Solve an Equation of Matrices Using Inversions

[BraedenP]

Homework Statement

(C-CB)^{-1}=B^{-1}E

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

Homework Equations

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

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:
C^{-1}-C^{-1}B^{-1}=B^{-1}E

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

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

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

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!

First I applied the inversion to everything inside the function:

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:

(C-CB)=E^{-1}B

They then somehow jump to:

C=(C+E^{-1})B

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]

Hi Braeden!

it's ordinary algebra …

C - CB = E^-1B

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

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, let's change the same question a bit. find the solution where no inverse term appears.

 

[BraedenP]

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

Okay, that just has me totally confused... I don't even know where to begin to now remove the E^{-1} 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]

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 ?

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?

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

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 …

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

(C-CB)=E^{-1}B

… 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.

.. 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.

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 E^{-1} 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 :zzz: …)

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?

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).