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 $A$?

  • $\begingroup$ If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$
    – user
    Jan 8, 2018 at 22:42

1 Answer 1


I think we can only find by Laplace expansion:

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


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

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

  • $\begingroup$ what is $B'$? the $B$ transpose? $\endgroup$
    – M a m a D
    Jan 5, 2018 at 19:25
  • $\begingroup$ sorry for the notation, B' is the matrix you obtain from A eliminating the firs row and the second column accordin to Laplace expansion en.wikipedia.org/wiki/Laplace_expansion $\endgroup$
    – user
    Jan 5, 2018 at 19:26
  • $\begingroup$ 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]: math.stackexchange.com/questions/575748/… $\endgroup$
    – M a m a D
    Jan 5, 2018 at 19:37
  • $\begingroup$ @Drupalist It seems to confirm that in general it is not possible to simplify further $\endgroup$
    – user
    Jan 5, 2018 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.