For which value(s) of λ,

Solution:

The given pair of linear equations is

λx + y = λ2 and x + λy = 1

a1 = λ, b1= 1,  c1 = – λ2

a2 =1,  b2=λ        c2=-1

(i) For no solution,

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\Rightarrow$ $\frac{\lambda}{1}=\frac{1}{\lambda} \neq \frac{-\lambda^{2}}{-1}$

$\Rightarrow \quad \lambda^{2}-1=0$

$\Rightarrow \quad(\lambda-1)(\lambda+1)=0$

$\Rightarrow \quad \lambda=1,-1$

Here, we take only $\lambda=-1$ because at $\lambda=1$ the system of linear equations has infinitely many solutions.

(ii) For infinitely many solutions,

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \quad \frac{\lambda}{1}=\frac{1}{\lambda}=\frac{\lambda^{2}}{1}$

$\Rightarrow \quad \frac{\lambda}{1}=\frac{\lambda^{2}}{1}$

$\Rightarrow \quad \lambda(\lambda-1)=0$

When $\lambda \neq 0$, then $\lambda=1$

(iii) For a unique solution.

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{\lambda}{1} \neq \frac{1}{\lambda}$

$\Rightarrow \quad \lambda^{2} \neq 1 \quad \Rightarrow \quad \lambda \neq \pm 1$

So, all real values of $\lambda$ except $\pm 1$.