Using Kirchhoff's rules find the the current I$\_(2)$

Use GaussPivot.m darr$$ script to solve this problem. function x $=$ GaussPivot$($A$,$b$)$

$\%$ GaussPivot: Gauss elimination pivoting

$\%$ x $=$ GaussPivot$($A$,$b$)$: Gauss elimination with pivoting.

$\%$ input:

$\%$ A $=$ coefficient matrix

$\%$ b $=$ right hand side vector

$\%$ output:

$\%$ x $=$ solution vector

$[$m$,$n$]=$size$($A$)$;

if m~$=$n$,$ error$(\text{'}$Matrix A must be square'$)$; end

nb$=$n$+1$;

Aug$=[$A b$]$;

$\%$ forward elimination

for k $=1$:n$-1$

$\%$ partial pivoting

$[$big$,$i$]=$max$($abs$($Aug$($k:n$,$k$)))$;

ipr$=$i$+$k$-1$;

if ipr~$=$k

Aug$([$k$,$ipr$],$:$)=$Aug$([$ipr$,$k$],$:$)$;

end

for i $=$ k$+1$:n

factor$=$Aug$($i$,$k$)/$Aug$($k$,$k$)$;

Aug$($i$,$k:nb$)=$Aug$($i$,$k:nb$)-$factor$*$Aug$($k$,$k:nb$)$;

end

end

$\%$ back substitution

x$=$zeros$($n$,1)$;

x$($n$)=$Aug$($n$,$nb$)/$Aug$($n$,$n$)$;

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

x$($i$)=($Aug$($i$,$nb$)-$Aug$($i$,$i$+1$:n$)*$x$($i$+1$:n$))/$Aug$($i$,$i$)$;

end