Develop a python implementation of Gaussian elimination using partial pivoting.

Develop a python code reducing echelon matrices to their reduced echelon forms.

You can use the python functions in the attached file. The shorter and smarter the python codes, the higher the grades.

Finding the bugs and the shortcomings of the attached example code will increase your grade.

import numpy as np

from scipy import linalg

def eliminateCol$($A$,$ k$=0)$:

# guarantee that A is float type

if A$.$dtype.kind $!=\text{'}$f$\text{'}$ and A$.$dtype.kind $!=\text{'}$c$\text{'}$: return None

# load dimensions

dim$0,$dim$1=$ np$.$shape$($A$)$

# initialize outputs

swap $=$ np$.$identity$($dim$0,$dtype$=$float$)$

oper $=$ np$.$identity$($dim$0,$dtype$=$float$)$

# k $<$ dim$1$

k $=$ k$\%$dim$1$

# if the column k is zero column, return

if A$[$:$,$k$].$any$()==$ False: return swap,oper

i$0=0$

if k $!=0$:

# find a row i$0$ whose left most k$-1$ entries are all zero

while i$0<$ dim$0$:

if A$[$i$0,0$:k$].$any$()==$ False: break

i$0+=1$

# find a nonzero element A$[$i$,$k$]$ to use as a pivot

i $=$ i$0$;

while i $<$ dim$0$:

if A$[$i$,$k$]!=0$: break

i$+=1$

if i $==$ dim$0$: return swap,oper

# swap the row i with the row i$0$

if i $!=$ i$0$: A$[[$i$0,$i$],]=$ A$[[$i$,$i$0],]$;swap$[[$i$0,$i$],]=$ swap$[[$i$,$i$0],]$

# row i$0$ is the pivot row and A$[$j$,$k$]$ is zero for i$0=0$:

if A$[$i$,$j$]!=0$:

quot $=$ A$[$i$,$j$]/$ A$[$k$,$j$]$

A$[$i$,$:$]=$ A$[$i$,$:$]-$ quot$*$A$[$k$,$:$]$

oper$[$i$,$k$]=-$quot

i$-=1$

return oper

def eliminateScale$($A$)$:

# guarantee that A is float type

if A$.$dtype.kind $!=\text{'}$f$\text{'}$ and A$.$dtype.kind $!=\text{'}$c$\text{'}$: return None

# load dimensions

dim$0,$dim$1=$ np$.$shape$($A$)$

# initialize outputs

oper $=$ np$.$identity$($dim$0,$dtype$=$float$)$

# find pivots

i $=0$

while i $<$ dim$0$:

j $=0$

while j $<$ dim$1$:

if A$[$i$,$j$]==0$: j$+=1$; continue

else:

oper$[$i$,$i$]=1/$A$[$i$,$j$]$

A$[$i$,$j:$]=$ A$[$i$,$j:$]/$A$[$i$,$j$]$

break

j$+=1$

i$+=1$

return oper