Question

Consider the system of equations
\(x + y + z = 6\)
\(x + 2y + 5z = 18\)
\(2x + 2y + \lambda z = \mu\)
If the system of equations has infinitely many solutions, then the value of \((\lambda + \mu)\) is equal to

Answer 1

Correct Answer: 14

Step 1: Write the system of equations.
The system of equations is:
\(x + y + z = 6\)
\(x + 2y + 5z = 18\)
\(2x + 2y + \lambda z = \mu\)

Step 2: Write the augmented matrix for the system.
The augmented matrix for the system of equations is:
\[ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 1 & 2 & 5 & 18 \\ 2 & 2 & \lambda & \mu \end{bmatrix} \]

Step 3: Apply the condition for infinitely many solutions.
For a system to have infinitely many solutions, the determinant of the coefficient matrix must be zero. We will first compute the determinant of the coefficient matrix: \[ \text{Determinant} = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 2 & \lambda \end{vmatrix} \] Expanding the determinant: \[ = 1 \cdot \begin{vmatrix} 2 & 5 \\ 2 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 5 \\ 2 & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix} \] \[ = 1 \cdot (2\lambda - 10) - 1 \cdot (\lambda - 10) + 1 \cdot (2 - 4) \] \[ = (2\lambda - 10) - (\lambda - 10) - 2 \] \[ = 2\lambda - 10 - \lambda + 10 - 2 = \lambda - 2 \] For the system to have infinitely many solutions, the determinant must be zero: \[ \lambda - 2 = 0 \] Thus, \(\lambda = 2\).

Step 4: Solve for \(\mu\) using the augmented matrix.
Now, substitute \(\lambda = 2\) into the third equation: \[ 2x + 2y + 2z = \mu \] Divide through by 2: \[ x + y + z = \frac{\mu}{2} \] Since \(x + y + z = 6\) from the first equation, we get: \[ \frac{\mu}{2} = 6 \quad \Rightarrow \quad \mu = 12 \] Step 5: Compute \(\lambda + \mu\).
Now that we have \(\lambda = 2\) and \(\mu = 12\), we find: \[ \lambda + \mu = 2 + 12 = 14 \] Step 6: Conclusion.
The value of \(\lambda + \mu\) is 14.