Step 1: Think of \( A \) as diagonalizable.
Any square matrix with distinct eigenvalues \( \lambda_1, \lambda_2, \lambda_3 \) can be written as \( A = PDP^{-1} \), where \( D \) is the diagonal matrix of eigenvalues. Once \( A \) is in this form, any polynomial or rational combination of \( A \) is also diagonal in the same basis, so we can work with the diagonal entries directly.
Step 2: Build the diagonal form of the combination.
For \( B = A^{-1} + 2I + A \), writing everything in terms of \( D \) gives
\[
B = PD^{-1}P^{-1} + 2PIP^{-1} + PDP^{-1} = P\left(D^{-1} + 2I + D\right)P^{-1}
\]
So \( B \) has the same eigenvectors as \( A \), and its eigenvalues are simply \( \frac{1}{\lambda_i} + 2 + \lambda_i \) for each \( i \).
Step 3: Check the pattern with a quick numeric example.
Take a simple diagonal matrix with eigenvalues \( 1, 2, 3 \), say \( D = \text{diag}(1,2,3) \). Then
\[
D^{-1} + 2I + D = \text{diag}\left(1+2+1,\ \tfrac{1}{2}+2+2,\ \tfrac{1}{3}+2+3\right) = \text{diag}(4,\ 4.5,\ 5.33)
\]
Each entry is exactly \( \frac{1}{\lambda} + 2 + \lambda \) for that eigenvalue, with no sign change anywhere.
So the eigenvalues of \( A^{-1} + 2I + A \) are
\[
\boxed{\frac{1}{\lambda_1} + 2 + \lambda_1,\ \ \frac{1}{\lambda_2} + 2 + \lambda_2,\ \ \frac{1}{\lambda_3} + 2 + \lambda_3}
\]
which is option (D).