Matrix Multiplication is Associative

Theorem
Matrix multiplication (conventional) is associative.

Proof

 * Let $\mathbf{A} = \left[{a}\right]_{m n}, \mathbf{B} = \left[{b}\right]_{n p}, \mathbf{C} = \left[{c}\right]_{p q}$ be matrices over a ring $\left({R, +, \circ}\right)$.

From inspection of the subscripts, we can see that both $\left({\mathbf{A} \mathbf{B}}\right) \mathbf{C}$ and $\mathbf{A} \left({\mathbf{B} \mathbf{C}}\right)$ are defined:

$\mathbf{A}$ has $n$ columns and $\mathbf{B}$ has $n$ rows, while $\mathbf{B}$ has $p$ columns and $\mathbf{C}$ has $p$ rows.


 * Consider $\left({\mathbf{A} \mathbf{B}}\right) \mathbf{C}$.

Let $\mathbf{R} = \left[{r}\right]_{m p} = \mathbf{A} \mathbf{B}, \mathbf{S} = \left[{s}\right]_{m q} = \left({\mathbf{A} \mathbf{B}}\right) \mathbf{C}$.

Then:

Thus $\displaystyle s_{i j} = \sum_{k=1}^p \sum_{l=1}^n a_{i l} \circ b_{l k} \circ c_{k j}$


 * Now consider $\mathbf{A} \left({\mathbf{B} \mathbf{C}}\right)$.

This time, let $\mathbf{R} = \left[{r}\right]_{n q} = \mathbf{B} \mathbf{C}, \mathbf{S} = \left[{s}\right]_{m q} = \mathbf{A} \left({\mathbf{B} \mathbf{C}}\right)$.

Then:

Thus $\displaystyle s_{i j} = \sum_{l=1}^n a_{i l} \circ \sum_{k=1}^p b_{l k} \circ c_{k j}$.


 * As addition is commutative, we can see that $s_{i j}$ is the same in both cases.

Hence $\left({\mathbf{A} \mathbf{B}}\right) \mathbf{C} = \mathbf{A} \left({\mathbf{B} \mathbf{C}}\right)$.