Find the determinant of a symmetric matrix How can we find the determinant of the following matrix?
$$A = \begin{pmatrix}
x_1y_1 & x_1y_2 & x_1y_3 & \cdots & x_1y_{n-1} & x_1y_n \\ 
x_1y_2 & x_2y_2 & x_2y_3 & \cdots & x_2y_{n-1} & x_2y_n \\ 
x_1y_3 & x_2y_3 & x_3y_3 & \cdots & x_3y_{n-1} & x_3y_n \\ 
\vdots &  \vdots &  \vdots &  \ddots &  \vdots &  \vdots \\
x_1y_{n-1} & x_2y_{n-1} & x_3y_{n-1} & \cdots & x_{n-1}y_{n-1} & x_{n-1}y_n \\ 
x_1y_n & x_2y_n & x_3y_n & \cdots & x_{n-1}y_n & x_ny_n
\end{pmatrix}$$
That is, the entry of $A$ is $a_{ij}=x_iy_j$ for $i\leq j$; and $a_{ij}=a_{ji}$ for $i>j$.
I do not have anything new. And I find also it is difficult to find its eigenvalues.
 A: A hint, a quick computer algebra calculation gives
for the determinant of the matrix $A_n$
\begin{align}
\text{det}A_1 &= +x_1y_1\\
\text{det}A_2 &= -x_1y_2(x_1y_2-x_2y_1)\\
\text{det}A_3 &= +x_1y_3(x_1y_2-x_2y_1)(x_2y_3-x_3y_2)\\
\text{det}A_4 &= -x_1y_4(x_1y_2-x_2y_1)(x_2y_3-x_3y_2)(x_3y_4-x_4y_3)\\
\text{det}A_5 &= +x_1y_5(x_1y_2-x_2y_1)(x_2y_3-x_3y_2)(x_3y_4-x_4y_3)(x_4y_5-x_5y_4) 
\end{align}
So this suggests to show first the formula for $\text{det}A_2$, and then to somehow show the recursion
\begin{equation}
y_n \text{det}A_{n+1}  = -y_{n+1}(x_ny_{n+1}-x_{n+1}y_n) \text{det}A_n,
\quad\text{with}\quad n > 2.
\end{equation}
Then one has to see what kind of coefficients $x_k$ and $y_k$ one has.
A: I don't have a solution method, but the determinant appears as problem *222 (the star/asterisk apparently indicating a difficult problem) in Problems in Higher Algebra by D. K. Faddeev and I. S. Sominskii [translated by J. L. Brenner] (W. H. Freeman, 1965).  The answer is given as $x_1 y_n \prod_{i = 1}^{n-1} (x_{i + 1} y_i -  x_i y_{i + 1})$.
