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.

The matrix can be expressed as the sum of two rank $1$ matrices, each of which is the outer product of vectors
$$M=\mathbf{11}^T+\mathbf{xy}^T$$
where $\mathbf{1}=[1,1,1,\cdots,1]^T\in\mathbb{R}^{n\times 1}$, $\mathbf{x}=[x_1,x_2,\cdots,x_n]^T\in\mathbb{R}^{n\times 1}$ and $\mathbf{y}=[y_1,y_2,\cdots,y_n]^T\in\mathbb{R}^{n\times 1}$.
Using the subadditive property of the rank, where $\text{rank}(A+B)\leq\text{rank}(A)+\text{rank}(B)$, we have
$$\text{rank}(M)\leq\text{rank}(\mathbf{11}^T)+\text{rank}(\mathbf{xy}^T)\Rightarrow \text{rank}(M)\leq 2$$
Thus, only when $n=2$ the matrix can be full rank, and so potentially have a non-zero determinant, which is easily evaluated as $(1+x_1y_1)(1+x_2y_2)-(1+x_2y_1)(1+x_1y_2)$. However for $n\geq3$, the matrix will be rank deficient, in which case the determinant will be $0$.  
