Step 1: Basis Fundamentals:
A basis for a vector space is a set of linearly independent vectors that span the space.
For \( \mathbb{R}^3 \), a 3-dimensional space, a basis consists of 3 linearly independent vectors. Fewer than 3 vectors cannot span \( \mathbb{R}^3 \), and more than 3 vectors are linearly dependent.
Step 2: Method:
Eliminate options based on the number of vectors. For sets of 3 vectors, check linear independence using the determinant of a matrix formed by the vectors. A non-zero determinant indicates linear independence and a basis for \( \mathbb{R}^3 \).
Step 3: Detailed Analysis:
Option A: \( S = \{(1, 1, 1), (1, 0, 1)\} \). This set has 2 vectors; it cannot span \( \mathbb{R}^3 \). Not a basis.
Option C: \( S = \{(1, 2, 3), (1, 3, 5), (1, 0, 1), (2, 3, 0)\} \). This set has 4 vectors in a 3-dimensional space; it must be linearly dependent. Not a basis.
Option B: \( S = \{(1, 1, 1), (1, 2, 3), (2, -1, 1)\} \). This set has 3 vectors. Check linear independence using the determinant:
\[ A_B = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{pmatrix} \]\
\[ \det(A_B) = 1(2 - (-3)) - 1(1 - 6) + 1(-1 - 4) = 1(5) - 1(-5) + 1(-5) = 5 + 5 - 5 = 5 \]\
Since the determinant is \( 5 eq 0 \), the vectors are linearly independent and form a basis for \( \mathbb{R}^3 \).
Option D: \( S = \{(1, 1, 2), (1, 2, 5), (5, 3, 4)\} \). This set has 3 vectors. Calculate the determinant:
\[ A_D = \begin{pmatrix} 1 & 1 & 2
1 & 2 & 5
5 & 3 & 4 \end{pmatrix} \]\
\[ \det(A_D) = 1(8 - 15) - 1(4 - 25) + 2(3 - 10) = 1(-7) - 1(-21) + 2(-7) = -7 + 21 - 14 = 0 \]\
Since the determinant is 0, the vectors are linearly dependent. Not a basis.
Step 4: Conclusion:
Only option (B) provides a linearly independent set of 3 vectors, forming a basis for \( \mathbb{R}^3 \).