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(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta =

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exactly three values of $\lambda$

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(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta = 0$.The coefficient matrix is:$\Delta = \begin{vmatrix} 1 & \lambda & -1 \ \lambda & -1 & -1 \ 1 & 1 & -\lambda \end{vmatrix} = 0$Expanding the determinant along the first row:$1((-1)(-\lambda) - (-1)(1)) - \lambda((\lambda)(-\lambda) - (-1)(1)) - 1((\lambda)(1) - (-1)(1)) = 0$$1(\lambda + 1) - \lambda(-\lambda^2 + 1) - 1(\lambda + 1) = 0$$(\lambda + 1) - \lambda(1 - \lambda^2) - (\lambda + 1) = 0$$-\lambda(1 - \lambda)(1 + \lambda) = 0$$\lambda(1 - \lambda)(1 + \lambda) = 0$Solving for $\lambda$,we get $\lambda = 0, 1, -1$.Thus,there are exactly three values of $\lambda$ for which the system has a non-trivial solution.

The system of linear equations $x + \lambda y - z = 0, \lambda x - y - z = 0, x + y - \lambda z = 0$ has a non-trivial solution for:

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