How to workout the determinant of the matrix $D_n(\alpha, \beta, \gamma)$.

Question

I am going through an example in my lecture notes. This is it:

Let's introduce the matrix $D_n(\alpha, \beta, \gamma)$, which looks like this:

$$\pmatrix{\beta & \gamma & 0 & 0 & ... & 0 \\ \alpha & \beta & \gamma& 0 & ... & 0 \\ 0 & \alpha & \beta & \gamma & ... & 0 \\ : & : & : &: & ... & : \\0 & 0 & 0 & 0 & ... & \beta}$$

To calculate the determinant, $d_n$, lets first decompose by row 1. Here, the first element is $\beta$. Removing this gives us the same matrix again but slightly smaller. We can therefore start with $d_n = \beta d_{n-1}$.

Now look at $\gamma$. As it is in row $1$ and column $2$, it has sign $(-1)$. Matrix now has new first element, $\alpha$ and so for the algebraic complement of $\gamma$, we decompose by column $1$. This gives us $d_n = \beta d_{n-1} - \gamma \alpha _{r - n}$. Let's call this $(*)$.

By now removing columns $\beta$ and $\gamma$ and rows $\beta$ and $\alpha$, we get the originial matrix again, but smaller. We then write $(*)$ as $d_n = \beta d_{n -1} - \gamma \alpha d_{n-2}$

I have a few questions with this. Firstly, when we start looking at column $\gamma$, how does decomposing with column $1$ give us that determinant bit? I am thinking I have written something down wrong but I am not sure what. Also, removing row $\alpha$ won't give us the same matrix again will it as we will now have $a_{11} = \alpha$ when it should be $\beta$, wouldn't we?

EDIT: Sorry, we do get the matrix again as removing column $\gamma$ takes out that $\alpha$. I still don't get the first bit on how they calculated the determinant using column $\gamma$.

Answer 1

The first term of your first cofactor expansion (with respect to $\beta$) gives you a term of the form $\beta d_{n-1}$ where $d_{n-1}$ is the determinant of the matrix with the first row and column removed, i.e. the $n-1$ by $n-1$ tridiagonal matrix. I assume that this part is clear.

If we move onto the second term in the expansion, then we have the term $(-1)^{1+2}\gamma d'_{n-1}$ where I use $d'_{n-1}$ to denote the determinant of the matrix with the first row and second column removed. i.e. $$d'_{n-1} = \begin{vmatrix}\alpha & \gamma & 0 & 0 & 0 & \cdots & 0 \\ 0 & \beta & \gamma & 0 & 0 & \cdots & 0\\0 & \alpha & \beta & \gamma & 0 & \cdots & 0 \\ 0 & 0 & \alpha & \beta &\gamma & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0& \cdots & 0 \end{vmatrix}$$

Now to find this determinant, we cofactor expand this smaller matrix along the first column. All entries are zero except for $\alpha$ so we have quite a bit of reduction. This gives $$d'_{n-1} = \alpha d_{n-2}$$ where $d_{n-2}$ is the $n-2$ by $n-2$ tri-diagonal determinant obtained by removing the first row and column of $d'_{n-1}$. In summary, we have $$d_n = \beta d_{n-1} + (-1)\gamma d'_{n-1} = \beta d_{n-1} - \gamma\alpha d_{n-2}$$

Answer 2

Example: $$ \underbrace{\left|\begin{matrix} \beta&\gamma&0&0\\ \alpha&\beta&\gamma&0\\ 0&\alpha&\beta&\gamma\\ 0&0&\alpha&\beta \end{matrix}\right|}_{d_4} \,=\,\beta\underbrace{\left|\begin{matrix} \beta&\gamma&0\\ \alpha&\beta&\gamma\\ 0&\alpha&\beta \end{matrix}\right|}_{d_3} -\gamma\left|\begin{matrix} \alpha&\gamma&0\\ 0&\beta&\gamma\\ 0&\alpha&\beta \end{matrix}\right| \,=\,\beta d_3 -\gamma\underbrace{\alpha\left|\begin{matrix} \beta&\gamma\\ \alpha&\beta \end{matrix}\right|}_{d_2}. $$ The first equality is obtained by Laplace expansion along the first row, while the second one is obtained by Laplace expansion along the first column.