Hello, I need fix the following code for scaled partial pivoting Gauss elimination.

This the main code

function$[$x$]=$GEPP$($A$,$b$)$

$\%[$x$]=$GEPP$($A$,$b$),$ returns column vector x

$\%$ Solves the linear system Ax$=$b using Gaussian Elimination with

$\%$ Partial Pivoting $($GEPP$)$

$[$N$,$~$]=$ size$($A$)$; $\%$ extracting number of rows

for k$=1$:N $\%$ loop over rows of A

A$\_$max $=$ A$($k$,$k$)$; $\%$ maximum A is the pivot

p $=$ k; $\%$ storing the existing row number into p

$\%$Next we test the nearest row that it is pivot larger than A$($k$,$k$)$

$\%-------$Finding the row if exist$-------$

for i$=$k$+1$:N $\%$ working under the pivot element all rows k$+1$ to N

if abs$($A$($i$,$k$))>$abs$($A$\_$max$)\%$Is $|$A$\_\{$i$,$k$\}|>|$A$\_$max$|$ if the element under the pivot larger than the pivot

A$\_$max $=$ A$($i$,$k$)$; $\%$ if yes, then Amax$=$the larger element

p $=$ i; $\%$ let p is the smallest integer $($row$)$in which condition is satisfied

end

end

$\%-----$if A$-$max$=0,$ then we can not evaluate its inverse$-----$

if abs$($A$\_$max$)==0$

msg$=$warning$(\text{'}$Matrix is not invertible!$\text{'})$;

throw$($msg$)$

end

$\%\%\%\%--------$Switching the two rows$---------$

$\%$ if p not equal to k$,$ I do the switch of rows in b

$\%$ PN $($if p$=$k$,$ I do not need to switch, pivot is the largest$)$

if p~$=$k

temp $=$ b$($p$)$; $\%$ the b$($p$)$ is stored into temporary name

b$($p$)=$ b$($k$)$;$\%$ b$($p$)$ is replaced by b$($k$)$

b$($k$)=$ temp;$\%$b$($k$)$ becomes b$($b$)$

for j$=1$:N $\%$ Similarly I switch the two rows in A

temp $=$ A$($p$,$j$)$;

A$($p$,$j$)=$ A$($k$,$j$)$;

A$($k$,$j$)=$ temp;

end

end

$\%---------$ Next, do the multiplier formula, to make all elements under

$\%$the present diagonal $($pivot$)$ element to be equal to $0$

for i$=$k$+1$:N $\%$ The constant vector b as well

b$($i$)=$ b$($i$)-$A$($i$,$k$)*$b$($k$)/$A$($k$,$k$)$;

for j$=$k$+1$:N

A$($i$,$j$)=$ A$($i$,$j$)-$A$($i$,$k$)*$A$($k$,$j$)/$A$($k$,$k$)$; $\%$ A$\_\{$i$,$k$\}/$A$\{$k$,$j$\}=$m$\_\{$i$,$j$\}$ multiplier

end

A$($i$,$k$)=0$;

end

end

$\%-----------$Back substitution$-----------$

x $=$ zeros$($N$,1)$; $\%$ the solution is stored into x

for i$=$N:$-1$:$1$

sum $=0$;

for j$=$i$+1$:N

sum $=$ sum$+$A$($i$,$j$)*$x$($j$)$;

end

x$($i$)=($b$($i$)-$sum$)/$A$($i$,$i$)$;

end

end

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

Here My trying to fix it for scaled $,$ thankyou

function$[$x$]=$GEPPS$($A$,$b$)$

$\%[$x$]=$GEPP$($A$,$b$),$ returns column vector x

$\%$ Solves the linear system Ax$=$b using Gaussian Elimination with

$\%$ Partial Pivoting $($GEPP$)$

$[$N$,$~$]=$ size$($A$)$; $\%$ extracting number of rows

for M$=1$:N

S$($M$)=$max$($abs$($A$($M$,$:$)))$;

end

for k$=1$:N $\%$ loop over rows of A

$\%$Scaled Partial pivoting

for a$=$k:N

d$($a$)=$abs$($A$($a$,$k$))/$S$($a$)$;

end

A$\_$max $=$ A$($k$,$k$)$; $\%$ maximum A is the pivot

p $=$ k; $\%$ storing the existing row number into p

$\%$Next we test the nearest row that it is pivot larger than A$($k$,$k$)$

$\%-------$Finding the row if exist$-------$

for i$=$k$+1$:N $\%$ working under the pivot element all rows k$+1$ to N

if abs$($A$($i$,$k$))>$abs$($A$\_$max$)\%$Is $|$A$\_\{$i$,$k$\}|>|$A$\_$max$|$ if the element under the pivot larger than the pivot

A$\_$max $=$ A$($i$,$k$)$; $\%$ if yes, then Amax$=$the larger element

p $=$ i; $\%$ let p is the smallest integer $($row$)$in which condition is satisfied

end

end

$\%-----$if A$-$max$=0,$ then we can not evaluate its inverse$-----$

if abs$($A$\_$max$)==0$

msg$=$warning$(\text{'}$Matrix is not invertible!$\text{'})$;

throw$($msg$)$

end

$\%\%\%\%--------$Switching the two rows$---------$

$\%$ if p not equal to k$,$ I do the switch of rows in b

$\%$ PN $($if p$=$k$,$ I do not need to switch, pivot is the largest$)$

if p~$=$k

temp $=$ b$($p$)$; $\%$ the b$($p$)$ is stored into temporary name

b$($p$)=$ b$($k$)$;$\%$ b$($p$)$ is replaced by b$($k$)$

b$($k$)=$ temp;$\%$b$($k$)$ becomes b$($b$)$

for j$=1$:N $\%$ Similarly I switch the two rows in A

temp $=$ A$($p$,$j$)$;

A$($p$,$j$)=$ A$($k$,$j$)$;

A$($k$,$j$)=$ temp;

end

end

$\%---------$ Next, do the multiplier formula, to make all elements under

$\%$the present diagonal $($pivot$)$ element to be equal to $0$

for M$=1$:N

S$($M$)=$max$($abs$($A$($M$,$:$)))$;

end

for i$=$k$+1$:N $\%$ The constant vector b as well

b$($i$)=$ b$($i$)-$A$($i$,$k$)*$b$($k$)/$A$($k$,$k$)$;

for j$=$k$+1$:N

A$($i$,$j$)=$ A$($i$,$j$)-$A$($i$,$k$)*$A$($k$,$j$)/$A$($k$,$k$)$; $\%$ A$\_\{$i$,$k$\}/$A$\{$k$,$j$\}=$m$\_\{$i$,$j$\}$ multiplier

end

A$($i$,$k$)=0$;

end

end

$\%-----------$Back substitution$-----------$

x $=$ zeros$($N$,1)$; $\%$ the solution is stored into x

for i$=$N:$-1$:$1$

sum $=0$;

for j$=$i$+1$:N

sum $=$ sum$+$A$($i$,$j$)*$x$($j$)$;

end

x$($i$)=($b$($i$)-$sum$)/$A$($i$,$i$)$;

end

end