Let $P=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]$  and Q$=\left[q_{ij}\right]$   be two $3X3$ matrices such that $Q-P^5=I_3$ . Then  $\frac{q_{21}+q_{31}}{q_{32}}$ is equal to

$P_2=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 6& 1& 0\\ 27& 6& 1\end{array}\right]$

$\Rightarrow P_4=\left[\begin{array}{ccc}1& 0& 0\\ 6& 1& 0\\ 27& 6& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 6& 1& 0\\ 27& 6& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 12& 1& 0\\ 90& 12& 1\end{array}\right]$

$\Rightarrow P_5=\left[\begin{array}{ccc}1& 0& 0\\ 12& 1& 0\\ 90& 12& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 15& 1& 0\\ 135& 15& 1\end{array}\right]$

$\because Q-P_5=I_3$

$\therefore Q=I_3+P_3=\left[\begin{array}{ccc}2& 0& 0\\ 15& 2& 0\\ 135& 15& 2\end{array}\right]$

$\frac{q_{21}+q_{31}}{q_{32}}=\frac{15+135}{15}=10$