With this code for an LU factorizaztion of square matrix A $(100*100)$ Print the diagonal entry of the U factor with smallest absolute value and compute the number of zero pivots in U$.$ Kindly give solution using python code $.$Code : import numpy as np

def lu$\_$factorization$\_$complete$\_$pivoting$($A$)$:

$"""$

Performs LU factorization with complete pivoting on a square matrix A$.$

Args:

A: A square numpy array.

Returns:

P: The permutation matrix for row exchanges.

Q: The permutation matrix for column exchanges.

L: The lower triangular matrix.

U: The upper triangular matrix.

d: A vector containing the multipliers mi$.$

$"""$

n $=$ len$($A$)$

P $=$ np$.$eye$($n$)$

Q $=$ np$.$eye$($n$)$

L $=$ np$.$zeros$(($n$,$ n$))$

U $=$ np$.$copy$($A$)$

d $=$ np$.$zeros$($n$)$

for k in range$($n $-1)$:

# Find the pivot element $($maximum absolute value$)$

pivot$\_$row $=$ k

pivot$\_$col $=$ k

max$\_$value $=$ abs$($U$[$k$,$ k$])$

for i in range$($k$,$ n$)$:

for j in range$($k$,$ n$)$:

if abs$($U$[$i$,$ j$])>$ max$\_$value:

max$\_$value $=$ abs$($U$[$i$,$ j$])$

pivot$\_$row $=$ i

pivot$\_$col $=$ j

# Perform row and column exchanges if necessary

if pivot$\_$row $!=$ k:

U$[[$k$,$ pivot$\_$row$]]=$ U$[[$pivot$\_$row, k$]]$ # Swap rows in U

P$[[$k$,$ pivot$\_$row$]]=$ P$[[$pivot$\_$row, k$]]$ # Update permutation matrix P

if pivot$\_$col $!=$ k:

U$[$:$,[$k$,$ pivot$\_$col$]]=$ U$[$:$,[$pivot$\_$col, k$]]$ # Swap columns in U

Q$[$:$,[$k$,$ pivot$\_$col$]]=$ Q$[$:$,[$pivot$\_$col, k$]]$ # Update permutation matrix Q

# Calculate multipliers and update L and U

for i in range$($k $+1,$ n$)$:

L$[$i$,$ k$]=$ U$[$i$,$ k$]/$ U$[$k$,$ k$]$

U$[$i$,$ k:n$]-=$ L$[$i$,$ k$]*$ U$[$k$,$ k:n$]$

return P$,$ Q$,$ L$,$ U$,$ dQProgram $($in a Python function; hint: extend the function from Assignment $1)$ a LU factorization

$($arka$)$

algorithm $($using complete plvoting$)$ for a square matrix A and compute the PQLU factors. The factorization

is obtained by elimination steps $k=1,$dots,$m-1$ so that $$

where the multipliers $m_i$ are given by $m_{i,k}=\frac{A_u^a}{A_{ij}^w},$ When $k=m-1,A_{i,j}^{(m)}$ is upper triangular, that is$,U_{i,j}=A_{ij}^{(m)}.$

The lower triangular matrix is obtained using the multipliers $m_{i,},$ that is $L_{ij}=m_{ij}$AAi$>j,L_{ij}=1,$ and

$k{A}_{L-1}^{(k)}PQ{L}_{i,j}=0$AAi. However, every $k-$step selects a pivot $A_{L-1}^{(k)}of$ maximum $absolute$ value via row exchanges

recorded $in$ the permutation matrix $P,$ and column exchanges recorded $in$ the permutation matrix $Q.$