1. Key Concepts and Formulas

For a system of linear equations in three variables, say $AX=B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix, we use determinants to analyze the nature of solutions. Let $\Delta = \det(A)$, and $\Delta_x, \Delta_y, \Delta_z$ be the determinants formed by replacing the respective columns of $A$ with the constant matrix $B$.

• Condition for No Solutions (Inconsistent System): The system has no solutions if and only if $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.
• Condition for Unique Solution: If $\Delta \ne 0$, the system has a unique solution given by Cramer's Rule: $x = \Delta_x/\Delta$, $y = \Delta_y/\Delta$, $z = \Delta_z/\Delta$.
• Condition for Infinitely Many Solutions: If $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$, the system has infinitely many solutions.

2. Step-by-Step Solution

Step 1: Formulate the Coefficient and Constant Matrices The given system of linear equations is:

$$\begin{align*} 2x - 3y + 5z &= 9 \\ x + 3y - z &= -18 \\ 3x - y + (\lambda^2 - |\lambda|)z &= 16 \end{align*}$$

We represent this system in matrix form $AX=B$:

$$A = \begin{pmatrix} 2 & -3 & 5 \\ 1 & 3 & -1 \\ 3 & -1 & \lambda^2 - |\lambda| \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 9 \\ -18 \\ 16 \end{pmatrix}$$

For convenience, let $K = \lambda^2 - |\lambda|$. Since $\lambda$ is a real number, $\lambda^2 = |\lambda|^2$, so $K = |\lambda|^2 - |\lambda|$.

Step 2: Calculate the Determinant of the Coefficient Matrix ($\Delta$) We need to calculate $\Delta = \det(A)$ for the condition $\Delta=0$.

$$\Delta = \left| \begin{matrix} 2 & -3 & 5 \\ 1 & 3 & -1 \\ 3 & -1 & K \end{matrix} \right|$$

Expanding along the first row:

$$\begin{align*} \Delta &= 2 \left| \begin{matrix} 3 & -1 \\ -1 & K \end{matrix} \right| - (-3) \left| \begin{matrix} 1 & -1 \\ 3 & K \end{matrix} \right| + 5 \left| \begin{matrix} 1 & 3 \\ 3 & -1 \end{matrix} \right| \\ &= 2(3K - (-1)(-1)) + 3(1K - (-1)(3)) + 5(1(-1) - 3(3)) \\ &= 2(3K - 1) + 3(K + 3) + 5(-1 - 9) \\ &= 6K - 2 + 3K + 9 - 50 \\ &= 9K - 43 \end{align*}$$

So, $\Delta = 9K - 43$.

Step 3: Find Values of $\lambda$ for which $\Delta = 0$ For the system to have no solutions (or infinitely many solutions), we must have $\Delta = 0$. Setting $\Delta = 0$: $9K - 43 = 0 \implies K = \frac{43}{9}$ Now, substitute back $K = \lambda^2 - |\lambda|$: $\lambda^2 - |\lambda| = \frac{43}{9}$ Let $t = |\lambda|$. Since $\lambda$ is a real number, $t \ge 0$. Also, $\lambda^2 = |\lambda|^2 = t^2$. The equation becomes a quadratic in $t$: $t^2 - t = \frac{43}{9} \\ 9t^2 - 9t - 43 = 0$ Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$t = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(9)(-43)}}{2(9)} \\ t = \frac{9 \pm \sqrt{81 + 1548}}{18} \\ t = \frac{9 \pm \sqrt{1629}}{18}$$

We have two possible values for $t$: $t_1 = \frac{9 + \sqrt{1629}}{18} \quad \text{and} \quad t_2 = \frac{9 - \sqrt{1629}}{18}$ Since $\sqrt{1629} \approx 40.36$, $t_1 \approx \frac{9+40.36}{18} \approx 2.74$, which is positive. However, $t_2 \approx \frac{9-40.36}{18} \approx -1.74$, which is negative. Since $t = |\lambda|$, $t$ must be non-negative ($t \ge 0$). Therefore, $t_2$ is not a valid solution for $|\lambda|$. The only valid value for $|\lambda|$ is $t_1 = \frac{9 + \sqrt{1629}}{18}$. Since this value is positive, it implies two distinct real values for $\lambda$: $\lambda = \pm \left( \frac{9 + \sqrt{1629}}{18} \right)$ These are the two real values of $\lambda$ for which $\Delta=0$.

