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(A) The given system of equations can be rewritten as:$(1 - \lambda)x - 2y - 2z = 0$$x + (2 - \lambda)y + z = 0$$-x - y - \lambda z = 0$For the system

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is a singleton

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(A) The given system of equations can be rewritten as:$(1 - \lambda)x - 2y - 2z = 0$$x + (2 - \lambda)y + z = 0$$-x - y - \lambda z = 0$For the system to have non-zero solutions,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 1 - \lambda & -2 & -2 \ 1 & 2 - \lambda & 1 \ -1 & -1 & -\lambda \end{vmatrix} = 0$Expanding the determinant along the first row:$(1 - \lambda) [(2 - \lambda)(-\lambda) - (-1)(1)] - (-2) [1(-\lambda) - (-1)(1)] + (-2) [1(-1) - (-1)(2 - \lambda)] = 0$$(1 - \lambda) [-2\lambda + \lambda^2 + 1] + 2 [-\lambda + 1] - 2 [-1 + 2 - \lambda] = 0$$(1 - \lambda)(\lambda - 1)^2 + 2(1 - \lambda) - 2(1 - \lambda) = 0$$-(\lambda - 1)^3 = 0$$\lambda = 1$Since there is only one value for $\lambda$,the set of all values is a singleton set.

The set of all values of $\lambda$ for which the system of linear equations $x - 2y - 2z = \lambda x$,$x + 2y + z = \lambda y$,and $-x - y = \lambda z$ has non-zero solutions.

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