Show if A,B,C are invertible matrices of same size

[Saladsamurai]

Show if A,B,C are invertible matrices of same size...

I know, I know. I should be awesome at these by now...

Homework Statement

Show if A,B,C are invertible matrices of the same size, than $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$

The Attempt at a Solution

Given some matrix A,$AA^{-1}=I$
If:
$AA^{-1}=I$
$BB^{-1}=I$
$CC^{-1}=I$

I am not sure where to go from here. I don't think I have any more definitions or product rules to incorporate.

It almost seems as if I would FIRST have to show that (ABC) is invertible to begin with. Then I can use the fact that (ABC)(ABC)^{-1}=I to discover how (ABC)^{-1} MUST be arranged in order for the product of the two to yield I.

Does that sound like a good place to start? Proving if A,B, and C are invertible, then (ABC) is too?

 

[rock.freak667]

Let $D=(ABC)^{-1}$
$\times (ABC)$
$\Rightarrow ABCD=I$

and then just multiply by $A^{-1}$ and so forth

 

[Saladsamurai]

Let $D=(ABC)^{-1}$
$\times (ABC)$
$\Rightarrow ABCD=I$

and then just multiply by $A^{-1}$ and so forth

I don't follow? What does the second line mean?--->$\times (ABC)$

 

[rock.freak667]

I don't follow? What does the second line mean?--->$\times (ABC)$

Multiply both sides by the matric ABC

 

[Saladsamurai]

So
$D=(ABC)^{-1}$
$\Rightarrow D(ABC)=(ABC)^{-1}(ABC)$

Well, I think I see where this is going. And I think that the only reason this works is because we are assuming that the product (ABC) IS invertible. Which brings me back to my original point. In order to show how the multiplication MUST be carried out, we must first SHOW or assume without proof that (ABC) is in fact invertible since our argument will be based on the fact that (ABC)*(ABC)^{-1}=Id.

 

[rock.freak667]

Well if A,B,C are nxn matrices then ABC is an nxn matrix.

and for the matrix ABC to be invertible $det(ABC) \neq 0$

and det(ABC)=det(A)*det(B)*det(C)

but the matrices A,B,C are invertible.

 

[Defennder]

Another way you could show that a product of two matrices A and B are invertible is by showing that there exists some matrix which when multiplied to AB on the left and on the right gives the identity matrix:

Suppose A and B are invertible, then:

$$AB(B^{-1}A^{-1}) = I$$ for multiplying on the right
$$B^{-1}A^{-1}AB = I$$ for multiplying on the left.

In both cases this reduces to I, so $$B^{-1}A^{-1}$$ is the inverse of AB.

Now make use of this result to prove your question.

 

[HallsofIvy]

It is not nnecessary to assume that ABC is invertible.

You are given that A, B, C separately are invertible so A^-1, B^-1, and C^-1 exist. Thus C^-1B^-1A^-1 exists. What do you get if you multiply (ABC)(C^-1B^-1C^-1) and (C^-1B^-1A^-1(ABC)?

That will prove that ABC is invertible.

 

[matt grime]

The same observation I wrote in your other thread about orthogonal matrices applies here.

Let's start with the definitions:

X is invertible if there is a Y with XY=YX=Id.

It isn't that I want to show ABC is invertible, but someone's been really helpful and just asked me to verify what the inverse is! I don't have to think at all, I just have to chuck the nominal inverse into the definition and see what happens. So do it...

 

[Defennder]

Hey yeah, I didn't see it that way. Makes it a lot easier.

 

[JinM]

This looks like a question that could spring up on my midterm tomorrow. Assuming invertibility, does this prove the result?

$$ABCC^{-1}B^{-1}A^{-1} = A(BI_{n})B^{-1}A^{-1}= A(BB^{-1})A^{-1} = AA^{-1} = I_n$$

Now, the other side,

$$C^{-1}B^{-1}A^{-1}ABC = C^{-1}B^{-1}BC = C^{-1}C = I_n$$

Thus $$(ABC)(C^-1B^-1A^-1}) = (C^-1B^-1A^-1)(ABC) = I_n$$.

Is this enough? No claim of originality, this was how a result was proved in my notes, and what everyone is getting at!

 

[Defennder]

That's basically what HallsOfIvy said.

 

[Saladsamurai]

I don't like it. I don't know why yet...but I just don't.

 

[matt grime]

You aren't assuming invertibility of ABC. You're just multiplying together 6 matrices, and showing that they give the identity, which is *proving* it is invertible. Just because you wrote "assuming invertibility of ABC" doen't mean you actually did assume it, or that you needed toi.

 

[b0it0i]

i think the problem you're having with this proof is understanding what the question is asking.
there are many other proofs with the same structure

the problem is saying show that

(ABC)-1 = C-1B-1A-1

in other words, they want you to show that the inverse of ABC is actually (behaves like) inverse C times inverse B times inverse A

an example would be... Show that

A Source of water = Rain

you must show that rain acts/ behaves like a source of water

apply the properties of "a source of water" to rain

now back to your problem...

Show that C-1B-1A-1 "behaves" like the (ABC)-1
how do you do that?... you check the property of (ABC)-1

that is... show that C-1B-1A-1 (ABC) = (ABC) C-1B-1A-1 = I

An important fact is to note that "the inverse of a matrix is unique"
since C-1B-1A-1 "behaves" like the inverse of ABC,
and that the inverse of a matrix is unique,
C-1B-1A-1 MUST be (ABC)-1

this is corny, but my past professor constantly called these types of proofs
"walks like a duck, quacks like a duck, MUST be a duck..." proof

 

[Saladsamurai]

Thanks people! I like it now