# Determinants of symmetric tridiagonal matrix after removing first row and column

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 $A$?

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

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.

• what is $B'$? the $B$ transpose? Jan 5, 2018 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 en.wikipedia.org/wiki/Laplace_expansion
– user
Jan 5, 2018 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]: math.stackexchange.com/questions/575748/… Jan 5, 2018 at 19:37
• @Drupalist It seems to confirm that in general it is not possible to simplify further
– user
Jan 5, 2018 at 20:17