Question

The existence of the unique solution of the system \[ x + y + z = \lambda, \quad 5x - y + \mu z = 10, \quad 2x + 3y - z = 6 \] depends on:

• A. \( \mu \) only
• B. \( \lambda \) only
• C. \( \lambda \) and \( \mu \) both
• D. neither \( \lambda \) nor \( \mu \)

Hint:

For systems of linear equations, use the determinant of the coefficient matrix to determine the existence of a unique solution. If the determinant is zero, the system may have no solution or infinitely many solutions.

Answer 1

The Correct Option is A

Step 1: Represent the system as a matrix equation.
The system of equations is:
\[ x + y + z = \lambda \] \[ 5x - y + \mu z = 10 \] \[ 2x + 3y - z = 6 \] This system can be written as: \[ \begin{pmatrix} 1 & 1 & 1 5 & -1 & \mu 2 & 3 & -1 \end{pmatrix} \begin{pmatrix} x y z \end{pmatrix} = \begin{pmatrix} \lambda 10 6 \end{pmatrix} \]

Step 2: Use the determinant to find conditions for a unique solution.
For the system to have a unique solution, the determinant of the coefficient matrix must be non-zero: \[ \text{Determinant} = \begin{vmatrix} 1 & 1 & 1 5 & -1 & \mu 2 & 3 & -1 \end{vmatrix} \] We calculate the determinant: \[ \text{Determinant} = 1 \begin{vmatrix} -1 & \mu 3 & -1 \end{vmatrix} - 1 \begin{vmatrix} 5 & \mu 2 & -1 \end{vmatrix} + 1 \begin{vmatrix} 5 & -1 2 & 3 \end{vmatrix} \] \[ \text{Determinant} = 1((-1)(-1) - (3)(\mu)) - 1((5)(-1) - (2)(\mu)) + 1((5)(3) - (2)(-1)) \] \[ \text{Determinant} = 1(1 - 3\mu) - 1(-5 - 2\mu) + 1(15 + 2) \] \[ \text{Determinant} = (1 - 3\mu) + (5 + 2\mu) + 17 = 23 - \mu \]

Step 3: Conclusion.
For a unique solution, the determinant must be non-zero: \[ 23 - \mu \neq 0 \quad \implies \quad \mu \neq 23 \] Thus, the existence of a unique solution depends on \( \mu \), which corresponds to option (A).