Incorrect result of $\det(A^{-1})$ by using the formulae $A^{-1} = \frac{1}{\det(A)}adj(A)$

Question

I know, this can be proved easily by:
1. $AA^{-1} = I$
2. $ f_{A^{-1}} = (f_A)^{-1}$

But, I also want to prove it by using $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$$ and here I go

$$\begin{align} \det(A^{-1}) & = \frac{1}{\det(A)} \det(\text{adj}(A)) \\ \det(A^{-1}) & = \det(A)^{n-2} \ \ \ (\because \det(\text{adj}(A)) = \det(A)^{n-1}) \\ \end{align}$$

I know it's me who is doing wrong. Any hint would be appreciated!

Answer 1

$\det(kB) = k^n \det(B)$ where $k$ is any scalar. Here $B=A^{-1}$, $k=1/\det(A)$. Your mistake was you did $\det(kB)=k\det(B)$