# Avoid for loop in multi-dimensional problem

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Howard Wilton on 21 Nov 2022
Commented: Howard Wilton on 21 Nov 2022
I am looking to create a 3D matrix the achieves the following:
A(:,:,1)=[0 -2*pi; 2*pi 0]; A(:,:,2)=[0 -4*pi; 4*pi 0];
Ultimately, I am trying to solve , which in this instance will evaluate to 0. But for now, I just want to figure out how to build the matrix .
I have the following code:
l=0:1; q=0:1; p=2*pi*(1:2);
sub = l(:)-q(:).' % gives a 2x2 matrix
sub2 = repmat(sub,[1 1 2]) % repeats the 2x2 matrix to give a 2x2x2 matrix
for p_ = 1:length(p)
A(:,:,p_) = p(p_)*sub2(:,:,p_);
end
I am looking to conduct a multiplation that yields without having to use a for loop. Any suggestions would be appreciated.

Stephen23 on 21 Nov 2022
Edited: Stephen23 on 21 Nov 2022
Some ideas:
format compact
A = cat(3,[0,-2*pi;2*pi,0],[0,-4*pi;4*pi,0]) % use CAT()
A =
A(:,:,1) = 0 -6.2832 6.2832 0 A(:,:,2) = 0 -12.5664 12.5664 0
A = pi*cat(3,[0,-2;2,0],[0,-4;-4,0]) % take out the commom PI
A =
A(:,:,1) = 0 -6.2832 6.2832 0 A(:,:,2) = 0 -12.5664 -12.5664 0
A = pi*cat(3,2,4).*[0,-1;1,0] % multiply two arrays
A =
A(:,:,1) = 0 -6.2832 6.2832 0 A(:,:,2) = 0 -12.5664 12.5664 0
A = cat(3,2,4).*[0,-pi;pi,0] % multiply two arrays
A =
A(:,:,1) = 0 -6.2832 6.2832 0 A(:,:,2) = 0 -12.5664 12.5664 0
Howard Wilton on 21 Nov 2022
thank you sir!

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