作者randymyliu (randy)
看板NTUfin01
标题管数(谢老师班)小考第四题另解
时间Sat Nov 7 17:34:16 2009
Proof.
Suppose to the contrary that A is singular, but adjA is nonsingular.
=> det(A)= 0 & inv(adjA) exists ( inv(adjA) is the inverse matrix of adjA )
=> A= det(A)inv(adjA)= 0inv(adjA)= O (zero matrix)
=> adjA= O (zero matrix) (-> <-)
This contradicts the hypothesis that adjA is nonsingular.
Hence, if A is singular, then adjA is singular.
这是矛盾法的证明方式
比较像考古题解答的证明
跟考古题做法一样的同学 实习课拿考卷跟我要分数
T.A.
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