Let $A$ be an invertible $n\times n$ matrix. Prove that $\det(\operatorname{adj}(A^{-1})) = (\det(A))^{1-n}$
I tried starting with $A^{-1} = 1/\det(A) \cdot \operatorname{adj}(A)$
I tried everything to reduce it to the proof..
the closest I could get was substituting $A^{-1}$ for $B$ and $A$ for $B^{-1}$ and wind up with the proof with $A$ replaced by $B$… but that can't be right…
Best Answer
If you substitute $A^{-1}$ for $A$ you get $\text{adj}(A^{-1})=(\det(A^{-1}))A$, so now we have
$\det(\text{adj}(A^{-1}))=(\det(A^{-1}))^{n}\det(A)=(\det(A))^{-n}\det(A)=(\det(A))^{1-n}$
$\;\;\;\;\;$since $\det(cA)=c^{n}\det(A)$ and $\det(A^{-1})=(\det(A))^{-1}$.