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(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's de

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contains exactly two elements.

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(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's determinants $(D_x, D_y, D_z)$ must be non-zero.The determinant $D$ is given by:$D = \begin{vmatrix} 2 & -1 & 2 \ 1 & -2 & \lambda \ 1 & \lambda & 1 \end{vmatrix} = 2(-2 - \lambda^2) + 1(1 - \lambda) + 2(\lambda + 2) = -4 - 2\lambda^2 + 1 - \lambda + 2\lambda + 4 = -2\lambda^2 + \lambda + 1$Setting $D = 0$:$-2\lambda^2 + \lambda + 1 = 0 \Rightarrow 2\lambda^2 - \lambda - 1 = 0 \Rightarrow (2\lambda + 1)(\lambda - 1) = 0$Thus,$\lambda = 1$ or $\lambda = -\frac{1}{2}$.Now,check $D_x$ for these values:$D_x = \begin{vmatrix} 2 & -1 & 2 \ -4 & -2 & \lambda \ 4 & \lambda & 1 \end{vmatrix} = 2(-2 - \lambda^2) + 1(-4 - 4\lambda) + 2(-4\lambda + 8) = -4 - 2\lambda^2 - 4 - 4\lambda - 8\lambda + 16 = -2\lambda^2 - 12\lambda + 8$For $\lambda = 1$,$D_x = -2 - 12 + 8 = -6 
eq 0$.For $\lambda = -\frac{1}{2}$,$D_x = -2(\frac{1}{4}) - 12(-\frac{1}{2}) + 8 = -0.5 + 6 + 8 = 13.5 
eq 0$.Since $D=0$ and $D_x 
eq 0$ for both values,the system has no solution for both $\lambda = 1$ and $\lambda = -\frac{1}{2}$.Therefore,$S = \{1, -\frac{1}{2}\}$,which contains exactly two elements.

Let $S$ be the set of all $\lambda \in \mathbb{R}$ for which the system of linear equations
$2x - y + 2z = 2$
$x - 2y + \lambda z = -4$
$x + \lambda y + z = 4$
has no solution. Then the set $S$

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Let $S$ be the set of all $\lambda \in \mathbb{R}$ for which the system of linear equations$2x - y + 2z = 2$$x - 2y + \lambda z = -4$$x + \lambda y + z = 4$has no solution. Then the set $S$