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To solve this problem, we need to determine if the vectors \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} \), and \( \mathbf{v}_3 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} \) are linearly independent or dependent.

### Part (a): Determine if the set \(\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}\) is linearly independent

To check for linear independence, we need to see if the equation 
\[
c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = 0
\]
has only the trivial solution \(c_1 = c_2 = c_3 = 0\).

We can set up the matrix equation:
\[
\begin{bmatrix}
1 & 4 & 2 \\
2 & 5 & 1 \\
3 & 6 & 0 \\
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2 \\
c_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}
\]

Next, we perform row reduction (Gaussian elimination) on the augmented matrix:
\[
\left[\begin{array}{ccc|c}
1 & 4 & 2 & 0 \\
2 & 5 & 1 & 0 \\
3 & 6 & 0 & 0 \\
\end{array}\right]
\]

After performing row reduction, if the matrix has a row of all zeros, it indicates that the vectors are linearly dependent. Otherwise, they are linearly independent.

Let's carry out the row operations:

1. **Row 2** \( R_2 = R_2 - 2R_1 \):
   \[
   \left[\begin{array}{ccc|c}
   1 & 4 & 2 & 0 \\
   0 & -3 & -3 & 0 \\
   3 & 6 & 0 & 0 \\
   \end{array}\right]
   \]

2. **Row 3** \( R_3 = R_3 - 3R_1 \):
   \[
   \left[\begin{array}{ccc|c}
   1 & 4 & 2 & 0 \\
   0 & -3 & -3 & 0 \\
   0 & -6 & -6 & 0 \\
   \end{array}\right]
   \]

3. **Row 3** \( R_3 = R_3 - 2R_2 \):
   \[
   \left[\begin{array}{ccc|c}
   1 & 4 & 2 & 0 \\
   0 & -3 & -3 & 0 \\
   0 & 0 & 0 & 0 \\
   \end{array}\right]
   \]

Since we ended up with a row of all zeros, the rank of the matrix is less than 3, indicating that the vectors are linearly dependent.

### Part (b): Find a linear dependence relation

Since the vectors are linearly dependent, there exists a non-trivial solution \(c_1, c_2, c_3\) such that
\[
c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = 0
\]

From the row-reduced form:
\[
c_1 + 4c_2 + 2c_3 = 0 \quad \text{(Row 1)}
\]
\[
-3c_2 - 3c_3 = 0 \quad \text{(Row 2)}
\]

Solving \( -3c_2 - 3c_3 = 0 \) gives \( c_2 = -c_3 \). Substituting \( c_2 = -c_3 \) in the first equation:
\[
c_1 + 4(-c_3) + 2c_3 = 0 \implies c_1 - 2c_3 = 0 \implies c_1 = 2c_3
\]

Thus, a linear dependence relation is:
\[
2c_3 \mathbf{v}_1 - c_3 \mathbf{v}_2 + c_3 \mathbf{v}_3 = 0
\]
or
\[
2 \mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 = 0
\]

This confirms that the vectors are linearly dependent.

Would you like more details or have any questions?

### Relative Questions:
1. How is the rank of a matrix related to the linear independence of vectors?
2. What does it mean for a set of vectors to span a space?
3. How can you determine if a matrix is invertible using row reduction?
4. What is the geometric interpretation of linearly dependent vectors in \(\mathbb{R}^3\)?
5. Can vectors in a set be linearly dependent if there are fewer vectors than components in each vector?

### Tip:
When checking for linear independence, the number of vectors should not exceed the dimension of the space. If it does, the vectors will always be linearly dependent.

Determine if the set {v1, v2, v3} is linearly independent. If possible, find a linear dependence relation among v1, v2, and v3.

Determine if the set {v1, v2, v3} is linearly independent by using row reduction techniques. Explore the solution using Gaussian elimination and learn how to find a linear dependence relation if the vectors are dependent.

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