# Homework Help: Matrix Proof

1. Sep 24, 2012

### pyroknife

Show that if A is invertible, then At is invertible
and (At )-1
= (A-1)t .

If A is invertible, then det(A)≠0
det(A)=det(A^t)
thus det(A^t)≠0

I'm not really sure how to prove the second part. It's an identity that I remembered, but don't know how to prove.

I'll take a crack at it though:

Multiply A^t to both sides
gives I=A^t (A^-1)^t
For the RHS, can i multiply A^t to A^-1 or does the ^t prevent me from doing that?

2. Sep 24, 2012

### jbunniii

If A is invertible, then there exists a matrix B such that AB = I and BA = I. What is the natural thing to do here, to both sides of each equation?

3. Sep 24, 2012

### pyroknife

I'm not sure what you are asking.

(At )-1 = (A-1)t .

If AB=I, then obviously, A is the inverse of B or vise versa.

So In this case I let A=(A^t)^-1
Thus A^=((A^t)^-1)^-1 ? <<<<multiply this to both sides? If so, the left said reduces to the I matrix, but the right side still looks complicated.

Last edited: Sep 24, 2012
4. Sep 24, 2012

### jbunniii

If A is invertible, then there exists a matrix B such that AB = I and BA = I. Try transposing both sides of each of these equations.

5. Sep 24, 2012

### Dick

This can be a little hard to see. You already have I=A^t (A^-1)^t. This tells you that A^t has an inverse. What is it? Now take a deep breath. The inverse of A^t is called (A^t)^(-1).

Now that I look back I'm not sure you even got that far. Take junniii's advice. Find the transpose of A^(-1)A=I.

Last edited: Sep 24, 2012
6. Sep 24, 2012

### pyroknife

OH!

So in this form

I=A^t [(A^-1)^t]

To get to the identity matrix A^T has to be multiplied by its inverse (A^T)^-1. The quantity [(A^-1)^t] must equal (A^T)^-1.

So simple, yet hard.

7. Sep 24, 2012

### Dick

Yes, really simple. But it can be completely opaque if you don't remember what inverse means and that inverses are unique.