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.
Best Answer
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 $$\det (A+\textbf{1}\textbf{1})=\det(A)(1+\textbf{1}^\textsf{T} (-\text{Id}-{\textbf{x}\textbf{y}^T\over 1 - \textbf{y}^\textsf{T}\textbf{x}})\textbf{1})=$$ $$=(-1)^{n+1}(1-\textbf{y}^T\textbf{x})(n-1+\frac{(\textbf{1}^\textsf{T}\textbf{x}\textbf{y}^T\textbf{1})}{1-\textbf{y}^T\textbf{x}})=(-1)^{n+1}((n-1)(1-\textbf{y}^T\textbf{x})+\textbf{1}^\textsf{T}\textbf{x}\textbf{1}^\textsf{T}\textbf{y}) $$