Weird notation in a matrix

Question

I am trying to solve this problem, but I am unsure what the problem is even asking. Also I have never seen such a matrix before:

Consider the matrix $$D_n=\begin{bmatrix}a&b&0&\dots&0\\b&a&b&\ddots&\vdots\\0&b&a&\ddots&0\\\vdots&\ddots&\ddots&a&b\\0&\dots&0&b&a\end{bmatrix}$$ Show that $$\det D_n=a\det D_{n-1}-b^2\det D_{n-2}$$ for $n=2,3,\dots$

Answer 1

It just means that the matrix carries on with the same pattern. So if $n=3$, $$ D_3=\begin{pmatrix}a&b&0\\b&a&b\\0&b&a\end{pmatrix} $$ If $n=5$, $$ D_5=\begin{pmatrix}a&b&0&0&0\\b&a&b&0&0\\0&b&a&b&0\\0&0&b&a&b\\0&0&0&b&a\end{pmatrix}$$ etc.

Answer 2

On the principal diagonal all the elements are equal to $a$, on the diagonals just above and below they are all equal to $b$, and zero everywhere else. I guess the $n$ stands for the dimension of the matrix.