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Is there any fast way to compute the determinant of this matrix

$$ \left( \begin{array}{ccccc} a & b & 0 &0 &0 \\ b & a & b &0 &0 \\ 0 & b & a &b &0 \\ 0 & 0 & b &a &b \\ 0 & 0 & 0 &b &a \end{array} \right) $$

And can you say anything about $$ \frac{det(M_{k+1})}{det(M_{k})} $$

where $M_{k}$ is a toeplitz matrix?

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2 Answers

up vote 3 down vote accepted

If $T_n$ is the determinant for $n$x$n$ matrix, then we have by expanding using the first row (and first column in the second matrix) that

$$T_n = a T_{n-1} - b^2 T_{n-2}$$

with $T_1 = a$ and $T_2 = a^2 - b^2$.

You should now be able to compute a closed form easily, using the characteristic equation.

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Can you please explain how you got this recurrence function? –  neticin Feb 9 '12 at 19:37
    
@neticin: See: mathworld.wolfram.com/DeterminantExpansionbyMinors.html. We pick the first row to get two minors, in the second minor, we pick the first column. –  Aryabhata Feb 9 '12 at 19:41
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See the paper Analytical Inversion of Symmetric Tridiagonal Matrices, and in particular the formulas (6) and (7). To apply this it is only necessary to remove a factor of $b$ from each row of your matrix, with the accompanying factor of $b^n$ from the determinant.

Of course if $b = 0$...

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