Let $$S$$ be the set of all $$\lambda \in R$$ 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 $$S$$:

We have a system of three linear equations in the variables $$x,\;y,\;z$$:

$$ \begin{aligned} 2x - y + 2z &= 2,\\ x - 2y + \lambda z &= -4,\\ x + \lambda y + z &= 4. \end{aligned} $$

Write the coefficient matrix $$A$$ and the augmented matrix $$[A|b]$$:

$$ A= \begin{bmatrix} 2 & -1 & 2\\ 1 & -2 & \lambda\\ 1 & \lambda & 1 \end{bmatrix}, \qquad [A|b]= \begin{bmatrix} 2 & -1 & 2 & \bigm| & 2\\ 1 & -2 & \lambda & \bigm| & -4\\ 1 & \lambda & 1 & \bigm| & 4 \end{bmatrix}. $$

A system of linear equations has

• a unique solution when $$\det(A)\neq 0,$$
• either no solution or infinitely many solutions when $$\det(A)=0.$$
For the latter case we compare the ranks: if $$\operatorname{rank}(A)<\operatorname{rank}([A|b])$$ we get no solution, whereas equal ranks give infinitely many solutions.

First compute $$\det(A)$$. Using cofactor expansion along the first row,

$$ \det(A)= 2\begin{vmatrix}-2 & \lambda\\ \lambda & 1\end{vmatrix} -(-1)\begin{vmatrix}1 & \lambda\\ 1 & 1\end{vmatrix} +2\begin{vmatrix}1 & -2\\ 1 & \lambda\end{vmatrix}. $$

Evaluate each $$2\times2$$ determinant:

$$ \begin{aligned} \begin{vmatrix}-2 & \lambda\\ \lambda & 1\end{vmatrix}&=(-2)(1)-\lambda\cdot\lambda=-2-\lambda^{2},\\[2mm] \begin{vmatrix}1 & \lambda\\ 1 & 1\end{vmatrix}&=1\cdot1-\lambda\cdot1=1-\lambda,\\[2mm] \begin{vmatrix}1 & -2\\ 1 & \lambda\end{vmatrix}&=1\cdot\lambda-(-2)\cdot1=\lambda+2. \end{aligned} $$

So

$$ \begin{aligned} \det(A)&=2(-2-\lambda^{2})+1(1-\lambda)+2(\lambda+2)\\ &=-4-2\lambda^{2}+1-\lambda+2\lambda+4\\ &=-2\lambda^{2}+\lambda+1. \end{aligned} $$

Set the determinant to zero to find the values of $$\lambda$$ that can possibly give inconsistency:

$$ -2\lambda^{2}+\lambda+1=0 \;\Longrightarrow\; 2\lambda^{2}-\lambda-1=0. $$

Solve this quadratic equation. The discriminant is

$$ \Delta=(-1)^{2}-4(2)(-1)=1+8=9, \qquad \sqrt{\Delta}=3. $$

Hence

$$ \lambda=\frac{1\pm3}{4}\;\Longrightarrow\; \lambda_{1}=1,\qquad \lambda_{2}=-\dfrac12. $$

The determinant is non-zero for every other real $$\lambda$$, giving a unique solution there. Thus, only the two values $$\lambda=1$$ and $$\lambda=-\dfrac12$$ need further investigation. We now check whether the system is inconsistent at those two values.

Case 1: $$\lambda=1$$.

The equations become

$$ \begin{aligned} 2x-y+2z&=2,\\ x-2y+z&=-4,\\ x+y+z&=4. \end{aligned} $$

The augmented matrix is

$$ \begin{bmatrix} 2 & -1 & 2 & | & 2\\ 1 & -2 & 1 & | & -4\\ 1 & 1 & 1 & | & 4 \end{bmatrix}. $$

Perform elementary row operations:

Subtract twice Row 2 from Row 1:

$$ R_{1}\leftarrow R_{1}-2R_{2}:\; [0\;\;3\;\;0\;|\;10]. $$

Subtract Row 2 from Row 3:

$$ R_{3}\leftarrow R_{3}-R_{2}:\; [0\;\;3\;\;0\;|\;8]. $$

Now subtract the new Row 3 from the new Row 1:

$$ R_{1}\leftarrow R_{1}-R_{3}:\; [0\;\;0\;\;0\;|\;2]. $$

This gives the equation $$0=2,$$ an impossibility. Thus $$\operatorname{rank}(A)=2$$ and $$\operatorname{rank}([A|b])=3,$$ so the system has no solution when $$\lambda=1.$$

Case 2: $$\lambda=-\dfrac12$$.

The equations now read

$$ \begin{aligned} 2x-y+2z&=2,\\ x-2y-\dfrac12z&=-4,\\ x-\dfrac12y+z&=4. \end{aligned} $$

The augmented matrix is

$$ \begin{bmatrix} 2 & -1 & 2 & | & 2\\ 1 & -2 & -\dfrac12 & | & -4\\ 1 & -\dfrac12 & 1 & | & 4 \end{bmatrix}. $$

Again perform row operations. First, eliminate the leading $$1$$ of Row 2 from Row 3:

$$ R_{3}\leftarrow R_{3}-R_{2}:\; \Bigl[0,\;\dfrac32,\;\dfrac32\;|\;8\Bigr]. $$

Next, eliminate the leading $$2$$ of the original Row 1 with twice Row 2:

$$ R_{1}\leftarrow R_{1}-2R_{2}:\; [0,\;3,\;3\;|\;10]. $$

Now compare the new Row 1 and Row 3. Half of Row 1 is $$[0,\;1.5,\;1.5\;|\;5].$$ Subtract this half from Row 3:

$$ R_{3}\leftarrow R_{3}-\frac12R_{1}:\; [0,\;0,\;0\;|\;3]. $$

Once again we reach an impossible equation $$0=3.$$ Therefore $$\operatorname{rank}(A)=2$$ and $$\operatorname{rank}([A|b])=3,$$ giving no solution when $$\lambda=-\dfrac12$$.

We have shown that the system is inconsistent precisely for the two values

$$ \lambda=1 \quad\text{and}\quad \lambda=-\dfrac12. $$

Hence the set $$S$$ of all real $$\lambda$$ that make the system unsolvable contains exactly two elements.

Hence, the correct answer is Option D.