MATLAB gets the inverse of the matrix wrong

Hello everyone. I write a code like this for Newton-Raphson; however system is converges. Thanks in advance to those who are interested.

clc;
clear all;
format shortG;
a=0.1;
b=0.3;
c=0.2;
d=0.2;
e=0.3;
f=0.25;
g=0.3;
h=0.2;
j=0.3;
k=0.2;
m=0.3;
n=0.2;
p=0.1;
u1=acosd(0.75);
v1=acosd(0.75);
v2=acosd(0.75);
y2=acosd(0.75);
y1=180-2*u1;
u2=180-2*u1;
max_error=0.000001;
%Initial guesses:
teta1=50;
teta2=40;
teta3=30;
teta4=60;
teta5=20;
teta6=15;
x1=0.2;
y=0.65;
for q=0:1:1
    for i=1:10000
        mat1=[(p+n)*cosd(teta1) e*cosd(teta2) 0 0 0 0 0 -1;
            -(p+n)*sind(teta1) -e*sind(teta2) 0 0 0 0 -1 0;
            0  e*cosd(teta2) 0 -c*cosd(teta1) -f*cosd(teta5) b*cosd(teta2) 0 0;
            0 -e*sind(teta2) 0 c*sind(teta4) f*sind(teta5) -b*sind(teta6) 0 0;
            (m-n)*cosd(teta1) 0 j*cosd(teta3) (-g-c)*cosd(teta4) -f*cosd(teta5) 0 0 0;
            (-m+n)*sind(teta1) 0 -j*sind(teta3) (g+c)*sind(teta4) f*sind(teta5) 0 0 0;
            k*cosd(teta1) -e*cosd(teta2) j*cosd(teta3) -h*cosd(teta4) 0 0 0 0;
            -k*sind(teta1) e*sind(teta2) -j*sind(teta3) h*sind(teta4) 0 0 0 0];
        
        f1=(p+n)*sind(teta1)+e*sind(teta2)-a*sind(q)-y;
        f2=(p+n)*cosd(teta1)+e*cosd(teta2)-a*cosd(q)-x1;
        f3=b*sind(teta6)-c*sind(teta4)-f*sind(teta5)+e*sind(teta2);
        f4=b*cosd(teta6)-c*cosd(teta4)-f*cosd(teta5)+e*cosd(teta2);
        f5=m*sind(teta1+u1)+j*sind(teta3)-g*sind(180+teta4-y2)-c*sind(teta4)-f*sind(teta5)-n*sind(teta1);
        f6=m*cosd(teta1+u1)+j*cosd(teta3)-g*cosd(180+teta4-y2)-c*cosd(teta4)-f*cosd(teta5)-n*cosd(teta1);
        f7=k*sind(teta1+180-y1)+j*sind(teta3)-h*sind(teta4+u2)+a*sind(q)-e*sind(teta2);
        f8=k*cosd(teta1+180-y1)+j*cosd(teta3)-h*cosd(teta4+u2)+a*cosd(q)-e*cosd(teta2);
        mat2=[f1;f2;f3;f4;f5;f6;f7;f8];
        s=[teta1;teta2;teta3;teta4;teta5;teta6;x1;y] -inv(mat1)*mat2;
        error1=abs((s(1)-teta1)/s(1));
        error2=abs((s(2)-teta2)/s(2));
        error3=abs((s(3)-teta3)/s(3));
        error4=abs((s(4)-teta4)/s(4));
        error5=abs((s(5)-teta5)/s(5));
        error6=abs((s(6)-teta6)/s(6));
        error7=abs((s(7)-x1)/s(7));
        error8=abs((s(8)-y)/s(8));
        %fprintf("mat1: %f \n",mat1(5,3))
        %fprintf("s1: %f \n",s(1))
        fprintf("mat1: %f \n", mat1(5,4))
        fprintf("s1: %f \n",s(1))
        fprintf("y: %f \n",teta1)
        fprintf("error 1: %f \n",error1)
        teta1=s(1);
        teta2=s(2);
        teta3=s(3);
        teta4=s(4);
        teta5=s(5);
        teta6=s(6);
        x1=s(7);
        y=s(8);
        if(error1<max_error)
            if(error2<max_error)
                if(error3<max_error)
                    if(error4<max_error)
                        if(error5<max_error)
                            if(error6<max_error)
                                if(error7<max_error)
                                    if(error8<max_error)
                                        fprintf("q=%d değeri için yardımcı koordinatlar: \n" + ...
                                            "teta1= %f \n teta2=%f teta3= %f \n teta4=%f teta5= %f \n teta6=%f \n" + ...
                                            "x1=%f y=%f \n ",q,teta1,teta2,teta3,teta4,teta5,teta6,x1,y)
                                        break;
                                    end
                                end
                            end
                        end
                    end
                end
            end
        end
        
    end
end

When i was searching about problem, i found that :

