- For \(n\geq 2\), consider the ring of \(n \times n\) matrices over a commutative unital ring \(R\), denoted \(\operatorname{Mat}_{n}(R)\). Recall the determinant map \(\det:\operatorname{Mat}_{n}(R)\to R\). For \(A \in \operatorname{Mat}_{n}(R)\) also recall the definition of the classical adjoint \(A^{a}\) of \(A\). Prove that:
\(\operatorname{det}\left(A^{a}\right)=\operatorname{det}(A)^{n-1}\)
\(\left(A^{a}\right)^{a}=\operatorname{det}(A)^{n-2} A\)
Solution
Write \(\operatorname{adj}(A)=A^a\).
1. It is enough to prove both identities for the generic matrix over \(S=\ZZ[x_{ij}]\).
Each entry of \(\operatorname{adj}(A)\) is a polynomial with integer coefficients in the entries of \(A\), and determinants are polynomial expressions as well. Thus both desired identities are polynomial identities with integer coefficients in the entries of \(A\). If they hold for the generic matrix \[ X=(x_{ij})\in M_n(S), \] then for any commutative unital ring \(R\) and any \(A=(a_{ij})\in M_n(R)\), the specialization homomorphism \(S\to R\), \(x_{ij}\mapsto a_{ij}\), carries those identities to the corresponding identities for \(A\).
2. Over the fraction field \(K=\operatorname{Frac}(S)\), the generic matrix \(X\) is invertible and\[ \operatorname{adj}(X)=\det(X)X^{-1}. \]
The polynomial \(\det(X)\) is nonzero in the domain \(S\), so it is a nonzero element of \(K\). Hence \(X\) is invertible over \(K\). The standard adjugate identity \[ X\operatorname{adj}(X)=\operatorname{adj}(X)X=\det(X)I_n \] then gives the displayed formula after multiplying by \(X^{-1}\).
3. One has\[ \det(\operatorname{adj}(X))=\det(X)^{n-1}. \]
Using 2 and multiplicativity of the determinant over \(K\), \[ \det(\operatorname{adj}(X)) =\det(\det(X)X^{-1}) =\det(X)^n\det(X^{-1}) =\det(X)^{n-1}. \] Both sides lie in \(S\), so the equality holds already in \(S\).
4. One has\[ \operatorname{adj}(\operatorname{adj}(X))=\det(X)^{n-2}X. \]
By 2, \(\operatorname{adj}(X)\) is invertible over \(K\). Therefore, applying the same adjugate formula to \(\operatorname{adj}(X)\) and then using 3, \[ \begin{aligned} \operatorname{adj}(\operatorname{adj}(X)) &=\det(\operatorname{adj}(X))\operatorname{adj}(X)^{-1}\\ &=\det(X)^{n-1}\bigl(\det(X)X^{-1}\bigr)^{-1}\\ &=\det(X)^{n-2}X. \end{aligned} \] Since \(n\ge2\), both sides are matrices whose entries lie in \(S\). Hence the identity holds in \(S\).
5. Specializing the generic identities gives, for every \(A\in M_n(R)\),\[ \det(A^a)=\det(A)^{n-1}, \qquad (A^a)^a=\det(A)^{n-2}A. \]
Apply 1 to the identities established in 3 and 4.