分析

简答题:设A为二阶矩阵,P=(,A),其中是非零向量且不是A的特征向量.
(I)证明P为可逆矩阵;
(Ⅱ)若A2+A-6=0,求P-1AP,并判断A是否相似于对角矩阵.

正确答案
(I)因为≠0且不是A的特征向量,所以A
与A线性无关,
则r(,A)=2,
则P可逆.
(11)解法一
因为A2+A-6=0,即A2=-A+6


解得=-3,2,
所以A的特征值为-3,2.于是A可相似对角化.
解法二
P-1AP同解法一.
由A2+A-6=0,
得(A2+A-6E)=0,
即(A+3E)(A-2E)=0,
≠0得(A2+A-6E)x=0有非零解,
故|(A+3E)(A-2E)|=0,
得|A+3E|=0或|A-2E|=0.
若|A+3E|≠0,则(A一2E)=0,故A=2与题意矛盾,
故|A+3E|=0,同理可得|A-2E|=0.
于是A的特征值为-3,2,
A有2个不同特征值,故A可相似对角化.
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