• When i call to inverse of first matrix in iteration; it gives me as:

0   1.7764e-15      -4.6837      -13.863   3.7184       11.211      -6.4978      -16.025
0            0      -4.1957      -12.418   1.6467        5.415      0.61168      -1.5034
0            0      -1.7856       -5.285   0.26549       1.1085       7.6963       6.0151
0            0      -1.0181      -3.0133   1.6857       4.8475      0.23662       -1.518
0   4.4409e-16      -2.1729      -6.4312    -4.7396     -0.86534       6.4819       3.8831
1.1102e-16  -7.9502e-16       5.7564       4.1585  -5.5487      -3.5879       6.1467       4.6237
0           -1       1.8855       5.5805   -1.1721      -3.6205       1.3753       3.9725
-1  -1.1102e-16      -1.8674      -5.5271    1.0955       3.4062      -1.1124      -3.4356

When i write the matrix on another page and calculated inverse of it, results are given as:

            0   1.7764e-15       -4.684      -13.863 3.7188       11.211      -6.4977      -16.025
            0   8.8818e-16      -4.1956      -12.418 1.6468        5.415       0.6116      -1.5034
            0            0      -1.7853      -5.2839  0.26533       1.1079       7.6962       6.0153
            0            0      -1.0181      -3.0132  1.6857       4.8476      0.23658       -1.518
            0   4.4409e-16      -2.1726      -6.4304  -4.7398     -0.86578        6.482       3.8834
            0   8.8818e-16       5.7566       4.1589 -5.5489      -3.5882       6.1468       4.6239
            0           -1       1.8855       5.5806  -1.1722      -3.6207       1.3753       3.9725
           -1   4.4409e-16      -1.8675      -5.5271   1.0956       3.4064      -1.1125      -3.4357

• As shown above, even if the value at (2,2) is 8.8818e-16, MATLAB gives 0 value. I continued about searching values, i writed 2nd matrix in iteration and take inverse of in code, it gives:

     0   8.8818e-16      -4.1229       -12.51  3.1709       10.036      -5.3542      -13.878
            0   1.1102e-16      -3.8266      -11.611  1.3591       5.1984      0.66694      -1.6326
            0            0      -1.7972      -5.4531  0.4079       1.8427       6.8106       4.1142
            0  -1.8041e-16     -0.96966      -2.9422  1.7444       4.9556     0.032418      -1.8284
            0   8.8818e-16      -2.0086      -6.0945 -4.8922     -0.70095        6.241       3.0747
            0            0        5.702       3.6962 -5.565      -3.3401       5.8793       3.8931
            0           -1       1.7726       5.3784 -1.0455      -3.4888        1.171       3.7098
           -1   3.3307e-16      -1.5747       -4.778  0.85786       2.9151     -0.78798      -2.7921

However when i write the 2nd matrix in iteration on another page and take the inverse of it, it gives:

     0            0      -4.1228      -12.509  3.1707       10.035       -5.354      -13.878
            0  -5.5511e-16      -3.8267      -11.611  1.3591       5.1984      0.66677      -1.6327
            0            0      -1.7973      -5.4534  0.408       1.8429       6.8101       4.1136
            0   1.9602e-16     -0.96963       -2.942   1.7444       4.9555     0.032273      -1.8284
            0   8.8818e-16      -2.0087      -6.0949  -4.8922     -0.70068       6.2408       3.0743
            0   4.4409e-16        5.702       3.6959     -5.565      -3.3399       5.8791       3.8928
            0           -1       1.7726       5.3784   -1.0454      -3.4887        1.171       3.7098
           -1  -4.4409e-16      -1.5747       -4.778    0.85782        2.915     -0.78798      -2.7921

• I repeated this situation about 3rd matrix in iteration, and given inverse matrix value in iteration is:

   0  -2.2204e-16      -3.8011      -11.647  2.8509       9.3197      -4.7219      -12.603
            0   5.5511e-17      -3.5913      -11.004 1.148       5.0202       0.7532      -1.6835
            0  -8.8818e-16      -1.7913      -5.4885 0.47991        2.277       6.3404       2.9598
            0   6.9389e-18     -0.94242      -2.8876   1.8163       5.0277    -0.095593      -2.0078
            0            0      -1.8857      -5.7779  -5.0559     -0.64089       6.1634       2.5847
            0            0       5.7376       3.4044 -5.6734      -3.2471       5.7999        3.469
            0           -1       1.7151       5.2551 -0.96483      -3.4177       1.0457       3.5597
           -1   1.1102e-16      -1.3791      -4.2256  0.70029       2.5632     -0.58609      -2.3628

