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• $\det(C_{k}(A))=(\det A)^{\binom {n-1}{k-1}}$ (Sylvester-Franke Theorem)

As in , introduce the sign matrix $S=$ diagonal matrix with entries alternating $\pm 1$ with $S_{11}=1$. And the reversal matrix $J$ with 1's on the antidiagonal and zeros elsewhere.

• $C_{k}(A)^{-1}=\det(A)^{-1}J(C_{n-k}(SAS))^{T}J$ (see below)

Compound matrices and adjugates

[See for a classical discussion related to this section.]

Recall the adjugate matrix is the transpose of the matrix of cofactors, signed minors complementary to single entries. Then we can write

$\operatorname {adj} (A)=JC_{n-1}(SAS)^{T}J$ (1)

with $T$ denoting transpose.

The basic property of the adjugate is the relation

$A\operatorname {adj} (A)=\det(A)I$,

hence $C_{k}(A)C_{k}(\operatorname {adj} (A))=\det(A)^{k}I$ while

$C_{k}(A)\operatorname {adj} (C_{k}(A))=\det(C_{k}(A))I$ (2)

Comparing these and using the Sylvester-Franke theorem yields the identity

• $\operatorname {adj} (C_{k}(A))=\det(A)^{{\binom {n-1}{k-1}}-k}C_{k}(\operatorname {adj} (A))$

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Jacobi's Theorem on the Adjugate

Jacobi's Theorem extends (1) to higher-order minors :

$C_{k}(\operatorname {adj} (A))=(\det A)^{k-1}J(C_{n-k}(SAS))^{T}J$

expressing minors of the adjugate in terms of complementary signed minors of the original matrix.

Substituting into the previous identity and going back to (2) yields

$C_{k}(A)\,J(C_{n-k}(SAS))^{T}J=\det(A)I$ and hence the formula for the inverse of the compound matrix given above.