If your matrix is of form [A:x] (as is usual when solving systems), items of A and x both have to be divisible by items of A but not the other way around. Thus, you could, for example, use floats for A and vectors for x. Example:

mtx = [[1.0, 1.0, 1.0, Vec3(0.0,  4.0, 2.0), 2.0],

       [2.0, 1.0, 1.0, Vec3(1.0,  7.0, 3.0), 3.0],

       [1.0, 2.0, 1.0, Vec3(15.0, 2.0, 4.0), 4.0]]

if gauss_jordan(mtx):

  print mtx

else:

  print "Singular!"

# Prints out (approximately):

#

# [[1.0, 0.0, 0.0, (  1.0,  3.0,  1.0),  1.0],

#  [0.0, 1.0, 0.0, ( 15.0, -2.0,  2.0),  2.0],

#  [0.0, 0.0, 1.0, (-16.0,  3.0, -1.0), -1.0]]

Auxiliary functions contributed by Eric Atienza (also released in Public Domain):

def solve(M, b):

  """

  solves M*x = b

  return vector x so that M*x = b

  :param M: a matrix in the form of a list of list

  :param b: a vector in the form of a simple list of scalars

  """

  m2 = [row[:]+[right] for row,right in zip(M,b) ]

  return [row[-1] for row in m2] if gauss_jordan(m2) else None

def inv(M):

  """

  return the inv of the matrix M

  """

  #clone the matrix and append the identity matrix

  # [int(i==j) for j in range_M] is nothing but the i(th row of the identity matrix

  m2 = [row[:]+[int(i==j) for j in range(len(M) )] for i,row in enumerate(M) ]

  # extract the appended matrix (kind of m2[m:,...]

  return [row[len(M[0]):] for row in m2] if gauss_jordan(m2) else None

def zeros( s , zero=0):

    """

    return a matrix of size `size`

    :param size: a tuple containing dimensions of the matrix

    :param zero: the value to use to fill the matrix (by default it's zero )

    """

    return [zeros(s[1:] ) for i in range(s[0] ) ] if not len(s) else zero