Question

设A,B均为n阶可逆矩阵,求证:(AB)^*=B*A*

Answer 1

证明:因为A,B可逆,故 A^-1,B^-1 存在,AB 可逆早或,
且有陆弯伍 A* = |A|A^-1,B* = |B|B^-1.
故闹蚂 (AB)* = |AB|(AB)^-1
= |A||B|B^-1A^-1
= (|B|B^-1)(|A|A^-1)
= B*A*.