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Given $N$ $n \times n$ matrices $\mathsf{A}^{1}, \dots, \mathsf{A}^{N}$, \begin{align} \det \left( \sum_{i = 1}^{N} \mathsf{A}^{i} \right) = \sum_{\sigma \in S} \det \mathsf{A}^{\sigma}, \end{align} where $S = \{ \sigma \colon \{ 1, \dots, n \} \mapsto \{ 1, \dots, N \} \}$ and $(\mathsf{A}^{\sigma})_{ij} = \mathsf{A}_{ij}^{\sigma(i)}$.

Where is this beautiful result published?

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does this apply to complex matrices also? –  Benjamin Sep 10 at 17:44
    
also, is $S$ the symmetric group? –  Benjamin Sep 10 at 17:58

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This is a straight-forward application of the linearity of the determinant on the rows of the matrix. Call the sum of the matrices $\Sigma_A$. Notice that each row of $\Sigma_A$ is a sum of the corresponding rows of each $A^i$. If we split the determinant along the first row, then we have $$\det\left(\Sigma_A\right) = \sum_{i=1}^N\det\left(\Sigma_A^{(i)}\right)$$ where each $\Sigma_A^{(i)}$ is the matrix obtained from $\Sigma_A$ by replacing the first row with the first row of $A^i$. Notice now that the second row (and on wards) of each $\Sigma_A^{(i)}$ remains a sum of rows. Splitting each $\Sigma_A^{(i)}$ along the second row now gives $$\det\left(\Sigma_A^{(i)}\right) = \sum_{j=1}^N\det\left(\Sigma_A^{(i,\,j)}\right)$$ where $\Sigma_A^{(i,\,j)}$ has first row of $A^i$ and second row of $A^j$. Substituting this new expression into our original summation, we have that $$\det\left(\Sigma_A\right) = \sum_{i=1}^N\sum_{j=1}^N\det\left(\Sigma_A^{(i,\,j)}\right)$$ or more compactly $$\det\left(\Sigma_A\right) = \sum_{(k_1,\ k_2)}\det\left(\Sigma_A^{(k_1,\,k_2)}\right)$$ Where the sum ranges over all $2$-tuples with $1\le k_1,\ k_2 \le N$. Continuing in the obvious fashion, we eventually reach $$\det\left(\Sigma_A\right) = \sum_{(k_1,\ \cdots, k_n)}\det\left(\Sigma_A^{(k_1,\,\cdots,k_n)}\right)$$ where the sum ranges over all $n$-tuples. Of course $\Sigma_A^{(k_1,\,\cdots,k_n)}$ is the matrix with row $i$ from $A^{k_i}$. This is exactly your desired sum.

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