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I have read about the Thomas algorithm. Right now, I am trying to use it to solve the following linear system

$$\begin{bmatrix} b_1 & c_1 & 0 & \dots & 0 & a_1\\ a_2 & b_ 2 & c_2 & 0 & \dots & 0\\ 0 & a_3 & b_3 & c_3 & 0 & \dots\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ c_n & 0 & \dots & 0 & a_n & b_n\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\ \vdots\\x_n\end{bmatrix} = \begin{bmatrix}r_1\\r_2\\r_3\\ \vdots\\r_n\end{bmatrix}$$ I have not been able to work out an algorithm for this matrix. Can someone point me to some reference that I can read about this? Thank you.

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  • $\begingroup$ Does your matrix have any special properties? Is (strictly) diagonally dominant by rows or symmetric positive definite? $\endgroup$ – Carl Christian May 29 '16 at 20:56
  • $\begingroup$ No. It's pretty much just a tridiagonal system of equations matrix except that the first and last row have an extra element like above. $\endgroup$ – Kane Billiot May 29 '16 at 20:58
  • $\begingroup$ Can you guarantee that the $a_i$, $b_i$, $c_i$ are not zero? $\endgroup$ – mvw May 29 '16 at 21:08
  • $\begingroup$ Yes. All three are non-zeroes. $\endgroup$ – Kane Billiot May 29 '16 at 21:12
  • $\begingroup$ My opinion (maybe I am wrong) is that you have too few zeros (25%) to expect a "miracle" solution compared to standard methods for solving a linear system. $\endgroup$ – Jean Marie May 29 '16 at 22:06
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The Thomas algorithm can be modified to solve this "periodic" case, as explained here.

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    $\begingroup$ The link does not work anymore. Could you provide a different one? $\endgroup$ – Kane Billiot Feb 22 '17 at 15:46

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