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


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


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

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