How can I Perform bitxor operation in matrix?

Assume You have two matrix A and B with even number of rows as shown below. I want the first column in A to undergo bitxor with the second column in B, and the second column in A with bitxor with first column in B. The same approach be applied with column 3 in A with column 4 in B as well as column 4 in A with third column in B.Assume you n even equal number of rows in both matrix. Is it possible to use for loop?

A=[1 2 3 4] and B=[11 22 33 44;3 4 5 6]

example bitxor(1,22),bitxor(1,4),bitxor(2,11),bitxor(3,44),bitxor(3,66),bitxor(4,33)

[Chunru]

"A" has a single row.

[dani elias]

Yes,A has only single row but B can have many rows and columns (the number of columns in both A and B are the same,(last column % 2=0))

Answer 1

[Bruno Luong]

A=[1 2 3 4], B=[11 22 33 44;3 4 5 6],

A = 1×4

1 2 3 4

B = 2×4

11 22 33 44 3 4 5 6

C=zeros(max(size(A),size(B)))

C = 2×4

0 0 0 0 0 0 0 0

C(:,1:2:end)=bitxor(A(:,1:2:end),B(:,2:2:end));
C(:,2:2:end)=bitxor(A(:,2:2:end),B(:,1:2:end));

C

C = 2×4

23 9 47 37 5 1 5 1

[dani elias]

Thank you for the answer,this work better as i want.Thank you once again