Prove this determinant identity combinatorially 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?
 A: I am very surprised that no solution was given yet.
Denote by $\mathrm{Com}(A)$ the comatrix of $A$. The determinant to be computed is the determinant of the product $D \times \mathrm{Com}(A) \times D$, where $D$ is the diagonal matrix with diagonal entries $((-1)^i)_{1 \le i \le n}$. Since $D^2=I_n$, this matrix is similar to $\mathrm{Com}(A)$, so it as the same determinant as $\mathrm{Com}(A)$.
Since $A \times \mathrm{Com}(A)^\top = (\det A) I_n$, we have $\det(A) \times \det(\mathrm{Com}(A)) = \det((\det A) I_n) = (\det A)^n$.
If $A$ is invertible, we derive $\det(\mathrm{Com}(A)) = (\det A)^{n-1}$.
If $A$ is not invertible and not null, then $\mathrm{Com}(A)$ cannot be invertible since $A \times \mathrm{Com}(A)^\top$ is null, so $\det(\mathrm{Com}(A))$ is null.
If $A$ is null and $n \ge 2$, then $\mathrm{Com}(A)$ is null.
If $A$ is null and $n=1$, then $\mathrm{Com}(A)$ is the $1 \times 1$ matrix with unique entry equal to $1$.
In all cases, we derive $\det(\mathrm{Com}(A)) = (\det A)^{n-1}$.
