If $\det(A) = 1$, what are possible values of $\det(\mbox{adj}(A))$?

Question

If A is square matrix (3×3), and $det(A)=1$, is it true that $det(adj(A))=1, -1$?

Answer 1

Recall the general formula $A\,{\rm adj}(A) = \det(A)\,{\rm Id}$. Apply $\det$ to get $$\det(A)\det({\rm adj}(A))= \det(A\,{\rm adj}(A)) = \det(\det(A)\,{\rm Id}) = \det(A)^n \det({\rm Id}) = \det(A)^n.$$So $\det({\rm adj}(A)) = \det(A)^{n-1}$, by cancelling $\det(A)$ (when non-zero) on both sides. If $\det(A) = 0$, the conclusion follows from continuity of $A \mapsto {\rm adj}(A)$ and denseness of non-singular matrices. In your case, we get $\det({\rm adj}(A)) = 1$.

Answer 2

No. $\det(\operatorname{adj}(A)) = (\det A)^{n-1} = 1$.

See here: https://en.wikipedia.org/wiki/Adjugate_matrix#Properties