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

For a system of $n$ linear equations in $n$ variables, represented in matrix form as $AX=B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix, there are specific conditions for it to have no solution. This situation is known as an inconsistent system.

Using Cramer's Rule, the conditions for a system to have no solution are:

1. The determinant of the coefficient matrix, $\Delta = \det(A)$, must be equal to zero ($\Delta = 0$).
2. At least one of the determinants $\Delta_x, \Delta_y, \Delta_z$ (formed by replacing a column of $A$ with the constant matrix $B$) must be non-zero. That is, at least one of $\Delta_x \ne 0$, $\Delta_y \ne 0$, or $\Delta_z \ne 0$.

In essence, if $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system is inconsistent and has no solution. This corresponds to a contradictory statement like $0 = k$ where $k \ne 0$ when solving by elimination.

2. Setting up the System in Matrix Form

The given system of linear equations is:

1. $x + y + z = 6$
2. $3x + 5y + 5z = 26$
3. $x + 2y + \lambda z = \mu$

We can express this system in the matrix form $AX=B$, where: $A = \begin{pmatrix} 1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 6 \\ 26 \\ \mu \end{pmatrix}$

3. Calculating the Determinant of the Coefficient Matrix ($\Delta$)

For the system to have no solution, the first condition is that the determinant of the coefficient matrix $A$ must be zero ($\Delta = 0$).

Let's calculate $\Delta = \det(A)$: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda \end{vmatrix}$ To simplify the calculation, we can perform row operations. Applying $R_2 \to R_2 - 3R_1$ and $R_3 \to R_3 - R_1$: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 3 - 3(1) & 5 - 3(1) & 5 - 3(1) \\ 1 - 1(1) & 2 - 1(1) & \lambda - 1(1) \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 1 & \lambda - 1 \end{vmatrix}$ Now, we expand the determinant along the first column, as it contains two zeros, simplifying the calculation: $\Delta = 1 \cdot \begin{vmatrix} 2 & 2 \\ 1 & \lambda - 1 \end{vmatrix} - 0 \cdot (\dots) + 0 \cdot (\dots)$ $\Delta = 1 \cdot [2(\lambda - 1) - 2(1)]$ $\Delta = 2\lambda - 2 - 2$ $\Delta = 2\lambda - 4$ For the system to have no solution, we must have $\Delta = 0$: $2\lambda - 4 = 0$ $2\lambda = 4$ $\lambda = 2$ This gives us the value of $\lambda$ required for the first condition of no solution.

4. Calculating the Determinants for Variables ($\Delta_x, \Delta_y, \Delta_z$)

With $\lambda = 2$, we now need to calculate $\Delta_x, \Delta_y,$ and $\Delta_z$. For no solution, at least one of these must be non-zero.

First, let's calculate $\Delta_x$ by replacing the first column of $A$ with the constant matrix $B$: $\Delta_x = \begin{vmatrix} 6 & 1 & 1 \\ 26 & 5 & 5 \\ \mu & 2 & \lambda \end{vmatrix}$ Substitute $\lambda = 2$: $\Delta_x = \begin{vmatrix} 6 & 1 & 1 \\ 26 & 5 & 5 \\ \mu & 2 & 2 \end{vmatrix}$ Observe that the second column ($C_2$) and the third column ($C_3$) are identical. A property of determinants states that if two columns (or rows) are identical, the determinant is zero. Therefore, $\Delta_x = 0$. This means $\Delta_x$ does not contribute to the "non-zero" condition for no solution, so we must check other determinants.

Next, let's calculate $\Delta_y$ by replacing the second column of $A$ with the constant matrix $B$: $\Delta_y = \begin{vmatrix} 1 & 6 & 1 \\ 3 & 26 & 5 \\ 1 & \mu & \lambda \end{vmatrix}$ Substitute $\lambda = 2$: $\Delta_y = \begin{vmatrix} 1 & 6 & 1 \\ 3 & 26 & 5 \\ 1 & \mu & 2 \end{vmatrix}$ We expand this determinant along the first row: $\Delta_y = 1 \cdot \begin{vmatrix} 26 & 5 \\ \mu & 2 \end{vmatrix} - 6 \cdot \begin{vmatrix} 3 & 5 \\ 1 & 2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 26 \\ 1 & \mu \end{vmatrix}$ $\Delta_y = 1(26 \cdot 2 - 5\mu) - 6(3 \cdot 2 - 5 \cdot 1) + 1(3\mu - 26 \cdot 1)$ $\Delta_y = (52 - 5\mu) - 6(6 - 5) + (3\mu - 26)$ $\Delta_y = 52 - 5\mu - 6(1) + 3\mu - 26$ $\Delta_y = 52 - 5\mu - 6 + 3\mu - 26$ Combine the constant terms and the terms involving $\mu$: $\Delta_y = (52 - 6 - 26) + (-5\mu + 3\mu)$ $\Delta_y = (46 - 26) - 2\mu$ $\Delta_y = 20 - 2\mu$ For the system to have no solution (given $\Delta = 0$), we require at least one of $\Delta_x, \Delta_y, \Delta_z$ to be non-zero. Since $\Delta_x = 0$, we must ensure $\Delta_y \ne 0$: $20 - 2\mu \ne 0$ $20 \ne 2\mu$ $\mu \ne 10$ Thus, for the system to have no solution, $\mu$ must not be equal to 10.

