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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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What happens algebraically if the two $n$s are distinct? –  Phira Nov 7 '11 at 12:24
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I'm not sure if I understand your comment. Do you mean if $A$ is an $m\times n$ matrix with $m\neq n$? In that case the determinant does not make sense and you would have to reformulate the equation. –  wircho Nov 7 '11 at 19:30
    
There is one $n$ which is the size of the square matrix $A$ and one $n$ which is the size of the square matrix of determinants. I am sure that there is an analogue identity for the sums of subdeterminants of a square matrix, but that is another story. –  Phira Nov 7 '11 at 19:32
    
You could consider the matrix of $k\times k$ minors (instead of $(n-1)\times(n-1)$ minors). This is an $\binom nk\times\binom nk$ matrix (instead of an $n\times n$ matrix). The determinant of this matrix of minors is also a power of $A$. I read that somewhere, but I don't remember the exact exponent. –  wircho Nov 7 '11 at 21:55
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Actually from the degree, the exponent is $\frac kn\binom nk$, isn't it? –  wircho Nov 8 '11 at 0:50

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