Re: [理工] 线代 反矩阵

楼主: JKLee (J.K.Lee)   2018-07-03 03:02:07
※ 引述《AAQ8 ()》之铭言:
: https://i.imgur.com/roKGvQG.jpg
: https://i.imgur.com/X26fYT3.jpg
: 这一题还可以理解
: 不太懂的地方是在找(I-Mn)的反矩阵时
: 是从哪里判断要先设(I+aMn)的
: 麻烦各位了 谢谢
另一种解法:
(M_n)^2 = nk*M_n
=> (M_n)^2 - nk*M_n = O
令 f(x) = x^2 - nk*x
则 f(M_n) = O
又 f(x) = (1-x)[-x - (1-nk)] + (1-nk)
=> O = f(M_n) = (I - M_n)[-M_n - (1-nk)I] + (1-nk)I
=> (I - M_n) [-M_n - (1-nk)I] = -(1-nk)I
=> (I - M_n) [M_n + (1-nk)I]/(1-nk) = I
=> (I - M_n) [M_n/(1-nk) + I] = I
=> (I - M_n)^(-1) = M_n/(1-nk) + I
作者: AAQ8 (不要就是要)   2018-07-03 17:53:00
谢谢J大

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