Given the basis vectors e1 = [1, 0, 0], [0, 1, 1], e3 = [0, 0, 1] for E3
. Which vectors can be removed form the 
basis and be replaced by b = [4, 3, 6] while still remaining a basis? Illustrate this geometrically.

To determine which vectors can be removed from the given basis \(\mathbf{e_1} = [1, 0, 0]\), \(\mathbf{e_2} = [0, 1, 1]\), \(\mathbf{e_3} = [0, 0, 1]\) in \( \mathbb{R}^3 \) and replaced by \(\mathbf{b} = [4, 3, 6]\) while still maintaining a basis, we need to check the linear independence of the vectors after substitution.

### Steps:

1. **Understand the Basis Requirement:**
   - A basis in \( \mathbb{R}^3 \) requires three linearly independent vectors.
   - The given vectors \(\mathbf{e_1}\), \(\mathbf{e_2}\), and \(\mathbf{e_3}\) form a basis since they span \( \mathbb{R}^3 \).

2. **Substitution and Linear Independence:**
   - Replace \(\mathbf{e_1}\), \(\mathbf{e_2}\), or \(\mathbf{e_3}\) with \(\mathbf{b}\), and check if the new set of vectors remains linearly independent.

### Case 1: Replace \(\mathbf{e_1}\) with \(\mathbf{b}\)
- New set: \(\mathbf{b} = [4, 3, 6]\), \(\mathbf{e_2} = [0, 1, 1]\), \(\mathbf{e_3} = [0, 0, 1]\).
- To check linear independence, compute the determinant of the matrix formed by these vectors:
  
  \[
  \text{det} \begin{bmatrix} 4 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 1 & 1 \end{bmatrix} = 4(1 \cdot 1 - 0 \cdot 1) - 0 + 0 = 4 \neq 0
  \]
  
- Since the determinant is non-zero, the set \(\{\mathbf{b}, \mathbf{e_2}, \mathbf{e_3}\}\) is linearly independent.

### Case 2: Replace \(\mathbf{e_2}\) with \(\mathbf{b}\)
- New set: \(\mathbf{e_1} = [1, 0, 0]\), \(\mathbf{b} = [4, 3, 6]\), \(\mathbf{e_3} = [0, 0, 1]\).
- Compute the determinant:
  
  \[
  \text{det} \begin{bmatrix} 1 & 4 & 0 \\ 0 & 3 & 0 \\ 0 & 6 & 1 \end{bmatrix} = 1(3 \cdot 1 - 0 \cdot 6) - 4(0 \cdot 1 - 0 \cdot 6) + 0 = 3 \neq 0
  \]

- Since the determinant is non-zero, the set \(\{\mathbf{e_1}, \mathbf{b}, \mathbf{e_3}\}\) is linearly independent.

### Case 3: Replace \(\mathbf{e_3}\) with \(\mathbf{b}\)
- New set: \(\mathbf{e_1} = [1, 0, 0]\), \(\mathbf{e_2} = [0, 1, 1]\), \(\mathbf{b} = [4, 3, 6]\).
- Compute the determinant:
  
  \[
  \text{det} \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 3 \\ 0 & 1 & 6 \end{bmatrix} = 1(1 \cdot 6 - 1 \cdot 3) = 1(6 - 3) = 3 \neq 0
  \]

- Since the determinant is non-zero, the set \(\{\mathbf{e_1}, \mathbf{e_2}, \mathbf{b}\}\) is linearly independent.

### Conclusion:
- Any of the vectors \(\mathbf{e_1}\), \(\mathbf{e_2}\), or \(\mathbf{e_3}\) can be replaced by \(\mathbf{b}\) to still form a valid basis in \( \mathbb{R}^3 \).

### Geometric Illustration:
- Geometrically, the replacement of any of the original basis vectors with \(\mathbf{b}\) changes the direction of one of the axes, but the three vectors will still span the three-dimensional space, forming a new basis.

Would you like a more detailed explanation or a visual illustration of this result?

Explore how to replace basis vectors in R^3, specifically e1 = [1, 0, 0], e2 = [0, 1, 1], e3 = [0, 0, 1], with a new vector b = [4, 3, 6], while ensuring the resulting set remains linearly independent. Understand the geometric implications of such replacements in three-dimensional space.

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