How to select specific entries of a matrix?

Let the matrix A be

A=[a_11 a_12 a_13 a_14;a_21 a_22 a_23 a_24;a_31 a_32 a_33 a_34; a_41 a_42 a_43 a_44];

how to select the following entries

a_12, a_14, a_21, a_23, a_32, a_34, a_41, a_43

In general, how to select the entries in the above pattern for large size of matrix.

The above entries can be obtaind by

for odd rows and even columns

B=A(1:2:end,2:2:end)

for even rows and odd columns

C=A(2:2:end,1:2:end)

but I want the combine results of B and C i.e.,

Ans1=[a_12 a14;a_21 a_23;a_32 a_34;a_41 a_43]

Ans2=[a_12 a14;a_23 a_21;a_32 a_34;a_43 a_41]

how can we obtained Ans 1 and Ans 2 for large square matrix.

Answer 1

[Awais Saeed]

I am writing a generic code so you can get an idea how to do what you need (which is how do you pick elements in particular manner).

clc;clear all;close all;
A = sym('A', [4 4])
% A = (symbolic 4×4 matrix)
% 
%   ⎡A₁₁  A₁₂  A₁₃  A₁₄⎤
%   ⎢                  ⎥
%   ⎢A₂₁  A₂₂  A₂₃  A₂₄⎥
%   ⎢                  ⎥
%   ⎢A₃₁  A₃₂  A₃₃  A₃₄⎥
%   ⎢                  ⎥
%   ⎣A₄₁  A₄₂  A₄₃  A₄₄⎦

result = sym('result', [4 2])
for row = 1:1:size(A,1)
	if (mod(row,2) ~= 0)    % check if row number is even or odd
		result(row,:) = [A(row,2:2:4)]
	else
		result(row,:) = [A(row,1:2:3)]
	end
end

% result = (sym 4×2 matrix)
% 
%   ⎡A₁₂  A₁₄⎤
%   ⎢        ⎥
%   ⎢A₂₁  A₂₃⎥
%   ⎢        ⎥
%   ⎢A₃₂  A₃₄⎥
%   ⎢        ⎥
%   ⎣A₄₁  A₄₃⎦

[Awais Saeed]

I think this should do the job or atleast you can have an idea

clc;clear all;close all;
A = [11 12 13 14 15 16; 21 22 23 24 25 26; 31 32 33 34 35 36; 41 42 43 44 45 46]

for row = 1:1:size(A,1)
	if (mod(row,2) ~= 0 ) % odd rows
		Ans1(row,:) = [A(row,2:2:end)];
	else
		Ans2(row,:) = [A(row,1:2:end)];
	end
end


% To remove all zeros rows from Ans1 and Ans2
ind = find(sum(Ans1,2)==0) ;
Ans1(ind,:) = []
ind = find(sum(Ans2,2)==0) ;
Ans2(ind,:) = []

[Ammy]

@Awais Saeed thank you very much , but I want a comblined answer.

[Awais Saeed]

clc;clear all;close all;
A = [11 12 13 14 15 16; 21 22 23 24 25 26; 31 32 33 34 35 36; 41 42 43 44 45 46]
for row = 1:1:size(A,1)
	if (mod(row,2) ~= 0 ) % odd rows
		Ans1(row,:) = [A(row,2:2:end)];
	else
		Ans1(row,:) = [A(row,1:2:end)];
	end
end

[Ammy]

Thank you very much, it works.

please help me to accept this answer, there is no option of acceptance on this answer.

[Awais Saeed]

You are welcome. All three snippets are actually doing the same thing. Its just the form of answers that are different (separate or combined or generic etc). Accepting this thread means you have been answered.