Determinant of Tridiagonal matrix 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
 A: 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:
$f_1 = |a_1|, f_0 = 1, f_{-1} =0$
$f_n = a_n f_{n-1} - c_{n-1} b_{n-1} f_{n-2}$
A: 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
$$
\begin{align}
\Delta_{n}=&
65(3^{n-2}-2^{n-2})+19(3 \cdot 2^{n-2} -2 \cdot 3^{n-2}) \\ =&
3^{n+1} - 2^{n+1}
\end{align}
$$
