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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 no solution?
Given:
The given system of equations is:
\( \lambda x+y=\lambda^{2} \) and \( x+\lambda y=1 \)
To do:
We have to find the value of $\lambda$ for which the given system of equations have
(i) no solution.
(ii) infinitely many solutions
(iii) a unique solution
Solution:
The given system of equations can be written as:
$\lambda x + y -\lambda^2=0$
$x + \lambda y -1=0$
The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.
Comparing the given system of equations with the standard form of equations, we have,
$a_1=\lambda, b_1=1, c_1=-\lambda^2$ and $a_2=1, b_2=\lambda, c_2=-1$
(i) 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}} \ $
Therefore,
$\frac{\lambda}{1}=\frac{1}{\lambda}≠\frac{-\lambda^2}{-1}$
$\lambda=\frac{1}{\lambda}≠\lambda^2$
$\lambda=\frac{1}{\lambda}$ and $\frac{1}{\lambda}≠\lambda^2$
$\lambda \times \lambda=1$ and $\lambda^2 \times \lambda≠1$
$\lambda^2=1$ and $\lambda^3≠1$
$\lambda=1$ or $\lambda=-1$ and $\lambda≠1$
Therefore,
$\lambda=-1$
The value of $\lambda$ for which the given system of equations has no solution is $-1$.
(ii) 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}} \ $
Therefore,
$\frac{\lambda}{1}=\frac{1}{\lambda}=\frac{-\lambda^2}{-1}$
$\lambda=\frac{1}{\lambda}=\lambda^2$
$\lambda=\frac{1}{\lambda}$ and $\frac{1}{\lambda}=\lambda^2$
$\lambda \times \lambda=1$ and $\lambda^2 \times \lambda=1$
$\lambda^2=1$ and $\lambda^3=1$
$\lambda=1$ or $\lambda=-1$ and $\lambda=1$
Therefore,
$\lambda=1$
The value of $\lambda$ for which the given system of equations has infinitely many solutions is $1$.
(iii) The condition for which the above system of equations has a unique solution is
$\frac{a_{1}}{a_{2}} \ ≠ \frac{b_{1}}{b_{2}} \ $
Therefore,
$\frac{\lambda}{1}≠ \frac{1}{\lambda}$
$\lambda≠ \frac{1}{\lambda}$
$\lambda \times \lambda≠ 1$
$\lambda^2≠ 1$
$\lambda≠ 1$ or $\lambda≠ -1$
Therefore, the values of $\lambda$ for which the given system of equations have a unique solution is "All real values except $-1$ and $1$".
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