是否可使A2+I=0?
证明:A为一n*n矩阵,当n 为奇数时,是否可使A2+I=0?当n为偶数时呢?
证明: 由已知, AA' = E 所以 |E-A|=|AA'-A| = |A(A'-E)| = |A||A'-E| = 1* |(A-E)'| = |A-E| = |-(E-A)| = (-1)^n|E-A| = - |E-A|. 故 |E-A| = 0.
答:|A-E|=|A-AA^T|=|A(E-A^T)|=|A|*|E-A^T| =|(E-A^T)^T|=|E-A|=(-1)^n|A-E|=-|A-E| 所以 2...详情>>
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