For which value$(s)$ of $\lambda$,do the pair of linear equations $\lambda x + y = \lambda^2$ and $x + \lambda y = 1$ have:
$(i)$ no solution?
$(ii)$ infinitely many solutions?
$(iii)$ a unique solution?

(A) The given pair of linear equations is $\lambda x + y = \lambda^2$ and $x + \lambda y = 1$.
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we have $a_1 = \lambda, b_1 = 1, c_1 = -\lambda^2$ and $a_2 = 1, b_2 = \lambda, c_2 = -1$.
$(i)$ For no solution,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
$\frac{\lambda}{1} = \frac{1}{\lambda} \neq \frac{-\lambda^2}{-1} \Rightarrow \lambda^2 = 1$ and $\lambda^2 \neq 1$. This is impossible for $\lambda = 1$. For $\lambda = -1$,$\frac{-1}{1} = \frac{1}{-1} \neq \frac{-(-1)^2}{-1} \Rightarrow -1 = -1 \neq 1$. Thus,for $\lambda = -1$,there is no solution.
$(ii)$ For infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$\frac{\lambda}{1} = \frac{1}{\lambda} = \frac{-\lambda^2}{-1} \Rightarrow \lambda^2 = 1$ and $\lambda = \lambda^3$. This holds for $\lambda = 1$.
$(iii)$ For a unique solution,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
$\frac{\lambda}{1} \neq \frac{1}{\lambda} \Rightarrow \lambda^2 \neq 1 \Rightarrow \lambda \neq \pm 1$. Thus,all real values of $\lambda$ except $\pm 1$ yield a unique solution.