Answer:
(u3,u4,u5)
Step-by-step explanation:
For finding a subset x of the set {u1, u2, u3, u4, u5} that is a basis for R3 we need to prove that x is linear independent and it's a generator of R3.
(0,0,0)=a(1,2,-1)+b(2,6,6)+c(-1,-3,-3)
(0,0,0)=(a+2b-c,2a+6b-3c,-a+6b-3c)
the system is not linear independent so (u1,u2,u3) can not be a basis of R3.
If we do the same procedure with u3 = (−1, −3, −3), u4 = (0, 2, 8), and u5 = (3, 7, −3)
(0,0,0)=a(-1,-3,-3)+b(0,2,8)+c(3,7,-3)
(0,0,0)=(-a+0b+3c,-3a+2b+7c,-3a+8b-3c)
If we solve the system of equations we have that: a=0, b=0, c=0. So the subset is linear independent.
Then we need to prove that it's a generator of R3
(x,y,z)=a(-1,-3,-3)+b(0,2,8)+c(3,7,-3)
(x,y,z)=(-a+0b+3c,-3a+2b+7c,-3a+8b-3c)
If we solve the system of equations we have that: x=-a+3c, y=-3a+2b+7c, z=-3a+8b-3c. So the subset is linear combination into R3 so it generates R3.
