Sum and Difference across row and column

Let say there is square matrix of order n-by-n, and n is odd integer

A = rand(3)
A =

    0.4218    0.9595    0.8491
    0.9157    0.6557    0.9340
    0.7922    0.0357    0.6787

I want to add the entries of second column and take difference of entries in second row to the middle number, i.e.

0.6557+0.9595+0.0357-0.9157-0.9340 = -0.1988

How can I do?

Answer 1

[Stephen Cobeldick]

>> sum(A(:,2))-A(2,1)-A(2,3)
ans = -0.19880

[Muhammad Usman]

I need a general form of this formula, like for a matrix of 11-by-11?

[Stephen Cobeldick]

For any square matrix where N is odd:

>> N = 3; % e.g. size(A,1)
>> X = 1+(N-1)/2; % the middle
>> sum(A(:,X))-sum(A(X,[1:X-1,X+1:N]))
ans = -0.19880

[Muhammad Usman]

Thanks

Answer 2

[Steven Lord]

You can do this with two simple sum calls and two basic arithmetic operations if you use the inclusion-exclusion principle. Since this sounds like it may be a homework assignment I'm not going to give the complete answer.