The problem involves finding the possible values of $\lambda$ for which the given system of linear equations has a unique solution.

Step-by-Step Solution:

1. Matrix Representation: The equation can be represented as:

   3 - \lambda & 2 \\ 2 & 3 - \lambda \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}.$$ The system has a unique solution if the determinant of the coefficient matrix is non-zero.

2. Determinant of the Coefficient Matrix: Compute the determinant of the matrix:

   3 - \lambda & 2 \\ 2 & 3 - \lambda \end{vmatrix}.$$ Using the determinant formula: $$\text{Det} = (3 - \lambda)(3 - \lambda) - (2)(2).$$ Simplify: $$\text{Det} = (3 - \lambda)^2 - 4.$$ Expand: $$\text{Det} = 9 - 6\lambda + \lambda^2 - 4.$$ $$\text{Det} = \lambda^2 - 6\lambda + 5.$$

3. Condition for a Unique Solution: A unique solution exists when the determinant is non-zero: $\lambda^2 - 6\lambda + 5 \neq 0.$

4. Solve for $\lambda$: Factorize the quadratic equation: $\lambda^2 - 6\lambda + 5 = (\lambda - 5)(\lambda - 1).$ The determinant is zero when: $\lambda = 5 \quad \text{or} \quad \lambda = 1.$ For a unique solution, these values must be excluded: $\lambda \neq 5 \quad \text{and} \quad \lambda \neq 1.$

Final Answer:

The system has a unique solution for: $\lambda \in \mathbb{R} \setminus \{1, 5\}.$

Let me know if you need further clarification or details!

Related Questions:

1. What happens when $\lambda = 1$ or $\lambda = 5$?
2. How does the determinant condition relate to the invertibility of a matrix?
3. Can this method be extended to larger systems of equations?
4. How do you solve such systems when $\lambda = 1$ or $\lambda = 5$?
5. How can we geometrically interpret the determinant being zero?

Tip:

Always check the determinant of the coefficient matrix when analyzing the solvability of a linear