InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
For which value (s) of `lambda`, do the pair of linear equations `lambdax + y = lambda^(2)` and `x + lambda y = 1 ` have (i) no solution ? (ii) infinitely many solutions ? (iii) a unique solution ? |
|
Answer» The given pair of linear equations is `lambdax + y = lambda^(2)` and `x + lambda y = 1` Here, `" " a_(1) = lambda, b_(1) = 1, c_(1) = - lambda^(2)` `a_(2) = 1, b_(2) = lambda, c_(2) = -1` (i) For no solution, `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) != (c_(1))/(c_(2))` `rArr " " (lambda)/(1) = (1)/(lambda) != (-lambda^(2))/(-1)` `rArr " " lambda^(2) - 1 = 0` `rArr " " (lambda - 1) (lambda + 1) = 0` `rArr " " lambda = 1, -1` Here, we take only `lambda = -1` because at `lambda = 1` the system of linear equations has infinitely many solutions. (ii) For infinitely many solutions, `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) = (c_(1))/(c_(2))` `rArr " " (lambda)/(1) = (1)/(lambda) = (lambda^(2))/(1)` `rArr " " (lambda)/(1) = (lambda^(2))/(1)` `rArr " " lambda (lambda - 1) = 0` when `lambda != 0, ` then ` lambda = 1` For a unique solution, ` (a_(1))/(a_(2))!=(b_(1))/(b_(2)) rArr (lambda)/(1) != (1)/(lambda)` `rArr " " lambda^(2) != 1 rArr lambda != +- 1` So, all real values of `lambda ` except `+-1`. |
|