Determinants of tridiagonal matrices 
A square matrix $A = [a_{ij}]$ is called ${\bf tridiagonal}$ if
$a_{ij}=0$ for $|i-j|>1$. Try to guess a formula for the determinant
of tridiagonal matrix, say $a_i = a_{ii}$ for $i=1,...,n$, $b_i =
 a_{i,i+1}$ and $c_i = a_{i+1,i}$ for $i=1,...,n-1$.

Attempt
So, I was thinking on reducing to smaller matrix. Say for $n=1$, we det A = $a_1$. Not in the case $n=2$, we have just the matrix with rows $[a_1, b_1$] and $[c_1,a_2]$. Thus,
$$ det A = a_1 a_2 - c_1 b_1 $$
Now, for the $n=3$ case, we start to see the zeros appear, but it becomes cumbersome to compute determinant after $n>3$. Is there a way to find closed nice for this problem? Or do I have to keep doing it expressing the actual determinant in terms of the previous as it is evident in the case $n=3$ since if we call $D_n$ to be the determinant on the nth case (for instnace, we saw that $D_2 = a_1 a_2 - c_1 b_1$ so that for the $n=3$ case I see that
$$ D_3 =a_3 D_2 - c_2 b_2 a_1 $$
Is this the right way to approach this problem? Moreover, why are tridiagonal matrices so important? Can someone give intuition into what they do? or in what situations we use them
 A: Recursion is the best way to solve this problem. As a hint, you showed that $$D_3 = a_3D_2-c_2b_2a_1 = a_3D_2 - c_2b_2D_1.$$ Can you generalize this to a formula for $D_n$ in terms of $D_{n-1}$, $D_{n-2}$, and a few of the entries of the matrix?
As for why they are important, many eigenvalue algorithms for symmetric/Hermitian matrices will first use similarity transforms to reduce the matrix to a tridiagonal form, and then find the eigenvalues of a tridiagonal matrix. 
Also, tridiagonal matrices come up when solving differential equations via discretization. Suppose we want to solve the $u''(x) = f(x)$ on the interval $[0,1]$. Pick a positive integer $N$, and let $v_n = u(\tfrac{n}{N})$ for $n = 0,1,\ldots,N$. Then, using an approximation of the second derivative, we have $$f(\tfrac{n}{N}) = u''(\tfrac{n}{N}) \approx \dfrac{u(\tfrac{n+1}{N})-2u(\tfrac{n}{N})+u(\tfrac{n-1}{N})}{(\tfrac{1}{N})^2} = N^2(v_{n+1}-2v_n+v_{n-1}).$$ If we do this for all $n = 1, 2, \ldots, N-1$, and then include equations for whatever boundary conditions we might have, we'll get a tridiagonal system of equations. Note, this was a fairly trivial example, but there are more complicated differential equations and PDEs that can be handled this way.
