Let M=beginpmatrix1&2&0\\0&1&2\\0&0&1endpmatrix and let n be a positive integer.

Write $M=I+N$ with $N=\begin{pmatrix}0&2&0\\0&0&2\\0&0&0\end{pmatrix}$, so $N^2$ has the single nonzero entry $4$ at position $(1,3)$ and $N^3=O$. Hence $M^n=I+nN+\binom n2N^2$ and, since the binomial series terminates, $M^{-n}=I-nN+\binom{-n}{2}N^2$. (A) The $(1,3)$ entry of $M^{n}$ is $4\binom n2=2n(n-1)$, which is $180$ for $n=10$: correct. (B) $M^{-1}=I-N+N^2$ has $(1,3)$ entry $+4$, not $-4$: incorrect. (C) Adding, the $N$ terms cancel and $\binom n2+\binom{-n}2=n^2$, so $M^n+M^{-n}=2I+n^2N^2$ is upper triangular with diagonal $2,2,2$ and determinant $8$: correct. (D) The only eigenvalue is $1$, and $M\neq I$, so $M$ is not diagonalizable (one Jordan block): incorrect. Answer: A and C.