It's not that tough.

Consider the given matrix as $X$ and split it into two as given below, where $I$ is the unit matrix of order $3$ .

$X\quad =\quad \left[ \begin{matrix} 1 & 2 & A \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{matrix} \right] =\quad \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] +\quad \left[ \begin{matrix} 0 & 2 & A \\ 0 & 0 & 4 \\ 0 & 0 & 0 \end{matrix} \right] =I+B$

Now,

${ B }^{ 2 }=\left[ \begin{matrix} 0 & 0 & 8 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \Rightarrow { B }^{ r }=0\quad \forall r>2$

We can now use binomial expansion to find $X^n$ as all the powers of $B$ greater than $2$ can be neglected.

$\Rightarrow { X }^{ n }={ (I+B) }^{ n }=I+nB+\frac { n(n-1) }{ 2 } { B }^{ 2 }\\ \Rightarrow { X }^{ n }=\left[ \begin{matrix} 1 & 2n & nA+4n(n-1) \\ 0 & 1 & 4n \\ 0 & 0 & 1 \end{matrix} \right]$

Now just put in given values to get answer as $\boxed{200}$

Claim: $\left[ \begin{matrix} 1&2&A\\ 0&1&4\\ 0&0&1 \end{matrix} \right]^N = \left[ \begin{matrix} 1&2N&AN+4N(N-1)\\ 0&1&4N\\ 0&0&1 \end{matrix} \right]$ .

Proof by induction:

For $N=1$ the equality is clear. Suppose it holds for certain $N\in\mathbb{N}$ , then we have:

$\begin{aligned} \left[ \begin{matrix} 1&2&A\\ 0&1&4\\ 0&0&1 \end{matrix} \right]^{N+1} &= \left[ \begin{matrix} 1&2&A\\ 0&1&4\\ 0&0&1 \end{matrix} \right]^N \left[ \begin{matrix} 1&2&A\\ 0&1&4\\ 0&0&1 \end{matrix} \right] \\ &= \left[ \begin{matrix} 1&2N&AN+4N(N-1)\\ 0&1&4N\\ 0&0&1 \end{matrix} \right] \left[ \begin{matrix} 1&2&A\\ 0&1&4\\ 0&0&1 \end{matrix} \right] \\ &= \left[ \begin{matrix} 1&2+2N&A+8N+AN+4N(N-1)\\ 0&1&4+4N\\ 0&0&1 \end{matrix} \right] \\ &= \left[ \begin{matrix} 1&2(N+1)&A(N+1)+4N(N+1)\\ 0&1&4(N+1)\\ 0&0&1 \end{matrix} \right] \end{aligned}$

as desired.

Hence we are looking for $A$ and $N$ such that $2N=18$ , hence $N=9$ , and $A+4N(N-1)=2007$ , giving $A=191$ and $A+N=191+9=\boxed{200}$ .

Since the matrix is upper diagonal it's likely to have 'magical' properties. With hopes of exploiting them I simply try forming successive powers.

I don't deserve to be so lucky! Having matched the other elements all I need to do is to solve for $A$ in the expression in the upper right element.