Question

Let $\alpha, \lambda, \mu \in R$. Consider the system of linear equations
$\alpha x+2 y=\lambda$
$3 x-2 y=\mu$
Which of the following statements(s) is (are) correct?

• A. If $\alpha=-3$, then the system has infinitely many solutions for all values of $\lambda$ and $\mu$
• B. If $\alpha \neq-3$, 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 $\alpha=-3$
• D. If $\lambda+\mu \neq 0$, then the system has no solution for $\alpha=-3$

Answer 1

Answer: (B), (C), (D)

$\Delta=\begin{vmatrix}\alpha & 2 \\3 & -2\end{vmatrix}=-2 \alpha-6$
for unique solution $\Delta \neq 0 \Rightarrow \alpha \neq-3$
$\Delta_{ x }=\begin{vmatrix}\lambda & 2 \\\mu & -2\end{vmatrix}=-2(\lambda+\mu) $
$\Delta_{y}=\begin{vmatrix}\alpha & \lambda \\3 & \mu\end{vmatrix}=\alpha \mu-3 \lambda$
If $\alpha=-3$ then $\Delta_{y}=-3(\lambda+\mu)$
so if $\lambda+\mu=0$ then system has infinite solution
and If $\alpha=-3$ and $\lambda+\mu \neq 0$ then no solution