# For any two elements A and B that form AB, neither A nor B have to be

1. Nov 4, 2009

### Gear300

For any two elements A and B that form AB, neither A nor B have to be invertible for (AB)-1 to exist, right?

2. Nov 4, 2009

### defunc

Re: Inverses

Both have to be invertible. Inv(AB) = inv(B)inv(A).

3. Nov 4, 2009

### Staff: Mentor

Re: Inverses

Right. For example..
$$\left[ \begin{array}{c c c} 1&0&0\\ 0&1&0 \end{array} \right] \left[ \begin{array}{c c} 1&0\\ 0&1\\ 0&0 \end{array} \right]~=~\left[ \begin{array}{c c} 1&0\\ 0&1\\ \end{array} \right]$$

The matrix on the right is clearly invertible, while the two matrices in the product aren't event square, let alone invertible.

4. Nov 4, 2009

### Staff: Mentor

Re: Inverses

See my counterexample.

5. Nov 4, 2009

### Gear300

Re: Inverses

Thanks for the reply. So that would mean that Inv(AB) = inv(B)inv(A) iff inv(B) and inv(A) exist (whereas Inv(AB) may exist without a defined inv(A) or inv(B)), right?

6. Nov 4, 2009

### pbandjay

Re: Inverses

Looks correct to me. The condition that I am familiar with is:

If A and B are invertible matrices of the same size, then AB is invertible and (AB)-1 = B-1A-1.

This is easy to prove by showing that (AB)(B-1A-1) = A(BB-1)A-1 = AIA-1 = AA-1 = I.

As far as the other way of your "iff", if (AB)-1 = B-1A-1 is given then it seems to me that the existence of B-1 and A-1 would directly follow since they are used in the initial condition.

7. Nov 5, 2009

### Gear300

Re: Inverses

I see. Thanks again for the replies. I have an additional question:

For a matrix C that is not a square matrix, there is no defined inverse; however, it is possible that there is a left inverse A and a right inverse B, in which A =/= B, for the matrix C, right?

8. Nov 5, 2009

### Staff: Mentor

Re: Inverses

Yes. See the wikipedia article here, under the section titled Matrices.

9. Nov 5, 2009

Re: Inverses