Solution: As $A$ is a $3\times 3$ matrix with distinct eigenvalues, it is diagonalizable, that is, there exists a matrix $P$ such that $PDP^{-1}=A$, where $D$ is a diagonal matrix with diagonal entries $1,-1$ and $0$. So we have \begin{align*} I + A^{100} & = I + \left( PDP^{-1} \right)^{100} \\ & = I + PD^{100}P^{-1} \\ & = PP^{-1} + PDP^{-1} \\ & = P\left(I+D^{100}\right)P^{-1}. \end{align*}

Thus, the determinant of the matrix $I+A^{100}$ will be \begin{align*} \det \left( I+A^{100} \right) & = \det \left( P\left(I+D^{100}\right)P^{-1} \right) \\ & = \det(P) \det \left( I+D^{100} \right) \det \left( P^{-1} \right) \\ & = \det(P) \det \left( P^{-1} \right) \det \left( I+D^{100} \right) \\ & = \det \left(P P^{-1} \right) \det \left( I+D^{100} \right) \\ & = \det \left(I \right) \det \left( I+D^{100} \right) \\ & = \det \left( I+D^{100} \right). \end{align*}

Now note that \[ I+D^{100} = \mathop{\mathrm{diag}}(1, 1, 1) + \mathop{\mathrm{diag}}(1^{100}, (-1)^{100}, 0^{100} ) = \mathop{\mathrm{diag}}(2,2,1). \] Thus, the determinant of the matrix $I+A^{100}$ will be $4$.