The given system of linear equations is:

1. ax + 2y = λ
2. 3x - 2y = μ

We can analyze the nature of the solutions using the determinant of the coefficient matrix.

Step 1: Find the determinant of the coefficient matrix.

The coefficient matrix A for the system is: $A = \begin{pmatrix} a & 2 \\ 3 & -2 \end{pmatrix}$

The determinant of A is: $det(A) = (a)(-2) - (2)(3) = -2a - 6$

Step 2: Analyze the condition for a unique solution.

A system of linear equations has a unique solution if and only if the determinant of its coefficient matrix is non-zero, i.e., det(A) ≠ 0. $-2a - 6 ≠ 0$ $-2a ≠ 6$ $a ≠ -3$ So, if a ≠ -3, the system has a unique solution for any values of λ and μ.

• Evaluating Option B: "If a ≠ -3, then the system has a unique solution for all values of λ and μ." This statement is correct based on our analysis.

Step 3: Analyze the case when det(A) = 0.

If det(A) = 0, the system will have either infinitely many solutions or no solution. This occurs when: $-2a - 6 = 0$ $a = -3$

Let's substitute a = -3 into the system of equations:

1. -3x + 2y = λ
2. 3x - 2y = μ

Step 4: Determine the conditions for infinitely many solutions or no solution.

We can add the two equations to eliminate the variables: $(-3x + 2y) + (3x - 2y) = λ + μ$ $0x + 0y = λ + μ$ $0 = λ + μ$

This equation provides the condition for the consistency of the system.

• Case 4.1: Infinitely many solutions The system is consistent and has infinitely many solutions if the condition 0 = λ + μ is satisfied. If λ + μ = 0, then μ = -λ. The second equation 3x - 2y = μ becomes 3x - 2y = -λ, which is equivalent to -1(-3x + 2y) = -1(λ), so it is the same as the first equation. The two equations are dependent, representing the same line, and thus have infinitely many solutions.

  • Evaluating Option C: "If λ + μ = 0, then the system has infinitely many solutions for a = -3." This statement is correct.
  • Evaluating Option A: "If a = -3, then the system has infinitely many solutions for all values of λ and μ." This statement is incorrect because it requires the specific condition λ + μ = 0.

• Case 4.2: No solution The system is inconsistent and has no solution if the condition 0 = λ + μ leads to a contradiction. This happens if λ + μ ≠ 0. In this case, we have 0 = (a non-zero value), which is impossible. Geometrically, the two lines are parallel and distinct.

  • Evaluating Option D: "If λ + μ ≠ 0, then the system has no solution for a = -3." This statement is correct.

Conclusion

Based on the step-by-step analysis, the correct statements are B, C, and D.