Key Concepts and Formulas

To analyze the nature of solutions for a system of three linear equations in three variables, we use determinants, often associated with Cramer's Rule. For a system: $\begin{aligned} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{aligned}$ We define the following determinants:

• $D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ (Determinant of the coefficient matrix)
• $D_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}$ (Replacing $x$-coefficients with constant terms)
• $D_y = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}$ (Replacing $y$-coefficients with constant terms)
• $D_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}$ (Replacing $z$-coefficients with constant terms)

The nature of the solutions is determined by these determinants:

• Unique Solution: The system has a unique solution if $D \neq 0$.
• No Solution (Inconsistent System): The system has no solution if $D=0$ AND at least one of $D_x, D_y, D_z$ is non-zero.
• Infinitely Many Solutions (Consistent, Dependent System): The system has infinitely many solutions if $D=0$ AND $D_x=0$ AND $D_y=0$ AND $D_z=0$.

We are looking for the conditions under which the system has no solution.

Step-by-Step Solution

Given the system of equations: $\begin{aligned} & x+y+z=6, \\ & x+2 y+5 z=9, \\ & x+5 y+\lambda z=\mu, \end{aligned}$

Step 1: Calculate the Determinant of the Coefficient Matrix ($D$) The first step is to calculate $D$. If $D \neq 0$, there is a unique solution, which is not what we want. If $D=0$, we proceed to check for no solution or infinitely many solutions.

The coefficient matrix is: $D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & \lambda \end{vmatrix}$ To simplify the calculation, we perform row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$. These operations do not change the value of the determinant: $D = \begin{vmatrix} 1 & 1 & 1 \\ 1-1 & 2-1 & 5-1 \\ 1-1 & 5-1 & \lambda-1 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 4 \\ 0 & 4 & \lambda-1 \end{vmatrix}$ Now, expand the determinant along the first column, which has two zeros: $D = 1 \cdot \left( (1)(\lambda-1) - (4)(4) \right) - 0 + 0$ $D = (\lambda-1) - 16$ $D = \lambda - 17$ For the system to have no solution (or infinitely many solutions), we must have $D=0$. Setting $D=0$: $\lambda - 17 = 0 \implies \lambda = 17$

Step 2: Calculate $D_z$ to Determine the Nature of Solutions when $D=0$ Since $D=0$ when $\lambda=17$, the system either has no solution or infinitely many solutions. To distinguish between these two cases, we need to calculate at least one of $D_x, D_y, D_z$. If any of these is non-zero, the system has no solution. If all are zero, it has infinitely many solutions. We'll calculate $D_z$ as it involves the constant term $\mu$.

$D_z$ is formed by replacing the $z$-coefficients column with the constant terms $(6, 9, \mu)$: $D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 9 \\ 1 & 5 & \mu \end{vmatrix}$ Again, use row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ to simplify: $D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1-1 & 2-1 & 9-6 \\ 1-1 & 5-1 & \mu-6 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 6 \\ 0 & 1 & 3 \\ 0 & 4 & \mu-6 \end{vmatrix}$ Expand along the first column: $D_z = 1 \cdot \left( (1)(\mu-6) - (3)(4) \right) - 0 + 0$ $D_z = (\mu-6) - 12$ $D_z = \mu - 18$

Step 3: Apply the Condition for "No Solution" For the system to have no solution, we must satisfy two conditions:

1. $D=0$
2. At least one of $D_x, D_y, D_z$ must be non-zero.

From Step 1, we found $D=0 \implies \lambda = 17$. From Step 2, we found $D_z = \mu - 18$. For $D_z$ to be non-zero, we must have: $\mu - 18 \neq 0 \implies \mu \neq 18$ Combining these conditions, the system has no solution if $\lambda = 17$ and $\mu \neq 18$.

Step 4: Match with Options Comparing our derived conditions with the given options:

• (A) $\lambda=17, \mu=18$: Here $D=0$ and $D_z=0$. Further checks (for $D_x, D_y$) would show they are also zero, leading to infinitely many solutions.
• (B) $\lambda=17, \mu \neq 18$: This exactly matches our derived conditions for no solution ($D=0$ and $D_z \neq 0$).
• (C) $\lambda=15, \mu \neq 17$: Here $D = 15-17 = -2 \neq 0$, so there is a unique solution.
• (D) $\lambda \neq 17, \mu \neq 18$: Here $D \neq 0$, so there is a unique solution.

Therefore, the system has no solution if $\lambda=17$ and $\mu \neq 18$.

Common Mistakes & Tips

• Confusing No Solution with Infinite Solutions: This is the most frequent error. Remember that $D=0$ is a necessary condition for both, but the values of $D_x, D_y, D_z$ differentiate them: non-zero for no solution, all zero for infinitely many solutions.
• Incorrect Determinant Calculation: Be careful with signs and arithmetic when expanding determinants, especially after applying row/column operations.
• Ignoring Row/Column Operations: Using row or column operations to introduce zeros before expanding a determinant can significantly simplify calculations and reduce errors. For instance, making all but one element zero in a row or column is a powerful technique.

Summary

To determine the conditions for a system of linear equations to have no solution, we first calculate the determinant of the coefficient matrix ($D$). For no solution, $D$ must be zero. If $D=0$, we then calculate $D_x$, $D_y$, and $D_z$. For no solution, at least one of these determinants must be non-zero. In this problem, we found $D = \lambda - 17$, so $D=0$ implies $\lambda=17$. We also found $D_z = \mu - 18$. For the system to have no solution, we need $D=0$ and $D_z \neq 0$ (or $D_x \neq 0$ or $D_y \neq 0$). Thus, $\lambda=17$ and $\mu \neq 18$ are the conditions for no solution. This corresponds to option (B).

The final answer is $\boxed{\text{(B)}}$.