Let $A$ be a square matrix with diagonal entries $a_{ii} = x_i y_i$ and off-diagonal entries $a_{ij} = x_{i} y_{j} + 1$ when $i \neq j$. Here, $x_{i}, y_{j}$, for $i, j \in \{ 1,\ldots, n \}$, are $2n$ real numbers. I would like to calculate $\det (A)$.
I tried to calculate the determinant of the simpler matrix where $a_{ij}=1+x_{i}y_{j}$, and I see it equals zero. But I got nothing more when I tried to expand the new determinant for $A$.
Many authors use $\mathbf{J_n}$ to represent the $n\times n$ matrix of all ones entries. Since the size $n\times n$ is unchanged throughout this problem, we will omit the subscript:
$$ \mathbf J = \begin{pmatrix} 1 & 1 & \cdots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{pmatrix} $$
Let $\mathbf{x} = \begin{pmatrix} x_1 & x_2 & \cdots & x_n \end{pmatrix}^T$, $\mathbf{y} = \begin{pmatrix} y_1 & y_2 & \cdots & y_n \end{pmatrix}^T$, and $\mathbf{1} = \begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix}^T$, so that $\mathbf J = \mathbf 1 \mathbf 1^T$ and:
$$ \mathbf A = \mathbf J - \mathbf I + \mathbf x \mathbf y^T $$
where $\mathbf I$ is the $n\times n$ identity matrix.
The well-known matrix determinant lemma tells us:
$$ \det \mathbf A = \det(\mathbf J - \mathbf I) (1 + \mathbf y^T (\mathbf J - \mathbf I)^{-1} \mathbf x ) $$
Fortunately the key "ingredients" here have already been worked out in this highly up-voted previous Question, namely:
$$ \det(\mathbf J - \mathbf I) = (-1)^{n-1} (n-1) $$
$$ (\mathbf J - \mathbf I)^{-1} = \left( \frac{1}{n-1} \right) \mathbf J - \mathbf I $$
The interested Reader might care to work out the first of these either from the aforementioned matrix determinant lemma or from evaluation of the characteristic polynomial of $\mathbf J$. The second of these can be verified by multiplying things out, and it can be derived by the Woodbury matrix identity.
Substituting these ingredients in the preceding expression gives:
$$ \begin{align*} \det A &= \det(\mathbf J - \mathbf I ) (1 + \mathbf y^T (\mathbf J - \mathbf I)^{-1} \mathbf x ) \\ &= (-1)^{n-1} (n-1) \left(1 + \mathbf y^T \left[\left(\frac{ 1}{n-1} \right) \mathbf J - \mathbf I \right] \mathbf x \right) \\ &= (-1)^{n-1} \left(n-1 + \mathbf y^T \left[ \mathbf J - (n-1)\mathbf I\right] \mathbf x \right) \\ &= (-1)^{n-1} \left(n-1 + \mathbf y^T \left[\left(\sum x_i \right) \mathbf 1 - (n-1) \mathbf x \right] \right) \\ &= (-1)^{n-1} \left(n-1 + \left[\left(\sum x_i \right) \left(\sum y_i \right) - (n-1) \mathbf y^T \mathbf x \right] \right) \\ &= (-1)^{n-1} \left[\left(\sum x_i \right) \left(\sum y_i \right) + (n-1) (1 -\mathbf y^T \mathbf x) \right] \end{align*} \\ $$
Let
$$\mathrm A := 1_n 1_n^\top - \mathrm I_n + \mathrm x \mathrm y^\top = - \mathrm I_n + \begin{bmatrix} | & |\\ \mathrm x & 1_n\\ | & |\end{bmatrix} \begin{bmatrix} | & |\\ \mathrm y & 1_n\\ | & |\end{bmatrix}^\top = - \left( \mathrm I_n - \begin{bmatrix} | & |\\ \mathrm x & 1_n\\ | & |\end{bmatrix} \begin{bmatrix} | & |\\ \mathrm y & 1_n\\ | & |\end{bmatrix}^\top \right)$$
Using the Weinstein-Aronszajn determinant identity,
$$\begin{array}{rl} \det (\mathrm A) &= (-1)^n \cdot \det \left( \mathrm I_n - \begin{bmatrix} | & |\\ \mathrm x & 1_n\\ | & |\end{bmatrix} \begin{bmatrix} | & |\\ \mathrm y & 1_n\\ | & |\end{bmatrix}^\top \right)\\ &= (-1)^n \cdot \det \left( \mathrm I_2 - \begin{bmatrix} | & |\\ \mathrm y & 1_n\\ | & |\end{bmatrix}^\top \begin{bmatrix} | & |\\ \mathrm x & 1_n\\ | & |\end{bmatrix}\right)\\ &= (-1)^n \cdot \det \begin{bmatrix} (1 - \mathrm y^\top \mathrm x) & - \mathrm y^\top 1_n\\ -1_n^\top \mathrm x & (1-n)\end{bmatrix}\\ &= (-1)^n \cdot \left( (1 - \mathrm y^\top \mathrm x) (1-n) - \mathrm y^\top 1_n 1_n^\top \mathrm x \right)\\ &= (-1)^n \cdot \left( \mathrm x^\top \left( (n-1) \, \mathrm I_n - 1_n 1_n^\top \right) \mathrm y + 1-n \right)\end{array}$$
