1. Key Concepts and Formulas

• System of Linear Equations: A set of equations with multiple variables. For a system of three linear equations in three variables ($x, y, z$): $a_1x + b_1y + c_1z = d_1$ $a_2x + b_2y + c_2z = d_2$ $a_3x + b_3y + c_3z = d_3$

• Cramer's Rule and Determinants: We define the following determinants:

  • Coefficient Determinant ($D$): The determinant of the matrix formed by the coefficients of the variables. $D = \left| \begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} \right|$
  • Determinants $D_1, D_2, D_3$: Obtained by replacing the $j$-th column of $D$ with the constant terms $d_1, d_2, d_3$. $D_1 = \left| \begin{matrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{matrix} \right|, \quad D_2 = \left| \begin{matrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{matrix} \right|, \quad D_3 = \left| \begin{matrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{matrix} \right|$

• Consistency Conditions:

  • Unique Solution (Consistent): If $D \neq 0$, the system has a unique solution.
  • Inconsistent System (No Solution): If $D = 0$ AND at least one of $D_1, D_2, D_3$ is non-zero, the system has no solution.
  • Infinitely Many Solutions (Consistent): If $D = 0$ AND $D_1 = 0$ AND $D_2 = 0$ AND $D_3 = 0$, the system has infinitely many solutions.

2. Step-by-Step Solution

The given system of linear equations is:

1. $2x_1 - 4x_2 + \lambda x_3 = 1$
2. $x_1 - 6x_2 + x_3 = 2$
3. $\lambda x_1 - 10x_2 + 4x_3 = 3$

We are looking for the value(s) of $\lambda$ for which the system is inconsistent (i.e., has no solution). This requires $D=0$ and at least one of $D_1, D_2, D_3$ to be non-zero.

Step 1: Calculate the Determinant of the Coefficient Matrix, $D$

First, we calculate $D$. If $D \neq 0$, the system has a unique solution and is consistent, so we only need to investigate values of $\lambda$ for which $D=0$.

$D = \left| \begin{matrix} 2 & -4 & \lambda \\ 1 & -6 & 1 \\ \lambda & -10 & 4 \end{matrix} \right|$ Expanding along the first row: $D = 2((-6)(4) - (1)(-10)) - (-4)((1)(4) - (1)(\lambda)) + \lambda((1)(-10) - (-6)(\lambda))$ $D = 2(-24 + 10) + 4(4 - \lambda) + \lambda(-10 + 6\lambda)$ $D = 2(-14) + 16 - 4\lambda - 10\lambda + 6\lambda^2$ $D = -28 + 16 - 14\lambda + 6\lambda^2$ $D = 6\lambda^2 - 14\lambda - 12$

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

To find potential values for inconsistency or infinitely many solutions, we set $D=0$: $6\lambda^2 - 14\lambda - 12 = 0$ Divide by 2 to simplify: $3\lambda^2 - 7\lambda - 6 = 0$ Factor the quadratic equation: $3\lambda^2 - 9\lambda + 2\lambda - 6 = 0$ $3\lambda(\lambda - 3) + 2(\lambda - 3) = 0$ $(3\lambda + 2)(\lambda - 3) = 0$ This yields two critical values for $\lambda$: $\lambda = 3 \quad \text{or} \quad \lambda = -\frac{2}{3}$

Step 3: Calculate $D_1, D_2, D_3$

Next, we calculate the determinants $D_1, D_2, D_3$ to distinguish between inconsistent systems and systems with infinitely many solutions for the critical values of $\lambda$.

$D_1 = \left| \begin{matrix} 1 & -4 & \lambda \\ 2 & -6 & 1 \\ 3 & -10 & 4 \end{matrix} \right|$ Expanding $D_1$ along the first row: $D_1 = 1((-6)(4) - (1)(-10)) - (-4)((2)(4) - (1)(3)) + \lambda((2)(-10) - (-6)(3))$ $D_1 = 1(-24 + 10) + 4(8 - 3) + \lambda(-20 + 18)$ $D_1 = -14 + 4(5) + \lambda(-2)$ $D_1 = -14 + 20 - 2\lambda = 6 - 2\lambda$

$D_2 = \left| \begin{matrix} 2 & 1 & \lambda \\ 1 & 2 & 1 \\ \lambda & 3 & 4 \end{matrix} \right|$ Expanding $D_2$ along the first row: $D_2 = 2((2)(4) - (1)(3)) - 1((1)(4) - (1)(\lambda)) + \lambda((1)(3) - (2)(\lambda))$ $D_2 = 2(8 - 3) - 1(4 - \lambda) + \lambda(3 - 2\lambda)$ $D_2 = 2(5) - 4 + \lambda + 3\lambda - 2\lambda^2$ $D_2 = 10 - 4 + 4\lambda - 2\lambda^2 = -2\lambda^2 + 4\lambda + 6$

