B=[1 2 3
     2 2 4
     3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
  K=null(B'); % the columns of K are a basis of the Kernel
  V=K(:,1); % in fact, in this case, V and K are identical
  a=rand(1,3);
  A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
  A*B,%test : must be zero