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: Set up the uniqueness condition.
A 3x3 linear system has a unique solution exactly when its coefficient determinant is nonzero.
Coefficient matrix: \(\begin{bmatrix}1 & 1 & 1\\ 5 & -1 & \mu\\ 2 & 3 & -1\end{bmatrix}\)

Step 2: Reduce rows before expanding, to keep the arithmetic light.
\(R_2 \to R_2 - 5R_1\) gives \((0,-6,\mu-5)\), and \(R_3 \to R_3 - 2R_1\) gives \((0,1,-3)\).
The matrix becomes \(\begin{bmatrix}1 & 1 & 1\\ 0 & -6 & \mu-5\\ 0 & 1 & -3\end{bmatrix}\), and expanding along the first column:
\[ \det = 1\cdot\big[(-6)(-3)-(\mu-5)(1)\big] = 18-(\mu-5) = 23-\mu \]

Step 3: Apply the uniqueness condition.
Unique solution needs \(23-\mu \neq 0\), so \(\mu \neq 23\).
\(\lambda\) never enters the coefficient determinant, since it only appears on the right-hand side, so it can be any real number.
\[ \boxed{\mu \neq 23 \text{ and } \lambda \in \mathbb{R}} \]