Question

Let $\alpha ,\lambda ,\mu \in R$. Consider the system of linear equations
$\alpha x+2y=\lambda$
$3x-2y=\mu$
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 $\alpha \ne -3$, then the system has a unique solution for all values of $\lambda$ and $\mu$
• C. If $\lambda +\mu =0$, then the system has infinitely many solutions for $\alpha =-3$
• D. If $\lambda +\mu \ne 0$, then the system has no solution for $\alpha =-3$

Answer 1

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

$\Delta =\left|\begin{array}{cc}\alpha & 2\\ 3& -2\end{array}\right|=-2\alpha -6$
for unique solution $\Delta \ne 0\Rightarrow \alpha \ne -3$
${\Delta }_x=\left|\begin{array}{cc}\lambda & 2\\ \mu & -2\end{array}\right|=-2(\lambda +\mu )$
${\Delta }_y=\left|\begin{array}{cc}\alpha & \lambda \\ 3& \mu \end{array}\right|=\alpha \mu -3\lambda$
If $\alpha =-3$ then ${\Delta }_y=-3(\lambda +\mu )$
so if $\lambda +\mu =0$ then system has infinite solution
and If $\alpha =-3$ and $\lambda +\mu \ne 0$ then no solution