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I want to ask a question about the associative property of matrices.

I was given an introductory course into vectors, specifically relating to matrices and then watched a video on 3Blue1Brown.

In his video, he gives a proof that the order of multiplying matrices matter.

He shows a shear and then a rotation in two different orders and shows that the basis vectors $i$ and $j$ do not match up.

$$M_1M_2 \neq M_2M_1$$

This made sense to me through his animations.

However, he later gives a proof for matrix multiplication being associative i.e.

$$(AB)C = A(BC)$$

and this didn't make sense to me.

I thought in both cases he was doing matrix multiplication, and he says in his first proof that order does matter, and then says that they're associative i.e. the order doesn't matter - and this has left me confused.

What am I missing here that's leaving me confused?

vik1245
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  • How does it not make sense, you just linked a proof of it? The proof for the associative property of matrix multiplication is easy to find and you should be able to make a counterexample for commutativity. –  Oct 10 '19 at 21:56
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    The phrase "order matters" is being used in two different senses. – mr_e_man Oct 10 '19 at 21:57
  • @mr_e_man can you explain this further please? – vik1245 Oct 10 '19 at 22:27
  • The two senses are commutativity and associativity. Another example of the difference between these (apart from matrix multiplication) is distance: $|x-y|=|y-x|$ (commutative), but $||x-y|-z|\neq|x-|y-z||$ (not associative). – mr_e_man Oct 10 '19 at 22:38

2 Answers2

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Order does matter, in that matrix multiplication is not commutative: $$AB \neq BA, \text{in general}.$$ It is easy to come up with examples to show this. Most choices of matrices will do the trick, just avoid multiples of the identity, etc.

However, order does not matter in that matrix multiplication is associative: $$ A(BC) = (AB)C.$$ That said, in proving this, you cannot assume the result. You have to assume order does matter until proven otherwise.

This is all summarized neatly in the observation that $\text{GL}_n(\mathbb{C})$ is a non-abelian group under multiplication, but don't worry if you have not come across these objects/terms yet.

DanLewis3264
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The order in which the matrices appear matters. But given that order, the order in which we multiply them in pairs does not. (To put it another way, the order of operations is irrelevant.) In other words, you don't need brackets to clarify what $ABC$ means, but you do need to avoid conflating $AB$ with $BA$.

J.G.
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