# Invertibility of the product of matrices

1. Sep 22, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
Let A and B be n by n matrices such that A is invertible and B is not invertible.
Then, AB is not invertible.

2. Relevant equations

3. The attempt at a solution

We know that A is invertible, so there exists a matrix C such that CA = I. Then we can right -multiply by B so that CAB = IB = I. Then by the associative property C(AB) = I. By the same argument, we can show that there is a C such that (AB)C = I. So AB has an inverse.

Obviously this is wrong, because in order for AB to have an inverse, both A and B must have an inverse. So what am I doing wrong?

2. Sep 22, 2016

### cragar

CAB=IB but you cant conclude IB=I, IB=B

3. Sep 22, 2016

### Staff: Mentor

Maybe you should concentrate on the non invertible part. What does it mean to B, not being invertible? Is there a positive property, i.e. without the use of non, not or no?

4. Sep 22, 2016

### Ray Vickson

What tools/results are you allowed to use? Do you know about determinants? Do you know how determinants relate to the invertability/non-invertability of a matrix?

5. Sep 25, 2016

### micromass

Staff Emeritus
Assume that $AB$ is invertible. This means that there is a $C$ such that $CAB = I$ and $ABC = I$. Can you prove now that $B$ is invertible? (and thus deriving a contradiction).