Question

The system of equations \[ x + y + z = 6, \] \[ x + 2y + 5z = 9, \] \[ x + 5y + \lambda z = \mu, \] has no solution if:

• A. \( \lambda = 17, \mu \neq 18 \)
• B. \( \lambda \neq 17, \mu \neq 18 \)
• C. \( \lambda = 15, \mu \neq 17 \)
• D. \( \lambda = 17, \mu = 18 \)

Hint:

For systems of linear equations, use substitution or elimination to simplify and solve. Inconsistent systems occur when the equations are parallel or contradictory.

Answer 1

The Correct Option is A

We are given the system of equations: \[ x + y + z = 6, \] \[ x + 2y + 5z = 9, \] \[ x + 5y + \lambda z = \mu. \] 
- We can solve this system by using elimination or substitution to obtain the conditions under which the system has no solution. For a system to have no solution, the determinant of the coefficient matrix must be zero, or the equations must be inconsistent. 
- After solving the system, we find that the system will have no solution when \( \lambda = 17 \) and \( \mu \neq 18 \). 

Conclusion: The correct answer is (1), as the system has no solution when \( \lambda = 17 \) and \( \mu \neq 18 \).

Answer 2

Problem:

Consider the system of equations:

\[ \begin{aligned} & (1)\quad x + y + z = 6 \\ & (2)\quad x + 2y + 5z = 9 \\ & (3)\quad x + 5y + \lambda z = \mu \end{aligned} \]

Step 1: Express \( x \) from Equation (1)

From equation (1), solve for \( x \):

\[ x = 6 - y - z \]

Step 2: Substitute into Equations (2) and (3)

Substitute into (2):

\[ (6 - y - z) + 2y + 5z = 9 \\ \Rightarrow 6 + y + 4z = 9 \\ \Rightarrow y + 4z = 3 \tag{4} \]

Substitute into (3):

\[ (6 - y - z) + 5y + \lambda z = \mu \\ \Rightarrow 6 + 4y + (\lambda - 1)z = \mu \tag{5} \]

Step 3: Use Equation (4) in Equation (5)

From Equation (4):

\[ y = 3 - 4z \]

Substitute into Equation (5):

\[ 6 + 4(3 - 4z) + (\lambda - 1)z = \mu \\ \Rightarrow 6 + 12 - 16z + (\lambda - 1)z = \mu \\ \Rightarrow 18 + (\lambda - 17)z = \mu \]

Step 4: Condition for No Solution

This equation is consistent for all \( z \), unless the coefficient of \( z \) becomes zero. That is: \[ \lambda = 17 \] In that case, the equation becomes: \[ 18 = \mu \] So, a contradiction arises when: \[ \lambda = 17 \quad \text{and} \quad \mu \ne 18 \]

✅ Final Answer:

The system has no solution if and only if:
\[ \boxed{\lambda = 17 \quad \text{and} \quad \mu \ne 18} \]