My Math Forum Coefficient of bivariate polynomial as a determinant of matrix

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 May 11th, 2016, 11:09 PM #1 Newbie   Joined: Oct 2014 From: China Posts: 12 Thanks: 2 Coefficient of bivariate polynomial as a determinant of matrix Given $$\begin{bmatrix} a\\ b\\ c\\ d\\ \end{bmatrix}=\begin{bmatrix} a_0t^3+a_1st^2+a_2s^2t+a_3s^3\\ a_4t^2+a_5st+a_6s^2\\ a_7t+a_8s\\ a_9\\ \end{bmatrix}$$ the following equation holds: $$-\frac{1}{d} \begin{vmatrix} d&c&b&a&0\\ 0&d&c&b&a\\ 3d&2c&b&0&0\\ 0&3d&2c&b&0\\ 0&0&3d&2c&b\\ \end{vmatrix} = \frac{1}{d}\begin{vmatrix} cd&2bd&3d\\ 2bd&3ad+bc&2c\\ 3ad&2ac&b\\ \end{vmatrix}\\ =\left(a_4^2a_7^2-4a_0a_7^3-4a_4^3a_9+18a_0a_4a_7a_9-27a_0^2a_9^2 \right)t^6 \\ +\left(2a_4a_5a_7^2-4a_1a_7^3+2a_4^2a_7a_8-12a_0a_7^2a_8-12a_4^2a_5a_9+18a_1a_4a_7a_9+18a_0a_5a_7a_9+18a_0a _4a_8a_9-54a_0a_1a_9^2 \right)st^5+\\ +(...)s^2t^4+(...)s^3t^3+(...)s^4t^2+(...)s^5t+(.. .)s^6$$ The resulting bivariate polynomial looks ugly, however, I find the coefficient of $t^6$ equals (that of $s^6$ has the same form): $$-\frac{1}{a_9}\begin{vmatrix} a_9&a_7&a_4&a_0&0\\ 0&a_9&a_7&a_4&a_0\\ 3a_9&2a_7&a_4&0&0\\ 0&3a_9&2a_7&a_4&0\\ 0&0&3a_9&2a_7&a_4\\ \end{vmatrix}$$ Thus I guess the coefficient of $st^5$, $s^2t^4$, etc could also be written as a determinant of matrix. But I tried rewrite that of $st^5$ in Mathematica for days in vain. Look forward for any clues. Last edited by whitegreen; May 11th, 2016 at 11:27 PM.

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