How to solve the following homogenous equation?

I want to solve the following homogenous equation AX = 0, where my A is as follows,

A =

   1.0e+08 *

   -3.1265   -1.0000         0
   -1.0000   -4.5841   -1.0000
         0   -1.0000   -0.2345

Rank of this matrix is 1. I want to solve such that X = [x1,x2,x3], and assuming x3 = 1.

Is there any other way in which I could solve other than going for dynamic variable? Please help!

Thank you in advance

Answer 1

[John D'Errico]

A =1.0e+08 *[   -3.1265   -1.0000         0
    -1.0000   -4.5841   -1.0000
    0   -1.0000   -0.2345];

First, let me verify your claim.

rank(A)

ans = 3

So you are wrong. The rank is actually 3, NOT 1. Is it at least close to being rank deficient?

format long g
svd(A)

ans = 3×1

1.0e+00 * 525349097.530803 269161622.121456 719.652258357894

And that suggsts it is not in fact even that close to being rank deficient. The third singular vlaue is small, but not that close to zero. However, then I need to consider if your matrix is REALLY what you claim it is. So very likely, you have provided only approximate values for the elements of the matrix. What they are more accurately, we can have no idea.

If your matrix is not rank deficient, then you CANNOT find any non-zero solution to the problem. And this matrix is not even remotely close to being rank 1, as you claim. You could make an argument that it MIGHT be close to a rank 2 matrix. But rank 1? You would be dreaming.

What is the smallest perturbation to A, such that it is even rank deficient, thus rank 2?

[U,S,V] = svd(A);
Adel = U(:,3)*S(3,3)*V(:,3)'

Adel = 3×3

3.81730227540681 -11.9348230353614 50.893205890217 -11.9348230353614 37.314310109281 -159.117974469869 50.893205890217 -159.117974469869 678.520645973206

Ahat = A - Adel

Ahat = 3×3

1.0e+00 * -312650003.817302 -99999988.065177 -50.893205890217 -99999988.065177 -458410037.31431 -99999840.8820255 -50.893205890217 -99999840.8820255 -23450678.520646

rank(Ahat)

ans = 
     2

As you can see, I had to make some significant changes to A even to find a matrix that was singular. And that is the SMALLEST such perturbation possible.

Again, unless A is rank deficient, there is NO non-zero solution.

But suppose you wanted to find the solutino to the perturbed problem? That is, solve for x such that Ahat*x = 0, with x(3)==1? That part is now trivial.

x = null(Ahat);
x = x/x(3)

x = 3×1

0.075006127215511 -0.234507196522584 1

Does this solve the problem? Of course it does.

Ahat*x

ans = 3×1

1.0e+00 * 2.07764117021725e-09 0 -3.72529029846191e-09

And to within floating point trash for this particular probem, that is what we see. (Remember that the elements of Ahat were pretty large. So trash on the order of 1e-9 is about correct here.)

[Subaharan Rajenthirakumar]

I was wrong, the rank isn't 1 for the matrix I attached. I made a change in the code and the rank was A for the previous version of my code (wrong ccode). Sorry about that.

However, irrespective of the rank of A, I need to find the variables (X), such that the last element equal to 1, so the rest would be relative to that (1).

The problem is as follows,

Initially, I would get w (modal frequency), and for this particular problem, I would get three values for w. I have checked that with manual calculations, and this code is correct.

Using each w, I have to calculate modal matrix (X), such that X(:,1) is for w(1), X(:,2) is for w(2) and so on, from AX = 0. Where A is defined in a particular way. Also, the last row of X should have all ones.

The resultant X matrix that I am getting from the code is NaN :(. Could you solve the problem please?

The following is my code, and I scripted as it would work the way I wanted. Though, I am not getting what I wanted. There could be a possiblity where I could get rank 2 but not likely rank 1, in that case, this code may not work (I guess)! but I don't have to worry about that at the moment.

clear all
syms w            %w-modal frequency
%defined properties M - mass and K - stiffness
M1 = [110e+03;160e+03;30e+03]; K = [40e+06,100e+06,100e+06];
DoF = length(M1);

%mass matrix
for i = 1:DoF
    M(i,i) = M1(i);
end
%standard stiffness matrix of a spring element
k = [1 -1;
    -1 1];          
kg = zeros([DoF,DoF]);  
kg(1,1) = K(1);        %stiffness matrix of the first element

%assemblage of the stiffness matrix to obtain the stiffness matrix of the
%entire system and solve for modal frequencies
for i = 1:DoF-1
    kg(i:i+1,i:i+1) = kg(i:i+1,i:i+1) + k*K(i+1)
end
detM = det(kg-M.*w) ==0;             %to determine the modal frequencies w
w = vpa(sqrt(solve(detM,w)),5);      %modal frequency of each mode 
w = double(w);
B = zeros([DoF,1]);
for i = 1:DoF
    A = (kg-M.*(w(i)^2));
    x = linsolve(A,B);
    for j = 1:length(x)
        x(j) = x(j)/x(length(x));
    end
    X(:,i) = x                      %vector of different mode shapes for each mass
end

[John D'Errico]

Read my answer. I added the extra point on how to solve for the vector x. (Actually, I did that before you responded. But you probably did not see it.) Look at the end of my answer, once you have a singular matrix for A.

[Subaharan Rajenthirakumar]

Yes, I got that.

Thank you for the guidance and learnt a lot!

[John D'Errico]

I almost forgot to say. You can scale the vector x that comes out, because it is not unique. Clearly if x is a solution to the problem

A*x = 0

then

A*(x*k) = 0

is also a solution. And that means you can scale x by dividing by the last element to make it 1, as long as that element was not exactly zero in the solution. If that were the case, then no solution would exist with x(3) == 1, since you would need to divide by zero.