{ij},$ and show that $((AB)C)_{ij}=(A(BC))_{ij}.$

For the first, you have that

$\displaystyle (AB)_{ij}=\sum_{k=1}^{n}A_{ik}B_{kj}=\sum_{k=1}^{n }(1\cdot 1)=n.$

For the second, plug into your formula:

$\displaystyle((AB)C)_{il}=\sum_{j}\left(\sum_{k}\ a_{ik}b_{kj}\right)\ c_{jl}\ \overset{?}{=}\sum_{k}\ a_{ik}\left(\sum_j\ b_{kj}c_{jl}\right)=(A(BC))_{il},$ yielding the test

$\displaystyle((AB)C)_{il}=\sum_{j}\left(\sum_{k}1\ right)2\overset{?}{=}\sum_{k}1\left(\sum_j 2\right)=(A(BC))_{il},$ which yields

$\displaystyle((AB)C)_{il}=\sum_{j}2n\overset{?}{=} \sum_{k}1\cdot 2n=(A(BC))_{il}.$

Can you finish?

5. If all I needed to do was compute both sides I believe you've already finished. I could probably even build a 2x2 that supports this. Thank you!!

6. You're welcome. Have a good one!

7. Thought I would let you know Ackbeet that I found the solution to be 2n^2, which made so much more sense after looking at your work, and learned something new about summations and matrices. I would like to thank you once again for giving me a new perspective between proofs, associative law, and computations with summations and matrices.

8. You're welcome!