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

12 views (last 30 days)
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];

Accepted Answer

Chetan Bhavsar
Chetan Bhavsar on 8 Nov 2021
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

More Answers (2)

Sean de Wolski
Sean de Wolski on 8 Nov 2021
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

Chunru
Chunru on 8 Nov 2021
Edited: Chunru on 8 Nov 2021
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
  2 Comments
Chunru
Chunru on 8 Nov 2021
MATLAB excels on the matrix computations. "c = sqrt(a.^2 + b.^2)" is much neater and more efficient that the codes with loops.

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by