Final answer:
To show that S = {(0, -1, 0, 1, 1, 0), (1, 0, 1, 1, 1, 0)} is a linearly independent subset of V, we need to show that no nontrivial linear combination of the vectors in S equals the zero vector. To extend S to a basis for V, we need to find one additional vector that is linearly independent from the vectors in S and spans the entire space V.
Step-by-step explanation:
To show that S = {(0, -1, 0, 1, 1, 0), (1, 0, 1, 1, 1, 0)} is a linearly independent subset of V, we need to show that no nontrivial linear combination of the vectors in S equals the zero vector. In other words, we want to find values of a and b such that a(0, -1, 0, 1, 1, 0) + b(1, 0, 1, 1, 1, 0) = (0, 0, 0, 0, 0, 0) does not hold.
If we set up the system of equations, we get:
a + b = 0
-a + b = 0
2b = 0
We can easily solve this system of equations:
b = 0
a = 0
Since the only solution is a = 0 and b = 0, we can conclude that S is a linearly independent subset of V.
To extend S to a basis for V, we need to find one additional vector that is linearly independent from the vectors in S and spans the entire space V. One way to do this is by finding a solution to the system of linear equations and choosing a vector that is not in the span of S. Once we find such a vector, we can add it to S to create a basis for V.