For which value(s) of $λ$, do the pair of linear equations $λx + y = λ^2$ and $x + λy = 1$ have no solution?

Given: 

The given system of equations is:

$λx + y = λ^2$ and $x + λy = 1$

To do: 

We have to find the value of $λ$ for which the given system of equations has no solution.

Solution:

The given system of equations can be written as:

$λx + y -λ^2=0$

$x + λy -1=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

The condition for which the above system of equations has no solution is

$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} ≠ \frac{c_{1}}{c_{2}} \ $

Comparing the given system of equations with the standard form of equations, we have,

$a_1=λ, b_1=1, c_1=-λ^2$ and $a_2=1, b_2=λ, c_2=-1$

Therefore,

$\frac{λ}{1}=\frac{1}{λ}≠\frac{-λ^2}{-1}$

$λ=\frac{1}{λ}≠λ^2$

$λ=\frac{1}{λ}$ and $\frac{1}{λ}≠λ^2$

$λ \times λ=1$ and $λ^2 \times λ≠1$

$λ^2=1$ and $λ^3≠1$

$λ=1$ or $λ=-1$ and $λ≠1$

Therefore,

$λ=-1$

The value of $λ$ for which the given system of equations has no solution is $-1$.   

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Updated on: 10-Oct-2022

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