# Matrix Notation and Matrix Multiplication

• February 8th 2011, 08:05 PM
Erich
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.
• February 8th 2011, 08:07 PM
Ackbeet
Could you please insert some punctuation into your big equation there? It's impossible to parse as is.
• February 8th 2011, 08:10 PM
Erich
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.
• February 9th 2011, 07:17 AM
Ackbeet
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?
• February 9th 2011, 07:33 AM
Erich
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!!
• February 9th 2011, 07:36 AM
Ackbeet
You're welcome. Have a good one!
• February 10th 2011, 08:55 PM
Erich
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.
• February 11th 2011, 04:57 AM
Ackbeet
You're welcome!