If $A^{-1}$ exists for the matrix $A =\begin{bmatrix}1&λ&-1\\-1&1&0\\λ&1&1\end{bmatrix}$ then

$λ≠ -1$

The correct answer is Option (2) → $λ≠ -1$

Given matrix:

$A = \begin{bmatrix} 1 & \lambda & -1 \\ -1 & 1 & 0 \\ \lambda & 1 & 1 \end{bmatrix}$

For $A^{-1}$ to exist, $\det(A) \ne 0$.

Compute determinant:

$\det(A) = 1\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \lambda\begin{vmatrix} -1 & 0 \\ \lambda & 1 \end{vmatrix} + (-1)\begin{vmatrix} -1 & 1 \\ \lambda & 1 \end{vmatrix}$

$= 1(1 - 0) - \lambda(-1 - 0) - [(-1)(1) - (\lambda)(1)]$

$= 1 + \lambda + (1 + \lambda)$

$= 2 + 2\lambda$

$\Rightarrow \det(A) = 2(1 + \lambda)$

For $A^{-1}$ to exist, $\det(A) \ne 0$

$\Rightarrow 1 + \lambda \ne 0$

$\Rightarrow \lambda \ne -1$

Therefore, $A^{-1}$ exists if and only if $\lambda \ne -1$.