用牛顿迭代法解三元三次方程的MATLAB程序!

用fsolve可解出来：先构造函数：function output=solveproblem(X)c=X(1);m=X(2);y=X(3);output(1)=(1-c)*(1-y)*(1-m)*10。61+y*(1-c)*(1-m)*30。56+c*(1-y)*(1-m)*60。67+m*(1-c)*(1-y)*0。95+y*m*(1-c)*28。45+y*c*(1-m) *8。19+c*m*(1-y) *2。51+c* y*m* 76。69-3。21output(2)=(1-c)*(1-y)*(1-m)*15。21+y*(1-c)*(1-m)*14。58+c*(1-y)*(1-m)*65。2+m*(1-c)*(1-y)*0。96+y*m*(1-c)*14。23+y*c*(1-m) *17。41+c*m*(1-y) *1。62+m*y *c*80。32output(3)=(1-c)*(1-y)*(1-m)*44。77+y*(1-c)*(1-m)*9。9+c*(1-y)*(1-m)*3。97+m*(1-c)*(1-y)*0。78+y*m*(1-c)*0。8+y*c*(1-m) *3。3+c*m*(1-y) *7。95+ m*c*y*66。79然后求x0=[1 1 1]; %初值ans1 = ones(20,3);options=optimset('fsolve');for k=1:20 [x,fval,exitflag,output]=fsolve(@solveproblem,x0,options); x0=x; ans1(k,:)=x;endans1 = -0。2892 0。9143 0。1893带入三个方程验证,分别等于0。0012,4。8363e-004,-0。0016 再问： 能看下我的问题补充，非常感谢~ 再答： function output=solveproblem(X1,Y1) c=X1(1); m=X1(2); y=X1(3); X=Y1(1); Y=Y1(2); Z=Y1(3); output(1)=(1-c)*(1-y)*(1-m)*10。61 + y*(1-c)*(1-m)*30。56 + c*(1-y)*(1-m)*60。67 + m*(1-c)*(1-y)*0。95 + y*m*(1-c)*28。45+y*c*(1-m) *8。19 + c*m*(1-y)*2。51 + c*y*m*76。69 - X output(2)=(1-c)*(1-y)*(1-m)*15。21 + y*(1-c)*(1-m)*14。58 + c*(1-y)*(1-m)*65。2 + m*(1-c)*(1-y)*0。96 + y*m*(1-c)*14。23+y*c*(1-m) *17。41+ c*m*(1-y)*1。62 + m*y*c*80。32 - Y output(3)=(1-c)*(1-y)*(1-m)*44。77 + y*(1-c)*(1-m)*9。9 + c*(1-y)*(1-m)*3。97 + m*(1-c)*(1-y)*0。78 + y*m*(1-c)*0。8+y*c*(1-m) *3。3 + c*m*(1-y)*7。95 + m*c*y*66。79 - Z ----------------------------------------- XYZ=[38。98 38。38 32。29; 7。07 9。73 10。97; 0。67 0。85 0。79; 69。85 68。99 56。01; 1。11 1。7 1。6; 52。05 54。56 49。08; 3。21 2。15 3。76; 44。6 46。98 41。77; 1。21 1。56 1。46; 63。46 63。94 51; 3 4。05 4。41; 13。01 13。6 11。03]; [m,n]=size(XYZ); x0=[1 1 1]; %初始值 ans1 = [zeros(m,3) XYZ]; %每行1~3列保存结果，4~6列保存X,Y,Z值 options = optimset('fsolve'); for k=1:m [x,fval,exitflag,output]=fsolve(@solveproblem,x0,options,XYZ(k,:)); ans1(k,1:3)=x; end