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.
這是矛盾法的證明方式
比較像考古題解答的證明
跟考古題做法一樣的同學 實習課拿考卷跟我要分數
T.A.
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