How to find pythagorean triangle's hypotenuse in a matrix?

Question

Ashfaq Ahmed 2021-11-8

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https://ww2.mathworks.cn/matlabcentral/answers/1581569-how-to-find-pythagorean-triangle-s-hypotenuse-in-a-matrix

回答： Sean de Wolski 2021-11-8

采纳的回答： C B

Hey guys,

I have two 4x4 matrices a and b. I want a third c matrix of 4x4 size where all the values will be c = sqrt(a.^2+b.^2).

For example, the first row of c will be [13 10 13 5];

How can I do this using a for loop?

a = [5 8 12 3; 8 12 63 20; 9 84 144 20; 24 11 15 180];
b = [12 6 5 4; 15 35 16 21; 40 13 17 99; 7 60 112 19];

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Answer 1

C B 2021-11-8

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a = [5 8 12 3; 8 12 63 20; 9 84 144 20; 24 11 15 180];
b = [12 6 5 4; 15 35 16 21; 40 13 17 99; 7 60 112 19];
c=[];
for i =1 : size(a,1)
c(i,:) = sqrt(a(i,:).^2+b(i,:).^2)
end
c = 1×4
    13    10    13     5
c = 2×4
    13    10    13     5
    17    37    65    29
c = 3×4
    13    10    13     5
    17    37    65    29
    41    85   145   101
c = 4×4
    13    10    13     5
    17    37    65    29
    41    85   145   101
    25    61   113   181

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Answer 2

Sean de Wolski 2021-11-8

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https://ww2.mathworks.cn/matlabcentral/answers/1581569-how-to-find-pythagorean-triangle-s-hypotenuse-in-a-matrix#answer_826484

This will be the most numrically stable.

a = [5 8 12 3; 8 12 63 20; 9 84 144 20; 24 11 15 180];
b = [12 6 5 4; 15 35 16 21; 40 13 17 99; 7 60 112 19];
hypot(a,b)
ans = 4×4
    13    10    13     5
    17    37    65    29
    41    85   145   101
    25    61   113   181

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Answer 3

Chunru 2021-11-8

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https://ww2.mathworks.cn/matlabcentral/answers/1581569-how-to-find-pythagorean-triangle-s-hypotenuse-in-a-matrix#answer_826439

编辑：Chunru 2021-11-8

a = [5 8 12 3; 8 12 63 20; 9 84 144 20; 24 11 15 180];
b = [12 6 5 4; 15 35 16 21; 40 13 17 99; 7 60 112 19];
c = sqrt(a.^2 + b.^2)
c = 4×4
    13    10    13     5
    17    37    65    29
    41    85   145   101
    25    61   113   181
% Using for loops should be avoid for such problem in matlab
c = zeros(size(a));
for i=1:size(a,1)
    for j=1:size(a, 2)
        c(i,j) = sqrt(a(i,j)^2 + b(i,j)^2);
    end
end
c
c = 4×4
    13    10    13     5
    17    37    65    29
    41    85   145   101
    25    61   113   181