row addition on matrix

Hello everyone, i got error in my coding , can you please help me.

A=[-0.5000, 2.5000, -4.0000, 5.0000, -3.5000,  2.5000]

B=[2.4217, -0.4930, 0, -0.4217, 0, 0
   0, 3.0265, -2.1892, 0, -2.3001, 0
   -2.1892, 0, 0, 0, -0.4217, 0.4696
   0, 0, 2.4217, -2.3001, 0, -3.6869]

here i want to add row element with each other and gives single row matrix. i.e. C11= 0-2.1892+Aij =-2.1892-0.500=-0.2676 (i have to avoid the element belongs to current value of i and j)

C=[-0.2676, 5.0334, -3.7676, 2.2783, -6.2217, -0.7173]

how can i get this single row matrix using mat lab code? can anyone help me please.

    for i=1:4
  for j=1:6
  sum=0;
  for jj=1:6
  if H(jj,j)==1 && jj~=i
  sum=sum+B(jj,j)
  end 
  end 
  if H(jj,j)==1 && jj~=i
  C(i,j)=sum+A(i,j)
  end 
  end 

H=[1 1 0 1 0 0;
   0 1 1 0 1 1;
   1 0 0 0 1 1;
   0 0 1 1 0 1]

where, jj=set of row locations of the H matrix equation

[Guillaume]

Please use the formatting tools offered by this forum. There is a {}Code button that will format anything you paste as code.

I've formatted the post for you. Didn't you see that all your matrices appeared all as one line and therefore appeared to be row vectors? Obviously, the answer to your question is different for row vectors and for matrices.

Also, do not use sum for a variable name as it stops you from using the sum function.

[Nora]

Thanks, for doing this for me. i am beginner and very new to this site and still on learning phase.

Answer 1

[Andrei Bobrov]

C = sum([A;B]);

Answer 2

[Guillaume]

Your jj should go from 1 to 4 or better from 1 to size(H, 1), not 1 to 6. The loops are a complete waste of time in any case.

If you still have the sum variable in your workspace, please clear it otherwise my answer will not work:

clear sum

You say that you want to "avoid the element belongs to current value of i and j" which I understands to mean you don't want the elements on the diagonal (that's what your code appears to do). In that case:

C = A + sum(B .* H .* ~eye(size(B)))

However, your example of C does not do that and is simply:

C = A + sum(B .* H)

Basically, multiply B with H element-wise so only the elements of B you want to keep are non-zero, then use the sum function which sums all the rows together by default and leaves you with just one row. Add that to A.

[Nora]

"avoid the element belong to current value of i and j" exactly means if i am calculating the C11, then i cannot use the element of first row,first column i.e. 2.4217 and for C12; avoid the element of first row , second col i.e. -0.4930 and do summation with rest of row element.

[Guillaume]

But, yet the C11 you showed is equal to -0.2676 which is A + B11 + B31 = 0.5 + 2.4217 + -2.1892, so I'm confused.

I you want to ignore the first row of B:

C = A + sum(B(2:end, :) .* H(2:end, :))