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Clarence
 one year ago
Just looking for confirmation for the following statement:
If matrix B is invertible and AB^1 = B^1A then AB = BA.
That should be True?
Clarence
 one year ago
Just looking for confirmation for the following statement: If matrix B is invertible and AB^1 = B^1A then AB = BA. That should be True?

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2you're given \[\large AB^{1} = B^{1}A\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2left multiply \(\large \color{red}{B}\) both sides to get : \[\large \color{red}{B}AB^{1} = \color{red}{B}B^{1}A\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that becomes \[\large \color{red}{B}AB^{1} = IA\] which is same as \[\large \color{red}{B}AB^{1} = A\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2next, right multiply both sides by \(\color{red}{B}\)

clarence
 one year ago
Best ResponseYou've already chosen the best response.1Then BAB^1B = AB which would be the same as BAI = AB and thus BA = AB

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Awesome, this is really the best answer but I thought I would throw in a little extra for you @Clarence A matrix always commutes with its inverse, \[BB^{1} = B^{1}B\] So you can kinda make a little shortcut if you know that \(A\) and \(B^{1}\) commute, then it's basically the transitive rule I guess you could call it: \(B\) commutes with \(B^{1}\) and \(B^{1}\) commutes with \(A\) therefore \(B\) commutes with \(A\).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2interesting, so if \(x\) commutes with \(y\) and \(z\) does that necessarily mean \(y\) commutes with \(z\) ?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2You know what I might be totally wrong here. if AB=BA and BC=CB does this mean AC=CA? I don't actually think we can show this is true after all.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2i dont know haha, i think it is true only if one of \(y\) or \(z\) is the inverse of \(x\)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I have to prove this or find a specific counter example now haha. Hmmmm

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2given : ``` xy = yx xz = zx ``` prove/disprove : ``` yz = zy ```

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I've found a counter example. x = putting on a hat y = putting on shoes z = putting on socks Reading xy left to right means put on hat first then put on shoes second. yx means put on your shoes first and then put your hat on second. xy=yx makes perfect sense, both result in the same outcome. They commute. Also we can clearly see xz=zx Now here comes the good part: \(yz \ne zy\) Right? Clearly putting your shoes on first and then your socks is not the same outcome as putting on your socks first and then your shoes! QED

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2basically i can do the tasks \(x,y\) and \(x,z\) in any order but il get different results if i change the order of \(y,z\) thats really a quick nice example :)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah simple to understand and fun. it's sorta how commuting was explained to me when I took quantum mechanics and it sorta stuck with me the idea of putting on socks and shoes in different orders haha.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nice explanation, your analogies are one of a kind @Kainui

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, they start out with me saying completely false information as fact and then realizing I'm totally wrong and forced to come up with some nifty response #_#;;

clarence
 one year ago
Best ResponseYou've already chosen the best response.1And I'm left wondering whether or not I was right in the first place... But I do understand that particular example and thoroughly enjoyed it so thanks for that!

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2You were right in the first place, you just can't use that "trick" I was describing. Everything before I posted is completely right. The first thing I said was false and after that was us figuring out that I was false basically haha.

clarence
 one year ago
Best ResponseYou've already chosen the best response.1Oh I see, I understand now! :) Aha, I really did quite like that example of yours though so thanks for that

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Awesome! Yeah you can find less weird examples of commuting too. For instance, in 3D we can say: A = rotation around xaxis B = stretching C = rotation around yaxis So A and B commute, B and C commute, but that doesn't mean A and C commute! In fact they don't, you can try this out by playing with a book! http://www.lightandmatter.com/html_books/genrel/ch07/figs/noncommutingrotations.png That's the picture I have in mind when I say this.

clarence
 one year ago
Best ResponseYou've already chosen the best response.1Aha, that sure is an awesome way to think about it in 3D! Though 3D questions in mechanics is confusing as it is already, especially with moments and stuff, but I'm pretty sure that's a bad thing to hear coming from an engineering freshman...

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Nah! You have to develop it and you will over time. Just keep in mind the physical picture is the important aspect and the numbers are just capturing the observed reality in a precise way that words can't do as well. In a way you can think of math as being a very precise form of language. But it also feels like so much more than that so I don't know haha. I think everyone gets bothered by rotational motion anyways that crap's hard haha

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2But yeah I called it crap but rotational motion is probably one of the most important things. I would really hate not having night time if the world didn't rotate. I think it's definitely worth trying to understand! haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol what, the earth is flat

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2yeah the world is rotating in the plane tho
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