How to consider elements from two different column matrix and perform set of arithmetical operation?

I am having two 7*1 column matrix. One matrix consists of 3 bit binary numbers and another matrix consists of 1 bit binary number. Now I have to perform some set of arithmetical equations by considering elements in both matrix. I have to repeat the arithmetical calculation 7 times. First time I will choose 1st element from matrix A and assume that only this element is active and remaining elements are inactive. The arithmetic operation performed is B-1. For example at first 110 is selected and assumed this number only is active and first element from B is subtracted with 1. Remaining elements are assumed inactive, so the value to B is assumed as 0.

Example:

A= [110; 101; 011; 111; 100; 001; 010]

B= [1; 1; 0; 0; 1; 0; 1]

Expected output:

1st choice - 110 assumed active, so corresponding B element is 1, hence (B-1) 1-1=0, remaining elements are assumed inactive, so B is considered as 0 for remaining 6 elements in matrix A. Hence expected results are [(1-1=0), (0-1=-1), (0-1=-1), (0-1=-1), (0-1=-1), (0-1=-1), (0-1=-1)].

2nd choice - 101 assumed active, so corresponding B element is 1, hence (B-1) 1-1=0, remaining elements are assumed inactive, so B is considered as 0 for remaining 6 elements in matrix A. Hence expected results are [(0-1=-1), (1-1=0), (0-1=-1), (0-1=-1), (0-1=-1), (0-1=-1), (0-1=-1)].

3rd choice - 011 assumed active, so corresponding B element is 0, hence (B-1) 0-1=0, remaining elements are assumed inactive, so B is considered as 0 for remaining 6 elements in matrix A. Hence expected results are [(0-1=-1), (0-1=-1), (0-1=-1), (0-1=-1), (0-1=-1), (0-1=-1), (0-1=-1)].

C= 1st choice 2nd choice 3rd choice 4th choice 5th choice 6th choice 7th choice

0 -1 -1 -1 -1 -1 -1

-1 0 -1 -1 -1 -1 -1

-1 -1 -1 -1 -1 -1 -1

-1 -1 -1 -1 -1 -1 -1

-1 -1 -1 -1 0 -1 -1

-1 -1 -1 -1 -1 -1 -1

-1 -1 -1 -1 -1 -1 0

Answer 1

[Voss]

% A = [110; 101; 011; 111; 100; 001; 010]; % final result doesn't depend on the 
                                           % value of A (is that right?)
B = [1; 1; 0; 0; 1; 0; 1];

N = numel(B);
C = eye(N) - 1;
C(1:N+1:end) = B-1

C = 7×7

0 -1 -1 -1 -1 -1 -1 -1 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 0

[ASHA PON]

Thank you for the reply. This is what I needed.