proposition: Lemma mulmx1C n (A B : 'M[R]_n) : A *m B = 1%:M -> B *m A = 1%:M.
proof: Proof. move=> AB1; pose A' := \det B *: \adj A. suffices kA: A' *m A = 1%:M by rewrite -[B]mul1mx -kA -(mulmxA A') AB1 mulmx1. by rewrite -scalemxAl mul_adj_mx scale_scalar_mx mulrC -det_mulmx AB1 det1. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat seq choice) mathcomp(fintype finfun bigop finset fingroup perm order) mathcomp(div prime binomial ssralg countalg finalg zmodp)