Linear Algebra ABC Proof

1. Feb 8, 2009

Sheneron

1. The problem statement, all variables and given/known data
Prove that if A,B,and C are square matrices and ABC = I, then B is invertible and B^-1 = CA.

3. The attempt at a solution

$$ABC = I$$
$$CABC = CI$$
$$CABC = C$$
$$CABCA = CA$$

so we have these two things:
$$(CAB)CA = CA$$
$$CA(BCA) = CA$$

so I thought that since CA times CAB = CA then CAB = I, and same for BCA. But, that is only true if the matrix is invertible, and the problem doesn't say whether C and A are invertible. Any suggestions? Thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 8, 2009

Staff: Mentor

If M and N are square matrices, and MN = I, then both M and N are invertible.

You're given that ABC = I, so A(BC) = I, and (AB)C = I, which says that A, BC, AB, and C are all invertible, and that A^(-1) = BC, and so on.

Your last equation is CA(BCA) = CA, which suggests to me that BCA = I, or that B(CA) = I. What's the relationship between B and CA?

3. Feb 9, 2009

Sheneron

Is an if you multiply invertible matrices together does it always yield an invertible matrix?

If so then CA(BCA) = CA implies that B(CA) = I, which means that B^-1=CA.

4. Feb 9, 2009

shaggymoods

Yes because det(AB) = det(A)*det(B) and if det(AB) is non-zero, then neither A nor B can have zero determinant, i.e., they must both be invertible. [if we're dealing with real or complex valued matrices at least]

5. Feb 9, 2009

Sheneron

Yes we are. Alright then, thank you both.