如何求带参数的矩阵的逆矩阵
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*
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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