$D_3 = \left| \begin{matrix} 2 & -4 & 1 \\ 1 & -6 & 2 \\ \lambda & -10 & 3 \end{matrix} \right|$ Expanding $D_3$ along the first row: $D_3 = 2((-6)(3) - (2)(-10)) - (-4)((1)(3) - (2)(\lambda)) + 1((1)(-10) - (-6)(\lambda))$ $D_3 = 2(-18 + 20) + 4(3 - 2\lambda) + (-10 + 6\lambda)$ $D_3 = 2(2) + 12 - 8\lambda - 10 + 6\lambda$ $D_3 = 4 + 12 - 10 - 2\lambda = 6 - 2\lambda$ Notice that $D_1 = D_3 = 6 - 2\lambda$.

Step 4: Analyze for Inconsistency for Each Critical $\lambda$ Value

We now test our critical values of $\lambda$ found in Step 2.

• Case A: $\lambda = 3$

  • $D = 0$ (from Step 2).
  • $D_1 = 6 - 2(3) = 0$.
  • $D_2 = -2(3)^2 + 4(3) + 6 = -2(9) + 12 + 6 = -18 + 12 + 6 = 0$.
  • $D_3 = 6 - 2(3) = 0$. Since $D = 0$ AND $D_1 = 0$ AND $D_2 = 0$ AND $D_3 = 0$, the system has infinitely many solutions for $\lambda = 3$. This means the system is consistent for $\lambda=3$.

• Case B: $\lambda = -\frac{2}{3}$

  • $D = 0$ (from Step 2).
  • $D_1 = 6 - 2\left(-\frac{2}{3}\right) = 6 + \frac{4}{3} = \frac{18+4}{3} = \frac{22}{3}$.
  • $D_2 = -2\left(-\frac{2}{3}\right)^2 + 4\left(-\frac{2}{3}\right) + 6 = -2\left(\frac{4}{9}\right) - \frac{8}{3} + 6 = -\frac{8}{9} - \frac{24}{9} + \frac{54}{9} = \frac{-8-24+54}{9} = \frac{22}{9}$.
  • $D_3 = 6 - 2\left(-\frac{2}{3}\right) = 6 + \frac{4}{3} = \frac{22}{3}$. Since $D = 0$ AND $D_1 \neq 0$ (and also $D_2 \neq 0$, $D_3 \neq 0$), the system is inconsistent (has no solution) for $\lambda = -\frac{2}{3}$.

Step 5: Conclusion

The system of linear equations is inconsistent only for $\lambda = -\frac{2}{3}$. This is exactly one negative value of $\lambda$.

3. Common Mistakes & Tips

• Calculation Errors: Determinant calculations can be lengthy and prone to arithmetic mistakes. Double-check each step, especially sign changes.
• Misinterpreting Conditions: Remember the exact conditions for unique solution ($D \neq 0$), no solution ($D=0$ and at least one $D_i \neq 0$), and infinitely many solutions ($D=0$ and all $D_i=0$).
• Quadratic Equation Solutions: Ensure you correctly solve for $\lambda$ when $D=0$. Factoring, quadratic formula, or completing the square are common methods.
• Gaussian Elimination Alternative: For complex systems, Gaussian elimination on the augmented matrix can sometimes be more robust, especially for distinguishing between no solution and infinitely many solutions. An inconsistent system will result in a row like $[0 \ 0 \ 0 \ | \ k]$ where $k \neq 0$.

4. Summary

To determine when a system of linear equations is inconsistent, we first calculate the determinant of the coefficient matrix, $D$. If $D \neq 0$, a unique solution exists, making the system consistent. If $D=0$, we then calculate $D_1, D_2,$ and $D_3$. The system is inconsistent if $D=0$ and at least one of $D_1, D_2, D_3$ is non-zero. Our calculations showed that $D=0$ for $\lambda=3$ and $\lambda=-2/3$. For $\lambda=3$, all $D_i$ were also zero, leading to infinitely many solutions (consistent). For $\lambda=-2/3$, we found that $D_1 \neq 0$, making the system inconsistent. Thus, the system is inconsistent for exactly one negative value of $\lambda$.

5. Final Answer

The final answer is $\boxed{B}$