Is this a well known determinant identity? Are there any generalizations? Let $A$ be a $3\times3$ matrix and for any $i,j\subseteq\{1,2,3\}$, let $A^{i,j}$ denote the $2\times2$ matrix resulting from removing row $i$ and column $j$ from $A$. Then:
$\det\left(\begin{array}{ccc}\det(A^{1,1})&\det(A^{1,2})&\det(A^{1,3})\\ \det(A^{2,1})&\det(A^{2,2})&\det(A^{2,3})\\ \det(A^{3,1})&\det(A^{3,2})&\det(A^{3,3})\end{array}\right)=\det(A)^2$
 A: Let $A$ be an $n \times n$ matrix. Let $C_{ij}$ be the $ij^{th}$ cofactor of $A$, defined by $C_{ij} = (-1)^{i+j} \det A^{ij}$ and let
$$
C = \begin{bmatrix}
    C_{11}  & C_{12} & \cdots &   C_{1n}   \\
    C_{21}  & C_{22} & \cdots &   C_{2n}   \\
  \vdots & \vdots & \ddots & \vdots \\ 
    C_{n1}  & C_{n2} & \cdots &  C_{nn}
\end{bmatrix}
$$
be the cofactor matrix. Also let $M$ be the matrix of minors; that is, the matrix that appears in the left hand side of the equation in the question. We will show that


*

*$\det C = (\det A)^{n-1}$, and

*$\det M = \det C$, 


which together establish the claim. 

Claim 1. From this wikipedia page, we know that 
$$
C^T A = A C^T = (\det A) I_n.
$$
If $A$ is invertible, then taking determinants on both sides, we get:
$$
\det (C^T) \cdot \det A = (\det A)^n \cdot 1.
$$
Since $\det (C^T) = \det C$ and $\det A \neq 0$, we get $\det C = (\det A)^{n-1}$, and we are done.
On the other hand, if $A$ is noninvertible, then $\det A = 0$ and 
$ C^T A = AC^T = 0$. In this case, (WLOG assuming $n \geq 1$) it suffices to show that $\det C = 0$, which is equivalent to showing that $C$ is noninvertible as well. Here, we have two subcases:


*

*If $A$ is the zero matrix, then $C$ is also the zero matrix, and hence non-invertible.

*Suppose $A$ is nonzero. Then if $C$ is invertible, then $C^T A = 0$ implies that $$A = \Big((C^{-1})^T C^T \Big) A = (C^{-1})^T \cdot 0 = 0 ,$$ which is a contradiction. Hence $C$ is noninvertible, and we are done. 

Claim 2. We now prove that $\det M = \det C$. We have the relation $C_{ij} = (-1)^{i+j} M_{ij} = (-1)^{i+j} \det A^{ij}$. So, if we take the matrix $M$, and multiply the $i^{\rm th}$ row by $(-1)^i$ and $j^{\rm th}$ column by $(-1)^j$, then we end up with the matrix $C$. In this process, the determinant gets multiplied by 
$$
\left(\prod_{i=1}^n (-1)^i \right) \left(\prod_{j=1}^n (-1)^j \right) =  \left(\prod_{i=1}^n (-1)^i \right)^2 = \prod_{i=1}^n (-1)^{2i}  = 1.
$$
In other words, $\det C = \det M$. 
