Could MATLAB please provide rref() without the rational conversion?

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 1984-2017 The MathWorks, Inc. [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 = zeros(1,0); 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) = 0; j = j + 1; else % Remember column index jb = [jb j]; %#ok<AGROW> % 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