Extra parenthesis
\begin{equation}\label{A-l-lambda}
A^{(\lambda)}_l =\sum_{I_l \subseteq\mathbf{n}}\per \mathbf{A}
^{(\lambda)}(I_l }I_l )\det\mathbf{A}^{((\lambda)}
(\overline I_{l}}\overline I_l ),}I_{l}}=l .
\end{equation}
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