The vectors u1 = (1, 2, −1), u2 = (2, 6, 6), u3 = (−1, −3, −3), u4 = (0, 2, 8), and u5 = (3, 7, −3) generate R 3 . Find a subset of the set {u1, u2, u3, u4, u5} that is a basis for R 3 . Clearly indicate and justify your final 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.

answer:

step-by-step explanation:

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step-by-step explanation: