Question

If the system of simultaneous linear equations \[ x+y+z = \lambda, \quad 5x - y + \mu z = 10, \quad 2x + 3y - z = 6 \] has a unique solution, then:

• A. \(\mu = 23 \text{ and } \lambda \in \mathbb{R}\)
• B. \(\mu \in \mathbb{R} \text{ and } \lambda \neq 23\)
• C. \(\mu \neq 23 \text{ and } \lambda \in \mathbb{R}\)
• D. \(\mu = 23 \text{ and } \lambda = 16\)

Hint:

For a system of three linear equations to have a unique solution, check that the determinant of the coefficient matrix is non-zero. Parameters not in the determinant can take any real value.

Answer 1

The Correct Option is C

Step 1: Condition for unique solution.
For a system of three linear equations in three variables to have a unique solution, the determinant of the coefficient matrix must be non-zero.
Coefficient matrix: \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 5 & -1 & \mu \\ 2 & 3 & -1 \end{bmatrix} \]

Step 2: Determinant calculation.
\[ \det(A) = 1\begin{vmatrix}-1 & \mu \\ 3 & -1\end{vmatrix} - 1\begin{vmatrix}5 & \mu \\ 2 & -1\end{vmatrix} + 1\begin{vmatrix}5 & -1 \\ 2 & 3\end{vmatrix} \]
Compute each minor:
1. \(1((-1)(-1) - (3)(\mu)) = 1(1 - 3\mu) = 1 - 3\mu\)
2. \(-1((5)(-1) - (2)(\mu)) = -1(-5 - 2\mu) = 5 + 2\mu\)
3. \(1((5)(3) - (2)(-1)) = 1(15 + 2) = 17\)
\[ \det(A) = (1 - 3\mu) + (5 + 2\mu) + 17 = 23 - \mu \]

Step 3: Non-zero determinant condition.
\[ \det(A) \neq 0 \implies 23 - \mu \neq 0 \implies \mu \neq 23 \]

Step 4: Conclusion.
\(\lambda\) does not appear in the determinant; hence \(\lambda \in \mathbb{R}\).
Thus, the unique solution exists if: \[ \boxed{\mu \neq 23 \text{ and } \lambda \in \mathbb{R}} \]