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# For which value(s) of $ \lambda $, do the pair of linear equations $ \lambda x+y=\lambda^{2} $ and $ x+\lambda y=1 $ have a unique 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 a unique 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 a unique solution is

$\frac{a_{1}}{a_{2}} \ ≠ \frac{b_{1}}{b_{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,

$\frac{λ}{1}≠ \frac{1}{λ}$

$λ≠ \frac{1}{λ}$

$λ \times λ≠ 1$

$λ^2≠ 1$

$λ≠ 1$ or $λ≠ -1$

Therefore, the values of $λ$ for which the given system of equations have a unique solution is "All real values except $-1$ and $1$".

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