Question

行列式证明设A为n阶矩阵,A*（A的转）=I,detA=-1,证明：det（I+A）=0

Answer 1

detA' = detA = -1

det(I+A)=det([I+A]']=det(A'+I)=det(A'+A'A)=det(A'(I+A))=det(A')det(I+A))=-det(I+A)

det(I+A)=0