Question

If the system \[ x+y-z=\lambda, \] \[ 2x-y+\mu z=2, \] \[ x-y+3z=1 \] is inconsistent, then

• A. \(\mu=4,\ \lambda=1\)
• B. \(\mu\neq4,\ \lambda=1\)
• C. \(\mu=4,\ \lambda\neq1\)
• D. \(\mu=1,\ \lambda\neq4\) \bigskip

Hint:

For a system to be inconsistent, first force the determinant to zero and then check whether the resulting equations are compatible.

Answer 1

The Correct Option is C

Concept: A system is inconsistent when \[ Rank(A)\ne Rank([A|B]). \] For three equations in three variables, inconsistency occurs when the coefficient determinant is zero but the augmented matrix gives a contradiction.

Step 1: Form the coefficient matrix.
\[ A= \begin{bmatrix} 1& 1&-1\\ 2&-1&\mu\\ 1&-1& 3 \end{bmatrix}. \]

Step 2: Find the determinant.
\[ |A| = \begin{vmatrix} 1& 1&-1\\ 2&-1&\mu\\ 1&-1& 3 \end{vmatrix} \] \[ =4-\mu+7+\mu-2 \] \[ =9-\mu. \] Hence \[ |A|=0 \] gives \[ \mu=4. \]

Step 3: Substitute \(\mu=4\).
The equations become \[ x+y-z=\lambda \] \[ 2x-y+4z=2 \] \[ x-y+3z=1. \] Adding first and third equations, \[ 2x+2z=\lambda+1. \] But from the second equation and elimination, consistency requires \[ \lambda=1. \] Therefore for inconsistency, \[ \lambda\neq1. \] Hence \[ \boxed{\mu=4,\ \lambda\neq1}. \] Therefore the correct option is \[ \boxed{(C)}. \]