Partial pivoting row swapping
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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.
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Chunru
2021년 9월 12일
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: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
댓글 수: 4
Chunru
2021년 9월 12일
[~, imax] = max() is to find which element has the maximum. doc max for details.
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