Simplifying the determinant of the matrix whose $(i,j)$-th entry is $b_i c_j$ for $i=j$ and $-b_i c_j$ for $i\neq j$ 
$A$ is a $n \times n$ real matrix.
$A_{ij} = \begin{cases} \phantom{-}b_{i}c_{j} & \text{if } i = j \\ 
-b_{i}c_{j} & \text{if } i \ne j  \end{cases}$
How to simplify $\det(A)$?


Update:
Can I simplify the determinant with elementary row and column operations as described at http://www.maths.nuigalway.ie/~rquinlan/MA203/section2-5.pdf?


*

*Divide each row by $b_{i}$ (elementary row operation)

*Divide each column by $c_{j}$ (elementary column operation)


$\det(A) = \left(\prod_{i=1}^{n} b_{i} \right) \left(\prod_{j=1}^{n} c_{j} \right) \det(S)$
where $S_{ij} = \begin{cases} + 1 & \text{if } i = j \\ 
-1 & \text{if } i \ne j  \end{cases}$

So, the problem reduces to finding $\det(S)$.

 A: For $n\ge 3$ there is a clear pattern. For $n=3$ we have
$$
\det(A)=- 4b_1b_2b_3c_1c_2c_3.
$$
For $n=4$ we have
$$
\det(A)=- 16b_1b_2b_3b_4c_1c_2c_3c_4.
$$
So we should have $\det(A)=-f(n)\prod_{i=1}^n b_i\prod_{i=1}^ n c_i$ for all $n\ge 3$ with a positive integer $f(n)$.
We have
$$
f(3)=4,\; f(4)=16,\; f(5)=48,\; f(6)=128,\; f(7)=320.
$$
Conjecture: $f(n)=2^n(n-1)$.
A: We can write $A = BSC$, where $S$ is the matrix you describe and
$$
B = \pmatrix{b_1\\ & b_2 \\ && \ddots\\ &&& b_n}, \quad 
C = \pmatrix{c_1\\ & c_2 \\ && \ddots\\ &&& c_n},
$$
so that as you noted, we have $\det(A) = \det(B)\det(C) \det(S)$.
In order to compute $\det(S)$, it suffices to note that
$$
S = 2I - \pmatrix{1 & \cdots & 1\\
\vdots & \ddots & \vdots \\
1 & \cdots & 1},
$$
which is to say that $S$ is a rank-one update of a scalar matrix.  One method to compute the determinant of such a matrix is by considering its eigenvalues: because $S$ has eigenvalues $2$ with multiplicity $n-1$ and $2-n$ with multiplicity $1$, we compute
$$
\det(S) = 2^{n-1}(2-n).
$$
A: Diedrich Burde's conjecture (with $n$ shifted by one) is confirmed.
After pulling out the $b_i$'s and $c_i$'s, what's left is a circulant matrix whose first row is $[1, -1, -1, \ldots, -1]$, and associated polynomial $f(x)=1-x-x^2-\cdots-x^{n-1}$.  The determinant of a circulant matrix is known to be $\prod_{j=0}^{n-1} f(\omega_j)$, where $\omega_j$ are the $n$ different complex $n$-th roots of unity, i.e. $\omega_j=\exp(\frac{2j\pi i}{n})$.
Now, $f(\omega_0)=f(1)=1-1-1-\cdots-1=2-n$.  For all other $j$, $\omega_j\neq 1$.  We have $$f(x)=2-(1+x+x^2+\cdots + x^{n-1})=2-\frac{x^n-1}{x-1}$$
and hence, for $j>0$, $f(\omega_j)=2-\frac{0}{\omega_j-1}=2$.
Substituting into our determinant formula, we get $$\prod_{j=0}^{n-1}f(\omega_j)=f(\omega_0)\prod_{j=1}^{n-1}f(\omega_j)=(2-n)\prod_{j=1}^{n-1} 2=(2-n)2^{n-1}$$
