Solving matrices of the form Ax = B ==> error: Matrix is close to singular or badly scaled

13
votes

I'm having trouble solving a system of the form Ax=B

The solution to the system should be 

x = inv(A)*B

However, this doesn't work.

I get the following error message when I try the above line of code:

Warning: Matrix is close to singular or badly scaled.
     Results may be inaccurate. RCOND = 1.156482e-018.

It seems that matlab is having trouble inverting the matrix that I've specified. I tried to verify that the inverse function was working properly by typing in inv(A)*A

This should give the identity matrix, however I got the same error and some garbage numbers.

This is the A matrix that I'm using:

A = [5/2   1/2  -1     0     0    -1/2  -1/2   0     0
     1/2   1/2   0     0     0    -1/2  -1/2   0     0 
    -1     0     5/2  -1/2  -1     0     0    -1/2   1/2
     0     0    -1/2   1/2   0     0     0     1/2  -1/2
     0     0    -1     0     3/2  -1/2   1/2   0     0
    -1/2  -1/2   0     0    -1/2   2     0    -1     0  
    -1/2  -1/2   0     0     1/2   0     1     0     0 
     0     0    -1/2   1/2   0    -1     0     2     0 
     0     0     1/2  -1/2   0     0     0     0     1]

Any ideas as to why this isn't working? I also tried to convert A to a sparse matrix (sparse(A)), and then run the inverse command. No dice.

4
matlablinear-algebra
This is fundamentally a conceptual, math problem, not a programming one. Your syntax is correct. Warnings are not errors, they're warnings. – Karl Knechtel
Thanks for your response Karl. I know that the inverse of a matrix multiplied by itself yields the identity matrix. For some reason this isn't working in matlab. This leads me to believe that there is an issue with the way I am inputting the data and not just a mathematics problem. – Tushar Garg
You don't seem to understand that not all matrices have an inverse. That is a mathematics problem. – Karl Knechtel
Even if a matrix is invertable, you shouldn't use inv(A) to solve a linear problem. Use mldivide instead. – user1512321

4 Answers

21
votes

The problem is indeed in your mathematics. The matrix you provided isn't of full rank, so it isn't invertible. You could verify that manually (haven't taken the time to do so), but MATLAB already points this out by showing that warning.

Since you are working with floating point numbers, this sometimes causes other subtle problems, one of which you can see in the result of det(A), which is in the order of 1e-16, i.e. machine precision or 0 in practice.

You can see that this Matrix is not of full rank by executing the rank function: rank(A) = 8. For a 9x9 matrix, this indeed means that the matrix is not invertible for doubles (as the rank function accounts for machine precision).

If you want to use MATLAB to get a result that corresponds to a manual calculation, you can use the Symbolic Toolbox and its vpa (variable precision arithmetic) to work around possible numerical problems at the cost of a slower calculation.

B = [5  1 -2  0  0 -1 -1  0  0;
     1  1  0  0  0 -1 -1  0  0;
    -2  0  5 -1 -2  0  0 -1  1;
     0  0 -1  1  0  0  0  1 -1;
     0  0 -2  0  3 -1  1  0  0;
    -1 -1  0  0 -1  4  0 -2  0;
    -1 -1  0  0  1  0  2  0  0;
     0  0 -1  1  0 -2  0  4  0;
     0  0  1 -1  0  0  0  0  2];
A = B/2;
size(A)    % = [9 9]
det(A)     % = -1.38777878078145e-17
rank(A)    % = 8
C = vpa(A);
det(C)     % = 0.0
rank(C)    % = 8

Both with VPA and floating points you will get that the rank is 8, the size is [9 9] and the determinant is practically 0, i.e. singular or not invertible. Changing a few entries might make your matrix regular (non-singular), but it is not guaranteed to work and it will solve a different problem.

To solve your actual problem A*x=b for x, you can try to use mldivide (a.k.a. the backslash operator) or a Moore-Penrose pseudo-inverse:

x1 = A\b;
x2 = pinv(A)*b;

But do remember that such a system does not have a unique solution, so both the pseudo-inverse and the backslash operator may (and in this case will) return very different solutions, whether any of them is acceptable really depends on your application.

11
votes

It means exactly what it says. The matrix is singular, which means it can't really be inverted. Not all matrices can.

In geometrical terms, you have a matrix that transforms one 9-dimensional object into another, but flattens one dimension out completely. That can't be undone; there's no way to tell how far to pull things out in that direction.

2
votes

The matrix is singular, consider B=2*A below:

B = [5  1 -2  0  0 -1 -1  0  0;
     1  1  0  0  0 -1 -1  0  0;
    -2  0  5 -1 -2  0  0 -1  1;
     0  0 -1  1  0  0  0  1 -1;
     0  0 -2  0  3 -1  1  0  0;
    -1 -1  0  0 -1  4  0 -2  0;
    -1 -1  0  0  1  0  2  0  0;
     0  0 -1  1  0 -2  0  4  0;
     0  0  1 -1  0  0  0  0  2]

det(B)

0
1
votes

bicgstab(A,b,tol,maxit), an iterative solver, was able to solve a singular linear system A*x=b for a singular matrix A: 

size(A)=[162, 162] 
rank(A)=14 
cond(A)=4.1813e+132

I used: 

tol=1e-10; 
maxit=100;

None of the above-mentioned (including svd, \, inv, pinv, gmres) worked for me but bicgstab did a good job. bicgstab converged at iteration 4 to a solution with relative residual 1.1e-11. It works fast for sparse matrices.

See documentation here: https://uk.mathworks.com/help/matlab/ref/bicgstab.html