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May 11th, 2016, 11:09 PM   #1
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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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