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
$\lambda x+y=\lambda 2andx+\lambda y=1a_1=\lambda ,b_1=1c_1=-{\lambda }^2{a}_2=1,b_2=\lambda ,c_2=-1$
For no solution.
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\Rightarrow \frac{\lambda }{1}=\frac{1}{\lambda }\ne \frac{-{\lambda }^2}{-1}\Rightarrow {\lambda }^2-1=0\Rightarrow (\lambda -1)(\lambda +1)=0\Rightarrow \lambda =1,-1$
Here, we take only $\lambda$ = - 1 because at $\lambda$ = 1 the system of linear equations has infinitely many solutions