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(D) The given system of linear equations is:$(2-\lambda)x_1 - 2x_2 + x_3 = 0$$2x_1 - (3+\lambda)x_2 + 2x_3 = 0$$-x_1 + 2x_2 - \lambda x_3 = 0$For a no

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contains two elements

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(D) The given system of linear equations is:$(2-\lambda)x_1 - 2x_2 + x_3 = 0$$2x_1 - (3+\lambda)x_2 + 2x_3 = 0$$-x_1 + 2x_2 - \lambda x_3 = 0$For a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 2-\lambda & -2 & 1 \ 2 & -(3+\lambda) & 2 \ -1 & 2 & -\lambda \end{vmatrix} = 0$Applying $R_1 	o R_1 + R_3$:$\begin{vmatrix} 1-\lambda & 0 & 1-\lambda \ 2 & -(3+\lambda) & 2 \ -1 & 2 & -\lambda \end{vmatrix} = 0$Taking $(1-\lambda)$ common from $R_1$:$(1-\lambda) \begin{vmatrix} 1 & 0 & 1 \ 2 & -(3+\lambda) & 2 \ -1 & 2 & -\lambda \end{vmatrix} = 0$Expanding along $R_1$:$(1-\lambda) [1(\lambda(3+\lambda) - 4) + 1(4 - (3+\lambda))] = 0$$(1-\lambda) [3\lambda + \lambda^2 - 4 + 4 - 3 - \lambda] = 0$$(1-\lambda) [\lambda^2 + 2\lambda - 3] = 0$$(1-\lambda)(\lambda+3)(\lambda-1) = 0$$-(1-\lambda)^2(\lambda+3) = 0$Thus,$\lambda = 1$ and $\lambda = -3$. The set of values is $\{1, -3\}$,which contains two elements.

The set of all values of $\lambda$ for which the system of linear equations $2x_1 - 2x_2 + x_3 = \lambda x_1$,$2x_1 - 3x_2 + 2x_3 = \lambda x_2$,and $-x_1 + 2x_2 = \lambda x_3$ has a non-trivial solution:

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