Key Concept: Conditions for a System of Linear Equations to Have No Solution

For a system of linear equations in three variables, say:

$$\begin{align*} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{align*}$$

We can represent this system in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix.

$$A = \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}$$

The determinant of the coefficient matrix, $\Delta = \det(A)$, plays a crucial role in determining the nature of the solution.

1. Unique Solution: If $\Delta \neq 0$, the system has a unique solution.
2. No Solution or Infinitely Many Solutions: If $\Delta = 0$, the system either has no solution or infinitely many solutions. To distinguish between these two cases, we examine the consistency of the equations.
   • No Solution: If $\Delta = 0$ and at least one of the determinants $\Delta_x, \Delta_y, \Delta_z$ (obtained by replacing a column of $A$ with $B$) is non-zero, then the system has no solution. This implies that the equations are inconsistent.
   • Infinitely Many Solutions: If $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$, then the system has infinitely many solutions. This implies that the equations are consistent and dependent (one or more equations can be derived from the others).

In this problem, we are looking for values of $\lambda$ for which the system has no solution. Therefore, we must satisfy two conditions:

1. The determinant of the coefficient matrix, $\Delta$, must be zero.
2. When $\Delta = 0$, the system of equations must be inconsistent (i.e., lead to a contradiction).

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Step 1: Write the System in Matrix Form and Calculate the Determinant of the Coefficient Matrix

The given system of linear equations is:

$$\begin{align*} 2x - y + 2z &= 2 \quad &(1) \\ x - 2y + \lambda z &= -4 \quad &(2) \\ x + \lambda y + z &= 4 \quad &(3) \end{align*}$$

The coefficient matrix $A$ for this system is:

$$A = \begin{pmatrix} 2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1 \end{pmatrix}$$

Now, we calculate the determinant $\Delta = \det(A)$:

$$\Delta = \left| \begin{matrix} 2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1 \end{matrix} \right|$$

To calculate the determinant, we expand along the first row:

$$\Delta = 2((-2)(1) - (\lambda)(\lambda)) - (-1)((1)(1) - (\lambda)(1)) + 2((1)(\lambda) - (-2)(1))$$ $$\Delta = 2(-2 - \lambda^2) + 1(1 - \lambda) + 2(\lambda + 2)$$ $$\Delta = -4 - 2\lambda^2 + 1 - \lambda + 2\lambda + 4$$ $$\Delta = -2\lambda^2 + \lambda + 1$$

Step 2: Find Values of $\lambda$ for which $\Delta = 0$

For the system to have no solution (or infinitely many solutions), the determinant $\Delta$ must be zero. So, we set $\Delta = 0$:

$$-2\lambda^2 + \lambda + 1 = 0$$

Multiplying by $-1$ to make the leading coefficient positive:

$$2\lambda^2 - \lambda - 1 = 0$$

This is a quadratic equation. We can solve it by factoring or using the quadratic formula. By factoring:

$$2\lambda^2 - 2\lambda + \lambda - 1 = 0$$ $$2\lambda(\lambda - 1) + 1(\lambda - 1) = 0$$ $$(2\lambda + 1)(\lambda - 1) = 0$$

This gives us two possible values for $\lambda$:

$$2\lambda + 1 = 0 \Rightarrow \lambda = -\frac{1}{2}$$ $$\lambda - 1 = 0 \Rightarrow \lambda = 1$$

These are the values of $\lambda$ for which the system might have no solution or infinitely many solutions. We now need to check each case to see if it leads to an inconsistency.

Step 3: Analyze the Case when $\lambda = 1$

Substitute $\lambda = 1$ into the original system of equations:

$$\begin{align*} 2x - y + 2z &= 2 \quad &(1) \\ x - 2y + (1)z &= -4 \quad &(2) \\ x + (1)y + z &= 4 \quad &(3) \end{align*}$$

Let's simplify and check for consistency:

$$\begin{align*} 2x - y + 2z &= 2 \quad &(1) \\ x - 2y + z &= -4 \quad &(2) \\ x + y + z &= 4 \quad &(3) \end{align*}$$

Consider equations (2) and (3). If we add equation (2) and equation (3):

$$(x - 2y + z) + (x + y + z) = -4 + 4$$ $$2x - y + 2z = 0$$

Now, we have a new equation $2x - y + 2z = 0$. Comparing this with equation (1), which is $2x - y + 2z = 2$, we see a clear contradiction:

$$0 = 2$$

This is impossible. Therefore, when $\lambda = 1$, the system of equations is inconsistent and has no solution. Thus, $\lambda = 1$ belongs to the set $S$.

Step 4: Analyze the Case when $\lambda = -\frac{1}{2}$

Substitute $\lambda = -\frac{1}{2}$ into the original system of equations:

$$\begin{align*} 2x - y + 2z &= 2 \quad &(1) \\ x - 2y + \left(-\frac{1}{2}\right)z &= -4 \quad &(2) \\ x + \left(-\frac{1}{2}\right)y + z &= 4 \quad &(3) \end{align*}$$

Let's simplify the equations:

$$\begin{align*} 2x - y + 2z &= 2 \quad &(1) \\ x - 2y - \frac{1}{2}z &= -4 \quad &(2) \\ x - \frac{1}{2}y + z &= 4 \quad &(3) \end{align*}$$

To check for consistency, let's try to manipulate equations to find a contradiction. Consider equation (3): $x - \frac{1}{2}y + z = 4$. If we multiply equation (3) by 2:

$$2 \times \left(x - \frac{1}{2}y + z\right) = 2 \times 4$$ $$2x - y + 2z = 8$$

Now, we have a new equation $2x - y + 2z = 8$. Comparing this with equation (1), which is $2x - y + 2z = 2$, we again see a clear contradiction:

$$8 = 2$$

This is impossible. Therefore, when $\lambda = -\frac{1}{2}$, the system of equations is inconsistent and has no solution. Thus, $\lambda = -\frac{1}{2}$ also belongs to the set $S$.

Summary of Findings:

Both values of $\lambda$ obtained from $\Delta = 0$, namely $\lambda = 1$ and $\lambda = -\frac{1}{2}$, lead to inconsistent systems of equations. Therefore, for both these values, the system has no solution. The set $S$ of all $\lambda \in R$ for which the system has no solution is $S = \left\{1, -\frac{1}{2}\right\}$.

Conclusion:

The set $S$ contains exactly two elements.

Relevant Tips and Common Mistakes:

• Don't Stop at $\Delta = 0$: A very common mistake is to assume that if $\Delta = 0$, the system automatically has no solution. Remember, $\Delta = 0$ only tells you that there is either no solution or infinitely many solutions. You must check the consistency of the equations for each value of $\lambda$ that makes $\Delta = 0$.
• Method for Checking Consistency: When $\Delta = 0$, look for linear combinations of the equations that result in a contradiction (e.g., $0 = k$ where $k \neq 0$) or identical left-hand sides with different right-hand sides. If such a contradiction is found, the system is inconsistent (no solution). If no such contradiction exists and the equations become dependent (e.g., one equation is a multiple of another or a sum of others), then there are infinitely many solutions.
• Algebraic Accuracy: Be careful with determinant calculations and solving quadratic equations. A small error can lead to incorrect values of $\lambda$.

The set $S = \left\{1, -\frac{1}{2}\right\}$ contains exactly two elements. Therefore, option (B) is the correct answer.

The final answer is $\boxed{A}$.