Bourbaki

Exercise 9, § 11

Algebra, III, 648

The adjoint of a square matrix XX of order nn over A is the matrix X~=(det⁡(Xji))\tilde{X} = (\det(X^{ji})) of minors of XX of order n−1n - 1. Show that if XX is invertible then det⁡(X~)=(det⁡X)n−1\det(\tilde{X}) = (\det X)^{n-1} and that every minor det⁡(X~H′,K′)\det(\tilde{X}_{H', K'}) of X~\tilde{X} of order pp is given by the formula

det⁡(X~H′,K′)=(det⁡X)p−1(XH′′,K′′) \det(\tilde{X}_{H', K'}) = (\det X)^{p-1}(X_{H'', K''})
where H′H' and K′K' are the complements of HH and KK respectively in {1,n}\{1, n\} ("Jacobi's identities"; use relation (80) of no. 11).

Solution

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Transcribed from the printed volume. Not translated.