Solving system of 9 nonlinear equaitons in 16 variables

Question

Sachchidanand Prasad 2022-11-28

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1865073-solving-system-of-9-nonlinear-equaitons-in-16-variables

回答： Torsten 2022-11-28

采纳的回答： Torsten

在 MATLAB Online 中打开

I have a system of equations as follows:

I am not able to use fsolve as it says in the documentaiton that the number of variables should be as same as the number of equations. I found this on the MathWorks which says that it can be done with fsolve. Please let me know if it can be solved by any other method or by using fsolve. It will also suffice if I can know the solution exists.

I am writing the MATLAB code that I have written using fsolve.

f = @(x) [x(1)*x(9)  + x(2)*x(12) + x(3)*x(15)  - 13;
          x(1)*x(10) + x(2)*x(13) + x(3)*x(16)  - 15;
          x(1)*x(11) + x(2)*x(14) - x(3)*(x(9)  + x(13)) + 1;
          x(4)*x(9)  + x(5)*x(12) + x(6)*x(15)  - 9;
          x(4)*x(10) + x(5)*x(13) + x(6)*x(16)  - 24;
          x(4)*x(11) + x(5)*x(14) - x(6)*(x(9)  + x(13));
          x(7)*x(9)  + x(8)*x(12) - x(15)*(x(1) + x(5)) - 7;
          x(7)*x(10) + x(8)*x(13) - x(16)*(x(1) + x(5)) -2;
          x(7)*x(11) + x(8)*x(14) + (x(1)+x(5))*(x(9)+x(13)) - 35];
A = zeros(1,9);
fsolve(f, A)

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Answer 1

Torsten 2022-11-28

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1865073-solving-system-of-9-nonlinear-equaitons-in-16-variables#answer_1114668

在 MATLAB Online 中打开

x0 = -10*ones(16,1);
AB = [13 15 -1;9 24 0;7 2 35];
options = optimset('TolFun',1e-16,'TolX',1e-16);
x = fmincon(@(x)fun(x,AB),x0,[],[],[],[],[],[],[],options);
Local minimum possible. Constraints satisfied.

fmincon stopped because the size of the current step is less than
the value of the step size tolerance and constraints are 
satisfied to within the value of the constraint tolerance.
A = [x(1) x(2) x(3);x(4) x(5) x(6);x(7) x(8) -(x(1)+x(5))]
A = 3×3
   -0.2352    1.4168   -3.4167
   -4.3530    2.0720   -3.1488
   -0.9938   -5.0327   -1.8368
B = [x(9) x(10) x(11);x(12) x(13) x(14);x(15) x(16) -(x(9)+x(13))]
B = 3×3
    0.7008   -2.2052   -1.3224
   -0.1068    1.3766   -5.9351
   -3.8974   -3.6675   -2.0774
A*B-AB
ans = 3×3
1.0e-06 *

   -0.0907   -0.1745   -0.0038
   -0.0097   -0.1094    0.0113
   -0.0376    0.0691   -0.1961
function obj = fun(x,AB)
  A = [x(1) x(2) x(3);x(4) x(5) x(6);x(7) x(8) -(x(1)+x(5))];
  B = [x(9) x(10) x(11);x(12) x(13) x(14);x(15) x(16) -(x(9)+x(13))];
  M = A*B - AB;
  M = M(:);
  obj = sum(M.^2);
end