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
Best Answer
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:
$f_1 = |a_1|, f_0 = 1, f_{-1} =0$
$f_n = a_n f_{n-1} - c_{n-1} b_{n-1} f_{n-2}$