# 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?

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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 infinitely many solutions.

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 infinitely many solutions 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 infinitely many solutions is $1$.

Updated on 10-Oct-2022 13:27:15