Circumscribes many notions.

For a symmetric matrix $\left[\begin{matrix} a & b\\ b & c \end{matrix}\right]$ we have:

$\left[\begin{matrix} a & b\\ b & c \end{matrix}\right] \left[\begin{matrix} 2 & 1\\ 1 & 0 \end{matrix}\right] = \left[\begin{matrix} 2a+b & a\\ a & b \end{matrix}\right].$

Hence $(a_{ij}(n))= \left[\begin{matrix} a_{n+1} & a_n\\ a_n & a_{n-1} \end{matrix}\right]$ where $a_0=0,a_1=1$ and $a_{n+1}=2a_n+a_{n-1}$ for $n\in\mathbb{N}$ . Setting $x:=\lim_{n\to\infty}\frac{a_{12}(n)}{a_{22}(n)}=\frac{a_n}{a_{n-1}}$ , we have $x=2+1/x$ , hence $x^2-2x-1=0$ giving $x=1+\sqrt2$ (the negative solution can be dismissed as $a_n>0$ for all $n$ ). Therefore $\left(\lim_{n\to\infty}\frac{a_{12}(n)}{a_{22}(n)}\right)^2=x^2=3+2\sqrt2=\sqrt8+\sqrt9$ , so $A+B=8+9=\boxed{17}$ .