The following system of equations:

$x+y-z=7$
$4x+λy-λz = 3$
$3x+2y-4z = 5$

does not possess a solution if the value of is:

4

The correct answer is Option (4) → 4

Given matrix:

$A = \begin{bmatrix} 1 & 1 & -1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{bmatrix}$

Compute determinant using the first row:

$\det(A) = 1 \cdot (\lambda \cdot -4 - (-\lambda) \cdot 2) - 1 \cdot (4 \cdot -4 - (-\lambda) \cdot 3) + (-1) \cdot (4 \cdot 2 - \lambda \cdot 3)$

$= 1(-4\lambda + 2\lambda) - 1(-16 + 3\lambda) - 1(8 - 3\lambda)$

$= -2\lambda + 16 - 3\lambda - 8 + 3\lambda$

$= -2\lambda + 8$

Set determinant = 0 for singularity:

$-2\lambda + 8 = 0$ ⟹ $\lambda = 4$

Hence, the system is inconsistent (i.e., does not possess a solution) when $\lambda = 4$.