A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 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.
anonymous
 3 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.

This Question is Closed

anonymous
 3 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??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What does it mean that it is invertible

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

anonymous
 3 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

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

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

anonymous
 3 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.

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1yes, multiply both sides by \(B^{1}\) and what do you get?

anonymous
 3 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
 3 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

anonymous
 3 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?

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

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

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

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

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

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

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

TuringTest
 3 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
 3 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

anonymous
 3 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
 3 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.

anonymous
 3 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!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.