Matrix with a certain pattern Consider the following matrix:
$$\left[\begin{array}{cccc} 1+x_1y_1 & 1+x_1y_2 & \ldots & 1+x_1y_n  \\
1+x_2y_1 & 1+x_2y_2 & \ldots & 1+x_2y_n \\
1+x_3y_1 & 1+x_3y_2 & \ldots & 1+x_3 y_n \\
\vdots & & & \\
1+x_ny_1 & 1+x_ny_2 & \ldots & 1+x_n y_n \\
\end{array}  \right]$$
with $x_k, y_k$ being real numbers. I suppose that there must be some neat way of computing the determinant of this matrix. Those ones are annoying to me.

Here are two approaches.
1) By inspecting the pattern of this matrix, one sees that it is of the form $A+x y^T$ where $A$ is the matrix of ones and $x,y$ are appropriate column vectors. The matrix determinant lemma then gives $\det(A+x y^T)=\det(A)+y^T \text{adj}(A)x$ where $\text{adj}(A)$ is the adjugate matrix of $A$. But the determinant of $A$ vanishes; moreover, if $n\geq 3$ then then so too does the the adjugate, so the determinant is equal to zero for $n\geq 3$.
2) If we imagine trying to compute this determinant by cofactors, we see that should be some degree polynomial of degree $2n$ in the variables $x_k,y_k$. Furthermore, said polynomial must be antisymmetric with respect to swaps of $x$- or $y$- variables. This suggests an answer of the form $\displaystyle A\prod_{i<j}(x_i-x_j)(y_i-y_j)$ where $A$ is some overall constant. To normalize this, we take $x_k=y_k=\omega^k$ where $\omega=e^{2\pi i/k}$. Then the matrix is circulant and for $n\geq 3$ has a zero eigenvalue, but the above polynomial doesn't vanish. Consequently both $A$ and the determinant itself must vanish if $n\geq 3$.
