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KatClaire
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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.
 one year ago
 one year ago
KatClaire Group Title
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

This Question is Closed

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
\[BX^{1}A=AB\]
 one year ago

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
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

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

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

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

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
say wuttt
 one year ago

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

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

joemath314159 Group TitleBest ResponseYou've already chosen the best response.0
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

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

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

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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

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

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

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

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

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

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

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

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

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

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

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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

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