用matlab 中solve求解线性方程组,求大神帮帮,..syms s3,x3,x4,se;[s3,x3,x4,se]=solve('s3*cos(x3)=0.125*cos(0.174)','s3*sin(x3)=0.275+0.125*sin(0.174)','0.6*cos(x3)+0.15*cos(x4)-se=0','0.6*sin(x3)+0.15*sin(x4)=0.3')结果出现:s3 =z2z2x3
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/20 15:36:52
![用matlab 中solve求解线性方程组,求大神帮帮,..syms s3,x3,x4,se;[s3,x3,x4,se]=solve('s3*cos(x3)=0.125*cos(0.174)','s3*sin(x3)=0.275+0.125*sin(0.174)','0.6*cos(x3)+0.15*cos(x4)-se=0','0.6*sin(x3)+0.15*sin(x4)=0.3')结果出现:s3 =z2z2x3](/uploads/image/z/15225276-12-6.jpg?t=%E7%94%A8matlab+%E4%B8%ADsolve%E6%B1%82%E8%A7%A3%E7%BA%BF%E6%80%A7%E6%96%B9%E7%A8%8B%E7%BB%84%2C%E6%B1%82%E5%A4%A7%E7%A5%9E%E5%B8%AE%E5%B8%AE%2C..syms+s3%2Cx3%2Cx4%2Cse%3B%5Bs3%2Cx3%2Cx4%2Cse%5D%3Dsolve%28%27s3%2Acos%28x3%29%3D0.125%2Acos%280.174%29%27%2C%27s3%2Asin%28x3%29%3D0.275%2B0.125%2Asin%280.174%29%27%2C%270.6%2Acos%28x3%29%2B0.15%2Acos%28x4%29-se%3D0%27%2C%270.6%2Asin%28x3%29%2B0.15%2Asin%28x4%29%3D0.3%27%29%E7%BB%93%E6%9E%9C%E5%87%BA%E7%8E%B0%EF%BC%9As3+%3Dz2z2x3)
用matlab 中solve求解线性方程组,求大神帮帮,..syms s3,x3,x4,se;[s3,x3,x4,se]=solve('s3*cos(x3)=0.125*cos(0.174)','s3*sin(x3)=0.275+0.125*sin(0.174)','0.6*cos(x3)+0.15*cos(x4)-se=0','0.6*sin(x3)+0.15*sin(x4)=0.3')结果出现:s3 =z2z2x3
用matlab 中solve求解线性方程组,求大神帮帮,..
syms s3,x3,x4,se;
[s3,x3,x4,se]=solve('s3*cos(x3)=0.125*cos(0.174)','s3*sin(x3)=0.275+0.125*sin(0.174)','0.6*cos(x3)+0.15*cos(x4)-se=0','0.6*sin(x3)+0.15*sin(x4)=0.3')
结果出现:
s3 =
z2
z2
x3 =
(3*cos(z3))/5 + (3*cos(z4))/20
(3*cos(z3))/5 + (3*cos(z4))/20
x4 =
z3
z3
se =
z4
为何呢?应该为数字吧...
用matlab 中solve求解线性方程组,求大神帮帮,..syms s3,x3,x4,se;[s3,x3,x4,se]=solve('s3*cos(x3)=0.125*cos(0.174)','s3*sin(x3)=0.275+0.125*sin(0.174)','0.6*cos(x3)+0.15*cos(x4)-se=0','0.6*sin(x3)+0.15*sin(x4)=0.3')结果出现:s3 =z2z2x3
如果你将[s3,x3,x4,se]作为 A*X = B 中的X,很显然这不科学,因为你左一个cos,右一个sin把 A 搞得早就不行了,早就不线性了,换句话说这根本就不是线性方程组都嘛.
直接求解得了吧,第一个方程除以第二个方程(当然s3肯定不等于0,你懂的),可得cot(x3) = 某个常数 解得x3,你懂的,x3得到了带入第4个得到x4,后面你也懂的...