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The problem asks to find two sets of \( y_1, y_2, y_3 \) such that the linear combination of the vectors \( r_1, r_2, r_3 \) results in the zero vector:

\[
y_1 \mathbf{r_1} + y_2 \mathbf{r_2} + y_3 \mathbf{r_3} = 0
\]

Where the vectors \( \mathbf{r_1}, \mathbf{r_2}, \mathbf{r_3} \) are given as:

\[
\mathbf{r_1} = \begin{pmatrix} 1 \\ 4 \\ 7 \end{pmatrix}, \quad \mathbf{r_2} = \begin{pmatrix} 2 \\ 5 \\ 8 \end{pmatrix}, \quad \mathbf{r_3} = \begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix}
\]

The equation to solve is:

\[
y_1 \begin{pmatrix} 1 \\ 4 \\ 7 \end{pmatrix} + y_2 \begin{pmatrix} 2 \\ 5 \\ 8 \end{pmatrix} + y_3 \begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\]

This gives us the following system of linear equations:

\[
\begin{aligned}
1. & \quad y_1 + 2y_2 + 3y_3 = 0 \\
2. & \quad 4y_1 + 5y_2 + 6y_3 = 0 \\
3. & \quad 7y_1 + 8y_2 + 9y_3 = 0 \\
\end{aligned}
\]

To solve this system, we can set up the augmented matrix and perform row reduction (Gaussian elimination) or recognize that the vectors are linearly dependent (since the third vector is a linear combination of the first two).

### Step 1: Form the matrix equation
\[
\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{pmatrix}
\begin{pmatrix}
y_1 \\
y_2 \\
y_3
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0
\end{pmatrix}
\]

### Step 2: Apply Gaussian elimination
Notice that row 3 can be obtained by adding row 1 and row 2 (times appropriate coefficients), confirming that the vectors are indeed linearly dependent.

### Step 3: Find the solutions
This system has an infinite number of solutions. To find two sets, let's express \( y_1, y_2, y_3 \) in terms of a free variable \( y_3 = t \).

From the first equation:
\[
y_1 = -2y_2 - 3t
\]

From the second equation:
\[
4(-2y_2 - 3t) + 5y_2 + 6t = 0
\]

Simplifying:
\[
-8y_2 - 12t + 5y_2 + 6t = 0 \implies -3y_2 - 6t = 0 \implies y_2 = -2t
\]

Substituting back:
\[
y_1 = -2(-2t) - 3t = 4t - 3t = t
\]

Thus, the general solution is:
\[
y_1 = t, \quad y_2 = -2t, \quad y_3 = t
\]

### Step 4: Two specific solutions
1. Set \( t = 1 \): 
\[
y_1 = 1, \quad y_2 = -2, \quad y_3 = 1
\]
2. Set \( t = 2 \):
\[
y_1 = 2, \quad y_2 = -4, \quad y_3 = 2
\]

### Final Answer
Two sets of solutions for \( y_1, y_2, y_3 \) are:

1. \( (1, -2, 1) \)
2. \( (2, -4, 2) \)

Would you like more details, or do you have any questions?

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Here are 5 questions related to this topic:
1. How do you determine if a set of vectors is linearly dependent?
2. What is the significance of the free variable in a system of linear equations?
3. How can Gaussian elimination be used to solve a system of equations?
4. What are the possible geometric interpretations of the solution to this problem?
5. What would happen if the vectors were linearly independent?

**Tip:** Always check for linear dependence by computing the determinant or by performing row operations to avoid unnecessary calculations.

The rows of a matrix M produce three vectors r1, r2, r3 (when taken as columns). Find two sets of y1, y2, y3 such that the linear combination y1r1 + y2r2 + y3r3 = 0.

Explore how to solve a problem involving linear combinations of vectors in a plane. Learn to find two sets of coefficients y1, y2, y3 such that the sum of vectors r1, r2, r3 equals the zero vector. This problem covers key concepts in linear algebra and is suitable for undergraduate students.

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