find 4 distinct vectors in F^3 that are linearly independent x+y+z=0

To find four distinct vectors in \(\mathbb{F}^3\) (where \(\mathbb{F}\) is a field) that are linearly independent and satisfy the equation \(x + y + z = 0\), note that it's impossible to find four such vectors in \(\mathbb{F}^3\). This is because the dimension of \(\mathbb{F}^3\) is 3, and any set of more than three vectors in \(\mathbb{F}^3\) must be linearly dependent.

However, I can help you find **three distinct linearly independent vectors** in \(\mathbb{F}^3\) that satisfy \(x + y + z = 0\).

Let's derive those vectors:

### Equation: \(x + y + z = 0\)

1. **Choose \(x\) and \(y\) freely**: Since the equation is linear, you can choose two of the variables freely and solve for the third.
   
2. **Solve for \(z\)**: \(z = -x - y\).

Let's find three distinct vectors:

- **Vector 1**: Let \(x = 1\) and \(y = 0\), then \(z = -1\). So, the first vector is \(\mathbf{v}_1 = (1, 0, -1)\).
  
- **Vector 2**: Let \(x = 0\) and \(y = 1\), then \(z = -1\). So, the second vector is \(\mathbf{v}_2 = (0, 1, -1)\).
  
- **Vector 3**: Let \(x = 1\) and \(y = 1\), then \(z = -2\). So, the third vector is \(\mathbf{v}_3 = (1, 1, -2)\).

### Linearly Independent Verification:
These vectors \(\mathbf{v}_1 = (1, 0, -1)\), \(\mathbf{v}_2 = (0, 1, -1)\), and \(\mathbf{v}_3 = (1, 1, -2)\) can be checked for linear independence by verifying that the only solution to \(c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{0}\) is \(c_1 = c_2 = c_3 = 0\).

Let's set up the equation:

\[
c_1(1, 0, -1) + c_2(0, 1, -1) + c_3(1, 1, -2) = (0, 0, 0)
\]

This gives us the following system of equations:

\[
c_1 + c_3 = 0 \quad (1)
\]
\[
c_2 + c_3 = 0 \quad (2)
\]
\[
-c_1 - c_2 - 2c_3 = 0 \quad (3)
\]

From (1), \(c_3 = -c_1\). From (2), \(c_3 = -c_2\). Thus, \(c_1 = c_2\).

Substituting into (3), we have:

\[
-c_1 - c_2 - 2(-c_1) = 0
\]
\[
-c_1 - c_1 + 2c_1 = 0 \implies 0 = 0
\]

Thus, \(c_1 = c_2 = c_3 = 0\), confirming that the vectors are linearly independent.

### Conclusion
The three distinct linearly independent vectors in \(\mathbb{F}^3\) that satisfy \(x + y + z = 0\) are:

\[
\mathbf{v}_1 = (1, 0, -1), \quad \mathbf{v}_2 = (0, 1, -1), \quad \mathbf{v}_3 = (1, 1, -2)
\]

Do you have any questions or need further details? 

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Here are five related questions:

1. What are the necessary and sufficient conditions for vectors to be linearly independent?
2. How would the problem change if the equation was \(x + y + z = 1\)?
3. Can we find four linearly dependent vectors in \(\mathbb{F}^3\) that satisfy \(x + y + z = 0\)?
4. What is the geometric interpretation of the vectors satisfying \(x + y + z = 0\) in \(\mathbb{F}^3\)?
5. How would you determine if a set of vectors in \(\mathbb{F}^n\) is linearly independent?

**Tip:** Remember that the maximum number of linearly independent vectors in \(\mathbb{F}^n\) is \(n\). Any set of more than \(n\) vectors in \(\mathbb{F}^n\) is linearly dependent.

Explore how to find four distinct vectors in F^3 that are linearly independent and satisfy the equation x + y + z = 0. Learn the method to derive three linearly independent vectors (1, 0, -1), (0, 1, -1), and (1, 1, -2) and verify their independence. Suitable for college-level students studying linear algebra and vector spaces.

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