Combination of rows of two different matrices

Sorry if this is a repeated question but I failed to find an answer myself. I have two matrices:

A = [-0.6, -0.2;
     -60,   2;
      6,   -20];
 
B = [-0.4, -0.8;
     -40,   8;
      4,   -80];

I want to find all the possible combinations of sum of each row (sum of individual elements of a row) of A with each row of B, i.e., my desired result is (order does not matter):

ans = [-1,     -1;
       -40.6,   7.8;
        3.4,   -80.2;
       -60.4,   1.2;
       -100,    10;
       -56,    -78;
        5.6,   -20.8;
       -34,    -12;
        10,    -100];

which is a matrix resulting from possible combinations of A and B.

Thanks in advance.

(Please no for loops. It is pretty trivial then.)

EDIT: I have used 2 columns and 3 rows as an example. Looking for more general solution, i.e., for n number of columns and m number of rows.

Answer 1

[Fangjun Jiang]

Feels non-ideal. Any better solution?

>> C=A(:,1)+B(:,1)';
D=A(:,2)+B(:,2)';
reshape([C(:),D(:)],[],2)
ans =
   -1.0000   -1.0000
  -60.4000    1.2000
    5.6000  -20.8000
  -40.6000    7.8000
 -100.0000   10.0000
  -34.0000  -12.0000
    3.4000  -80.2000
  -56.0000  -78.0000
   10.0000 -100.0000

better one

>> m=size(A,1);
ind1=repmat(1:m,1,m);
ind2=repelem(1:m,m);
A(ind1,:)+B(ind2,:)
ans =
   -1.0000   -1.0000
  -60.4000    1.2000
    5.6000  -20.8000
  -40.6000    7.8000
 -100.0000   10.0000
  -34.0000  -12.0000
    3.4000  -80.2000
  -56.0000  -78.0000
   10.0000 -100.0000

[J AI]

This works, but I should have been a bit more specific with my question - the number of columns I have used is 2, however, it may vary. So I was looking for a more general solution. Thanks anyway!

[J AI]

Great work! Thanks a lot

Answer 2

[Bruno Luong]

A = [-0.6, -0.2;
     -60,   2;
      6,   -20];
 
B = [-0.4, -0.8;
     -40,   8;
      4,   -80];

Single statement

  
reshape(permute(A,[3 1 2])+permute(B,[1 3 2]),[],size(A,2))

or a variation

reshape(reshape(A,1,size(A,1),[])+reshape(B,size(B,1),1,[]),[],size(A,2))

Gives

ans =

   -1.0000   -1.0000
  -40.6000    7.8000
    3.4000  -80.2000
  -60.4000    1.2000
 -100.0000   10.0000
  -56.0000  -78.0000
    5.6000  -20.8000
  -34.0000  -12.0000
   10.0000 -100.0000

>> 

[J AI]

These ones are actually faster for larger data - the variation being the fastest. Thanks a lot for your contribution!

[Bruno Luong]

Yes the variation version does no more no less than the required combination sums and puts at the results at the right place. Not a hair uneccesary arithmetic or memory moving (first version).

[Fangjun Jiang]

Brilliant, Bruno Luong! I learned implicit expansion in a whole new dimension.