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I'm working on an independent project where I am attempting to find what rotation will result in transforming from one Euclidean space to another, where they have the same origin. Assumption is that such a transform exists. I was having some trouble (haven't used my matrix algebra in a while), and I was assuming:

$$ AB = I $$ $$ A = IB^{-1} $$ $$ A = B^{-1}I $$ $$ BA = I $$ thus: $$ AB = BA $$ Since I was having trouble, I was relooking at the matrix operations and saw that this is not true in general.

In general, $AB \ne BA$, even if $A$ and $B$ are both square. If $AB = BA$, then we say that $A$ and $B$ commute.

Am I correct in my matrix operation assumptions?

Edit

@Onir brings up an example showing that they are not, however I'm confused as to what operation that I started above is invalid. Can someone explain?

Adrian
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    You proved that a matrix commutes with its inverse (which is true), but not that matrices commute in general (which is false). – eyeballfrog May 23 '21 at 13:55
  • in $\mathbb R^2$ rotating $\theta$ degrees and then reflecting over the $x$ axis is different to reflecting over the $x$ axis and then rotating $\theta$ degrees except when $\theta$ is $0$ or $180$ – Asinomás May 23 '21 at 13:57
  • @Onir, yeah, actually, your right. That is the same with translations and rotations. Or even two rotations on different rotation axes. But if there is an inverse matrix defined for A and B, then why are my operations above invalid? What rule am I violating? Can you only multiply an inverse on the right? – Adrian May 23 '21 at 21:59
  • I second eyeballfrog's comment – Asinomás May 23 '21 at 22:03
  • Oh. I understand now. I've not broken anything and my results are still valid. Thx. – Adrian May 23 '21 at 23:11

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So, according to the comments there doesn't seem to be anything wrong with my operations. All that I did was show the community of two matrices one of which is the inverse of the other. It being inverse is implied. This of course is not a general solution for all matrices. I cannot for instance show that $ABC=CBA$ because either $AB$ or $BC$ must appear in that order in whatever valid symbolic transformation I come up with.

Adrian
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