Matlab Gaussian elimination with partial pivoting

AI Thread Summary
The discussion focuses on implementing Gaussian elimination with partial pivoting in MATLAB, specifically addressing the challenge of determining the maximum row for pivoting during forward elimination. The user seeks clarification on whether the maximum value should be found in the entire matrix or specific to the current pivot row. They share their code and express difficulty in translating the pseudo code provided by their professor. Additional resources, including a relevant discussion and MATLAB code examples, are mentioned as potential aids. The conversation highlights the importance of correctly identifying the pivot row to ensure accurate results in solving linear equations.
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Homework Statement


Hi all, I'm writing a program to solve a system of linear algebraic equations using the method of Gaussian elimination. The only thing I can't figure out is how to perform the actual pivot.
In the %Forward elimination nest, I can't figure out how I am supposed to find the max_row. Is it just any row that has the maximum value or is it specific to the pivot row? Most of the function is translated from pseudo code posted by my professor and the only part I can't translate is how to get the max_row. Any help would be greatly appreciated!

Homework Equations


Matlab:
function x=Lab7gauss(A,b)
%Solves a given system of linear equations Ax=b using the Gauss Elimination
%method. Equations in format Ax=b.%Function code
A=input('Enter the coefficient matrix: ');
b=input('Enter the right hand side vector b: ');
[m,n]=size(A);
if m~=n
    error('Coefficient matrix A must be square')
end
Ab=[A,b];
nc=n+1;

%Forward elimination

for p=1:n-1     %p is current pivot row
    %partial pivoting
    max_row=max(A);
    if max_row~=p
        p=max_row;
        max_row=p;
    end
    for i=p+1:n     %operate on all rows below pivot
        factor = Ab(i,p)/Ab(p,p);
        for c=p:nc
        Ab(i,c)=Ab(i,c)-factor*Ab(p,c);
        end
    end
end

%back substitution
x(n)=Ab(n,nc)/Ab(n,n);
for k=n-1:-1:1
    x(k)=Ab(k,nc);
    for j=k+1:n
        x(k)=x(k)-Ab(k,j)*x(j);
    end
    x(k)=x(k)/Ab(k,k);
end
[C,I]=max(A)
 
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