Let $P \left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$  and I be  the identity matrix

of  order  3. If  $Q=\left[q_{ii}\right]$  ] is a matrix such that $P^{50}-Q=I$ then,

$\frac{9_{31}+q_{32}}{q_{21}}$ equal 

We have $P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$

$\begin{array}{l}\therefore P^2=PP=\left[\begin{array}{lll}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 8& 1& 0\\ 16+32& 8& 1\end{array}\right]\\ P^3=P^{2}P=\left[\begin{array}{ccc}1& 0& 0\\ 8& 1& 0\\ 16+32& 8& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 12& 1& 0\\ 16+32+48& 12& 1\end{array}\right]\end{array}$

By observing the symmetry, we obtain 

      $\begin{array}{l}{P}^{50}=\left[\begin{array}{cccc}4\times 50& 1& 0& \\ 16+32+48+\dots 50& \text{ terms }& 4\times 50& 1\end{array}\right]\\ \Rightarrow P^{50}=\left[\begin{array}{ccc}1& 0& 0\\ 200& 1& 0\\ \frac{16\times 50\times 51}{2}& 200& 1\end{array}\right]\end{array}$

     $\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{P}^{50}-Q=I\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}Q=P^{50}-I=\left[\begin{array}{ccc}0& 0& 0\\ 200& 0& 0\\ 20400& 200& 0\end{array}\right]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{q}_{21}=200,q_{31}=20400\text{ and }{q}_{32}=200\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{q_{31}+q_{32}}{q_{21}}=\frac{20400+200}{200}=\frac{20600}{200}=103\end{array}$