(We don't need to calculate $\Delta_z$ because we've already found a condition ($\Delta_y \ne 0$) that satisfies the second requirement for no solution.)

5. Combining Conditions and Final Answer

Based on our calculations, for the system of equations to have no solution, the values of $\lambda$ and $\mu$ must satisfy:

• $\lambda = 2$
• $\mu \ne 10$

Comparing these conditions with the given options: (A) $\lambda$ = 3, $\mu$ = 5 (B) $\lambda$ = 3, $\mu$ $\ne$ 10 (C) $\lambda$ $\ne$ 2, $\mu$ = 10 (D) $\lambda$ = 2, $\mu$ $\ne$ 10

Our derived conditions match option (D).

Alternative Method: Gaussian Elimination (Row Operations)

This problem can also be efficiently solved using Gaussian elimination on the augmented matrix. This method often provides a more intuitive understanding of why a system has no solution.

The augmented matrix for the system is: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 3 & 5 & 5 & | & 26 \\ 1 & 2 & \lambda & | & \mu \end{pmatrix}$ Perform row operations to reduce the matrix to row echelon form:

1. $R_2 \to R_2 - 3R_1$ (Eliminate $x$ from the second equation)
2. $R_3 \to R_3 - R_1$ (Eliminate $x$ from the third equation) $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 2 & 2 & | & 8 \\ 0 & 1 & \lambda - 1 & | & \mu - 6 \end{pmatrix}$
3. $R_2 \to \frac{1}{2} R_2$ (Simplify the second equation) $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 1 & | & 4 \\ 0 & 1 & \lambda - 1 & | & \mu - 6 \end{pmatrix}$
4. $R_3 \to R_3 - R_2$ (Eliminate $y$ from the third equation) $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 1 & | & 4 \\ 0 & 0 & (\lambda - 1) - 1 & | & (\mu - 6) - 4 \end{pmatrix}$ This simplifies to: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 1 & | & 4 \\ 0 & 0 & \lambda - 2 & | & \mu - 10 \end{pmatrix}$ The last row of this matrix corresponds to the equation: $0x + 0y + (\lambda - 2)z = \mu - 10$ For the system to have no solution, this equation must be a contradiction. This means the coefficient of $z$ must be zero, while the constant term on the right side must be non-zero.

• Coefficient of $z$: $\lambda - 2 = 0 \implies \lambda = 2$.
• Constant term: $\mu - 10 \ne 0 \implies \mu \ne 10$. Both methods consistently lead to the same conditions: $\lambda = 2$ and $\mu \ne 10$.

6. Tips and Common Mistakes

• Determinant Calculation: Be very careful with arithmetic and signs when calculating determinants. A single error can lead to incorrect conditions for $\lambda$ and $\mu$.
• Conditions for Solution Types: It's crucial to remember the precise conditions for each type of solution:
  • Unique Solution: $\Delta \ne 0$.
  • No Solution: $\Delta = 0$ AND (at least one of $\Delta_x, \Delta_y, \Delta_z \ne 0$).
  • Infinitely Many Solutions: $\Delta = 0$ AND ($\Delta_x = 0$ AND $\Delta_y = 0$ AND $\Delta_z = 0$).
• Gaussian Elimination as a Verification: Using row operations (Gaussian elimination) is an excellent way to cross-check your results from Cramer's Rule. It directly reveals the consistency or inconsistency of the system.

7. Summary and Key Takeaway

To determine the conditions for a system of linear equations to have no solution, we first find the value(s) of the parameters that make the determinant of the coefficient matrix ($\Delta$) equal to zero. Then, using these parameter values, we check if any of the other determinants ($\Delta_x, \Delta_y, \Delta_z$) are non-zero. If $\Delta=0$ and at least one $\Delta_i \ne 0$, the system has no solution. In this problem, this led to the conditions $\lambda=2$ and $\mu \ne 10$.