Question

Let $a,\lambda ,\mu \in R.$ Consider the system of linear equations
$ax+2y=\lambda 3x-2y=\mu$
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 $a\ne -3,$ 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 $a=-3$
• D. If $\lambda +\mu \ne 0,$ then the system has no solution for $a=-3$

Answer 1

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

The correct options are
B If $a\ne -3,$ 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 $a=-3$
D If $\lambda +\mu \ne 0,$ then the system has no solution for $a=-3$

$ax+2y=\lambda \cdots \left(i\right)3x-2y=\mu \cdots \left(ii\right)$
From $\left(i\right)\text{and}\left(ii\right),$ we get
$(a+3)x=(\lambda +\mu )\therefore x=\frac{\lambda +\mu }{a+3}$

Option (1):
If $a=-3,$ then $0\times x=\lambda +\mu$
If $\lambda \ne -\mu$, then there is no solution for all values of $\lambda$ and $\mu$

Option (2):
If $a\ne -3,$ then $x=\frac{\lambda +\mu }{a+3}$ has unique solution.

Option (3):
If $\lambda +\mu =0,$ then $x=\frac{\lambda +\mu }{a+3}$ has infinitely many solutions for $a=-3$

Option (4):
Similarly if $\lambda +\mu \ne 0,$ then the system has no solutions for $a=-3$