Vectors Linear Independent

by CSNabeel
Tags: independent, linear, vectors
 P: 12 1. The problem statement, all variables and given/known data Considering the following vectors R$$^{4}$$: v1 = (1,2,0,2) v2 = (2,3,1,4) v3 = (0,1,-1,0) Determine if these vectors are linearly independent. Let S be the linear span of the three vectors. Define a basis and the dimensions of S. Express the vector v=(3,5,1,6) as a linear combination of the three vectors. Can this be achieved in a unique way? Justify your answer? 2. Relevant equations I tried to put it into matrix form and reduce via row echolon but I'm not if this is the correct or proper way 3. The attempt at a solution [ 1 2 0 2 2 3 1 4 0 1 -1 0 3 5 1 6] [ 1 2 0 2 0 -1 1 0 0 1 -1 0 0 0 0 0 ] x +2y = 2 y - z = 0 -y + 2 = 0 therefore y=z making it linearly independent
 P: 365 You need to prove that p=q=r=0, for v1,v2,v3 to be linear independent: $$pv_1 + qv_2 +rv_3=0$$ $$p(1,2,0,2)+q(2,3,1,4)+r(0,1,-1,0)=0$$ You should express the vector v in same manner as linear combination of v1,v2,v3: i.e pv1+qv2+rv3=v p,q,r are random scalars. Regards.
 P: 12 so with that being said which of the two do I follow from below to work out the answer? a) 1p + 2q = 0 2p +3q +r = 0 q - r = 0 2p + 4q = 0 b) 1p + 2q = 3 2p +3q +r = 5 q - r = 1 2p + 4q = 6 and if I follow b I'm I right to think that p = 1 q =2 and r = 0
P: 365

Vectors Linear Independent

 Quote by CSNabeel so with that being said which of the two do I follow from below to work out the answer? a) 1p + 2q = 0 2p +3q +r = 0 q - r = 0 2p + 4q = 0 b) 1p + 2q = 3 2p +3q +r = 5 q - r = 1 2p + 4q = 6 and if I follow b I'm I right to think that p = 1 q =2 and r = 0