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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?

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  1. ganeshie8
    • one year ago
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    you're given \[\large AB^{-1} = B^{-1}A\]

  2. ganeshie8
    • one year ago
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    left multiply \(\large \color{red}{B}\) both sides to get : \[\large \color{red}{B}AB^{-1} = \color{red}{B}B^{-1}A\]

  3. ganeshie8
    • one year ago
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    that becomes \[\large \color{red}{B}AB^{-1} = IA\] which is same as \[\large \color{red}{B}AB^{-1} = A\]

  4. ganeshie8
    • one year ago
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    next, right multiply both sides by \(\color{red}{B}\)

  5. clarence
    • one year ago
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    Then BAB^-1B = AB which would be the same as BAI = AB and thus BA = AB

  6. Kainui
    • one year ago
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    Awesome, 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 \(A\).

  7. ganeshie8
    • one year ago
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    interesting, so if \(x\) commutes with \(y\) and \(z\) does that necessarily mean \(y\) commutes with \(z\) ?

  8. Kainui
    • one year ago
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    You 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.

  9. ganeshie8
    • one year ago
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    i dont know haha, i think it is true only if one of \(y\) or \(z\) is the inverse of \(x\)

  10. Kainui
    • one year ago
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    I have to prove this or find a specific counter example now haha. Hmmmm

  11. ganeshie8
    • one year ago
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    given : ``` xy = yx xz = zx ``` prove/disprove : ``` yz = zy ```

  12. Kainui
    • one year ago
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    I'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

  13. ganeshie8
    • one year ago
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    nice one ! xD

  14. ganeshie8
    • one year ago
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    basically 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 :)

  15. Kainui
    • one year ago
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    Yeah 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.

  16. anonymous
    • one year ago
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    Nice explanation, your analogies are one of a kind @Kainui

  17. Kainui
    • one year ago
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    Yeah, 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 #_#;;

  18. clarence
    • one year ago
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    And 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!

  19. Kainui
    • one year ago
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    You 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.

  20. clarence
    • one year ago
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    Oh I see, I understand now! :) Aha, I really did quite like that example of yours though so thanks for that

  21. Kainui
    • one year ago
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    Awesome! Yeah you can find less weird examples of commuting too. For instance, in 3D we can say: A = rotation around x-axis B = stretching C = rotation around y-axis 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/noncommuting-rotations.png That's the picture I have in mind when I say this.

  22. clarence
    • one year ago
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    Aha, 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...

  23. Kainui
    • one year ago
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    Nah! 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

  24. Kainui
    • one year ago
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    But 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

  25. anonymous
    • one year ago
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    Lol what, the earth is flat

  26. Kainui
    • one year ago
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    yeah the world is rotating in the plane tho

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