# Is there a standard solution for Gauss elimination in Python?

One function that can be worth checking is _remove_redundancy, if you wish to remove repeated or redundant equations:

import numpy as np
import scipy.optimize

a = np.array([[1.,1.,1.,1.],
[0.,0.,0.,1.],
[0.,0.,0.,2.],
[0.,0.,0.,3.]])
print(scipy.optimize._remove_redundancy._remove_redundancy(a, np.zeros_like(a[:, 0]))[0])


which gives:

[[1. 1. 1. 1.]
[0. 0. 0. 3.]]


As a note to @flonk answer, using a LU decomposition might not always give the desired reduced row matrix. Example:

import numpy as np
import scipy.linalg

a = np.array([[1.,1.,1.,1.],
[0.,0.,0.,1.],
[0.,0.,0.,2.],
[0.,0.,0.,3.]])

_,_, u = scipy.linalg.lu(a)
print(u)


gives the same matrix:

[[1. 1. 1. 1.]
[0. 0. 0. 1.]
[0. 0. 0. 2.]
[0. 0. 0. 3.]]


even though the last 3 rows are linearly dependent.

I finally found, that it can be done using LU decomposition. Here the U matrix represents the reduced form of the linear system.

from numpy import array
from scipy.linalg import lu

a = array([[2.,4.,4.,4.],[1.,2.,3.,3.],[1.,2.,2.,2.],[1.,4.,3.,4.]])

pl, u = lu(a, permute_l=True)


Then u reads

array([[ 2.,  4.,  4.,  4.],
[ 0.,  2.,  1.,  2.],
[ 0.,  0.,  1.,  1.],
[ 0.,  0.,  0.,  0.]])


Depending on the solvability of the system this matrix has an upper triangular or trapezoidal structure. In the above case a line of zeros arises, as the matrix has only rank 3.