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I wanna find a closed form of determinant of the following matrix

$$A(n) = \begin{pmatrix} B_{1} & B_{2} & \cdots & B_{n} & 1 \\ B_{n} & B_{1} & \cdots & B_{n-1} &1 \\ \vdots & \vdots & \ddots & \vdots \\ B_{2} & B_{3} & \cdots & B_{1} & 1\\ A_{1} & A_{2} & \cdots & A_{n} &1 \end{pmatrix}$$

For instance :

for $n=3$

$$A(3) = \begin{pmatrix} B_{1} & B_{2} & B_{3} &1 \\ B_{3} & B_{1} & B_{2}&1 \\ B_{2} & B_{3} & B_{1}&1 \\ A_{1} & A_{2} & A_{3} & 1 \end{pmatrix}$$

and

$\det \bigg[A(n)\bigg]= \left(-A_1-A_2-A_3+B_1+B_2+B_3\right)\left(B_1^2-\left(B_2+B_3\right) B_1+B_2^2+B_3^2-B_2 B_3\right) $

I used Mathematica for some $n$ and I guessed that determinant of $A(n)$ has a form $$\color{blue}{\det \bigg[A(n)\bigg]=\bigg(\sum_{i=1}^{n}B_i-\sum_{i=1}^{n}A_i\bigg)\cdot F(B_k)}$$

where $f(B_k)$ is a function of only $B_k$, it doesn't depend on $A_k$.

How to prove this and how to find closed form of $F$ ?

Thank you in advance to any one who may be able to give me some ideas

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closed as off-topic by Ryan Budney, Franz Lemmermeyer, Stefan Kohl, Alex Degtyarev, András Bátkai Jun 25 '16 at 10:20

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  • 1
    $\begingroup$ The formula (without identifying $F$) is essentially obvious since $\det A$ is a degree $1$ polynomial in the $A_j$'s, and it's easy to confirm that $A_1 = \sum B_j - \sum_{j\not=1} A_j$ is the correct zero. $\endgroup$ – Christian Remling Jun 23 '16 at 21:24
  • 2
    $\begingroup$ $(\sum B_i)F(B_k)$ is the part of $\det A(n)$ which does not contain $A_k$. This is exactly the $(n+1,n+1)$ minor (i.e. determinant of the submatrix with first $n$ rows and first $n$ columns). That submatrix is a circulant matrix whose determinant is given in wiki. $\endgroup$ – Yuzhou Gu Jun 23 '16 at 22:46
  • $\begingroup$ @YuzhouGu why do you think that $(\sum B)F(B)$ is $\textbf{exactly}$ $(n+1,n+1)$ minor? $\endgroup$ – vito-ვიტო Jun 24 '16 at 12:01
  • $\begingroup$ Expand the determinant. $\endgroup$ – Yuzhou Gu Jun 24 '16 at 12:06
  • $\begingroup$ @YuzhouGu $\det A(n)=(\sum B)F(B)-(\sum A)F(B)$, then $(\sum B)F(B)$ certainly doesn't contain $A_k$. but it doesn't follows that it is exactly $(n+1,n+1)$ minor. I can construct new matrix from $A(n)$ whose determinant doesn't contain $A_k$ and it isn't a circulant matrix. $\endgroup$ – vito-ვიტო Jun 24 '16 at 12:23