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To solve this system of linear equations and find the values of \(\lambda\) that give non-trivial solutions, we need to set up the augmented matrix and determine when the system has a non-zero solution.

The given system is:
\[
\begin{cases}
(\lambda + 3)x_1 + x_2 + 2x_3 = 0 \\
\lambda x_1 + (\lambda - 1)x_2 + x_3 = 0 \\
3(\lambda + 1)x_1 + \lambda x_2 + (\lambda + 3)x_3 = 0 
\end{cases}
\]

We convert this into an augmented matrix:

\[
\begin{pmatrix}
\lambda + 3 & 1 & 2 \\
\lambda & \lambda - 1 & 1 \\
3(\lambda + 1) & \lambda & \lambda + 3 
\end{pmatrix}
\]

For this system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

The determinant of the matrix is:
\[
\begin{vmatrix}
\lambda + 3 & 1 & 2 \\
\lambda & \lambda - 1 & 1 \\
3(\lambda + 1) & \lambda & \lambda + 3 
\end{vmatrix}
\]

We can calculate this determinant using cofactor expansion along the first row:

\[
\text{Det} = (\lambda + 3) \begin{vmatrix} \lambda - 1 & 1 \\ \lambda & \lambda + 3 \end{vmatrix} - 1 \begin{vmatrix} \lambda & 1 \\ 3(\lambda + 1) & \lambda + 3 \end{vmatrix} + 2 \begin{vmatrix} \lambda & \lambda - 1 \\ 3(\lambda + 1) & \lambda \end{vmatrix}
\]

We calculate each 2x2 determinant:

\[
\begin{vmatrix} \lambda - 1 & 1 \\ \lambda & \lambda + 3 \end{vmatrix} = (\lambda - 1)(\lambda + 3) - 1(\lambda) = \lambda^2 + 2\lambda - 3
\]

\[
\begin{vmatrix} \lambda & 1 \\ 3(\lambda + 1) & \lambda + 3 \end{vmatrix} = \lambda(\lambda + 3) - 1 \cdot 3(\lambda + 1) = \lambda^2 + 3\lambda - 3\lambda - 3 = \lambda^2 - 3
\]

\[
\begin{vmatrix} \lambda & \lambda - 1 \\ 3(\lambda + 1) & \lambda \end{vmatrix} = \lambda \cdot \lambda - (\lambda - 1) \cdot 3(\lambda + 1) = \lambda^2 - 3(\lambda^2 + \lambda - 1) = \lambda^2 - 3\lambda^2 - 3\lambda + 3 = -2\lambda^2 - 3\lambda + 3
\]

Substituting these back, we get:

\[
\text{Det} = (\lambda + 3)(\lambda^2 + 2\lambda - 3) - (\lambda^2 - 3) + 2(-2\lambda^2 - 3\lambda + 3)
\]

Simplifying:

\[
= (\lambda + 3)(\lambda^2 + 2\lambda - 3) - \lambda^2 + 3 - 4\lambda^2 - 6\lambda + 6
\]
\[
= (\lambda^3 + 2\lambda^2 - 3\lambda + 3\lambda^2 + 6\lambda - 9) - \lambda^2 + 3 - 4\lambda^2 - 6\lambda + 6
\]
\[
= \lambda^3 + 5\lambda^2 + 3\lambda - 9 - \lambda^2 + 3 - 4\lambda^2 - 6\lambda + 6
\]
\[
= \lambda^3 + 5\lambda^2 - \lambda^2 + 3\lambda - 6\lambda - 9 + 3 + 6 - 4\lambda^2
\]
\[
= \lambda^3 + (5 - 1 - 4)\lambda^2 + (3 - 6)\lambda - 9 + 9
\]
\[
= \lambda^3 - 3\lambda
\]

Set the determinant to zero:

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

Factor:

\[
\lambda(\lambda^2 - 3) = 0
\]

So, the solutions for \(\lambda\) are:

\[
\lambda = 0, \pm\sqrt{3}
\]

These are the values of \(\lambda\) for which the system has non-trivial solutions. 

Would you like further details or have any questions? Here are five questions you might want to ask:

1. Can you show step-by-step how to compute the determinants for each part?
2. How do you solve the system once \(\lambda\) values are found?
3. Can we verify the solutions for non-trivial cases of \(\lambda\)?
4. What is the geometric interpretation of this solution?
5. How does this result relate to eigenvalues and eigenvectors?

**Tip:** When solving determinants, double-check each step carefully to avoid errors in complex calculations.

Learn how to solve a system of linear equations with a parameter lambda, including determining when the system has non-trivial solutions using determinant calculations and Cramer's rule. Suitable for advanced high school students and those preparing for college-level mathematics.

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