What is $\det(I + A^{50})$ if $A$ has eigenvalues $1$, $-1$ and $0$? Since $3 \times 3$ matrix $A$ has $0$ as one of its eigenvalues, we know that $A$ is singular. From this we can conclude that $\det(A) = 0$ also $\mbox{tr} (A) = 0$. Further, $\det(A^{50}) = 0$. Now I think that $\mbox{tr} (I + A^{50}) = 3$ but how can I get the determinant from here?
 A: Since $A$ is a $3 \times 3$ matrix with distinct eigenvalues, it is diagonalizable. Hence, there exists $P$ such that $A = P D P^{-1}$ where $D = \operatorname{diag}(1,-1,0)$. Then 
$$ I + A^{50} = I + P D^{50} P^{-1} = P(I + D^{50}) P^{-1} $$
so $I + A^{50}$ is similar to 
$$I + D^{50} = \operatorname{diag}(1 + 1^{50}, 1 + (-1)^{50}, 1 + 0^{50}) = \operatorname{diag}(2, 2, 1). $$
Since the determinant of a diagonalizable matrix is the product of its eigenvalues, we have $\det(I + D^{50}) = 4$.
A: Let 
$$D=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}$$
Since all the eigenvalues have multiplicity $1$, we know that there is some matrix $P$ such that $A=P^{-1}DP$. Then $A^{50}=P^{-1}D^{50}P$. Can you continue?
A: Hint: Since $A$ is $3\times 3$ you know that there exists an invertable matrix $U$ such that $A=U^{-1}DU$ where $D=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}$.

Now you can conclude $A^{50}=U^{-1}D^{50}U$ and $D^{50}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$.



This yields $$\det(I+A^{50})=\det(U^{-1}U+(U^{-1}D^{50}U))=\det(U^{-1}(I+D^{50})U)=\det(I+D^{50})=4.$$

