Let `a,lambda,mu in R,` Consider the system of linear equations `ax+2y=lambda 3x-2y=mu` Which of the flollowing statement (s) is (are) correct?

Options

A. If `alpha =-3` then the system has infinitely many solutions for all values of `lambda " and "mu`
B. If `alpha ne -3` then the system has a unique solution for all values of `lambda " and " mu`
C. If `lambda +mu =0` then the system has infiniely many solutions for `alpha =-3`
D. if `lambda +mu ne 0` then the system has no solution for `alpha =-3`

Explanation

To solve the problem, we need to analyze the given system of linear equations:

1. \( \alpha x + 2y = \lambda \) (Equation 1)
2. \( 3x - 2y = \mu \) (Equation 2)

We will determine the conditions under which the system has unique solutions, infinitely many solutions, or no solutions based on the values of \( \alpha, \lambda, \) and \( \mu \).

### Step 1: Write the system in matrix form
The system can be represented in matrix form as:
\[ \begin{pmatrix} \alpha & 2 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \lambda \\ \mu \end{pmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
The determinant of the coefficient matrix is given by:
\[ D = \alpha \cdot (-2) - 2 \cdot 3 = -2\alpha - 6 \]

### Step 3: Analyze the determinant
- If \( D \neq 0 \), the system has a unique solution.
- If \( D = 0 \), the system may have either infinitely many solutions or no solutions.

Setting the determinant to zero to find the critical value of \( \alpha \):
\[ -2\alpha - 6 = 0 \implies -2\alpha = 6 \implies \alpha = -3 \]

### Step 4: Evaluate the statements
1. **Statement 1**: If \( \alpha = -3 \), then the system has infinitely many solutions for all values of \( \lambda \) and \( \mu \).
- When \( \alpha = -3 \), the determinant is zero. We need to check the consistency of the equations. Substituting \( \alpha = -3 \) into the equations gives:
- \( -3x + 2y = \lambda \)
- \( 3x - 2y = \mu \)
- Solving these equations will show that they do not necessarily have infinitely many solutions for all \( \lambda \) and \( \mu \). Thus, this statement is **false**.

2. **Statement 2**: If \( \alpha \neq -3 \), then the system has a unique solution for all values of \( \lambda \) and \( \mu \).
- Since \( D \neq 0 \) when \( \alpha \neq -3 \), this statement is **true**.

3. **Statement 3**: If \( \lambda + \mu = 0 \), then the system has infinitely many solutions for \( \alpha = -3 \).
- When \( \alpha = -3 \), we check if \( \lambda + \mu = 0 \) leads to consistent equations. This condition can lead to infinitely many solutions, so this statement is **true**.

4. **Statement 4**: If \( \lambda + \mu \neq 0 \), then the system has no solutions for \( \alpha = -3 \).
- If \( \alpha = -3 \) and \( \lambda + \mu \neq 0 \), the equations will be inconsistent, leading to no solutions. Thus, this statement is **true**.

### Conclusion
The correct statements are:
- Statement 2: True
- Statement 3: True
- Statement 4: True

### Summary of Results
- Statement 1: False
- Statement 2: True
- Statement 3: True
- Statement 4: True