Solve an Equation of Matrices Using Inversions

BraedenP

1. The problem statement, all variables and given/known data
[tex](C-CB)^{-1}=B^{-1}E[/tex]

Solve the system for B, with the assumption that C,B, and (C-CB) are invertible.

2. Relevant equations

The rules for matrix invertibility (but I've already come to the conclusion that all matrices in this equation are invertible.

3. The attempt at a solution

I attempted to get a solution, but I don't think it's correct:

First I applied the inversion to everything inside the function:
[tex]C^{-1}-C^{-1}B^{-1}=B^{-1}E[/tex]

Then I multiplied both sides by E, to cancel out the inverse on the right side:
[tex]C^{-1}B-C^{-1}=E[/tex]

Then I moved the [tex]C^{-1}[/tex] term to the right-hand side:
[tex]C^{-1}B=E+C^{-1}[/tex]

Finally, I multiplied both sides by [tex]C^{-1}[/tex] to isolate X:
[tex]B=CE[/tex]

This is the solution I got to, but it doesn't seem right. Have I missed anything, made an error in an assumption or calculation, or have I taken a completely wrong direction?

Any help would be greatly appreciated!

Thanks,
Braeden

tiny-tim

Hi Braeden!
Nope!

1/(3 + 5) isn't 1/3 + 1/5, and it doesn't work for matrices either …

(and btw, (CB)^-1 = B^-1C^-1, not C^-1B^-1)

start again, and try multiplying by something

gau_physics

in such questions, its better to use the rule that :- (A)(A^-1)=I (identity matrix)

BraedenP

Okay.. Thanks, guys -- but for some reason I still can't get it. I can get answers, but none of them are what the solution says is the answer.

The solution first takes the inverse of everything, resulting in:

[tex](C-CB)=E^{-1}B[/tex]

They then somehow jump to:

[tex]C=(C+E^{-1})B[/tex]

That's the step I'm not sure about.. Could you explain that step?

tiny-tim

Hi Braeden!!

it's ordinary algebra …

C - CB = E^-1B

so C = CB + E^-1B = (C + E^-1)B

BraedenP

Oh! Okay.. I'm just confused about the operations that are allowed on matrices, and which ones aren't.

But that makes perfect sense! Then to isolate and solve for B I just take the inverse of both sides, right?

gau_physics

Guys, lets change the same question a bit. find the solution where no inverse term appears.

BraedenP

Okay, that just has me totally confused... I don't even know where to begin to now remove the [tex]E^{-1}[/tex] term... I could get it out of the right-hand side, but then I'm going to have CE on the left-hand side..

gau_physics

C=(C + E^-1)B

C=C.B + E^-1.B

C(1-B)=E^-1.B

Multiplying both sides by E^-1 ;
EC(1-B)=B

Is it right ?

gau_physics

Also remember C.E and E.C are not the same.

gau_physics

Its better Braeden that you keep a list of all algebraic operations on matrices and formulae on a page while solving coz matrix algebra is confusing

BraedenP

Okay, that's right, but it still doesn't solve for B.. In order to isolate B, I will need an inverse, won't I?

And yeah.. Our textbook doesn't provide a concrete list of allowed operations and rules regarding the operations, so I've been scraping it off of websites and stuff, which could be adding to my confusion.

tiny-tim

Hi Braeden!

Looking back, it occurs to me that this was wrong …
… because the question does not tell you that E is invertible (ie, we're not told that E^-1 exists, we'll have to prove that later).

Start again with the original (C - CB)^-1 = B^-1E,

and get rid of the two inverses by multiplying them out.
Matrices have the usual rules for addition and multiplication, including use of brackets.

They don't have division … you can't divide by a matrix (though, if its inverse exists, you can of course multiply by that inverse, which has the same effect ).

BraedenP

Oh, sorry.. I forgot to mention that, in the previous question, we were asked to first prove that [tex]E^{-1}[/tex] is, in fact, invertible.

And yeah.. I used your method to reach the answer. It worked perfectly! Thanks!

Also, regarding the arithmetic rules:

Basically, addition is easy, but would something like this hold true?

B+C=C+A therefore B=A or is it more complex than that?

Thanks again for all of your help -- I appreciate it!

tiny-tim

Hi Braeden!

(just got up …)
Yes.

It doesn't work for multiplication of matrices, because multiplication isn't commutative (AB ≠ BA), but it works for addition, because addition is commutative (A + B = B + A).