Analysis of Solutions for System of Two Linear Equations

To analyze the system of linear equations: $ax + 2y = \lambda \quad \text{--- (1)}$ $3x - 2y = \mu \quad \text{--- (2)}$

where $a, \lambda, \mu \in \mathbb{R}$, we can evaluate the determinant of the coefficient matrix, $D$:

$$D = \begin{vmatrix} a & 2 \\ 3 & -2 \end{vmatrix} = -2a - 6 = -2(a + 3)$$

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Case 1: $a \neq -3$

When $a \neq -3$, the determinant $D \neq 0$.

By Cramer's Rule, a non-zero determinant implies that the system of linear equations is consistent and independent, possessing a unique solution for any real values of $\lambda$ and $\mu$.

$\implies \text{Option (B) is correct.}$

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Case 2: $a = -3$

When $a = -3$, the determinant $D = 0$. Substituting $a = -3$ into equations (1) and (2): $-3x + 2y = \lambda \quad \text{--- (3)}$ $3x - 2y = \mu \quad \text{--- (4)}$

Adding equations (3) and (4): $(-3x + 2y) + (3x - 2y) = \lambda + \mu$ $0 = \lambda + \mu$

• Subcase 2a: $\lambda + \mu = 0$ If $\lambda + \mu = 0$, then $\lambda = -\mu$. Equation (3) becomes $3x - 2y = \mu$, which is identical to equation (4). Both equations represent the same straight line, yielding infinitely many solutions.

  $\implies \text{Option (C) is correct.}$

• Subcase 2b: $\lambda + \mu \neq 0$ If $\lambda + \mu \neq 0$, the equation $0 = \lambda + \mu$ leads to a contradiction. The two lines are parallel and distinct, meaning the system has no solution (inconsistent).

  $\implies \text{Option (D) is correct.}$

• Analysis of Option (A): Option (A) asserts that for $a = -3$, the system has infinitely many solutions for all values of $\lambda$ and $\mu$. However, as shown in Subcase 2b, if $\lambda + \mu \neq 0$, the system has no solution. Thus, Option (A) is incorrect.

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Conclusion

The correct options are B, C, and D.