急！！！！哪位高手会用matlab解矩阵方程AXB+CXD=E，万分感激 求高手解答用Matlab 解复杂方程组的解
\u6025\uff01\uff01\uff01\uff01\uff01\u54ea\u4f4d\u5927\u54e5\u4f1amatlab \u5e2e\u5fd9\u5199\u4e2a\u7a0b\u5e8f\uff0c\u7528\u725b\u987f\u8fed\u4ee3\u6cd5\u6c42\u65b9\u7a0bX^3-3*x-1=0\u57282\u9644\u8fd1\u7684\u89e3syms x
x0=2;
f=x^3-3*x-1;
eps=1e-6;
maxcnt=1000;
fx=diff(f,x);
x1=x0;
cnt=1;
while cnt<=maxcnt
x2=x1-subs(f/fx,x,x1);
if abs(x1-x2)<eps
break;
end
[cnt,x1,x2] %\u8fed\u4ee3\u6b21\u6570\uff0c\u8fed\u4ee3\u524d\uff0c\u8fed\u4ee3\u540e
x1=x2;
cnt=cnt+1;
end
subs(f,x,x2)
syms x y z a b c
[x,y,z]=solve('72.8*(1+cos(a))=2*(sqrt(x*21.8)+sqrt(25.5*y)+sqrt(25.5*z))','50.8*(1+cos(b))=2*sqrt(x*50.8)','48*(1+cos(c))=2*(sqrt(x*29)+sqrt(47*y)+sqrt(1.92*z))','x,y,z')
x =
12.700000000000000000000000000000+25.400000000000000000000000000000*cos(b)+12.700000000000000000000000000000*cos(b)^2
y =
-.98416931170728906540241988580218*cos(c)+9.7643439805890752790240685157559*cos(a)*cos(b)-23.462398366292790782350185322144*cos(c)*cos(b)+7.1488996766620738069752563394849*cos(b)^2+3.3341639162497969191291895638814*cos(a)^2+.40958164398297044880227538559881*cos(a)+.59974496762043211062439587258181*cos(b)-16.023090169105698668480172257920*cos(c)*cos(a)+19.250659611845600192713968847131*cos(c)^2+.12578649948056747012125686189223e-1
z =
.15194565438596826916006412565125e-193*(.58477278352492431981436322974990e98+.58477278352492431981436322974990e98*cos(a)-.26731041191058757560106827855831e98*((cos(b)+1.)^2)^(1/2)-1.*(-.64771139107889760381210034820223e194*cos(c)+.64262081202967154457022801098613e195*cos(a)*cos(b)-.15441309237244940736312726425476e196*cos(c)*cos(b)+.47049056490306928474554525841306e195*cos(b)^2+.21943134403702282352778341593461e195*cos(a)^2+.82783874266682206962299704487537e192+.26955798481907364090410143645319e194*cos(a)+.39471018111316040430045885264655e194*cos(b)-.10545277016218098275353846992100e196*cos(c)*cos(a)+.12669437431192070703927236534687e196*cos(c)^2)^(1/2))^2
______________________________
\u5c31\u662f\u4ee3\u5165\u6c42\u503c\u5c31\u884c\u4e86\uff0c\u4f8b\u5982\uff1a\uff08\u63a5\u7740\u4e0a\u9762\u7684\u6c42X\uff09
x=subs(x,{a,b,c},{77.7,33.3,48.5})
x =
127/10+127/5*cos(333/10)+127/10*cos(333/10)^2
__________________________________
\u5176\u5b9e\u5728\u4e00\u5f00\u59cb\u5c31\u7ed9\u5b9aa,b,c\u7684\u503c\u4e5f\u662f\u53ef\u4ee5\u7684\uff0c\u5982\u679c\u4e0d\u8981\u6c42\u4e00\u5b9a\u8981\u6c42\u89e3\u6790\u89e3\u7684\u8bdd\uff0c\u697c\u4e0b\u8bf4\u7684fsolve\u5012\u4e5f\u662f\u4e00\u79cd\u9009\u62e9
aX+Xb=-c
这个是Sylvester方程,其中的a,b,c你自己转换吧，因为有矩阵逆的符号，我不好表示。（注意c前面有个负号）
然后你用MATLAB自带的工具箱中的lyap()函数求解。命令输入如下：
X=lyap(a,b,c)
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