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

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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??
What does it mean that it is invertible

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Other answers:

determinant is not equal to zero
Does that mean I can move them around any special way? I'm so confused
say wuttt
I still don't know how to do this :(
is that from linear algebra class in college?
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.
Okay I know that but I still want to know if what I wrote up there is on the right track
yes, multiply both sides by \(B^{-1}\) and what do you get?
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
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
sorry my laptop decided to restart lol. So you're not allowed to rearrange the letters at all, right?
or where did you get the B to? on the right side
no, matrix multiplication is not commutative (cant move the letters)
oh I see, I misread the question, sorry
\[BX^{-1}A=AB\] all matrices are invertible solve for \(X\) correct?
I tried switching them around in different ways but nothing works lol
yeah, there must be some trick we're missing :/
Like there's no way to rearrange it so that BA=AB ?
only if BA and AB are inverses of each other, which I am trying to prove... so far unsuccessfully.
I mean if A and B are inverses of each other, then AB=BA=I
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!
This will continue to bother me, please let me know the answer when you find out :) Sorry I couldn't really help.
I really appreciate that you tried! I'll post the answer when it's up!

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