[stingray27]

Originally Posted by Michael Hill

If you want to calculate OPR, especially on large sets of data, I STRONGLY suggest not trying to invert the matrix. It's an extremely computationally intensive operation. However, SOLVING, the matrix is much less so. You can do so by getting your matrix into reduced row echelon form. Here's the MATLAB code for it, I'm not entirely sure how one would do it in VB, but presumably, you would use the same algorithm:

function [A,jb] = rref(A,tol)
%RREF   Reduced row echelon form.
%   R = RREF(A) produces the reduced row echelon form of A.
%
%   [R,jb] = RREF(A) also returns a vector, jb, so that:
%       r = length(jb) is this algorithm's idea of the rank of A,
%       x(jb) are the bound variables in a linear system, Ax = b,
%       A(:,jb) is a basis for the range of A,
%       R(1:r,jb) is the r-by-r identity matrix.
%
%   [R,jb] = RREF(A,TOL) uses the given tolerance in the rank tests.
%
%   Roundoff errors may cause this algorithm to compute a different
%   value for the rank than RANK, ORTH and NULL.
%
%   Class support for input A:
%      float: double, single
%
%   See also RANK, ORTH, NULL, QR, SVD.

%   Copyright 1984-2005 The MathWorks, Inc. 
%   $Revision: 5.9.4.3 $  $Date: 2006/01/18 21:58:54 $

[m,n] = size(A);

% Does it appear that elements of A are ratios of small integers?
[num, den] = rat(A);
rats = isequal(A,num./den);

% Compute the default tolerance if none was provided.
if (nargin < 2), tol = max(m,n)*eps(class(A))*norm(A,'inf'); end

% Loop over the entire matrix.
i = 1;
j = 1;
jb = [];
while (i <= m) && (j <= n)
   % Find value and index of largest element in the remainder of column j.
   [p,k] = max(abs(A(i:m,j))); k = k+i-1;
   if (p <= tol)
      % The column is negligible, zero it out.
      A(i:m,j) = zeros(m-i+1,1);
      j = j + 1;
   else
      % Remember column index
      jb = [jb j];
      % Swap i-th and k-th rows.
      A([i k],j:n) = A([k i],j:n);
      % Divide the pivot row by the pivot element.
      A(i,j:n) = A(i,j:n)/A(i,j);
      % Subtract multiples of the pivot row from all the other rows.
      for k = [1:i-1 i+1:m]
         A(k,j:n) = A(k,j:n) - A(k,j)*A(i,j:n);
      end
      i = i + 1;
      j = j + 1;
   end
end

% Return "rational" numbers if appropriate.
if rats
    [num,den] = rat(A);
    A=num./den;
end

I definitely agree. Just at the time I didn't know a lot about matrices because I was a junior in high school. When I wrote my java code for it, I actually calculated the matrix.