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Let's break down the problem:

You are given a homogeneous linear system over \( \mathbb{R} \) with four equations and three unknowns. It is known that the vectors
\[
\mathbf{v_1} = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \quad \text{and} \quad \mathbf{v_2} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}
\]
are solutions to the system.

The questions are:

1. Can we determine if the vector \( \mathbf{v} = \begin{bmatrix} 1 \\ -5 \\ -2 \end{bmatrix} \) is a solution to the system?
2. Can we determine if the vector \( \mathbf{v} = \begin{bmatrix} 1 \\ -5 \\ 1 \end{bmatrix} \) is a solution to the system?

### Solution Approach

Since this is a homogeneous system, any solution vector \( \mathbf{v} \) must be a linear combination of \( \mathbf{v_1} \) and \( \mathbf{v_2} \) to satisfy the system. To determine if a vector is a solution, we need to check if it can be expressed as
\[
\mathbf{v} = a \mathbf{v_1} + b \mathbf{v_2}
\]
for some scalars \( a \) and \( b \).

1. **For** \( \mathbf{v} = \begin{bmatrix} 1 \\ -5 \\ -2 \end{bmatrix} \):
   - Set up the equation:
     \[
     a \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + b \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -5 \\ -2 \end{bmatrix}
     \]
   - This gives the system:
     \[
     a + b = 1
     \]
     \[
     -a + b = -5
     \]
     \[
     b = -2
     \]
   - Solving this system will reveal if such \( a \) and \( b \) exist.

2. **For** \( \mathbf{v} = \begin{bmatrix} 1 \\ -5 \\ 1 \end{bmatrix} \):
   - Set up the equation similarly:
     \[
     a \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + b \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -5 \\ 1 \end{bmatrix}
     \]
   - This gives the system:
     \[
     a + b = 1
     \]
     \[
     -a + b = -5
     \]
     \[
     b = 1
     \]
   - Solve this system to determine if such \( a \) and \( b \) exist.

Let me calculate these values for clarity.The solutions to the two questions are as follows:

1. For the vector \( \mathbf{v} = \begin{bmatrix} 1 \\ -5 \\ -2 \end{bmatrix} \):
   - We found that \( a = 3 \) and \( b = -2 \) satisfy the system of equations. Therefore, this vector **can be expressed** as a linear combination of \( \mathbf{v_1} \) and \( \mathbf{v_2} \), and it **is a solution** to the system.

2. For the vector \( \mathbf{v} = \begin{bmatrix} 1 \\ -5 \\ 1 \end{bmatrix} \):
   - There is no solution for \( a \) and \( b \) that satisfies the system of equations. Thus, this vector **cannot be expressed** as a linear combination of \( \mathbf{v_1} \) and \( \mathbf{v_2} \), and it **is not a solution** to the system.

Would you like further details or have any questions?

Here are some related questions to explore further:
1. How do you determine if a vector is a solution to a homogeneous system?
2. What are the properties of homogeneous systems in linear algebra?
3. Can a homogeneous system with more equations than unknowns have non-trivial solutions?
4. How does linear dependence between vectors affect the solution set of a system?
5. What are the implications of a solution being expressible as a linear combination of basis vectors?

**Tip:** In a homogeneous system, if you know some solutions, any linear combination of these solutions will also be a solution.

Given a homogeneous linear system over R with four equations and three unknowns. It is known that the vectors v1 = [1, -1, 0] and v2 = [1, 1, 1] are solutions to the system. Determine if the vectors v = [1, -5, -2] and v = [1, -5, 1] are solutions to the system.

Learn how to determine if specific vectors are solutions to a given homogeneous linear system. This problem covers key linear algebra concepts, including vector spaces, linear combinations, and solving for linear dependence. Suitable for undergraduate students.

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