Question

Question 1 (i)
For which value (s) $\lambda$, do the pair of linear equations $\lambda x+y={\lambda }^{2}andx+\lambda y=1$ have no solution.

Answer 1

The given pair of linear equations is
$\lambda x+y={\lambda }^{2}andx+\lambda y=1a_1=\lambda ,b_1=1c_1=-{\lambda }^2{a}_2=1,b_2=\lambda ,c_2=-1$
For no solution.
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

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

When $\frac{\lambda }{1}=\frac{1}{\lambda }$,

${\lambda }^2-1=0\Rightarrow (\lambda -1)(\lambda +1)=0\Rightarrow \lambda =1,-1$

When $\lambda =1$,

$\frac{-{\lambda }^2}{-1}=\frac{-\left(1^2\right)}{-1}=1$

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

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

When $\lambda =-1$,

$\frac{-{\lambda }^2}{-1}=\frac{-\left({(-1)}^2\right)}{-1}=1$

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

Thus, for $\lambda =-1$, the pair of equations will have no solution.