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I'm taking my first linear algebra course and one of the problem asks us to find the determinant of the Cartan matrix by row or column operations.

The entries of Cartan matrix is given as follow:

\begin{equation*} a_{ij} = \begin{cases} 2 \text{ if } i = j \\ -1 \text{ if } |i-j| = 1 \\ 0 \text{ otherwise} \end{cases} \end{equation*}

So the matrix is basically: \begin{pmatrix} 2 & -1 & 0 & \cdots & 0 \\ -1 & 2 & -1 & \ddots & \vdots \\ 0 & -1 & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \ddots & -1 \\ 0 & \cdots & 0 & -1 & 2 \end{pmatrix} I think I need an algorithm that can be repeated to simplify the matrix so that the determinant can be easier to find. I have tried smaller cases for 2x2 matrix and 3x3 matrix but cannot find anything that's not bruteforce-like. Therefore, I want to seek help from you, thanks very much.

Additionally, I must use row/column operations to finish the task but not the Laplace expansion or other ways.

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There's probably multiple options here, but what I would do is:

Row reduce the matrix to a diagonal one, without scaling. Then the determinant remains unchanged, and the determinant of a diagonal matrix is easily computed: it is simply the product of the diagonal entries. You should get something like this: \begin{pmatrix} 2 & -1 & 0 & \cdots & 0 \\ 0 & 3/2 & -1 & \ddots & \vdots \\ 0 & 0 & 4/3 & -1 & 0 \\ \vdots & \ddots & \ddots & \ddots & -1 \\ 0 & \ldots & \ldots & \ldots & \ldots \end{pmatrix} And then try to find a closed formula for the diagonal entries of the row reduced matrix. I'd think the diagonal entries of the reduced matrix are something like $$d_{ii} = 1+\frac{1}{i},$$ but you'll have to verify this to make sure. Then, the determinant would evaluate to $$\prod_{i = 1}^n d_{ii}.$$

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