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Let's solve the system of linear equations:

\[
\begin{aligned}
x_1 + 3x_2 - 4x_3 &= 0 \quad \text{(1)} \\
2x_1 - 5x_2 &= 0 \quad \text{(2)} \\
-2x_1 - 6x_2 + 8x_3 &= 0 \quad \text{(3)}
\end{aligned}
\]

### Step 1: Solve for \(x_1\) and \(x_2\) using equation (2)

From equation (2):
\[
2x_1 - 5x_2 = 0
\]
Solve for \(x_1\):
\[
x_1 = \frac{5}{2}x_2 \quad \text{(4)}
\]

### Step 2: Substitute \(x_1\) in equation (1)

Substitute \(x_1 = \frac{5}{2}x_2\) into equation (1):
\[
\frac{5}{2}x_2 + 3x_2 - 4x_3 = 0
\]
Simplifying:
\[
\left(\frac{5}{2} + 3\right)x_2 - 4x_3 = 0
\]
\[
\left(\frac{5}{2} + \frac{6}{2}\right)x_2 - 4x_3 = 0
\]
\[
\frac{11}{2}x_2 = 4x_3
\]
Solving for \(x_3\):
\[
x_3 = \frac{11}{8}x_2 \quad \text{(5)}
\]

### Step 3: Substitute \(x_1\) and \(x_3\) in equation (3)

Substitute \(x_1 = \frac{5}{2}x_2\) and \(x_3 = \frac{11}{8}x_2\) into equation (3):
\[
-2\left(\frac{5}{2}x_2\right) - 6x_2 + 8\left(\frac{11}{8}x_2\right) = 0
\]
Simplifying:
\[
-5x_2 - 6x_2 + 11x_2 = 0
\]
\[
0 = 0
\]
This equation is always true, so it provides no additional information. Thus, we only have the relationships between the variables as given in equations (4) and (5).

### Step 4: Conclusion

Let \(x_2 = t\), where \(t\) is a free parameter. Then the general solution is:
\[
x_1 = \frac{5}{2}t, \quad x_2 = t, \quad x_3 = \frac{11}{8}t
\]
Thus, the solution is:
\[
(x_1, x_2, x_3) = t\left(\frac{5}{2}, 1, \frac{11}{8}\right)
\]

The **trivial solution** occurs when \(t = 0\), leading to:
\[
(0, 0, 0)
\]

The **non-trivial solution** occurs for any non-zero value of \(t\).

---

Would you like more details or have any questions about this solution?

Here are some related questions:
1. How do you solve systems of linear equations using matrices?
2. What is the difference between trivial and non-trivial solutions in linear algebra?
3. How do you determine if a system of equations has infinite solutions?
4. What are the properties of homogeneous systems of linear equations?
5. How does Gaussian elimination help in solving systems of equations?

**Tip:** When dealing with systems of linear equations, always check if a system is homogeneous (all zeros on the right-hand side). Homogeneous systems always have at least the trivial solution.

Determine all trivial and non-trivial solutions of the following system: x1 + 3x2 - 4x3 = 0; 2x1 - 5x2 = 0; -2x1 - 6x2 + 8x3 = 0

Explore how to find both the trivial and non-trivial solutions of the system of linear equations: x1 + 3x2 - 4x3 = 0, 2x1 - 5x2 = 0, -2x1 - 6x2 + 8x3 = 0. Learn how to express the general solution in parametric form and understand the behavior of homogeneous systems.

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