This is probably really simple, and I'm missing something, but Thomas' algorithm doesn't seem to work.
6
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3$\begingroup$ Let's start with something simpler. How do you solve the number 6? $\endgroup$– Steven LandsburgCommented Jun 20, 2013 at 18:18
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1$\begingroup$ Presumably "solve" = invert. This matrix is invertible iff its order is not $2\bmod 3$, in which case it has determinant $\pm 1$ and experiments such as (in gp) 1/matrix(19,19,i,j,abs(i-j)<2) suggest the inverse matrix always has $0,\pm1$ entries with tractable patterns. BTW the determinant depends on the order mod 6; perhaps that's what Steven Landsburg was hinting at? $\endgroup$– Noam D. ElkiesCommented Jun 20, 2013 at 18:26
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$\begingroup$ Noam: You give me far too much credit! $\endgroup$– Steven LandsburgCommented Jun 20, 2013 at 18:33
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$\begingroup$ the formula for the inverse (for an arbitrary diagonal) is here: phys.lsu.edu/~amarti9/adfaerf/… $\endgroup$– Carlo BeenakkerCommented Jun 20, 2013 at 18:34
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1$\begingroup$ the formula for the inverse $M^{\rm inverse}$ of your $n\times n$ matrix $M$ is Eq. 10 of this 1996 paper phys.lsu.edu/~amarti9/adfaerf/… by Hu and O'Connell: $$M^{\rm inverse}_{ij}=-(-1)^{i+j}\frac{\cos[\frac{\pi}{3}(n+1-|j-i|)]-\cos[\frac{\pi}{3}(n+1-i-j)]}{2\sin\frac{\pi}{3}\sin[\frac{\pi}{3}(n+1)]}$$ this paper also gives more general expressions for the inverse when you add an arbitrary multiple of the unit matrix to your $M$. $\endgroup$– Carlo BeenakkerCommented Jun 21, 2013 at 7:23
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1 Answer
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Let
$$M=\pmatrix{0&1&-1\cr 1&-1&1\cr -1&1&0\cr} \quad N=\pmatrix{0&1&-1\cr 0&-1&1\cr 0&0&0\cr} \quad P=\pmatrix{0&0&0\cr 1&-1&0\cr -1&1&0\cr} $$
Then it's not hard to check that for $n=3k$, the inverse of your matrix is
$$\pmatrix{M&N&N&\cdots& N\cr P&M&N&\cdots &N\cr P&P&M&\ldots &N\cr \vdots&\vdots&\vdots&&\vdots\cr P&P&P&\cdots&M\cr}$$
For $n=3k+2$, your matrix is singular, as Noam Elkies observed in comments.
That leaves the case $n=3k+1$, which I believe is also easy, though I haven't worked it out.
answered Jun 20, 2013 at 21:24