Sum and Difference across row and column
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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?
采纳的回答
Stephen23
2020-5-8
>> sum(A(:,2))-A(2,1)-A(2,3)
ans = -0.19880
3 个评论
Muhammad Usman
2020-5-8
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
2020-5-8
更多回答（1 个）
Steven Lord
2020-5-8
0 个投票
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.
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