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The system of linear equations:$x + \lambda y - z = 0$ $\lambda x - y - z = 0$$x + y - \lambda z = 0Has a non-trivial solution for:A. infinitely many values of \lambda B. exactly one value of \lambda C. exactly two values of \lambda D. exactly three values of \lambda Last updated date: 15th Sep 2024 Total views: 439.5k Views today: 13.39k Answer Verified 439.5k+ views Hint:If we are given the system of equations like here we are given: x + \lambda y - z = 0 \lambda x - y - z = 0 x + y - \lambda z = 0 If all variable values are not equal to 0, then it is a non-trivial solution. \left| {\begin{array}{*{20}{c}} 1&\lambda &{ - 1} \\ \lambda &{ - 1}&{ - 1} \\ 1&1&{ - \lambda } \end{array}} \right| = 0 Get the determinant of coefficients and find the value of \lambda . Complete step-by-step answer: So here we are given the system of equations like here we are given: x + \lambda y - z = 0 \lambda x - y - z = 0 x + y - \lambda z = 0 And it is a non-trivial solution. It means that at least one of the variables must not be equal to 0. If all are equal to zero, then it is a trivial solution. So here the determinant of the coefficients as shown must be equal to 0 \left| {\begin{array}{*{20}{c}} 1&\lambda &{ - 1} \\ \lambda &{ - 1}&{ - 1} \\ 1&1&{ - \lambda } \end{array}} \right| = 0 Now solving it, we get 1(\lambda + 1) - \lambda ( - {\lambda ^2} + 1) - 1(\lambda + 1) = 0 (\lambda + 1) - \lambda (1 - {\lambda ^2}) - 1(\lambda + 1) = 0 We know identity (a-b) (a+b) = a^2-b^2 Applying the identity to 2nd term (\lambda + 1) + \lambda ({\lambda ^2}-1) - 1(\lambda + 1) = 0 (\lambda + 1) + \lambda (\lambda + 1) (\lambda - 1) - 1(\lambda + 1) = 0 (\lambda + 1) + \lambda (\lambda + 1) (\lambda - 1) - 1(\lambda + 1) = 0 Taking (\lambda + 1) and simplifying further we get, \lambda (\lambda - 1)(\lambda + 1) = 0 So we get \lambda$$ = 0,1, - 1$
So there are three values for which it is a non-trivial solution.

So, the correct answer is “Option D”.

Note:If set of the linear equations are given to be trivial then variables all are equal to zero but if it is a non-trivial solution then at least one must not be equal to zero.