Find the sum of all nine entries of M^-15.

Write $M=I+N$ with $N=\begin{pmatrix}0&2&0\\0&0&2\\0&0&0\end{pmatrix}$. Then $N^{2}$ has the single non-zero entry $4$ in position $(1,3)$ and $N^{3}=0$. For a nilpotent $N$, $(I+N)^{-n}=I-nN+\binom{n+1}{2}N^{2}$ (the binomial series for the negative exponent, which terminates). With $n=15$ this is $I-15N+120N^{2}$. The entry sums are: $I$ contributes $3$; $-15N$ contributes $-15\cdot4=-60$; $120N^{2}$ contributes $120\cdot4=480$. Total $=3-60+480=423$. Direct computation of the inverse power of $M$ with exact arithmetic gives the same sum, $423$. It is worth noting that the entry sum is not the same as the trace or determinant, so one must really track all three non-zero diagonals. A common mistake is to use $\binom{n}{2}$ instead of $\binom{n+1}{2}$ for the negative power, which would give $3-60+420=363$ and is therefore wrong. Verification with an exact rational inverse removes any doubt.