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Let $a, \lambda, \mu \in \mathbb{R}$. Consider the system of linear equations
$a x+2 y=\lambda$
$3 x-2 y=\mu$Which of the following statement($s$) is(are) correct?
($A$) If $a=-3$, then the system has infinitely many solutions for all values of $\lambda$ and $\mu$
($B$) If $a \neq-3$, then the system has a unique solution for all values of $\lambda$ and $\mu$
($C$) If $\lambda+\mu=0$, then the system has infinitely many solutions for $a=-3$
($D$) If $\lambda+\mu \neq 0$, then the system has no solution for $a=-3$
$A,C$
$B,C$
$B,C,D$
$B,C,A$
Solution
$\alpha x+2 y=\lambda$
$3 x-2 y=\mu$
$\Delta=\left|\begin{array}{ll}\alpha & 2 \\ 3 & -2\end{array}\right|=-2 \alpha-6$
$\Delta=0, \therefore, \alpha=-3$
$\Delta_1=\left|\begin{array}{ll}\lambda & 2 \\ \mu & -2\end{array}\right|=-2 \lambda-2 \mu=-2(\lambda+\mu)$
$\Delta_2=\left|\begin{array}{ll}-3 & \lambda \\ 3 & \mu\end{array}\right|=-3 \mu-3 \lambda=-3(\lambda-\mu)$
