Let $P=\begin{pmatrix}1&0&0\\3&1&0\\9&3&1\end{pmatrix}$ and $Q=[q_{ij}]$ be two $3\times3$ matrices satisfying $Q - P^5 = I_3$. Find $\dfrac{q_{21}+q_{31}}{q_{32}}$.
Notice $P = I + N$ where $N=\begin{pmatrix}0&0&0\\3&0&0\\9&3&0\end{pmatrix}$ (strictly lower triangular, so $N^3=0$).
$P^5 = (I+N)^5 = I + 5N + 10N^2$ (since $N^3=0$).
$N^2 = \begin{pmatrix}0&0&0\\0&0&0\\9&0&0\end{pmatrix}$... computing: $(N^2)_{31}=3\cdot3=9$, others 0.
$P^5 = I + 5N + 10N^2$. So $Q = P^5 + I = 2I + 5N + 10N^2$.
$q_{21} = 5\cdot3 = 15$, $q_{31} = 5\cdot9+10\cdot9=45+90=135$... wait: $q_{31}=(5N)_{31}+(10N^2)_{31}=5\cdot9+10\cdot9=45+90=135$. Wait $N_{31}=9, N^2_{31}=9$: $q_{31}=5(9)+10(9)=135$. $q_{32}=(5N)_{32}=5\cdot3=15$.
$\dfrac{q_{21}+q_{31}}{q_{32}} = \dfrac{15+135}{15} = \mathbf{10}$
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