已知A,B均为n阶方阵,B是可逆矩阵,且满足A²+AB+B²=0,证明A和A+B均可逆,且求出它们的逆矩阵.

Question

已知A,B均为n阶方阵,B是可逆矩阵,且满足A²+AB+B²=0,证明A和A+B均可逆,且求出它们的逆矩阵.

Answer 1

A^2+AB+B^2=0 =>A(A+B)=-B^2 =>|A(A+B)|=|A||A+B|=-|B||B|
B可逆=>|B|不等于0=>|A|和|A+B|均不等于0=>A,A+B为可逆矩阵
A^2+AB+B^2=0 =>A^2B^-2+AB^-1+E=0 =>A(-AB^-2-AB^-1)=E =>A^-1=(-AB^-2-AB^-1)
A^2+AB+B^2=0 =>A(A+B)=-B^2 =>B^2A(A+B)=E =>(A+B)^-1=B^2A