If A is square matrix (3×3), and $det(A)=1$, is it true that $det(adj(A))=1, -1$?
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1Possible duplicate of Prove that if A is ANY $n\times n$ matrix, then $det(adj(A)) = (det(A))^{n-1}$. (how to when A is singular?) – darij grinberg Nov 02 '19 at 20:23
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@Sebastiano: $\operatorname{adj} A$ is the better notation, since $A^*$ has too many different meanings. – darij grinberg Nov 02 '19 at 20:24
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No. $\det(\operatorname{adj}(A)) = (\det A)^{n-1} = 1$.
See here: https://en.wikipedia.org/wiki/Adjugate_matrix#Properties
amsmath
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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 $\det({\rm adj}(A)) = 1$.
Ivo Terek
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