Originally Posted by

**joll** I see... that way, the determinant can be written

$\displaystyle (-1)^{n+1}a_n\begin{vmatrix}-1&0&\cdots&0&0\\x&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots&&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}+ (-1)^{n+2}a_{n-1}\begin{vmatrix} x&0&\cdots&0&0\\0&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots &&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}\dots$.

The determinent term in (n,i)cofactor (I mean, the \vmatrix part of the equation above) of the matrix is given by

$\displaystyle (-1)^{n-i}x^{i-1}$

Therefore

$\displaystyle D_{n+1}=a_0x^n+\dots a_n$(am I correct?).

I haven't been learning linear algebra for long yet, so knowing many ways to solve one problem really helps me. Thank you very much.

yes, you've got the idea but you forgot that the original determinant is $\displaystyle (n+1) \times (n+1)$ not $\displaystyle n \times n$.

so, after expanding along the last row, you'll get that the determinant is equal to

$\displaystyle \sum_{i=1}^{n+1}(-1)^{n+1+i}a_{n+1-i}(-1)^{n+1-i}x^{i-1}=\sum_{i=1}^{n+1}a_{n+1-i}x^{i-1}=a_n + \ldots + a_0x^n.$