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pure maths
2011-12-07 5:24 am
回答 (2)
2011-12-07 6:06 am
✔ 最佳答案
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2011-12-07 12:00:17 補充：
感謝Rvy的指正
因為我無入個direct link 度睇,所以唔見有'xy>0', 我以為個題目suppose P^(-1) 存在。
補充:
∵xy>0
∴(x>0 and y>0) or (x<0 and y<0)
when x>0 and y>0
det(P) = - x - y < 0
when x<0 and y<0
det(P) = - x - y >0
∵det(P) is not equal to 0
∴P^(-1) exists
2011-12-07 9:31 am
小王 ans miss the checking part
|P|=-y-x=-(x+y)
As xy>0
(x<0 and y<0) or (x>0 and y>0)
i.e. x+y=/=0
|P|=/=0
P-1 exists.
then do the task followed by 小王's ans.
|P|=-y-x=-(x+y)
As xy>0
(x<0 and y<0) or (x>0 and y>0)
i.e. x+y=/=0
|P|=/=0
P-1 exists.
then do the task followed by 小王's ans.
收錄日期: 2021-04-13 18:23:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111206000051KK00879