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KatClaire
 2 years ago
Suppose A, B and X are invertible matrices such that BinverseXA = AB:
Find an expression for X in terms of A and B.
KatClaire
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0What does it mean that it is invertible

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

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

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

joemath314159
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1yes, multiply both sides by \(B^{1}\) and what do you get?

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

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

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

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

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

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

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

TuringTest
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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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