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This question already has an answer here:

What is the trick for evaluating the determinant of this matrix?

$$\begin{bmatrix} 2 & -1 \\ -1 & 2 & -1 \\ & -1 & 2 & -1 \\ && -1 & 2 & -1 \\ &&& -1 & 2 & -1 \\ &&&& -1 & 2 \end{bmatrix}$$

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marked as duplicate by Daniel Fischer Feb 15 '15 at 9:17

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    $\begingroup$ What do you mean to evaluate it? $\endgroup$ – Amzoti Feb 15 '15 at 1:51
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    $\begingroup$ Determinate? What do you mean by evaluate? $\endgroup$ – Chinny84 Feb 15 '15 at 1:52
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    $\begingroup$ there is a three term recursive formula. you can look up the determinant of a tridiagonal matrix. $\endgroup$ – abel Feb 15 '15 at 1:57
  • $\begingroup$ Apologies. Question updated. $\endgroup$ – sailor Feb 15 '15 at 1:57
  • $\begingroup$ This is not just any tridiagonal matrix, it's a very special one. I think we can find a basis of eigenvectors analytically. The $j$th component of eigenvector $m$ is something like $\sin(m \pi j/N) $. Knowing the eigenvalues, we get the determinant. $\endgroup$ – littleO Feb 15 '15 at 2:44
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let $f_n$ be the the determinant of the $n \times n$ tridiagonal matrix with diagonal elements all equal to $2$ and the sub and super diagonal has $-1$ on them. then by expanding the determinant by the first row, you get the recursive relation $$f_n = 2f_{n-1} - f_{n-2},\quad f_1 = 2,\ f_2 = 3.$$ you can verify that $$f_n = n+1$$ is in fact the solution.

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