• Aug 29th 2009, 04:25 AM
I need to prove that |adj(A)|=|A|^(n-1) for every non singular A in F^nxn, while n>2.

The solution says:

http://img190.imageshack.us/img190/1217/43950593.png
I can understand the first equality - |adj(A)|=||A|*A^(-1)|,
but I can't understand the rest.

Thank you! :)
• Aug 29th 2009, 04:32 AM
I'm sorry, there was a little mistake - only the first row is relevant to my question - I don't understand it...
• Aug 29th 2009, 05:11 AM
Defunkt
You need to use two facts to prove this:
1. If $\displaystyle det(A) \neq 0$ then $\displaystyle det(A^{-1}) = \frac{1}{det(A)}$
2. If c is a scalar in F, $\displaystyle det(cA) = c^ndet(A)$

• Aug 29th 2009, 05:19 AM
Well.. I got to this line:

but how can I continue from here?

I just can't find a way to expand this phrase...
• Aug 29th 2009, 05:26 AM
Defunkt
Quote:

Well.. I got to this line:

but how can I continue from here?

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

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

$\displaystyle det(adj(A)) = \frac{(det(A))^n}{det(A)} = (det(A))^{n-1}$
• Aug 29th 2009, 05:38 AM
Why:

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

?
• Aug 29th 2009, 05:44 AM
Defunkt
Quote:

Why:

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

?

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

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

Since $\displaystyle det(A)$ is a scalar in F. Also, $\displaystyle det(A^{-1}) = \frac{1}{det(A)} \Rightarrow det(det(A) * det(A^{-1})) = \frac{det(A)^n}{det(A)} = det(A)^{n-1}$
• Aug 29th 2009, 07:55 AM