Inverse question (matrices)

[TsAmE]

Let A and B be invertible n x n matrices.

Show that \(\displaystyle \mathbf{AB}\) is invertible, with \(\displaystyle \mathbf{(AB)^{-1}} = \mathbf{B^{-1}A^{-1}}\)

I am clueless with this question and dont know where to start.

 

[Deleted member 4993]

Let A and B be invertible n x n matrices.

Show that \(\displaystyle \mathbf{AB}\) is invertible, with \(\displaystyle \mathbf{(AB)^{-1}} = \mathbf{B^{-1}A^{-1}}\)

I am clueless with this question and dont know where to start.

Try pre and post multiplying with AB - and tell us what you find.

We are clueless as to what do you know.

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[TsAmE]

Pre multiplying with AB:

\(\displaystyle \mathbf{(AB)(AB)^{-1}} = \mathbf{(AB)B^{-1}A^{-1}}\)

\(\displaystyle 1 = \mathbf{(AB)B^{-1}A^{-1}}\)

Post multiplying with AB:

\(\displaystyle \mathbf{(AB)^{-1}(AB)} = \mathbf{ B^{-1}A^{-1}(AB)}\)

\(\displaystyle 1 = \mathbf{B^{-1}A^{-1}(AB) }\)

Is this right? What next are you suppose to do?

 

[Deleted member 4993]

Pre multiplying with AB:

\(\displaystyle \mathbf{(AB)(AB)^{-1}} = \mathbf{(AB)B^{-1}A^{-1}}\)

\(\displaystyle 1 = \mathbf{(AB)B^{-1}A^{-1}}\)

Post multiplying with AB:

\(\displaystyle \mathbf{(AB)^{-1}(AB)} = \mathbf{ B^{-1}A^{-1}(AB)}\)

\(\displaystyle 1 = \mathbf{B^{-1}A^{-1}(AB) }\)

Is this right? What next are you suppose to do?

Now use the fact that:

AA[sup:2ibqv0ds]-1[/sup:2ibqv0ds] = I = A[sup:2ibqv0ds]-1[/sup:2ibqv0ds]A

and

IA = AI = A

 

[TsAmE]

Thanks. I am just curious, I have never understood the rule IA = AI = A. Why is it that the identity matrix multiplied by the matrix (that has an inverse) just give you the matrix?

Here is what I got:

\(\displaystyle \mathbf{I} = \mathbf{(AB)B^{-1}A^{-1}}\)

\(\displaystyle \mathbf{IA} = \mathbf{(AB)B^{-1}A^{-1}I}\)

\(\displaystyle \mathbf{A} = \mathbf{(AB)B^{-1}A^{-1}I}\)