For any $n > 2$ and any $(n \times n)$-matrix $A$ over an arbitrary field, the adjugate of the adjugate of $A$ equals $\det(A)^{n - 2} A$. Is there a unified way, without dividing into two cases – $A$ invertible and $A$ non-invertible – to prove this result ?
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2See http://math.stackexchange.com/questions/92837/proof-mathrmadj-mathrmadja-mathrmdetan-2-cdot-a-for-a/92842#92842. – Davide Giraudo Jun 25 '12 at 19:21
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are you referring to the second proof (because as far as i can tell, the first proof is by dividing into the two cases) ? – tipshoni Jun 25 '12 at 19:36
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Exactly the same universal approach in the related problem below works on your similar problem.
Hint $ $ Denote the adjoint of $\rm\:A\:$ by $\rm\:A^*.\:$ Then
$$\rm\: A A^* = |A|\: I_n \,\Rightarrow\ |A|\, |A^*| = |A|^n \,\Rightarrow\ |A^*| = |A|^{n-1}\qquad$$
where the cancellation of $\rm\:|A|\:$ is done universally, i.e. consider matrix extries as indeterminates, so the determinant is a nonzero polynomial in a domain $\:\!\rm\Bbb Z[a_{\:\!i\:\!j}],\,$ so it is cancellable. For further discussion of such universal cancellation of "apparent singularities" see here and here and here.
Bill Dubuque
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