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To solve the system of linear equations for a non-trivial solution, we first need to write the system as a matrix and then find the value of the parameter \( \lambda \) that makes the determinant of the coefficient matrix equal to zero. This condition ensures a non-trivial solution. Here are the equations:

\[
\begin{aligned}
2x_1 + x_2 + 2x_3 &= 0, \\
4x_1 - x_2 + 7x_3 &= 0, \\
x_1 + \lambda x_2 + 2x_3 &= 0.
\end{aligned}
\]

### Step 1: Write the system as a matrix equation
The system can be represented as \( A \mathbf{x} = \mathbf{0} \), where \( A \) is the coefficient matrix and \( \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \). The matrix \( A \) is:

\[
A = \begin{pmatrix}
2 & 1 & 2 \\
4 & -1 & 7 \\
1 & \lambda & 2
\end{pmatrix}
\]

### Step 2: Find the determinant of matrix \( A \)
To ensure a non-trivial solution, the determinant of \( A \) must be zero. Let's compute the determinant of \( A \):

\[
\text{det}(A) = \begin{vmatrix} 2 & 1 & 2 \\ 4 & -1 & 7 \\ 1 & \lambda & 2 \end{vmatrix}
\]

I will calculate the determinant of this matrix.The determinant of the matrix \( A \) is \( -6\lambda - 3 \).

### Step 3: Set the determinant to zero
For the system to have a non-trivial solution, the determinant must be zero:

\[
-6\lambda - 3 = 0
\]

Solving for \( \lambda \):

\[
\lambda = -\frac{1}{2}
\]

### Step 4: Substitute \( \lambda = -\frac{1}{2} \) into the system
Now that we know \( \lambda = -\frac{1}{2} \), we substitute it back into the system of equations and solve for \( x_1 \), \( x_2 \), and \( x_3 \).

The system becomes:

\[
\begin{aligned}
2x_1 + x_2 + 2x_3 &= 0, \\
4x_1 - x_2 + 7x_3 &= 0, \\
x_1 - \frac{1}{2}x_2 + 2x_3 &= 0.
\end{aligned}
\]

I will now solve this system for a non-trivial solution.The general solution to the system is:

\[
x_1 = -1.5x_3, \quad x_2 = x_3, \quad x_3 = x_3
\]

For a non-trivial solution, \( x_3 \neq 0 \). Therefore, the solution can be written in terms of \( x_3 \), where \( x_3 \) is any non-zero scalar:

\[
\mathbf{x} = x_3 \begin{pmatrix} -1.5 \\ 1 \\ 1 \end{pmatrix}
\]

Thus, the solution is:

\[
\mathbf{x} = \begin{pmatrix} -1.5 \\ 1 \\ 1 \end{pmatrix} x_3
\]

where \( x_3 \neq 0 \). This gives a family of non-trivial solutions.

Would you like more details or have any questions?

Here are five related questions for practice:
1. What is the significance of a matrix having a zero determinant?
2. How do we identify whether a system of linear equations has a unique or infinite number of solutions?
3. Can a system have a trivial solution even if the determinant is non-zero?
4. How does changing a parameter affect the solvability of a linear system?
5. What are eigenvalues, and how are they related to the determinant of a matrix?

**Tip:** Always check the determinant when solving systems involving parameters to determine if the system has non-trivial solutions.

Determine the value of the parameter λ for which the system has a non-trivial solution, and find this solution.

Learn how to determine the parameter λ for which a system of linear equations has a non-trivial solution. This problem involves finding the determinant of a 3x3 matrix and setting it to zero. Suitable for college-level linear algebra students.

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