When i write the 3nd matrix in iteration on another page and take the inverse of it, it gives:

    0   1.5543e-15       -3.801      -11.646  2.8509       9.3194      -4.7218      -12.603
  -4.4409e-16   3.3404e-16      -3.5915      -11.004  1.1481       5.0203      0.75311      -1.6836
  -2.2204e-16    3.058e-16      -1.7914      -5.4887  0.48003       2.2772       6.3401       2.9594
  -1.1102e-16   2.3616e-16     -0.94243      -2.8875  1.8163       5.0277    -0.095579      -2.0077
  -2.2204e-16   7.4989e-16      -1.8858       -5.778 -5.0558     -0.64076       6.1632       2.5844
            0            0       5.7377       3.4042  -5.6735       -3.247       5.7999       3.4688
            0           -1       1.7151        5.255  -0.96484      -3.4176       1.0457       3.5595
           -1  -2.2204e-16      -1.3791      -4.2255 0.70031       2.5631     -0.58609      -2.3627

If i clarify openly my problem, my code is converges and i think it is about taken falsely of inverse of matrix in iteration. I think i write all of term correctly but don't know why it converges and i consider this is the problem's source.

[Bruno Luong]

Is there a question, or you just come to tell your (wrong) finding?

[Mustafa Duran]

I stated at the bottom of the text that "If i clarify openly my problem, my code is converges and i think it is about taken falsely of inverse of matrix in iteration. I think i write all of term correctly but don't know why it converges and i consider this is the problem's source."

Also clarified that inverse of matrix in iteration is different than calculated on the new code.

[Torsten]

@Mustafa Duran

Isn't

d/dx(sind(x)) = pi/180*cosd(x)

?

My advice: Never work with the degree trigonometric functions in the above context.

[Mustafa Duran]

Thanks, I changed degree parameters to radian, however divergence problem is still exist. I considered the problem is caused at the taking inverse of matrix. Sometimes MATLAB round so small number to zero as i shown above. But i realize that divergence problem is not about it. It doesn't effect value.

I think i found the problem that is about Newton-Raphson's divergence problem:

Warning: Matrix is close to singular or badly scaled. Results

may be inaccurate. RCOND = 2.308764e-23.

Answer 1

[Torsten]

%Initial guesses:
teta1=50;
teta2=40;
teta3=30;
teta4=60;
teta5=20;
teta6=15;
x1=0.2;
y=0.65;
x0 = [teta1, teta2, teta3, teta4, teta5, teta6]*pi/180;
x0 = [x0,x1,y];
q = 0.1;
sol = fsolve(@(x)fun(x(1),x(2),x(3),x(4),x(5),x(6),x(7),x(8),q),x0)

function ff = fun(teta1, teta2, teta3 ,teta4 ,teta5, teta6,x1,y,q)
  a=0.1;
  b=0.3;
  c=0.2;
  d=0.2;
  e=0.3;
  f=0.25;
  g=0.3;
  h=0.2;
  j=0.3;
  k=0.2;
  m=0.3;
  n=0.2;
  p=0.1;
  u1=acos(0.75);
  v1=acos(0.75);
  v2=acos(0.75);
  y2=acos(0.75);
  y1=pi-2*u1;
  u2=pi-2*u1;

  ff(1)=(p+n)*sin(teta1)+e*sin(teta2)-a*sin(q)-y;
  ff(2)=(p+n)*cos(teta1)+e*cos(teta2)-a*cos(q)-x1;
  ff(3)=b*sin(teta6)-c*sin(teta4)-f*sin(teta5)+e*sin(teta2);
  ff(4)=b*cos(teta6)-c*cos(teta4)-f*cos(teta5)+e*cos(teta2);
  ff(5)=m*sin(teta1+u1)+j*sin(teta3)-g*sin(pi+teta4-y2)-c*sin(teta4)-f*sin(teta5)-n*sin(teta1);
  ff(6)=m*cos(teta1+u1)+j*cos(teta3)-g*cos(pi+teta4-y2)-c*cos(teta4)-f*cos(teta5)-n*cos(teta1);
  ff(7)=k*sin(teta1+pi-y1)+j*sin(teta3)-h*sin(teta4+u2)+a*sin(q)-e*sin(teta2);
  ff(8)=k*cos(teta1+pi-y1)+j*cos(teta3)-h*cos(teta4+u2)+a*cos(q)-e*cos(teta2);
end