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This is for those of you who understand the Lindstrom-Gessel-Viennot lemma. I am looking for a proof of the following identity using paths and such:

Let $A$ be an $n\times n$ matrix, and for $i,j\in\{1,\ldots,n\}$, let $A^{ij}$ denote the matrix resulting from $A$ after removing row $i$ and column $j$, then:

$\det\left(\begin{array}{cccc}\det(A^{11})&\det(A^{12})&\cdots&\det(A^{1n})\\ \det(A^{21})&\det(A^{22})&\cdots&\det(A^{2n})\\ \vdots &\vdots &\ddots &\vdots\\ \det(A^{n1})&\det(A^{n2})&\cdots &\det(A^{nn})\end{array}\right)=\det(A)^{n-1}$

Read this for the algebraic proof:

Is this a well known determinant identity? Are there any generalizations?

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    You are right. For some reason, I assumed that we take consecutive minors and still miscalculated.2011-11-08

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