Proposition 5

Tag 064B, permanent. Cite it as [Bourbaki, Tag 064B].

If $X$ and $Y$ are two square matrices over a commutative ring $A$ with the same finite indexing set, then

\det(XY) = (\det X)(\det Y).

For X to be invertible, it is necessary and sufficient that det X be an invertible element of A, and then

(9)

\det(X^{-1}) = (\det X)^{-1}.

This follows immediately from no. 1, Proposition 1 and no. 2, Theorem 1.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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