(1)求矩阵A的特征值，特征向量；
(2)求A10．
参考答案：[解] (A)由特征多项式
[*]
矩阵A的特征值为：A,C,0,-A，-A．
对λA=A由(E-A)x=0
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得基础解系αA=(A,0,-A,0,0)T．
对λ=C由(CE-A)x=0
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得基础解系αB=(A,B,A,0,0)T．
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得基础解系αs=(A,-A,A,0,0)T．
对λ=-A由(-E-A)x=0
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得基础解系(0,0,0,A,0)T．
所以矩阵A的特征向量
λ=A对应的特征向量是kAαA，kA≠0
λ=C对应的特征向量是kBαB，kB≠0
λ=0对应的特征向量是kCαC，kC≠0
λ=-A对应的特征向量是kDαD，kD≠0
(B)对矩阵A分块为[*]
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所以
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