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.