Partial pivoting row swapping

a=[1 4 3 5;2 0 2 6;1 1 0 5;1 2 3 4];
b=[5;6;7;10]
ab=[a b]

for i=1,3
if ab(i,i)< abs(max(ab(:,i)))
     piv=ab(i,:)
      ab(i,:)=ab(i+1,:)
    ab(i+1,:)=piv

    for j=2:4
  ab(j,:)=ab(j,:)-ab(j,1)/ab(1,1)*ab(1,:)
end
end
end

Guys when I run the code I get

ab =

2 0 2 6 6

0 4 2 2 2

0 1 -1 2 4

0 2 2 1 7

I should make zero below the diagonal matrix. I did firs column but the others I could not. Can you help me? Thank you in advance.

Answer 1

[Chunru]

a=[1 4 3 5;2 0 2 6;1 1 0 5;1 2 3 4];
b=[5;6;7;10]

b = 4×1

5 6 7 10

ab=[a b]

ab = 4×5

1 4 3 5 5 2 0 2 6 6 1 1 0 5 7 1 2 3 4 10

for i=1:size(ab, 1)-1
    % pivoting
    [~, imax] = max(abs(ab(i:end, i)));
    %ab(i:end, i)
    %imax
    
    % exchange rows
    piv=ab(imax+(i-1),:);
    ab(imax+(i-1), :)=ab(i, :);
    ab(i,    :)=piv;
    ab
    
    % elimination
    for j=i+1:size(ab, 1)
        ab(j,:)=ab(j,:)-ab(j,i)/ab(i,i)*ab(i,:)
    end

end

ab = 4×5

2 0 2 6 6 1 4 3 5 5 1 1 0 5 7 1 2 3 4 10

ab = 4×5

2 0 2 6 6 0 4 2 2 2 1 1 0 5 7 1 2 3 4 10

ab = 4×5

2 0 2 6 6 0 4 2 2 2 0 1 -1 2 4 1 2 3 4 10

ab = 4×5

2 0 2 6 6 0 4 2 2 2 0 1 -1 2 4 0 2 2 1 7

ab = 4×5

2 0 2 6 6 0 4 2 2 2 0 1 -1 2 4 0 2 2 1 7

ab = 4×5

2.0000 0 2.0000 6.0000 6.0000 0 4.0000 2.0000 2.0000 2.0000 0 0 -1.5000 1.5000 3.5000 0 2.0000 2.0000 1.0000 7.0000

ab = 4×5

2.0000 0 2.0000 6.0000 6.0000 0 4.0000 2.0000 2.0000 2.0000 0 0 -1.5000 1.5000 3.5000 0 0 1.0000 0 6.0000

ab = 4×5

2.0000 0 2.0000 6.0000 6.0000 0 4.0000 2.0000 2.0000 2.0000 0 0 -1.5000 1.5000 3.5000 0 0 1.0000 0 6.0000

ab = 4×5

2.0000 0 2.0000 6.0000 6.0000 0 4.0000 2.0000 2.0000 2.0000 0 0 -1.5000 1.5000 3.5000 0 0 0 1.0000 8.3333

[Rabia Sonmez]

Hey got it.Thank you so much:)

[Rabia Sonmez]

hey I could not understand [~,imax]? Can you explain it

[Chunru]

[~, imax] = max() is to find which element has the maximum. doc max for details.

[Rabia Sonmez]

Okay thank you 🙏