Implement a function Gauss$($A$)$ that takes a square matrix $\backslash ($ A $\backslash )$ as input, performs Gaussian elimination with partial pivoting $($slide $36$ of Part$5.$pdf$),$ and returns the matrices $\backslash ($ P$,$ L$,$ U $\backslash )$ from the resulting $\backslash (\backslash$mathrm$\{$PA$\}=\backslash$mathrm$\{$LU$\}\backslash )$ factorization as output. Then, use your function to compute a $\backslash (\backslash$mathrm$\{$PA$\}=\backslash$mathrm$\{$LU$\}\backslash )$ factorization of the matrix $\backslash ($ A $\backslash )$ from Q$2.$

An incomplete MATLAB code is provided below.

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A $=[0,2,1,-1$;$1,1,3,3$;$4,4,0,7$; $2,1,1,1]$;

$[$P$,$L$,$U$]=$ Gauss$($A$)$

function $[$P$,$L$,$U$]=$ Gauss$($A$)\%$ Input: A in R R $\{$n$\backslash$timesn$\}$

$[$m$,$n$]=$size$($A$)$; $\%$ get the dimensions of A

if $($m ~ n$)$

error$(\text{'}$not a square matrix'$)$; s if A is not a square matrix, give an error message to the user

end

P $=$ eye$($n$)$; L $=$ eye$($n$)$; U $=$ A; & For now, set P and L as identity matrices and U as A$.$ We will overwrite their entries later,

for i $=1$:$($n$-1)$

$[$~$,$maxindex$]=$ max$($abs$($U$($i:n$,$i$)))$; s check where the largest element in absolute value in column i on or below the diagonal is located

r $=$ maxindex $+$ i $-1$; s r is the index s$.$t$.$ u$\_\{$ri$\}$ is the largest element in absolute value in column i $($diagonal and below$)$

if r ~ $=$ i

s if r is not equal to in$,$ we need to swap some rows

U$([$i$,$r$],$i:n$)=$ U$([$r$,$i$],$i:n$)$;

L$([$i$,$r$],1$:$($i$-1))=$ L$(???)$;

P$(???)=$ P$(???)$;

end

for j $=($i$+1)$:n

L$($j$,$i$)=?$l$?$;

$???=???$;

end

end

end

$```$