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 one year ago
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
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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KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0I 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??

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0What does it mean that it is invertible

modphysnoob
 one year ago
Best ResponseYou've already chosen the best response.0determinant is not equal to zero

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0Does that mean I can move them around any special way? I'm so confused

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0I still don't know how to do this :(

modphysnoob
 one year ago
Best ResponseYou've already chosen the best response.0is that from linear algebra class in college?

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.0If 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.

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0Okay I know that but I still want to know if what I wrote up there is on the right track

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1yes, multiply both sides by \(B^{1}\) and what do you get?

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0you 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

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1I'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

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0sorry my laptop decided to restart lol. So you're not allowed to rearrange the letters at all, right?

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0or where did you get the B to? on the right side

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1no, matrix multiplication is not commutative (cant move the letters)

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1oh I see, I misread the question, sorry

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1\[BX^{1}A=AB\] all matrices are invertible solve for \(X\) correct?

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0I tried switching them around in different ways but nothing works lol

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1yeah, there must be some trick we're missing :/

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0Like there's no way to rearrange it so that BA=AB ?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1only if BA and AB are inverses of each other, which I am trying to prove... so far unsuccessfully.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1I mean if A and B are inverses of each other, then AB=BA=I

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0I 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!

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1This will continue to bother me, please let me know the answer when you find out :) Sorry I couldn't really help.

KatClaire
 one year ago
Best ResponseYou've already chosen the best response.0I really appreciate that you tried! I'll post the answer when it's up!
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