Recurrence relation for the determinant of a tridiagonal matrix Let
$$f_n := \begin{vmatrix}
a_1 & b_1 \\
c_1 & a_2 & b_2 \\
& c_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-1} \\
& & & c_{n-1} & a_n
\end{vmatrix}$$
Apparently, the determinant of the tridiagional matrix above is given by the recurrence relation 
$$f_n = a_n f_{n-1} - c_{n-1} b_{n-1}f_{n-2}$$
with initial values $f_0 = 1$ and $f_{-1} = 0$ (according to Wikipedia). Can anyone please explain to me how they came to this recurrence relation? 
I don't really understand how to derive it.  
 A: The recurrence is obtained by developing the determinant along the last column (or, equivalently, along the last row).
A: For $n \ge 2$ using Laplace expansion on the last row gives
\begin{align}
f_n &= 
\begin{vmatrix}
a_1 & b_1 \\
c_1 & a_2 & b_2 \\
& c_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-3} \\
& & & c_{n-3} & a_{n-2} & b_{n-2} \\
& & & & c_{n-2} & a_{n-1} & b_{n-1} \\
& & & & & c_{n-1} & a_n
\end{vmatrix}
\\
&=
(-1)^{2n-1}
c_{n-1}
\begin{vmatrix}
a_1 & b_1 \\
c_1 & a_2 & b_2 \\
& c_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-3} \\
& & & c_{n-3} & a_{n-2} \\
& & & & c_{n-2} & b_{n-1}
\end{vmatrix}
+ (-1)^{2n}
a_n
\begin{vmatrix}
a_1 & b_1 \\
c_1 & a_2 & b_2 \\
& c_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-2} \\
& & & c_{n-2} & a_{n-1}
\end{vmatrix}
\\
&= 
- c_{n-1}
(-1)^{2(n-1)}
b_{n-1} 
\begin{vmatrix}
a_1 & b_1 \\
c_1 & a_2 & b_2 \\
& c_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-3} \\
& & & c_{n-3} & a_{n-2}
\end{vmatrix}
+ a_n f_{n-1} \\
&= a_n f_{n-1} - c_{n-1} b_{n-1} f_{n-2}
\end{align}
as recurrence relation. 
For the initial conditions:
From comparing the above formula with matrices for $n=1$ and $n=2$ we get:
$$
f_1 = a_1 f_0 - c_0 b_0 f_{-1} \overset{!}{=} a_1 \\
f_2 = a_2 f_1 - c_1 b_1 f_0 \overset{!}{=} a_1 a_2 - c_1 b_1
$$
The latter implies $f_0 = 1$ and the former $f_{-1} = 0$.
