Let $A \in K^{nxn}$ and $Cofac(A)$ be the cofactormatrix to A. I have to show
(1) $cofac(AB) = cofac(B)*cofac(A)$.
In fact I have:
$^t(cofac A) = cofac (^t A) = adj(A).$
Then I have (I have already proven it!!):
$adj(AB) = adj(B)*adj(A)$. So I get: $adj(AB)= cofac(^t AB) = ^t(cofac AB) = ^t(cofac B) * ^t(cofac A)$.
That is not exactly what I wanted to show, but it's very close to it!?! Thank you!
$\DeclareMathOperator{\adj}{adj}\DeclareMathOperator{\cofac}{cofac}$If I understand correctly, you have proven that $\adj(AB) = \adj(B) \adj(A)$, where $\adj(A) = \cofac(^tA)$. Then: $$\cofac(AB) = \adj(^t(AB)) = \adj(^tB ^tA) = \adj(^tA) \adj(^tB) = \cofac(A) \cofac(B)$$
Which is the correct equation. Another proof is to notice that $\cofac(A) = \det(A) ^t(A^{-1})$.