multiplying two matrices in an efficient way

Dear all,

 I have

A= [ 29     1;
  27     4;
  31     0;
  28     0;
  28     3;
  25     5;
  23     8;
  20    10;
  18    13;
  15    16;
  29     1;
  27     4;
  31     0;
  28     0;
  28     3;
  25     5;
  23     8;
  20    10;
  18    13;
  15    16]

and

 B=[0.4607    0.4400    0.9167    0.8701;
    0.4400    0.4432    0.8701    0.8808;
    0.4432    0.4419    0.8808    0.8809;
    0.4419    0.4607    0.8809    0.9156;
    0.4607    0.4547    0.9156    0.9039;
    0.4547    0.4377    0.9039    0.8679;
    0.4377    0.4461    0.8679    0.8839;
    0.4461    0.4396    0.8839    0.8740;
    0.4396    0.4409    0.8740    0.8752;
    0.4409    0.4409    0.8752    0.8752;
    0.8487    0.8742    1.6974    1.7484;
    0.8742    0.7616    1.7484    1.5232;
    0.7616    0.7522    1.5232    1.5044;
    0.7522    0.8492    1.5044    1.6984;
    0.8492    0.8986    1.6984    1.7971;
    0.8986    0.9138    1.7971    1.8275;
    0.9138    0.7911    1.8275    1.5821;
    0.7911    0.9008    1.5821    1.8016;
    0.9008    0.9285    1.8016    1.8570;
    0.9285    0.9285    1.8570    1.8570]

And I want to calculate

sum(A.*B(:,1:2),2)

sum(A.*B(:,3:4),2)

in one step because my B contains 120 columns and I want to multiply A with every 2 columns of B that is B(:,1:2) B(:,3:4) B(:,5:6)

S0, I am looking something like

sum(A.*[B(:,1:2) B(:,3:4) B(:,5:6) ],2)

 thanks

Answer 1

[Oleg Komarov]

squeeze(sum(bsxfun(@times, reshape(B,20,2,[]),A),2))

[Sabbas]

thank you both!

[Oleg Komarov]

@Azzi: "must" is a strong word when the final request is "something like" and in clear contraddiction with what he "wants" to calculate.

Answer 2

[Mon Beng]

5044

Answer 3

[Mon Beng]

if true
  % code
end 5044