Question

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

Answer 1

Step 1: Convert the equations to standard form

Given two equations are,

$\lambda x+y={\lambda }^2$ and $x+\lambda y=1$

The standard linear equation is given as,

$ax+by+c=0$

Hence, the two equations in the standard form can be represented as,

$\lambda x+y={\lambda }^2\Rightarrow \lambda x+y-{\lambda }^2=0$

$x+\lambda y=1\Rightarrow x+\lambda y-1=0$

Step 2: Compare the equations with the standard equation

Comparing the two equations with the standard equation we define,

$a_1=\lambda$, $b_1=1$ and $c_1=-{\lambda }^2$

$a_2=1$, $b_2=\lambda$, and $c_2=-1$

Step 3: Define the condition for infinite solutions

For infinitely many solutions, the condition is,

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

Thus,

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

Step 4: Solve for $\lambda$

Consider the first and last part of the equation,

$$\begin{gathered} \begin{array}{rcl}\lambda & =& {\lambda }^{2}and{\lambda }^2=1\\ & \Rightarrow & \lambda (\lambda -1)=0and\lambda =\pm 1\end{array} \\ \Rightarrow \begin{array}{rcl}& & \lambda \end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 1\end{array} \end{gathered}$$

Therefore, when $\lambda =1$, the set of equations has infinitely many solutions.