For the matrix $A = \begin{bmatrix} 2 & -1 & 1 \\ \lambda & 2 & 0 \\ 1 & -2 & 3 \end{bmatrix}$ to be invertible, the value of $\lambda$ is

$\mathbb{R} - \{-10\}$

The correct answer is Option (4) → $\mathbb{R} - \{-10\}$ ##

$A = \begin{bmatrix} 2 & -1 & 1 \\ \lambda & 2 & 0 \\ 1 & -2 & 3 \end{bmatrix}$

For invertible, the determinant of $A$ should not be equal to zero.

Therefore, $|A| = 2(6 - 0) - (-1)(3\lambda - 0) + 1(-2\lambda - 2)$

$|A| = 12 + 3\lambda - 2\lambda - 2$

$|A| = 10 + \lambda \neq 0$

$\lambda \neq -10$

Thus, $\lambda \in \mathbb{R} - \{-10\}$.