证明: (1)
由 AA* = |A|E
知 (A*)^-1 = (1/|A|)A
由 A^-1 (A^-1)* = |A^-1|E
知 (A^-1)* = |A^-1|A = (1/|A|)A
比较两式得
(A*)^-1 = (A^-1)*
(2) 由 A* (A*)* = |A*|E = |A|^(n-1) E
等式两边左乘A 得
AA* (A*)* = |A|^(n-1) A
由 AA* = |A|E
|A| (A*)* = |A|^(n-1) A
由A可逆得'
(A*)* = |A|^(n-2) A
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