A为三阶矩阵,E为三阶单位矩阵A的三个特征值分别为1,2,-3,则下列矩阵中是可逆矩阵的是:A.A-

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若要A+aE可逆,只需|A+aE|≠0,即a不是-A的特征值,亦即-a不是A的特征值.因此a≠-1,-2,3即可.观察选项,只有A+E可逆,选B.
======以下答案可供参考======
供参考答案1:
原因很简单，由Jordan标准化知，A可看作上三角阵，对角线值为1,2,-3.detA=-6.
所以det(A-E)=det(A+3E)=det(A-2E)=0，而det(A+E)=-12供参考答案2:
|A|的值为三个特征值的乘积，当|A|不为0，则矩阵的秩为3，可逆
|A-E|=（1-1）（2-1）（-3-1）=0
|A+E|=（1+1）（2+1）（-3+1）不为0，选B
供参考答案3:
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