inconsistent linear system of equation

Hello,

I am working on analysis of truss structure. This structure includes 386 link bodies and 132 joints. Every joint connects several link bodies. Links are substitute with axial forces, that are spread to components of each axis (x,y,z). So I wrote 3 equations for every joint (exept 4 joints, that have only plane components of forces, these have 2 equations). I wrote another 24 forces, which prevant structure movement. 6 of these forces should be solved, and 18 forces will be unsloved variables, which is okay, beacause i want them solve in differnt calculations.

So I have a 392 variables (they will be dependent to 18 unsloved variables) and 392 equations. But this linear system of equation is inconsistent and doesn't have solution. I modeled this truss structure in Ansys apdl, and in this programme, I got a results of every axial force and if I give this values to my system of equations, I'll get a pretty accurate solutions of each equation. This solution is not exactly zero, but it is very close to it (for example 10^-6).

So is there any solve method, which can solve this inconsistent system, for example some Iterative method and if its not, how can I find which collums are lineary dependent?

System of equations and Ansys results are in attachment.

[Matt J]

If you know that the system is inconsistent, why do you expect to find a simultaneous solution to all of the equations?

[Jakub Dohnal]

Because i have a solution from Ansys, and if I substitute those values into the equations, the results of every equations are almost zero (very small numbers like 10^-6). So i don't know how to modify this system to get some solution. Or at least I want to know those linear dependent collums and i don't know how to find them.

[John D'Errico]

One worry is you seem to be creating these equations in MATLAB using 9 digits of precision. But the system is singular.

Ad = double(A);
size(Ad)
ans =
   392   392

rank(Ad)
ans =
   383

Is the system inconsistent? Yes. But then b contains symbolic elements, not just numbers.

rank([Ad,b])
ans =
   384
   
b(1:10)
ans =
                                             0
 2876.8499999999985448084771633148 - 1.0*FS1ay
                                    -1.0*FS1az
                                             0
 2876.8499999999985448084771633148 - 1.0*FS1by
                                    -1.0*FS1bz
                                    -1.0*FS1cx
 2876.8499999999985448084771633148 - 1.0*FS1cy
                                             0
                                    -1.0*FS1dx

Which columns of A are linearly dependent on the others? Well, that is difficult to say, because we can only say that 9 columns are dependent on the others. Which exact set of 9 that should be eliminated is somewhat arbitrary.

[R,jb] = rref(Ad);
setdiff(1:392,jb)
ans =
   311   319   327   335   343   352   354   386   388

So rref chose that set of 9 columns.

I'm not entirely certain what solution you compare this to, since b is not numeric as we have it, nor do I see how it is you think that you got a solution that is correct to roughly 1e-6, since b is not numeric.

Answer 1

[Matt J]

You can find the least squares solution to a linear system of equation A*x=b using backslash

x_leastsquares = A\b