- #1
Philip Wong
- 95
- 0
hi guys,
I have two homogeneous systems S1 and S2. The solution for
NS(S1) = {[-2,1,0], [-1,0,1]},
NS(S2) = {[-2,1,0], [-3,0,1]}.
I know that in a system if u and v are vectors, the sum of u+v is also a solution in the homogeneous system. i.e. S1=span{[-2,1,0], [-1,0,1]} then [-2,1,0] + [-1,0,1] is also a solution in that system (close under addition if I am correct).
But what about if I want to find the set of vectors which are solution to both S1 and S2? Do I use the same methods to it?
i.e. Solution = span {[-2,1,0]+[-2,1,0] ; [-1,0,1]+[-3,0,1]}
I think this is wrong, but I'm not sure how
I have two homogeneous systems S1 and S2. The solution for
NS(S1) = {[-2,1,0], [-1,0,1]},
NS(S2) = {[-2,1,0], [-3,0,1]}.
I know that in a system if u and v are vectors, the sum of u+v is also a solution in the homogeneous system. i.e. S1=span{[-2,1,0], [-1,0,1]} then [-2,1,0] + [-1,0,1] is also a solution in that system (close under addition if I am correct).
But what about if I want to find the set of vectors which are solution to both S1 and S2? Do I use the same methods to it?
i.e. Solution = span {[-2,1,0]+[-2,1,0] ; [-1,0,1]+[-3,0,1]}
I think this is wrong, but I'm not sure how