Question

Which of the following set of vectors forms the basis for \( \mathbb{R}^3 \)?

• A. \( S = \{(1, 1, 1), (1, 0, 1)\} \)
• B. \( S = \{(1, 1, 1), (1, 2, 3), (2, -1, 1)\} \)
• C. \( S = \{(1, 2, 3), (1, 3, 5), (1, 0, 1), (2, 3, 0)\} \)
• D. \( S = \{(1, 1, 2), (1, 2, 5), (5, 3, 4)\} \)

Hint:

To quickly check for a basis of \( \mathbb{R}^n \), first count the vectors. You need exactly \(n\) vectors. If you have \(n\) vectors, form a matrix and find its determinant. A non-zero determinant means you have a basis.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A set of vectors forms a basis for a vector space if two conditions are met: 1. The vectors are linearly independent. 2. The vectors span the vector space. For the vector space \( \mathbb{R}^3 \), which has a dimension of 3, any set of 3 linearly independent vectors will form a basis. Any set with fewer than 3 vectors cannot span \( \mathbb{R}^3 \), and any set with more than 3 vectors must be linearly dependent.

Step 2: Key Formula or Approach:
We can eliminate options based on the number of vectors. For the options with exactly 3 vectors, we can check for linear independence by forming a matrix with the vectors as rows or columns and calculating its determinant. If the determinant is non-zero, the vectors are linearly independent and thus form a basis for \( \mathbb{R}^3 \).

Step 3: Detailed Explanation:

Option A: \( S = \{(1, 1, 1), (1, 0, 1)\} \). This set has only 2 vectors. It cannot span the 3-dimensional space \( \mathbb{R}^3 \). 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. This set must be linearly dependent. Not a basis.
Option B: \( S = \{(1, 1, 1), (1, 2, 3), (2, -1, 1)\} \). This set has 3 vectors. We check for linear independence by calculating the determinant of the matrix formed by these vectors. \[ A_B = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{pmatrix} \] \[ \det(A_B) = 1(2 - (-3)) - 1(1 - 6) + 1(-1 - 4) = 1(5) - 1(-5) + 1(-5) = 5 + 5 - 5 = 5 \] Since the determinant is \( 5 \neq 0 \), 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. Let's check 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: Final Answer:
Only the set of vectors in option (B) is linearly independent and has the correct number of vectors to form a basis for \( \mathbb{R}^3 \).