How do I find independent equations from a system of linear equations?

Question

Nidhish Jain 2017-12-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/373201-how-do-i-find-independent-equations-from-a-system-of-linear-equations

编辑： Nidhish Jain 2017-12-15

I am trying to find independent equations out of a system of linear equations. In total I have 24 equations, but when I checked the rank of the matrix formed by these equations I got 6. That means there are 6 independent equations, but how do I find them?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

James Tursa 2017-12-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/373201-how-do-i-find-independent-equations-from-a-system-of-linear-equations#answer_296484

See this link:

https://www.mathworks.com/matlabcentral/answers/108835-how-to-get-only-linearly-independent-rows-in-a-matrix-or-to-remove-linear-dependency-b-w-rows-in-a-m

7 个评论
显示 5更早的评论隐藏 5更早的评论

Nidhish Jain 2017-12-15

编辑：Nidhish Jain 2017-12-15

在 MATLAB Online 中打开

Here, I am using syms. After running the code, check Knew. It's a 24*6 matrix. These are actually 24 equations with 6 variables. I am trying to find 6 independent equations out of these 24.

  clc;
  clear;
  syms kn kt Rp Rq d kxd kyd kxyd
  klocal = [kn 0     0        -kn 0      0;...
            0  kt    kt*Rp    0   -kt    kt*Rq;...
            0  kt*Rp kt*Rp^2  0   -kt*Rp kt*Rp*Rq;...
           -kn 0     0        kn  0      0;...
            0  -kt   -kt*Rp   0   kt     -kt*Rq;...
            0  kt*Rq kt*Rp*Rq 0   -kt*Rq kt*Rq^2];
  angle = [pi/4 7*pi/4 pi/4 3*pi/4 pi/4 5*pi/4 3*pi/4 pi/4 7*pi/4];
  %----1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
  C = [1 2 3 4 5 6 0 0 0 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;... %1
       0 0 0 4 5 6 1 2 3 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;... %2
       0 0 0 0 0 0 1 2 3 4  5  6  0  0  0  0  0  0  0  0  0  0  0  0;... %3
       0 0 0 0 0 0 0 0 0 4  5  6  1  2  3  0  0  0  0  0  0  0  0  0;... %4
       0 0 0 0 0 0 0 0 0 0  0  0  1  2  3  4  5  6  0  0  0  0  0  0;... %5
       0 0 0 4 5 6 0 0 0 0  0  0  1  2  3  0  0  0  0  0  0  0  0  0;... %6
       0 0 0 4 5 6 0 0 0 0  0  0  0  0  0  0  0  0  1  2  3  0  0  0;... %7
       0 0 0 0 0 0 0 0 0 0  0  0  0  0  0  0  0  0  1  2  3  4  5  6;... %8
       0 0 0 0 0 0 0 0 0 0  0  0  1  2  3  0  0  0  0  0  0  4  5  6];   %9
  K = sym(zeros(24));
   for ii = 1:9
      Lc = cos(angle(ii));
      Ls = sin(angle(ii));
      p =  [ Lc Ls 0 0   0  0;...
            -Ls Lc 0 0   0  0;...
             0  0  1 0   0  0;...
             0  0  0 Lc  Ls 0;...
             0  0  0 -Ls Lc 0;...
             0  0  0 0   0  1];
       Kglobal = p*klocal*p';
       for j = 1:24
           for r = 1:24
               a = C(ii,j);
               b = C(ii,r);
               if a~=0 && b~=0
                   K(j,r) = Kglobal(a,b)+K(j,r);
                   Ksystem = K;
               end
           end
       end
   end
  q = 1i;
  Knew = sym(zeros(24,6));
  for ii = 1:24
      for j = 1:6
          Knew(ii,j) = Ksystem(ii,j) + kyd*Ksystem(ii,j+6) + kxyd*Ksystem(ii,j+12) + kxd*Ksystem(ii,j+18);
      end
  end

James Tursa 2017-12-15

Highlight the code, and then push the "{ } Code" button ...

Nidhish Jain 2017-12-15

在 MATLAB Online 中打开

This is when I am using numerical values for each variable. Here the rank of Knew is 3.

clc;
clear;
kn = 2.5197*1000;
kt = 0.1481*kn;
Rp = 5/1000;
Rq = 2.5/1000;
m1 = 1.05*10^3*pi*(Rp)^2*9/1000;
m2 = 2.15*10^3*pi*(Rq)^2*9/1000;
I1 = m1*Rp^2/2;
I2 = m2*Rq^2/2;
klocal = [kn 0     0        -kn 0      0;...
          0  kt    kt*Rp    0   -kt    kt*Rq;...
          0  kt*Rp kt*Rp^2  0   -kt*Rp kt*Rp*Rq;...
         -kn 0     0        kn  0      0;...
          0  -kt   -kt*Rp   0   kt     -kt*Rq;...
          0  kt*Rq kt*Rp*Rq 0   -kt*Rq kt*Rq^2];
angle = [pi/4 7*pi/4 pi/4 3*pi/4 pi/4 5*pi/4 3*pi/4 pi/4 7*pi/4];
%----1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
C = [1 2 3 4 5 6 0 0 0 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;... %1
     0 0 0 4 5 6 1 2 3 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;... %2
     0 0 0 0 0 0 1 2 3 4  5  6  0  0  0  0  0  0  0  0  0  0  0  0;... %3
     0 0 0 0 0 0 0 0 0 4  5  6  1  2  3  0  0  0  0  0  0  0  0  0;... %4
     0 0 0 0 0 0 0 0 0 0  0  0  1  2  3  4  5  6  0  0  0  0  0  0;... %5
     0 0 0 4 5 6 0 0 0 0  0  0  1  2  3  0  0  0  0  0  0  0  0  0;... %6
     0 0 0 4 5 6 0 0 0 0  0  0  0  0  0  0  0  0  1  2  3  0  0  0;... %7
     0 0 0 0 0 0 0 0 0 0  0  0  0  0  0  0  0  0  1  2  3  4  5  6;... %8
     0 0 0 0 0 0 0 0 0 0  0  0  1  2  3  0  0  0  0  0  0  4  5  6];   %9
K = zeros(24);
 for ii = 1:9
     Lc = cos(angle(ii));
    Ls = sin(angle(ii));
    p =  [ Lc Ls 0 0   0  0;...
          -Ls Lc 0 0   0  0;...
           0  0  1 0   0  0;...
           0  0  0 Lc  Ls 0;...
           0  0  0 -Ls Lc 0;...
           0  0  0 0   0  1];
     Kglobal = p*klocal*p';
     for j = 1:24
         for r = 1:24
             a = C(ii,j);
             b = C(ii,r);
             if a~=0 && b~=0
                 K(j,r) = Kglobal(a,b)+K(j,r);
                 Ksystem = K;
             end
         end
     end
 end
q = 1i;
d = (Rp+Rq)*cos(pi/4)*2;
kx = 0;
ky = 0;
kxd = exp(q*kx*d);
kyd = exp(q*ky*d);
kxyd = exp(q*(kx+ky)*d);
Knew = zeros(24,6);
for i = 1:24
    for j = 1:6
        Knew(i,j) = Ksystem(i,j) + kyd*Ksystem(i,j+6) + kxyd*Ksystem(i,j+12) + kxd*Ksystem(i,j+18);
    end
end