Need a hint on calculating this $n$-th order determinant

Question

I am trying to calculate the following determinant:

$$ \begin{vmatrix} a_0 & -1 & 0 & 0 & \ldots & 0 & 0 \\ a_1 & x & -1 & 0 & \ldots & 0 & 0 \\ a_2 & 0 & x & -1 & \ldots & 0 & 0 \\ a_3 & 0 & 0 & x & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ a_{n-1} &0 & 0 & 0 & \ldots & x & -1 \\ a_n & 0 & 0 & 0 & \ldots & 0 & x \\ \end{vmatrix} $$

but I don't really know how to start solving it. I've thought about multiplying the second column with $a_0$ and adding it to the first, and doing the same with the other columns (while multiplying them with $a_2,a_3,\ldots , a_n$) respectively but that doesn't seem to get me anywhere. Any hints would be appreciated.

Expand along the first column and use the properties of "determinant of triangular matrices" to evaluate the determinant. — mike, Jan 03 '22 at 17:30
Does this answer your question? Determinant of a companion matrix — Martin R, Jan 04 '22 at 02:48
@MartinR: This is a different matrix with a different determinant and a different way to compute it. — markvs, Jan 07 '22 at 05:46

markvs · Accepted Answer · 2022-01-06T04:11:55.497

3

Denote the determinant by $d(a_0,\ldots,a_n,x)$. Expand along the first row: $$d(a_0,\ldots,a_n,x)=a_0x^n+d(a_1,\ldots,a_n,x).$$ So by induction:

$$d(a_0,\ldots,a_n,x)=a_0x^n+a_1x^{n-1}+\ldots+a_n.$$

By the way, the matrix is of size $n+1$, not $n$.

edited Jan 06 '22 at 04:11

answered Jan 03 '22 at 17:23

markvs

19,653

Weird. My textbook says the result is 1. – downmath Jan 07 '22 at 16:06
Take $n=1$:to check. – markvs Jan 07 '22 at 16:11

Need a hint on calculating this $n$-th order determinant

1 Answers1