2
$\begingroup$

$$ D_n = \begin{vmatrix} a_1 & 1 & 0 & \cdots& 0 & 0\\ -1& a_2 & 1 & \cdots & 0 & 0 \\ 0 & -1 & a_3 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & a_{n-1} & 1 \\ 0 & 0 & 0 & \cdots & -1 & a_n \end{vmatrix} $$ I thought to multiply last column with $\frac{1}{a_n}$ and add it to the (n-1)-th column and so on but $a_n$ can be equal to $0$.

$\endgroup$
3
  • 1
    $\begingroup$ The answer has to be a polynomial in $a_1, a_2, \ldots, a_n$. If you get a solution that works for nonzero $a_n$ in the form of such a polynomial, it will be valid for $a_n=0$ as well. $\endgroup$ Jun 16 '16 at 22:40
  • 3
    $\begingroup$ You matrix is a tri-diagonal matrix and its determinant can be computed using a 3-term recurrence relation $D_n = a_n D_{n-1} + D_{n-2}$. see this. $\endgroup$ Jun 16 '16 at 22:45
  • $\begingroup$ @achille hui thank you! $\endgroup$
    – Gjekaks
    Jun 16 '16 at 22:47
0
$\begingroup$

The answer is $$ D_n = \prod_{k=1}^n a_n \left( 1+\sum_{i=1}^{n-1}\frac{1}{a_ia_{i-1}}\right) $$ The way to do this is to let $$ E_n = \frac{D_n}{\prod_{k=1}^n a_n} $$ and use Achille hui's insight to get $$ E_k = E_{k-1}+\frac{1}{a_{k-1}a_{k-2}} $$ for all $3 \leq k \leq n$ and sum up. You have to be careful dealing with the endpopints; I think I got them right but you should check this.

$\endgroup$

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.