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I have been solving matrices questions lately and this pattern keeps showing up:

Given matrix $ P = \begin{bmatrix} 1 & 0 & 0\\ 4 & 1 & 0\\ 16 & 4 & 1\\ \end{bmatrix}$ and $I$ be identity matrix of order 3. If $[Q] = [q_{ij}]$ is a matrix such that $P^{50}-Q= I$ then $\frac{q_{31}+q_{32}}{q_{21}} = ?$

While solving I found this pattern for a general matrix of the type: $$L = \begin{bmatrix} 1 & 0 & 0\\ a & 1 & 0\\ b & a & 1\\ \end{bmatrix}$$ As

$$L^n = \begin{bmatrix} 1 & 0 & 0\\ a×n & 1 & 0\\ K & a×n & 1\\ \end{bmatrix}$$

Where $K = (L_{31})_{{n-1}}+a^2×(n-1) + b$

If I have to figure out $L_{50}$ using this algorithm, I need to find $L_{49}, L_{48}$ and so on which basically defeats the purpose. How do you reiterate the value of $L_{n-1}$ so that I can use directly the value of $b$ to calculate $L_{31}$ of any power of matrix?

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By recursion you actually show that $$P^n=\begin{bmatrix} 1 & 0 & 0 \\ 4n & 1 & 0 \\ 16n + \frac{(n-1)n}{2}4^2 & 4n & 1\end{bmatrix} ,$$ so you necessarily have $$Q=\begin{bmatrix} 0 & 0 & 0 \\ 4n & 0 & 0 \\ 16n + \frac{(n-1)n}{2}4^2 & 4n & 0\end{bmatrix} $$ where $n=50$ (but we will substitute later), so we obtain $$\frac{q_{31}+q_{32}}{q_{21}}= \frac{16n + \frac{(n-1)n}{2}4^2 + 4n}{4n} = 4 + 2(n-1) + 1 =103.$$

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Note that $L=I+N$ with $N^3=0$. Therefore, $L^n = I+n N + \frac{n(n-1)}{2}N^2$. Since $$ N^2 = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ a^2 & 0 & 0\\ \end{bmatrix} $$ we get $$ L^n = \begin{bmatrix} 1 & 0 & 0\\ n a & 1 & 0\\ nb +\frac{n(n-1)}{2}a^2& n a & 1\\ \end{bmatrix} $$

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