Why won't my matrix invert??? I get FAIL every time. QUICK HELP NEEDED

I have a 15x15 matrix and it won't invert. What I have is

    % code
mc  = 2;         % Disk mass
mg  = 1;         % Bar mass
mb  = 1;         % Collar mass
L   = 2;         % Bar length
r   = 1;         % Disk radius
g   = 9.81;      % Gravity
Ig  = mg*L^2/12; % Bar moment of inertia
Ic  = mc*r^2/2;  % Disk moment of inertia
A=[  -mb  1    0    0   0    0    0    0    0     0   0   0   0   0   0 
      0   0    1    1   0    0    0    0    0     0   0   0   0   0   0
      0  -1    0    0  -mg   1    0    0    0     0   0   0   0   0   0
      0   0   -1    0   0    0    1   -mg   0     0   0   0   0   0   0
      0   yb/2 xa/2 0   0    yb/2 xa/2 0   -Ig    0   0   0   0   0   0
      0   0    0    0   0    1    0    0    0     mc  0   1   0   0   0
      0   0    0    0   0    0   -1    0    0     0   0   0   1   0   0
      0   0    0    0   0    r    0    0    0     0   Ic  r   0   0   0
      1   0    0    0   0    0    0    0    yb    0   0   0   0   0   1
      0   0    0    0   0    0    0    0    xa    0   0   0   0   1   0
     -1   0    0    0   1    0    0    0   -yb/2  0   0   0   0   0   0
      0   0    0    0   0    0    0    1    xa/2  0   0   0   0   0   0 
      0   0    0    0   0    0    0    0    0     1   r   0   0   0   0
      0   0    0    0  -1    0    0    0   -yb/2  0   0   0   0   0   1
      0   0    0    0   0    0    0   -1    xa/2  0   0   0   0   1   0
    ];
C = inv(A)

and I tried inverting it with C=inv(A) and C = A^-1 and neither worked. Can anyone see why this is happening?

[dpb]

>>  C = inv(A) 
Undefined function or variable 'yb'. 
>>

Looking farther I see there's no 'xa', either, at least...

Answer 1

[Yu Jiang]

The reason that the matrix is not invertible is because that it is a singular matrix. In fact, the matrix has a rank of 14, meaning there is one column depending on the other columns. I suggest you check the matrix to see why the dependency occurs.

[Andrew Reibold]

Yu Jiang is right! The matrix has a determinant of zero which means that there is no possible inverse.

You can see this yourself by using the determinant function like this:

det(A)

[John D'Errico]

NO NO NO. Do NOT use the determinant to determine if a matrix is singular!

Despite what they taught you in school, a determinant is just a poorly conditioned computation. Yes, it might be ok for a 2x2 matrix. But a 15x15 matrix is just bad mojo to compute a determinant on.

Use rank, cond, or svd to determine if a matrix is singular. NEVER rely on det.

I was going to go into more detail, but I have no idea what yb and xa are. I suppose one can go symbolic for those variables though, so...

syms yb xa

(...)

det(A)
ans =
0

rank(A)
ans =
14

We can go a bit deeper yet, and compute the singular values of A, to find that exactly one of those singular values is identically zero.

[John D'Errico]

I should add, very often matrices like this are singular for a good reason. Usually that singularity is a reflection of the fact that your problem is under-constrained. So it might be possible for example, to translate the entire system by some amount, and change nothing about the minimum energy state of that system. Or perhaps it might be possible to rotate the entire system without any change to the potential energy.