I need to prove that |adj(A)|=|A|^(n-1) for every non singular A in F^nxn, while n>2.

The solution says:

I can understand the first equality - |adj(A)|=||A|*A^(-1)|,
but I can't understand the rest.

Thank you!

2. I'm sorry, there was a little mistake - only the first row is relevant to my question - I don't understand it...

3. You need to use two facts to prove this:
1. If $det(A) \neq 0$ then $det(A^{-1}) = \frac{1}{det(A)}$
2. If c is a scalar in F, $det(cA) = c^ndet(A)$

4. Well.. I got to this line:

but how can I continue from here?

I just can't find a way to expand this phrase...

Well.. I got to this line:

but how can I continue from here?

I just can't find a way to expand this phrase...
$det(adj(A)) = det(det(A) * det(A^{-1})) = (det(A))^n * det(A^{-1}) = (det(A))^n * \frac{1}{det(A)}$

$det(adj(A)) = \frac{(det(A))^n}{det(A)} = (det(A))^{n-1}$

6. Why:

$det(det(A) * det(A^{-1})) = (det(A))^n * det(A^{-1})$

?

Why:

$det(det(A) * det(A^{-1})) = (det(A))^n * det(A^{-1})$

?
Using $det(\alpha A) = \alpha^n det(A)$ , we get:

$det(det(A) * det(A^{-1})) = det(A)^n * det(A^{-1})$

Since $det(A)$ is a scalar in F. Also, $det(A^{-1}) = \frac{1}{det(A)} \Rightarrow det(det(A) * det(A^{-1})) = \frac{det(A)^n}{det(A)} = det(A)^{n-1}$

8. Okay, I think I can understand the main idea... it's a little hard for me to 'digest' it right now (because intuitively it looks like this 'n' came out of nowhere), but I'll go over it until I'll understand it...

Thank you !