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I'm a bit confused with this determinant. We have the determinant

$$\Delta_n=\left\vert\begin{matrix} 5&3&0&\cdots&\cdots&0\\ 2&5&3&\ddots& &\vdots\\ 0&2&5&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&\ddots&0\\ \vdots& &\ddots&\ddots&\ddots&3\\ 0&\cdots&\cdots&0&2&5\end{matrix} \right\vert$$

I compute $\Delta_2=19$, $\Delta_3=65$.

Then I would like to find a relation for $n\geq 4$ which links $\Delta_n, \Delta_{n-1}$ and $\Delta_{n-2}$ and thus find an expression of $\Delta_n$. How could we do that for $n\geq 4$?

Thank you

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You have a tridiagonal matrix. A tridiagonal matrix has a nice form for the determinant. If the diagonal is $a_1,a_2, \ldots$, above diagonal $b_1,b_2,\ldots$ and below diagonal is $c_1,c_2,\ldots$, then the determinant of the $n$-th principal minor (i.e. the matrix formed by the top left $n \times n$ submatrix) is given by the following recursion:

$f_1 = |a_1|, f_0 = 1, f_{-1} =0$

$f_n = a_n f_{n-1} - c_{n-1} b_{n-1} f_{n-2}$

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Let prove the theorem. Suppose the determinant of tri-diagonal matrix as $\Delta_{n}$, and operate the following calculation.

$$ \begin{align} \Delta_{n}=& \det \begin{bmatrix} a_{1} & b_{1} & 0 & \cdots & 0 & 0 & 0 \\ c_{1} & a_{2} & b_{2} & \ddots & \vdots & \vdots & \vdots \\ 0 & c_{2} & a_{3} & \ddots & a_{n-2} & b_{n-2} & 0 \\ \vdots & \vdots & \vdots & \ddots & c_{n-2} & a_{n-1} & b_{n-1} \\ 0 & 0 & 0 & \cdots & 0 & c_{n-1} & a_{n} \end{bmatrix} \\ \\ =& \det \begin{bmatrix} a_{1} & b_{1} & 0 & \cdots & 0 & 0 & 0 \\ c_{1} & a_{2} & b_{2} & \ddots & \vdots & \vdots & \vdots \\ 0 & c_{2} & a_{3} & \ddots & a_{n-2} & b_{n-2}-c_{n-1}\frac{b_{n-1}}{a_{n-1}} & 0 \\ \vdots & \vdots & \vdots & \ddots & c_{n-2} & a_{n-1}-c_{n-1}\frac{b_{n-1}}{a_{n-1}} & 0 \\ 0 & 0 & 0 & \cdots & 0 & c_{n-1} & a_{n} \end{bmatrix} \\ \\ =& a_{n} \Delta_{n-1} - b_{n-1}c_{n-1} \Delta_{n-2} \end{align} $$

Based on this formula, it can be described as below using matrix and vector product. Then, we try to estimate $\Delta_{n}$.

$$ \begin{align} \begin{bmatrix} \Delta_{n} \\ \Delta_{n-1} \end{bmatrix} =& \begin{bmatrix} a_{n} & -b_{n-1}c_{n-1} \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \Delta_{n-1} \\ \Delta_{n-2} \end{bmatrix} \\ =& \prod_{k=4}^n \begin{bmatrix} a_{n+4-k} & -b_{n-k+3}c_{n-k+3} \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \Delta_{3} \\ \Delta_{2} \end{bmatrix} \end{align} $$

This problem's case, these elements are identity each diagonal factors like $a_{i}=5$ $b_{i}=3$, $c_{i}=2$. Therefore this equation can be simplified as follows.

$$ \begin{bmatrix} \Delta_{n} \\ \Delta_{n-1} \end{bmatrix} = \begin{bmatrix} 5 & -6 \\ 1 & 0 \end{bmatrix} ^{n-3} \begin{bmatrix} 65 \\ 19 \end{bmatrix} $$

After that, we get the eigenvalues, eigenvectors and diagonalization of the matrix.

$$ \begin{align} \begin{bmatrix} \Delta_{n} \\ \Delta_{n-1} \end{bmatrix} =& \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 3^{n-1} & 0 \\ 0 & 2^{n-1} \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} ^{-1} \begin{bmatrix} 65 \\ 19 \end{bmatrix} \\ =& \begin{bmatrix} 3^n-2^n & 3 \cdot 2^{n-2} -2 \cdot 3^{n-2} \\ 3^{n-3} - 2^{n-3} & 3 \cdot 2^{n-3} - 2 \cdot 3^{n-3} \end{bmatrix} \begin{bmatrix} 65 \\ 19 \end{bmatrix} \end{align} $$

Eventually, $\Delta_{n}$ is

$$ \begin{align} \Delta_{n}=& 65(3^{n-2}-2^{n-2})+19(3 \cdot 2^{n-2} -2 \cdot 3^{n-2}) \\ =& 3^{n+1} - 2^{n+1} \end{align} $$

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