Given that, the equations.
- $ kx + y = 3 \; \longrightarrow (1) $
- $ 4x + ky = 4 \; \longrightarrow (2) $
For a system of equations.
- $ a_{1}x + b_{1}y + c_{1} = 0 $
- $ a_{2}x + b_{2}y + c_{2} = 0 $
to have unique solution, the condition to be satisfied is
$$ \boxed{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}} $$
We can re-write the equation $(1)$ and $(2)$ as follows:
- $ kx + y – 3 = 0$
- $ 4x + ky – 4 =0 $
For unique solution $: \frac{k}{4} \neq \frac{1}{k} \neq \frac{-3}{-4} $
Taking first two terms,
$ \frac{k}{4} \neq \frac{1}{\text{k}} $
$ \Rightarrow k^{2} \neq 4 $
$ \Rightarrow k \neq \sqrt{4} $
$ \Rightarrow k \neq \pm 2 $
$ \Rightarrow \boxed{|k| \neq 2} $
Correct Answer$: \text{A} $
$\textsf{PS:}$
$1.$ For a system of equation $ a_{1}x + b_{1}y + c_{1} =0 ; \; a_{2}x + b_{2}y + c_{2} = 0 $ to have no solution, the condition to be satisfied is,
$$ \boxed{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}} $$
$2.$ For a system of equation $ a_{1}x + b_{1}y + c_{1} =0 ; \; a_{2}x + b_{2}y + c_{2} = 0 $ to have infinitely many solutions, we must have
$$ \boxed{\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}} $$