Final answer:
Set (A) of vectors of the form (a,0,0) is a subspace of ℝ^3 as it includes the zero vector and is closed under addition and scalar multiplication. Set (B) of vectors of the form (a,1,1) is not a subspace of ℝ^3 because it does not include the zero vector and is not closed under these operations.
Step-by-step explanation:
The student's question asks for the determination of which sets are subspaces of ℝ^3 (the set of all 3-dimensional real vectors) using a certain theorem. Specifically, they ask about the sets of vectors of the form (a,0,0) and (a,1,1), where 'a' can be any real number.
To determine if a set is a subspace, it must satisfy three conditions:
- It must contain the zero vector.
- It must be closed under vector addition.
- It must be closed under scalar multiplication.
For set (A), all vectors of the form (a,0,0), it includes the zero vector (0,0,0), is closed under vector addition ((a,0,0) + (b,0,0) = (a+b,0,0)), and is closed under scalar multiplication (k*(a,0,0) = (ka,0,0)). Therefore, set (A) is a subspace of ℝ^3.
For set (B), vectors of the form (a,1,1), it does contain the zero vector, as there is no 'a' such that (a,1,1) equals (0,0,0). Even if we ignore the first condition, this set is also not closed under vector addition or scalar multiplication as, for example, if we multiply any vector in set (B) by the scalar 0, we do not get a vector in set (B). Therefore, set (B) is not a subspace of ℝ^3.