Vtyjkyuk

It is given that A is a matrix with dimensions m x n, and C is a matrix with dimensions r x s.

In order for the product ABC to be defined, I assumed that matrix B has dimensions n x r.

I am now asked to write an expression for the (i,j) element of ABC in terms of only the elements of matrices A, B, and C. I don't know how to begin. It's not easy for me to wrap my head around the fact that the matrices have unknown dimensions respresented by variables. If they were defined, I could deduce an expression based off of matrix multiplication rules. Any tips on how to proceed?

Let $A$ be a $m$-by-$n$ matrix, $B$ a $n$-by-$p$ matrix, and $C$ a $p$-by-$r$ matrix. I'm sure that you're aware that the entry in the $j$th row and $k$th column of $AB$ is given by
$$(AB)_j,k = sum_r=1^n A_j,r B_r,k.$$
Notice this is the product of the $(j,r)$th entry of $A$ and $(r,k)$th entry of $B$. To represent the $(j,k)$th entry of $(AB)C$, we do so by taking the product of the $(j,r)$th entry of $AB$ and $(r,k)$th entry of $C$. The main difference is that our sum is no longer to $n$, but rather to $p$ (can you tell why)?

We then have the following (in a spoiler):

$$((AB)C)_j,k = sum_r=1^p (AB)_j,rC_r,k = sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k. $$

Also, as a quick extra tidbit, technically it's improper to call it quits after the above work. How do you know for instance that $(AB)C=ABC$? What if $A(BC) neq (AB)C$? Which one then would denote $ABC$?

Lucky for us, they're the same! Of course that takes a proof. Can you see how to rearrange the above sum to also show that $(AB)C=A(BC)$? (That is, matrix multiplication is associative!) Here is that in a spoiler:

beginalign* sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k &= sum_r=1^p left( sum_s=1^n A_j,s B_s,r C_r,k right) \ &= sum_s=1^n A_j,s left( sum_r=1^p B_s,r C_r,k right) \ &= sum_s=1^nA_j,s (BC)_s,k\ & = (A(BC))_j,k. endalign*