Findind the determinant of the square matrix $(a_{ij}) \in \mathbb{R}^{n \times n}$, where $a_{ij} = (1 - \delta_{ij}) + x_iy_j$ I am having difficulty finding the determinant of the following matrix:
$$
\begin{bmatrix}
x_1y_1&1+x_1y_2&\ldots&1+x_1y_n\\
1+x_2y_1&x_2y_2&\ldots&1+x_2y_n\\
\vdots&\vdots&\ddots&\vdots\\
1+x_ny_1&1+x_ny_2&\ldots&x_ny_n\\
\end{bmatrix}.
$$
I considered a related matrix
$$
\begin{bmatrix}
1+x_1y_1&1+x_1y_2&...&1+x_1y_n\\
1+x_2y_1&1+x_2y_2&...&1+x_2y_n\\
...&...&...&...\\
1+x_ny_1&1+x_ny_2&...&1+x_ny_n\\
\end{bmatrix}
$$
and I was able to show that the determinant of this matrix is $0$ if $n$ is at least $3$. Then I tried to see how this can be applied to find the determinant of the first matrix, but I haven't found any connection.
Any help is appreciated.
 A: This matrix is $\textbf{x}\textbf{y}^\textsf{T}+\textbf{1}\textbf{1}^\textsf{T}-\text{Id}$.
Sherman–Morrison formula

$$\left(A + uv^\textsf{T}\right)^{-1} = A^{-1} - {A^{-1}uv^\textsf{T}A^{-1} \over 1 + v^\textsf{T}A^{-1}u}$$
gives when $A=-\text{Id}$
$$(-\text{Id}+\textbf{x}\textbf{y}^\textsf{T})^{-1}=-\text{Id}-{\textbf{x}\textbf{y}^\textsf{T}\over 1 - \textbf{y}^\textsf{T}\textbf{x}}$$
Now since
\begin{equation}
\det(A + \mathbf{c}\mathbf{d}^T) = \det(A)(1 + \mathbf{d}^T A^{-1}\mathbf{c}) 
\end{equation}
we get for $A=-\text{Id}+\textbf{x}\textbf{y}^T$, firstly
$$\det A= \det(-\text{Id})(1-\textbf{y}^T\textbf{x})$$
Then
$$\det (A+\textbf{1}\textbf{1})=\det(A)(1+\textbf{1}^\textsf{T} (-\text{Id}-{\textbf{x}\textbf{y}^T\over 1 - \textbf{y}^\textsf{T}\textbf{x}})\textbf{1})=$$
$$=(-1)^{n+1}(1-\textbf{y}^T\textbf{x})(n-1+\frac{(\textbf{1}^\textsf{T}\textbf{x}\textbf{y}^T\textbf{1})}{1-\textbf{y}^T\textbf{x}})=(-1)^{n+1}((n-1)(1-\textbf{y}^T\textbf{x})+\textbf{1}^\textsf{T}\textbf{x}\textbf{1}^\textsf{T}\textbf{y}) $$
A: $$
{\rm A} := 
    \begin{bmatrix}
            x_1 y_1 & 1 + x_1 y_2 & \ldots & 1 + x_1 y_n \\
        1 + x_2 y_1 &     x_2 y_2 & \ldots & 1 + x_2 y_n \\
             \vdots &      \vdots & \ddots &     \vdots \\
        1 + x_n y_1 & 1 + x_n y_2 & \ldots &     x_n y_n
    \end{bmatrix}
    = \unicode{x1D7D9}_n \unicode{x1D7D9}_n^\top - {\rm I}_n + {\rm x} {\rm y}^\top
    = (- {\rm I}_n) +
    \begin{bmatrix}
        | & | \\
        {\rm x} & \unicode{x1D7D9}_n\\ | & |
    \end{bmatrix}
    \begin{bmatrix}
           |    &  | \\
        {\rm y} & \unicode{x1D7D9}_n \\ 
           |    &  |
    \end{bmatrix}^\top
$$
Using the (generalized) matrix determinant lemma,
$$
\det ({\rm A}) = \cdots = \color{blue}{(-1)^n \cdot \big( (1-n) ( 1 - {\rm x}^\top {\rm y} ) - ( \unicode{x1D7D9}_n^\top {\rm x} ) ( \unicode{x1D7D9}_n^\top {\rm y} ) \big)}$$