Step 4: Calculate $\Delta_x$ (or $\Delta_y$ or $\Delta_z$) to Check for Inconsistency To determine if the system has no solutions for these values of $\lambda$, we need to check if at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero when $\Delta=0$. Let's calculate $\Delta_x$.

$$\Delta_x = \left| \begin{matrix} 9 & -3 & 5 \\ -18 & 3 & -1 \\ 16 & -1 & K \end{matrix} \right|$$

Expanding along the first row:

$$\begin{align*} \Delta_x &= 9 \left| \begin{matrix} 3 & -1 \\ -1 & K \end{matrix} \right| - (-3) \left| \begin{matrix} -18 & -1 \\ 16 & K \end{matrix} \right| + 5 \left| \begin{matrix} -18 & 3 \\ 16 & -1 \end{matrix} \right| \\ &= 9(3K - (-1)(-1)) + 3((-18)K - (-1)(16)) + 5((-18)(-1) - 3(16)) \\ &= 9(3K - 1) + 3(-18K + 16) + 5(18 - 48) \\ &= 27K - 9 - 54K + 48 + 5(-30) \\ &= -27K + 39 - 150 \\ &= -27K - 111 \end{align*}$$

So, $\Delta_x = -27K - 111$.

Step 5: Check for Inconsistency Now, substitute the value $K = \frac{43}{9}$ (which makes $\Delta=0$) into the expression for $\Delta_x$:

$$\Delta_x = -27 \left( \frac{43}{9} \right) - 111 \\ \Delta_x = -3 \times 43 - 111 \\ \Delta_x = -129 - 111 \\ \Delta_x = -240$$

Since $\Delta_x = -240 \ne 0$, it confirms that for $K = \frac{43}{9}$ (i.e., for the two values of $\lambda$ found in Step 3), the system of equations has no solutions.

Step 6: Count the Number of Real Values of $\lambda$ We found two distinct real values of $\lambda$, namely $\lambda = \frac{9 + \sqrt{1629}}{18}$ and $\lambda = -\frac{9 + \sqrt{1629}}{18}$, for which $\Delta = 0$ and $\Delta_x \ne 0$. Therefore, for these two real values of $\lambda$, the system of linear equations has no solutions.

The number of such real values of $\lambda$ is 2.

3. Common Mistakes & Tips

• Confusing Conditions: A common error is to confuse the conditions for "no solutions" with "infinitely many solutions". Remember, $\Delta=0$ is necessary but not sufficient for either; the values of $\Delta_x, \Delta_y, \Delta_z$ distinguish between the two cases.
• Handling Absolute Value: When substituting $t = |\lambda|$, ensure you consider only non-negative values for $t$. If $t > 0$, then $|\lambda|=t$ yields two real values for $\lambda$ ($\pm t$). If $t=0$, it yields one value ($\lambda=0$). If $t < 0$, it yields no real values for $\lambda$.
• Determinant Calculation Errors: Be meticulous with signs and arithmetic when calculating determinants, especially when expanding $3 \times 3$ determinants. Using row/column operations to simplify determinants before expansion can help reduce errors.

4. Summary

We first established the conditions for a system of linear equations to have no solutions: the determinant of the coefficient matrix ($\Delta$) must be zero, and at least one of the determinants formed by replacing a column with the constant terms ($\Delta_x, \Delta_y, \Delta_z$) must be non-zero. We defined $K = \lambda^2 - |\lambda|$ for simplification. Calculating $\Delta = 9K - 43$, we set it to zero to find $K = 43/9$. This led to a quadratic equation in $|\lambda|$, which yielded two distinct real values for $\lambda$. Next, we calculated $\Delta_x = -27K - 111$. Substituting $K=43/9$ into $\Delta_x$ resulted in $\Delta_x = -240$, which is non-zero. Since $\Delta=0$ and $\Delta_x \ne 0$ for these two values of $\lambda$, the system has no solutions for both of them.

5. Final Answer

The number of real values of $\lambda$ for which the system of linear equations has no solutions is 2.

The final answer is $\boxed{2}$, which corresponds to option (C).