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  • (i)For which value(s) of lambda, do the pair of linear equations lambda x + y = lambda square and x +lambday =1 have no solution

For which value(s) of \lambda , do the pair of linear equations \lambdax + y = \lambda ^{2} and x +\lambday = 1 have

(i) no solution?
(ii) infinitely many solutions?
(iii) a unique solution?

Answers (1)

(i)Solution:
The given equations are  
\lambdax + y =\lambda ^{2} , x + \lambday = 1
In equation
\lambdax + y – \lambda ^{2} = 0
a1 = \lambda, b1 = 1, c1 = –\lambda ^{2}
In equation
x + \lambday – 1 = 0
a2 =1, b2 = \lambda, c2 = –1
For no solution \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}
\frac{a_{1}}{a_{2}}= \frac{\lambda }{1}
\frac{b_{1}}{b_{2}}= \frac{1}{\lambda }
\frac{c_{1}}{c_{2}}= \frac{-\lambda^{2}}{ -1}= \lambda^{2}
\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}
\frac{\lambda }{1}= \frac{1}{\lambda}\neq \lambda^{2}
\frac{\lambda }{1}= \frac{1}{\lambda}
\lambda ^{2}= 1
\lambda=1,-1
Hence value of \lambda is -1

\lambda \neq 1 because in this case \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}})
(ii)Solution:
The given equations are  
\lambdax + y =\lambda ^{2} , x + \lambday = 1
In equation
\lambdax + y – \lambda ^{2} = 0
a1 = \lambda, b1 = 1, c1 = –\lambda ^{2}
In equation
x + \lambday – 1 = 0
a2 =1, b2 = \lambda, c2 = –1
\frac{a_{1}}{a_{2}}=\frac{\lambda }{1}; \frac{b_{1}}{b_{2}}= \frac{1}{\lambda };\frac{c_{1}}{c_{2}}=\frac{-\lambda^{2} }{-1}= \lambda ^{2}
For infinite many solution
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}
\frac{\lambda }{1}= \frac{1}{\lambda }= \lambda ^{2}
Only one value of l satisfy all the three equations, that is \lambda = 1
(iii) Solution:
The given equations are
\lambdax + y = \lambda^{2}, x + \lambday = 1
In equation
\lambdax + y – \lambda^{2} = 0
a1 =\lambda , b1 = 1, c1 = -\lambda^{2}
In equation
x + \lambday  = 1
a2 =1, b2 = \lambda, c2 =–1

\frac{a_{1}}{a_{2}}= \frac{\lambda }{1};\frac{b_{1}}{b_{2}}= \frac{1}{\lambda};\frac{c_{1}}{c_{2}}= \frac{-\lambda^{2} }{-1}= \lambda ^{2}
For unique solution \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}

\frac{\lambda }{1}\neq \frac{1}{\lambda }
\lambda ^{2}\neq 1
\lambda \neq \pm 1
All real values of \lambda except \lambda \neq \pm 1

Posted by

infoexpert27

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