Question

If the system of linear equations : \[ x+y+z=4,\quad x+2y+3z=6,\quad 4x+5y+\lambda z=\mu \] has more than one solution, then the value of \( \lambda+\mu \) is equal to:

• A. \(18\)
• B. \(9\)
• C. \(12\)
• D. \(24\)

Hint:

For infinitely many solutions, one equation must be a linear combination of the others, including the constant terms.

Answer 1

The Correct Option is D

To determine the value of \( \lambda + \mu \) such that the given system of linear equations has more than one solution, we need to analyze the system:

The system of equations is: 

1. \( x + y + z = 4 \)
2. \( x + 2y + 3z = 6 \)
3. \( 4x + 5y + \lambda z = \mu \)

For the system to have more than one solution, the equations must be dependent, meaning that the third equation must be a linear combination of the first two equations.

We express the first two equations in a matrix form:

\(\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 4 & 5 & \lambda \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ 6 \\ \mu \end{bmatrix}\)

To check for dependency, the determinant of the coefficient matrix must be zero:

Calculate the determinant:

\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 4 & 5 & \lambda \end{vmatrix} = 1(2\lambda - 15) - 1(1\lambda - 12) + 1(5 - 8) \]

\[ = 2\lambda - 15 - \lambda + 12 + 5 - 8 \]

\[ = \lambda + (-6) \]

For dependence (determinant zero):

\( \lambda - 6 = 0 \)

\(\lambda = 6\)

Substitute \(\lambda = 6\) into the third equation:

\(4x + 5y + 6z = \mu \)

Since the system has more than one solution, the third equation is a linear combination of the first two equations:

1. Multiplying the first equation by 2: \(2x + 2y + 2z = 8\)
2. Subtracting from the second equation: \((x + 2y + 3z) - (2x + 2y + 2z) = 6 - 8\)
3. \(-x + z = -2 \rightarrow x - z = 2\)

If we replace \(x - z = 2\) in \(4x + 5y + 6z = \mu\):

Simplify using equation 1 and equation 2, we find \(\mu = 18\):

Thus, \( \lambda + \mu = 6 + 18 = 24 \).

Therefore, the correct answer is \(24\).