- #1

pyroknife

- 613

- 3

^{t}is invertible

and (A

^{t})

^{-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?