Can someone do this calculation without for loops ?

a = [1 2 3; 4 5 6];
b = [ 1 2 3];
for n = 1: size(a,1)
    for m = 1:size(a,2)
k(n,m,:)= b.*b*a(n,m)
    end
end

Answer 1

[José-Luis]

 k = bsxfun(@times, a , reshape(b.^2,1,1,[]))

 alt_k = bsxfun(@times, a , permute(b.^2,[3,1,2]))

[Amelos]

this one works also , thanks k = reshape(kron(b.*b,a),[size(a),numel(b.*b)]);

Answer 2

[Joakim Magnusson]

Do you mean like this?

fun=@(a,b) b.*b*a
k = bsxfun(fun,a,b)

Answer 3

[Azzi Abdelmalek]

a = [1 2 3; 4 5 6];
b = [ 1 2 3];
bb=reshape(b.*b,1,1,[])
out=bsxfun(@times,a,bb)