Ans is (C) 2
$Ax = λx$, where $λ$ is the eigen value of $A$ . Hence $(A−λI)x = 0$ or $|A−λI| = 0$
So, for our given matrix, we have
$\begin{bmatrix} A-\lambda I &I \\ I& A - \lambda I \end{bmatrix} = 0$
This is a $ 2 \times 2$ block matrix where the first and last and the second and third elements are the same. So, applying the formula for determinant of a block matrix as given here
When A = D and B = C, the blocks are square matrices of the same order and the following formula holds (even if A and B do not commute)
$det \begin{pmatrix} A & B\\ C& D \end{pmatrix} = det (A-B) \ det(A+B)$
(second last case) https://en.wikipedia.org/wiki/Determinant#Block_matrices
we get,
$|A−\lambda I− I| \times |A-\lambda I + I|= 0 $
$\implies |A-(\lambda+1)I| \times |A - (\lambda -1)I| = 0$
Each of the eigen value of $A$ is the solution of the equation $|A- \alpha I |= 0$ ($\alpha$ being the eigen value of $A$). So, we can equate $\lambda +1$ and $\lambda -1$ to any of the eigen value of $A$, and that will get our value of $\lambda$. If we take $\alpha = 1$, we get $\lambda = 2$, and that is one of the choice. For no other choice, this equation holds. So, (c) 2 is the answer.