A的转至矩阵乘以A=E,A的行列式=-1,证E+A=0
问题描述:
A的转至矩阵乘以A=E,A的行列式=-1,证E+A=0
答
由已知 A'A = E, |A| = -1 所以有 |E+A| = |A'A+A| = |(A'+E)A| = |A'+E||A| = -| (A'+E)'| = - |A+E| 所以 |E+A| = 0.
答
是证A+E的行列式等于0吧!
由已知 A'A = E,|A| = -1
所以有
|E+A| = |A'A+A| = |(A'+E)A| = |A'+E||A| = -| (A'+E)'| = - |A+E|
所以 |E+A| = 0.