Let P = [ cases 1 & 0 & 0 4 & 1 & 0 16 & 4 & 1 cases ] and I be the...

P = \left[\begin{cases}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{cases}\right] ⇒ {P}^{2} = \left[\begin{cases}1 & 0 & 0 \\ 8 & 1 & 0 \\ 16 + 32 & 8 & 1\end{cases}\right]
So, {P}^{3} = \left[\begin{cases}1 & 0 & 0 \\ 12 & 1 & 0 \\ 16 + 32 + 48 & 12 & 1\end{cases}\right]
Using symmetry, we can write
{P}^{50} = \left[\begin{cases}1 & 0 & 0 \\ 200 & 1 & 0 \\ \dfrac{16.50.51}{2} & 200 & 1\end{cases}\right]
As, ${P}^{50} - Q = I ⇒ Q = {P}^{50} - I = [$\begin{cases}0 & 0 & 0 \\ 200 & 0 & 0 \\ \dfrac{16.50.51}{2} & 200 & 0\end{cases}$],$
${q}_{31} = \dfrac{16.50.51}{2},$ ${q}_{32} = 200$ and ${q}_{21} = 200$
$∴$ $\dfrac{{q}_{31} + {q}_{32}}{{q}_{21}} = \dfrac{16.50.51}{2.200} + 1 = 102 + 1 = 103$