Repeated shear

Multiply out $A^2$ and $A^3$ and look for a pattern.

1. Direct multiplication gives $A^2=\left[\begin{array}{cc}1& 4\\ 0& 1\end{array}\right]$ and $A^3=\left[\begin{array}{cc}1& 6\\ 0& 1\end{array}\right]$: each extra factor adds $2$ to the top right entry.
2. By induction $A^n=\left[\begin{array}{cc}1& 2n\\ 0& 1\end{array}\right]$, since $\left[\begin{array}{cc}1& 2n\\ 0& 1\end{array}\right]\times \left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 2n+2\\ 0& 1\end{array}\right]$.
3. For $n=10$ the top right entry is $2\times 10=20$.

Answer: $20$