Question

若n阶矩阵A满足A的3次幂等于3A(A-I),试证I-A可逆，并求（I-A)的-1次幂

Answer 1

A＝0
－A^3＋3A^2－3A＋I＝I
(I－A)^3＝I
所以，(I－A)[(I－A)^2]＝I，即(I－A)(A^2－2A＋I)＝I，所以I－A可逆，且逆矩阵是A^2－2A＋I

Answer 2

由 A^3 = 3A(A-I)
得 (A^2-2A+ I)(I-A) = I
所以 I-A 可逆, 且
(I-A)^(-1) = A^2-2A+ I

过程是凑 (A-I) 因子:
由 A^3 = 3A(A-I)
得: A^3 -3A^2 +3A = 0
A^2(A-I) +A^2 - 3A^2 +3A = 0
A^2(A-I) -2A(A-I) -2A +3A = 0
A^2(A-I) -2A(A-I) + (A-I) + I = 0

(A^2-2A+ I)(A-I) = -I
(A^2-2A+ I)(I-A) = I