如何求带参数的矩阵的逆矩阵

ljh2010 · 发表于 2011-10-20 22:25

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下面是求帕克变换矩阵的逆矩阵，为什么会是这样的结果？如何求带参数的逆矩阵？我是低手，求高手解答,不甚感激！！
>> syms x
>> c=2/3*[cos(x) cos(x-2/3*pi) cos(x+2/3*pi);-sin(x) -sin(x-2/3*pi) -sin(x+2/3*pi);1/2 1/2 1/2]

c =

[       2/3*cos(x), -2/3*cos(x+1/3*pi), -2/3*sin(x+1/6*pi)]
[       -2/3*sin(x),  2/3*sin(x+1/3*pi), -2/3*cos(x+1/6*pi)]
[             1/3,             1/3,             1/3]

>> inv(c)

ans =

[                         3/2*(sin(x+1/3*pi)+cos(x+1/6*pi))/(cos(x)*sin(x+1/3*pi)+cos(x)*cos(x+1/6*pi)-sin(x)*cos(x+1/3*pi)+sin(x)*sin(x+1/6*pi)+cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi)),                         3/2*(cos(x+1/3*pi)-sin(x+1/6*pi))/(cos(x)*sin(x+1/3*pi)+cos(x)*cos(x+1/6*pi)-sin(x)*cos(x+1/3*pi)+sin(x)*sin(x+1/6*pi)+cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi)), 3*(cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi))/(cos(x)*sin(x+1/3*pi)+cos(x)*cos(x+1/6*pi)-sin(x)*cos(x+1/3*pi)+sin(x)*sin(x+1/6*pi)+cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi))]
[                               -3/2*(-sin(x)+cos(x+1/6*pi))/(cos(x)*sin(x+1/3*pi)+cos(x)*cos(x+1/6*pi)-sin(x)*cos(x+1/3*pi)+sin(x)*sin(x+1/6*pi)+cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi)),                               3/2*(cos(x)+sin(x+1/6*pi))/(cos(x)*sin(x+1/3*pi)+cos(x)*cos(x+1/6*pi)-sin(x)*cos(x+1/3*pi)+sin(x)*sin(x+1/6*pi)+cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi)),             3*(cos(x)*cos(x+1/6*pi)+sin(x)*sin(x+1/6*pi))/(cos(x)*sin(x+1/3*pi)+cos(x)*cos(x+1/6*pi)-sin(x)*cos(x+1/3*pi)+sin(x)*sin(x+1/6*pi)+cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi))]
[                               -3/2*(sin(x)+sin(x+1/3*pi))/(cos(x)*sin(x+1/3*pi)+cos(x)*cos(x+1/6*pi)-sin(x)*cos(x+1/3*pi)+sin(x)*sin(x+1/6*pi)+cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi)),                               -3/2*(cos(x)+cos(x+1/3*pi))/(cos(x)*sin(x+1/3*pi)+cos(x)*cos(x+1/6*pi)-sin(x)*cos(x+1/3*pi)+sin(x)*sin(x+1/6*pi)+cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi)),             3*(cos(x)*sin(x+1/3*pi)-sin(x)*cos(x+1/3*pi))/(cos(x)*sin(x+1/3*pi)+cos(x)*cos(x+1/6*pi)-sin(x)*cos(x+1/3*pi)+sin(x)*sin(x+1/6*pi)+cos(x+1/3*pi)*cos(x+1/6*pi)+sin(x+1/6*pi)*sin(x+1/3*pi))]

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[求助] 如何求带参数的矩阵的逆矩阵

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