Finding a Solution for Invertible Matrices

[The Student]

If A, B and C are n x n invertible matrices, does the equation C^(-1)(A + X)B^(-1) = I_n. If so, find it.

I think that I got the first part of the question right which is X = CB - A. But I have no idea how to find the solution from the information given in the question.

 

[JohnZ]

you need to know some basic operations of matrices, like multiplication and addition.

 

[HallsofIvy]

If A, B and C are n x n invertible matrices, does the equation C^(-1)(A + X)B^(-1) = I_n. If so, find it.

I think that I got the first part of the question right which is X = CB - A. But I have no idea how to find the solution from the information given in the question.

I have no idea what you are saying here. I would think the answer to "does the equation ...." would be "yes" or "no". Then you say "If so, find it". Find what?

Perhaps it was "does the equation have a solution" and "if so find the solution". Starting from \(\displaystyle C^{-1}(A+ X)B^{-1}= I_n\), multiply both sides of the equation, (on the left) by C and multiply both sides of the equation (on the right) by B. What do get? Do you understand what I mean by "on the left" and "on the right"?

 

[The Student]

I have no idea what you are saying here. I would think the answer to "does the equation ...." would be "yes" or "no". Then you say "If so, find it". Find what?

Perhaps it was "does the equation have a solution" and "if so find the solution".

Oh god, you're absolutely right; I left out "have a solution, X" after the I_n.

Starting from \(\displaystyle C^{-1}(A+ X)B^{-1}= I_n\), multiply both sides of the equation, (on the left) by C and multiply both sides of the equation (on the right) by B. What do get? Do you understand what I mean by "on the left" and "on the right"?

I understand that order matters when multiplying matrices, so I will keep that in mind for the following work.

CC^(-1)(A + X)B^(-1)B = CI_nB

I_n(A + X)I_n = CB

AI_nI_n + XI_nI_n = CB

A + X = CB

X = CB - A

 

[HallsofIvy]

Oh god, you're absolutely right; I left out "have a solution, X" after the I_n.


I understand that order matters when multiplying matrices, so I will keep that in mind for the following work.

CC^(-1)(A + X)B^(-1)B = CI_nB

I_n(A + X)I_n = CB

AI_nI_n + XI_nI_n = CB

A + X = CB

X = CB - A

Yes, very good.