Question

Question 1 (ii)
For which value(s) of $\lambda$ does the pair of linear equations $\lambda x+y={\lambda }^{2}andx+\lambda y=1$ have infinitely many solutions?

Answer 1

The given pair of linear equations is
$\lambda x+y={\lambda }^2$ and

$x+\lambda y=1$

Here,
$a_1=\lambda ,b_1=1c_1=-{\lambda }^2{a}_2=1,b_2=\lambda ,c_2=-1$

For infinitely many solutions,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

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

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

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

Thus, either $\lambda =0$ or $\lambda =1$

When $\lambda =0$,

$\frac{a_1}{a_2}=\frac{\lambda }{1}=\frac{0}{1}=0$

$\frac{b_1}{b_2}=\frac{1}{\lambda }=\frac{1}{0}$

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ Hence, the pair of linear equations will have one unique solution.

When $\lambda =1$,

$\frac{a_1}{a_2}=\frac{\lambda }{1}=\frac{1}{1}=1$

$\frac{b_1}{b_2}=\frac{1}{\lambda }=\frac{1}{1}=1$

$\frac{c_1}{c_2}=\frac{-{\lambda }^2}{1}=\frac{-1}{-1}=1$

Thus, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Hence, when $\lambda =1$, the given pair of linear equations will have infinitely many solutions.