Let P = [ cases 1 & 0 & 0 3 & 1 & 0 9 & 3 & 1 cases ] and Q = [ q _...

Given,

P = \left[\begin{cases}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{cases}\right]

⇒ {P}^{2} = \left[\begin{cases}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{cases}\right]\left[\begin{cases}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{cases}\right] = \left[\begin{cases}1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1\end{cases}\right]

⇒ {P}^{4} = {P}^{2} \times {P}^{2} = \left[\begin{cases}1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1\end{cases}\right]\left[\begin{cases}1 & 0 & 0 \\ 6 & 1 & 0 \\ 27 & 6 & 1\end{cases}\right] = \left[\begin{cases}1 & 0 & 0 \\ 12 & 1 & 0 \\ 90 & 12 & 1\end{cases}\right]

⇒ {P}^{5} = {P}^{4} \times P = \left[\begin{cases}1 & 0 & 0 \\ 12 & 1 & 0 \\ 90 & 12 & 1\end{cases}\right]\left[\begin{cases}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{cases}\right] = \left[\begin{cases}1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1\end{cases}\right]

Let,

Q = \left[\begin{cases}{q}_{11} & {q}_{12} & {q}_{13} \\ {q}_{21} & {q}_{22} & {q}_{23} \\ {q}_{31} & {q}_{31} & {q}_{33}\end{cases}\right]

Now, given $Q - {P}^{5} = {I}_{3}$

$⇒ {q}_{21} - 15 = 0, {q}_{31} - 135 = 0, {q}_{32} - 15 = 0$

$⇒ {q}_{21} = 15, {q}_{31} = 135, {q}_{32} = 15$

$∴ \dfrac{{q}_{21} + {q}_{31}}{{q}_{32}} = \dfrac{15 + 135}{15} = 10$.