= Solution
For two <upper triangular matrices> and $i>j$,
$$
(AB)_{ij}=\sum_{k=1}^3A_{ik}B_{kj}=0.
$$
Indeed a potentially nonzero factor $A_{ik}$ requires $k\geq i$, while a potentially nonzero $B_{kj}$ requires $k\leq j$. These requirements cannot both hold when $i>j$. Thus their <matrix product> is upper triangular.
Direct <matrix multiplication> for the given $A$ gives
$$
A^2=\begin{pmatrix}1&0&2\\0&1&-2\\0&0&1\end{pmatrix},\qquad
A^3=\begin{pmatrix}1&2&-2\\0&-1&3\\0&0&-1\end{pmatrix}.
$$
Adding $A^3+A^2-A$ yields $I$, so $A(A^2+A-I)=(A^2+A-I)A=I$. Hence
$$
\boxed{A^{-1}=A^2+A-I=\begin{pmatrix}1&2&2\\0&-1&-1\\0&0&-1\end{pmatrix}.}
$$
Every nonnegative <matrix power> is in $\operatorname{span}\{I,A,A^2\}$: reduce powers of degree at least three using $A^3=-A^2+A+I$. More formally, repeated reduction shows that products of any two polynomials of degree at most two in $A$ again lie in this span. Since $A^{-1}$ also lies in it, positive powers of $A^{-1}$ prove the same assertion for all negative integers. This is an instance of the <integer powers for an annihilating polynomial with roots one and minus one>, because $A^3+A^2-A-I=(A-I)(A+I)^2=0$.
To find the required entries for every integer, use the block upper-triangular structure. The top-left entry of $A^n$ is $1$ for positive and negative powers. Its lower-right block is
$$
J=-I_2+N=-(I_2-N),\qquad
N=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad N^2=0.
$$
The identity $(I_2-N)^n=I_2-nN$ holds for all integers: multiplication adds the coefficients of $N$, and $(I_2-N)^{-1}=I_2+N$ proves the negative cases. Therefore
$$
\boxed{(A^n)_{11}=1,\quad(A^n)_{22}=(-1)^n,\quad(A^n)_{23}=n(-1)^{n-1}.}
$$
Write $s=(-1)^n$ and $A^n=\alpha_nA^2+\beta_nA+\gamma_nI$. Comparing those three entries gives
$$
\alpha_n+\beta_n+\gamma_n=1,\qquad
\alpha_n-\beta_n+\gamma_n=s,\qquad
-2\alpha_n+\beta_n=-ns.
$$
Subtracting the first two equations finds $\beta_n$; the third then finds $\alpha_n$, and the first finds $\gamma_n$. The unique solution is
$$
\boxed{\alpha_n=\frac{1-s}{4}+\frac{ns}{2},\quad
\beta_n=\frac{1-s}{2},\quad
\gamma_n=\frac{1+3s}{4}-\frac{ns}{2},\qquad s=(-1)^n.}
$$
For clarity, substituting also gives the full <matrix power>
$$
\boxed{A^n=\begin{pmatrix}
1&1-s&(1-s)/2+ns\\0&s&-ns\\0&0&s
\end{pmatrix}\quad(n\in\mathbb Z).}
$$
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