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

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];

Answer 1

[Chetan Bhavsar]

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

Answer 2

[Sean de Wolski]

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

Answer 3

[Chunru]

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

[Ashfaq Ahmed]

Thank you so much. But is there any specific reason why should I avoid for loop?

[Chunru]

MATLAB excels on the matrix computations. "c = sqrt(a.^2 + b.^2)" is much neater and more efficient that the codes with loops.