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How do I calculate the following determinant? $$ \begin{vmatrix} a_n & -1 & 0 & \dots & 0 & 0 \\ a_{n-1} & x & -1 & \dots & 0 & 0 \\ a_{n-2} & 0 & x & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ a_1 & 0 & 0 & \dots & x & -1 \\ a_0 & 0 & 0 & \dots & 0 & x \end{vmatrix} $$

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  • $\begingroup$ using induction on $n$ $\endgroup$ – user368558 Mar 26 '17 at 17:19
  • $\begingroup$ Use induction on n and Laplace expansion on the first row $\endgroup$ – mlainz Mar 26 '17 at 17:25
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Let $F(a_0, a_1, \dots, a_n)$ be the determinant in your question. Expanding by minors across the first row, we get that $F(a_0, a_1, \dots, a_n)$ is equal to $$ a_n \begin{vmatrix} x & -1 & \dots & 0 & 0 \\ 0 & x & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & x & -1 \\ 0 & 0 & \dots & 0 & x \end{vmatrix} -(-1) \begin{vmatrix} a_{n-1} & -1 & \dots & 0 & 0 \\ a_{n-2} & x & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a_1 & 0 & \dots & x & -1 \\ a_0 & 0 & \dots & 0 & x \end{vmatrix} $$ or $a_n x^n + F(a_0, a_1, \dots, a_{n-1}).$

From there, you can get the answer by induction on $n$.

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