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This question asked for an intelligent way to find $$ \det \begin{bmatrix} 2 & -1 \\ -1 & 2 & -1 \\ & -1 & 2 & -1 \\ && -1 & 2 & -1 \\ &&& -1 & 2 & -1 \\ &&&& -1 & 2 \end{bmatrix}. $$ I did not attempt to answer it, but I ran a few things through R, and found the singular values: $$ 3.8019377,\quad 3.2469796,\quad 2.4450419,\quad 1.5549581,\quad 0.7530204,\quad 0.1980623 $$ When I plotted these on the $y$-axis with the sequence $1,2,3,4,5,6$ on the $x$-axis, I thought they were in a straight line, and a second later I thought that they were not. So I fitted the least-squares line and put $1,2,3,4,5,6$ on the $x$-axis and the residuals on the $y$-axis, and saw something. The residuals are: $$ -0.08315789,\quad 0.11592223,\quad 0.06802274,\quad -0.06802274,\quad -0.11592223,\quad 0.08315789 $$ The $k$th residual is $-1$ times the $(7-k)$th residual. Why is that?

(At least some of this is to be expected: the sum of all six residuals must be $0$ and their linear combination with coefficients $1,2,3,4,5,6$ must be $0$; we know that much before knowing anything about the six $y$ values.)

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Michael Hardy
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The eigenvalues of a sign-symmetric $n\times n$ tridiagonal matrix $\pmatrix{a&b\\ c&\ddots&\ddots\\&\ddots&\ddots}$ with constant diagonals are given by $a+2\sqrt{bc}\cos\frac{k\pi}{n+1}$ with $k=1,2,\ldots,n$ (see this thread, for instance). So, the residuals from linear regression are given by the cosine function. Hence your observation.

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user1551
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