Question

The system of linear equations \( x+y+z=6 \), \( x+2y+5z=10 \), \( 2x+3y+\lambda z=\mu \) has a unique solution, if

• A. \( \lambda \neq 16, \mu=6 \)
• B. \( \lambda=6, \mu=16 \)
• C. \( \lambda=6, \mu \neq 16 \)
• D. \( \lambda \neq 6, \mu \in \mathbb{R} \)

Hint:

For a square system of linear equations, the conditions for the number of solutions are determined by the rank of the coefficient matrix A and the augmented matrix [A|b].
\textbf{Unique solution:} \( \det(A) \neq 0 \).
\textbf{No solution:} \( \det(A) = 0 \) and \( \text{rank}(A)<\text{rank}([A|b]) \).
\textbf{Infinite solutions:} \( \det(A) = 0 \) and \( \text{rank}(A) = \text{rank}([A|b]) \). The condition for a unique solution only depends on the coefficient matrix A.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A system of linear equations \( Ax=b \) has a unique solution if and only if the determinant of the coefficient matrix A is non-zero. The values on the right-hand side of the equations (in this case, \( \mu \)) do not affect the uniqueness of the solution, only its existence.

Step 2: Key Formula or Approach:
We write the system in matrix form \( Ax=b \) and find the determinant of the coefficient matrix A. \[ A = \begin{pmatrix} 1 & 1 & 1
1 & 2 & 5
2 & 3 & \lambda \end{pmatrix}, \quad x = \begin{pmatrix} x
y
z \end{pmatrix}, \quad b = \begin{pmatrix} 6
10
\mu \end{pmatrix} \] For a unique solution, we must have \( \det(A) \neq 0 \).

Step 3: Detailed Explanation:
Let's calculate the determinant of A. \[ \det(A) = 1 \begin{vmatrix} 2 & 5
3 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 5
2 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 2
2 & 3 \end{vmatrix} \] \[ = 1(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) \] \[ = 2\lambda - 15 - \lambda + 10 - 1 \] \[ = \lambda - 6 \] For the system to have a unique solution, we need \( \det(A) \neq 0 \). \[ \lambda - 6 \neq 0 \implies \lambda \neq 6 \] If \( \lambda \neq 6 \), the system has a unique solution regardless of the value of \( \mu \). The value of \( \mu \) can be any real number.

Step 4: Final Answer:
The system has a unique solution if \( \lambda \neq 6 \) and \( \mu \in \mathbb{R} \).