Partial pivoting row swapping

2021 9 11
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回答已采纳
更新时间：2021 9 13
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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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 采纳的回答

Chunru
Chunru 2021-9-12
Ran in:

0 个投票

Ran in:
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

4 个评论
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Rabia Sonmez
Rabia Sonmez 2021-9-12
Hey got it.Thank you so much:)
Rabia Sonmez
Rabia Sonmez 2021-9-12
hey I could not understand [~,imax]? Can you explain it
Chunru
Chunru 2021-9-12
[~, imax] = max() is to find which element has the maximum. doc max for details.
Rabia Sonmez
Rabia Sonmez 2021-9-13
Okay thank you 🙏

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