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KatClaire

Suppose A, B and X are invertible matrices such that BinverseXA = AB: Find an expression for X in terms of A and B.

  • one year ago
  • one year ago

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  1. KatClaire
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    \[BX^{-1}A=AB\]

    • one year ago
  2. KatClaire
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    I have a question to see if I'm doing this right. Do I first times each side by the inverse of B to get rid of it then by X??

    • one year ago
  3. KatClaire
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    What does it mean that it is invertible

    • one year ago
  4. modphysnoob
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    determinant is not equal to zero

    • one year ago
  5. KatClaire
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    Does that mean I can move them around any special way? I'm so confused

    • one year ago
  6. modphysnoob
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    say wuttt

    • one year ago
  7. KatClaire
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    I still don't know how to do this :(

    • one year ago
  8. modphysnoob
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    is that from linear algebra class in college?

    • one year ago
  9. KatClaire
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    yes

    • one year ago
  10. joemath314159
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    If the matrix A is invertible, that means there exists a matrix\[A^{-1}\]such that:\[AA^{-1}=A^{-1}A=I\]where I is the identity matrix.

    • one year ago
  11. KatClaire
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    Okay I know that but I still want to know if what I wrote up there is on the right track

    • one year ago
  12. TuringTest
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    yes, multiply both sides by \(B^{-1}\) and what do you get?

    • one year ago
  13. KatClaire
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    you get \[x^{-1} A = B^{-1}AB\] so I went on and multiplied each side by X and then got \[A=XB^{-1}AB\] But I'm confused beacuse it doesn't look right

    • one year ago
  14. TuringTest
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    I'm having a little trouble too actually :/ while I work on it, I can say that you can use \(AB=B^{-1}AB\) to make that a little prettier, but I still can't solve for \(X\) yet

    • one year ago
  15. KatClaire
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    sorry my laptop decided to restart lol. So you're not allowed to rearrange the letters at all, right?

    • one year ago
  16. KatClaire
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    or where did you get the B to? on the right side

    • one year ago
  17. TuringTest
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    no, matrix multiplication is not commutative (cant move the letters)

    • one year ago
  18. TuringTest
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    oh I see, I misread the question, sorry

    • one year ago
  19. TuringTest
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    \[BX^{-1}A=AB\] all matrices are invertible solve for \(X\) correct?

    • one year ago
  20. KatClaire
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    Yup!

    • one year ago
  21. KatClaire
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    I tried switching them around in different ways but nothing works lol

    • one year ago
  22. TuringTest
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    yeah, there must be some trick we're missing :/

    • one year ago
  23. KatClaire
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    Like there's no way to rearrange it so that BA=AB ?

    • one year ago
  24. TuringTest
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    only if BA and AB are inverses of each other, which I am trying to prove... so far unsuccessfully.

    • one year ago
  25. TuringTest
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    I mean if A and B are inverses of each other, then AB=BA=I

    • one year ago
  26. KatClaire
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    I think I'll just leave it and wait for the solution to be posted online to understand it lol thanks a lot for your help though!

    • one year ago
  27. TuringTest
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    This will continue to bother me, please let me know the answer when you find out :) Sorry I couldn't really help.

    • one year ago
  28. KatClaire
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    I really appreciate that you tried! I'll post the answer when it's up!

    • one year ago
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