It is not possible to uniquely determine a matrix given its eigenvalues. For eigenvalues ${1, 2, 3}$, ${A}$ could be a lot of matrices e.g., any upper triangular matrix with ${1, 2, 3}$ as diagonal elements.

If you’re asking for the eigenvalues of the inverse of ${A}$, then it’s simply the reciprocal of the eigenvalues of ${A}$ i.e., ${1, }$ $\frac{1}{2}$ and $\frac{1}{3}$. (Note that ${A}^{-1}$ always exists because all of ${A}$’s eigenvalues are non-zero).

Eigenvalues for the inverse of A would be the reciprocal but here the question is not asking that. I had forgotten to add options as it is an MCQ question.

In that case, solve (x-1)(x-2)(x-3) = 0 and put A in place of x after expanding. It'll come as A^3 - 6A + 11A = 6I. After multiplying A^-1 on both sides, I got option C.