busterh2_qlgaw 线性代数证明题 设方阵A满足A^2-A=2E,证明A及A+2E可逆,并求它们的逆阵。A^2

wangccwl2022-10-04 11:39:541条回答

busterh2_qlgaw
线性代数证明题
设方阵A满足A^2-A=2E,证明A及A+2E可逆,并求它们的逆阵。
A^2-A=2E
即AA-A-2EE=0
(A-2E)(A+E)=0
解得:A=2E或A=-E
当A=2E时,A显然可逆,A+2E=4E也可逆
A^(-1)=(1/2)E,(A+2E)^(-1)=(1/4)E
当A=-E时,A显然可逆,A+2E=E也可逆
A^(-1)=-E,(A+2E)^(-1)=E
请问这样证明是否正确?若否,是否应根据行列式不等于0来证明?请指正。谢谢!

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美莎雪 共回答了20个问题 | 采纳率80%

你的做法不对,两个非零矩阵的乘积可以为0,应当如图证明。经济数学团队帮你解答,请及时采纳。谢谢!

1年前

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