= Solution
{c}
Write $A=I_3+N$, where
$$
N=\begin{pmatrix}0&a&b\\0&0&c\\0&0&0\end{pmatrix},\qquad
N^2=\begin{pmatrix}0&0&ac\\0&0&0\\0&0&0\end{pmatrix},\qquad N^3=0.
$$
The <binomial theorem for commuting matrices> applies because $I_3N=NI_3$. All terms of degree at least three vanish, so for every positive <integer> $n$,
$$
\boxed{A^n=I_3+nN+\binom n2N^2
=\begin{pmatrix}
1&na&nb+\dfrac{n(n-1)}2ac\\
0&1&nc\\
0&0&1
\end{pmatrix}.}
$$
The same expression directly gives $A$ when $n=1$, so no small exponent is excluded.
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