For any n>2 and any n by n matrix A over an arbitrary field, the adjugate of the adjugate of A equals the determinant of A to power n-2 times 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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–  Davide Giraudo Jun 25 '12 at 19:21
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

Hint $\$ Denote the adjoint of $\rm\:A\:$ by $\rm\:A^*.\:$ Then
$$\rm\: A A^* = |A|\: I_n \ \Rightarrow\ |A|\, |A^*| = |A|^n\ \Rightarrow\ |A^*| = |A|^{n-1}$$
where the cancellation of $\rm\:|A|\:$ is done universally, i.e. considering the matrix extries as indeterminates, so the determinant is a nonzero polynomial in a domain $\rm\:\mathbb Z[a_{ij}],$ hence is cancellable. For further discussion of such universal cancellation of "apparent singularities" see here and here and here.