TITLE: Correlation between polynomial equations and matrix determinants
QUESTION [5 upvotes]: Expanding
$p(x)=(ax-b)(cx+d)$
we get
$acx^2+(ad-bc)x-bd$.
Notice the determinant of the matrix
$\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
$
is $ad-bc$ exactly like the constant of $x$ in the polynomial expansion.
So I searched links between the equation and the matrix determinant. I only found the solution of the equation $p(x)=0$ to be $x \in \{b/a, -d/c\}$. Then I found numerous other things, but none of them were simple enough to be considered worthy remembering.
My question: What are the correlations between the polynomials and determinants of the matrices with entries being the coefficients of the polynomial?

REPLY [2 votes]: Conceptually, one nice way to "explain" this is as follows. We know from high-school that for a polynomial $\rm\: x^n - \color{#C00}{c_{n-1}}x^{n-1}\!+\,\cdots+c_0,\:$ the coefficient $\rm\,\color{#C00}{c_{n-1}}\,$ equals the sum of the roots. Hence $\rm\: \ell(f)\, :=\, \color{#C00}{c_{n-1}}$ satisfies a $\rm\color{blue}{logarithmic}$ law $\rm\ \ell(fg)\, =\, \ell(f)+\ell(g)\ $ (as is also easily verified directly). Your observation may be interpreted as a variation on this logarithmic property of root sums.
Define $\rm\,\ f\,\left[\begin{array}{cc}\rm a&\rm b\\ \rm c&\rm d\end{array}\right]\, :=\ \ell((ax\!-\!b)(cx\!+\!d))\, :=\, $ coefficient of $\rm\,x\,$ (the "generic" value of $\ell)$  

$\rm\,f\,$ is multilinear: $\, $ True, essentially, by the above $\rm\color{blue}{logarithmic}$ property of $\:\ell$.
$\rm\,f\,$ is alternating:  $\rm\, \ f\,\left[\begin{array}{cc}\rm a&\rm b\\ \rm a&\rm b\end{array}\right] =\ \ell((ax\!-\!b)(ax\!+\!b))\, =\, \ell(a^2x^2\!+\color{#C00}0\cdot x-\!b^2)\, =\, \color{#C00}0$
$\rm\,f\,$ is normalized: $\rm\ f(I) = 1\:$ by $\rm\: f\,\left[\begin{array}{cc}\rm 1&\rm 0\\ \rm 0&\rm 1\end{array}\right] =\ \ell(1\!\cdot\! x\!-\!0)(0\!\cdot\! x\!+\!1))\, =\, \ell(\color{#C00}1\cdot x) = \color{#C00}1$

However, as is well-known, $\rm\:det\,$ (determinant) is the unique solution of the above equations. Therefore, by uniqueness, we deduce that $\rm\ f = det.$