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The symmetric traditional matrix $A$ and its determinant is given.

$$ A = \begin{bmatrix} a_1&b_1&0&0&0&0& \cdots &0\\ b_1&a_2&b_2&0&0&0&\cdots&0\\ 0&b_2&a_3&b_3&0&0&\cdots&0\\ 0&0&b_3&a_4&b_4&0&\cdots&0\\ 0&0&0&b_4&a_5&b_5&\cdots&0\\ 0&0&0&\ddots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&0&0&b_{n-2}&a_{n-1}&b_{n-1}\\ 0&0&0&0&0&0&b_{n-1}&a_n\\ \end{bmatrix} $$

What is the determinant of matrix $B$ which exactly $A$ after removing first row and column?

$$ B = \begin{bmatrix} a_2&b_2&0&0&0&\cdots&0\\ b_2&a_3&b_3&0&0&\cdots&0\\ 0&b_3&a_4&b_4&0&\cdots&0\\ 0&0&b_4&a_5&b_5&\cdots&0\\ 0&0&\ddots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&0&b_{n-2}&a_{n-1}&b_{n-1}\\ 0&0&0&0&0&b_{n-1}&a_n\\ \end{bmatrix} $$

Is there any known way to calculate this from $A$?

M a m a D
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1 Answers1

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I think we can only find by Laplace expansion:

$$det A=a_1\cdot detB-b_1\cdot detB'$$

and

$$det B'=b_1\cdot detC-b_2\cdot detC'$$

and so on, but it seems not possible to simplify further.

user
  • 154,566
  • what is $B'$? the $B$ transpose? – M a m a D Jan 05 '18 at 19:25
  • sorry for the notation, B' is the matrix you obtain from A eliminating the firs row and the second column accordin to Laplace expansion https://en.wikipedia.org/wiki/Laplace_expansion – user Jan 05 '18 at 19:26
  • From [This question][1] We know for a tridiagonal matrix the following recursion relation is true (notations described in there) $$f_i = d_if_{i-1} - c_ia_{i-1}f_{i-2}$$

    Won't this help? [1]: https://math.stackexchange.com/questions/575748/determinant-of-symmetric-tridiagonal-matrices

     – M a m a D Jan 05 '18 at 19:37
  • @Drupalist It seems to confirm that in general it is not possible to simplify further – user Jan 05 '18 at 20:17