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Find a formula for the permanent of the following square matrix with $-1$ on the main diagonal and $1$ everywhere else.

$$ A = \begin{bmatrix} -1 & 1 & 1 & \dots & 1 \\ 1 & -1 & 1 & \dots & 1 \\ 1 & 1 & -1 & \dots & 1 \\ \vdots & \vdots &\vdots & \ddots & \vdots \\ 1 & 1 & 1 & \dots & -1 \end{bmatrix} $$

I was able to find a recursive formula for it, but I am looking for a closed-form expression. How would I go about this problem? I computed the permanent of small matrices of that form, but I do not see a pattern.

Rodrigo de Azevedo
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Noam
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1 Answers1

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Let $A_n$ be an $ n \times n $ matrix where all diagonal elements are $-1$ and all off-diagonal elements are $1$. We compute its permanent:

$$ \text{perm}(A_n) = \sum_{\sigma \in S_n} \prod_{i=1}^n a_{i,\sigma(i)}. $$

Each term depends on the number of fixed points $ k $ of $ \sigma $. If $ \sigma $ has $ k $ fixed points, the product contributes $ (-1)^k $. The number of such permutations is: $$ \binom{n}{k} \cdot D_{n-k}, $$ where $ D_{n-k} = (n-k)! \sum_{i=0}^{n-k} \frac{(-1)^i}{i!} $ is the number of derangements (no fixed points). Substituting: $$ \text{perm}(A_n) = \sum_{k=0}^n \binom{n}{k} D_{n-k} \cdot (-1)^k = n! \sum_{k=0}^n \sum_{i=0}^{n-k} \frac{(-1)^k}{k!} \cdot \frac{(-1)^i}{i!}. $$ Changing variables $ m = k+i $, and using $ \sum_{k=0}^m \frac{1}{k!(m-k)!} = \frac{2^m}{m!} $, we simplify: $$ \text{perm}(A_n) = n! \sum_{m=0}^n \frac{(-2)^m}{m!}. $$

Unless I made a mistake somewhere again.

kabenyuk
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