solving a linear system using Gaussian Elimination with python.

I need to implement two fucntions two do the Gaussian Elimation in python.

There are 3 steps

1: fowardElimination => code is given

2.inconsistemSystem => function that checks the system is whetehr consisten or inconsistent (input is assumed to be in echelon form)

3.backsubstitution(B) = > return the reudced row echelon form matrix of B

I have no clue how to implement 2,3 please help ... It says function 2 & 3 are simillar to fowardElimination Function

Provided codes :

import numpy as np import warnings def swapRows(A, i, j): """  interchange two rows of A  operates on A in place  """  tmp = A[i].copy() A[i] = A[j] A[j] = tmp def relError(a, b): """  compute the relative error of a and b  """  with warnings.catch_warnings(): warnings.simplefilter("error") try: return np.abs(a-b)/np.max(np.abs(np.array([a, b]))) except: return 0.0 def rowReduce(A, i, j, pivot): """  reduce row j using row i with pivot pivot, in matrix A  operates on A in place  """  factor = A[j][pivot] / A[i][pivot] for k in range(len(A[j])): # we allow an accumulation of error 100 times larger # than a single computation # this is crude but works for computations without a large # dynamic range if relError(A[j][k], factor * A[i][k]) < 100 * np.finfo('float').resolution: A[j][k] = 0.0 else: A[j][k] = A[j][k] - factor * A[i][k] # stage 1 (forward elimination) def forwardElimination(B): """  Return the row echelon form of B  """  A = B.copy() m, n = np.shape(A) for i in range(m-1): # Let lefmostNonZeroCol be the position of the leftmost nonzero value # in row i or any row below it leftmostNonZeroRow = m leftmostNonZeroCol = n ## for each row below row i (including row i) for h in range(i,m): ## search, starting from the left, for the first nonzero for k in range(i,n): if (A[h][k] != 0.0) and (k < leftmostNonZeroCol): leftmostNonZeroRow = h leftmostNonZeroCol = k break # if there is no such position, stop if leftmostNonZeroRow == m: break # If the leftmostNonZeroCol in row i is zero, swap this row # with a row below it # to make that position nonzero. This creates a pivot in that position. if (leftmostNonZeroRow > i): swapRows(A, leftmostNonZeroRow, i) # Use row reduction operations to create zeros in all positions # below the pivot. for h in range(i+1,m): rowReduce(A, i, h, leftmostNonZeroCol) return A