Question

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 ?

Answer 1

The given linear equations are
$$\begin{gathered} \lambda x+y={\lambda }^2 \\ x+\lambda y=1 \end{gathered}$$
(i) We know that the system of equations

will have no solution if

So,
$$\begin{gathered} \frac{\lambda }{1}=\frac{1}{\lambda }\ne \frac{{\lambda }^2}{1} \\ \Rightarrow {\lambda }^2=1 \\ \Rightarrow \lambda =\pm 1 \end{gathered}$$
Also,
$$\begin{gathered} \frac{1}{\lambda }\ne \frac{{\lambda }^2}{1} \\ \Rightarrow {\lambda }^3\ne 1 \\ \Rightarrow \lambda \ne 1 \end{gathered}$$
Thus, the given system of equations has no solutions when $\lambda =-1$.
(ii) We know that the system of equations

will have infinitely many solutions if

$$\begin{gathered} \therefore \frac{\lambda }{1}=\frac{1}{\lambda }=\frac{{\lambda }^2}{1} \\ \Rightarrow {\lambda }^2=1 \\ \Rightarrow \lambda =\pm 1 \end{gathered}$$
Also,
$$\begin{gathered} \frac{1}{\lambda }=\frac{{\lambda }^2}{1} \\ \Rightarrow {\lambda }^3=1 \\ \Rightarrow \lambda =1 \end{gathered}$$
Thus, the given system of equations has infinitely many solutions when $\lambda =1$.
(iii) We know that the system of equations

will have a unique solution if

$$\begin{gathered} \Rightarrow \frac{\lambda }{1}\ne \frac{1}{\lambda } \\ \Rightarrow {\lambda }^2\ne 1 \\ \Rightarrow \lambda \ne \pm 1 \end{gathered}$$
Thus, the given system of equations has a unique solution for all real values of $\lambda$ except $\pm 1$.