Is there any matrix representation for the second order derivation of a determinant? $\newcommand{\piff}[2]{\frac{\partial{#1}}{\partial{#2}}}\newcommand{\ppiff}[3]{\frac{\partial^2 {#1}}{\partial{#2}\partial{#3}}}\newcommand{\pa}[2]{\ppiff{|A|}{#1}{#2}}$
Suppose matrix
$$A = \begin{pmatrix}
x_{11} & x_{12} & \cdots & x_{1n}\\
x_{21} & x_{22} & \cdots & x_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
x_{n1} & x_{n2} & \cdots & x_{nn}
\end{pmatrix}$$
Consider $\det{A}$, denoted $|A|$, as a $n^2$-variable function of variable $x_{ij}(1\le i,j \le n)$.
I've found that
$$\piff{|A|}{A} = \begin{pmatrix}
\piff{|A|}{x_{11}} & \piff{|A|}{x_{12}} & \cdots & \piff{|A|}{x_{1n}}\\
\piff{|A|}{x_{21}} & \piff{|A|}{x_{22}} & \cdots & \piff{|A|}{x_{2n}}\\
\vdots & \vdots & \ddots & \vdots\\
\piff{|A|}{x_{n1}} & \piff{|A|}{x_{n2}} & \cdots & \piff{|A|}{x_{nn}}
\end{pmatrix} = C$$
where $C$ is the cofactor matrix.
Is there any matrix representation for follwing $n^2$ by $n^2$ matrix?
$$
\begin{pmatrix}
\pa{x_{11}}{x_{11}} & \pa{x_{11}}{x_{12}} & \cdots\cdots & \pa{x_{11}}{x_{nn}} \\
\pa{x_{12}}{x_{11}} & \pa{x_{12}}{x_{12}} & \cdots\cdots & \pa{x_{12}}{x_{nn}} \\
\vdots & \vdots & \ddots & \vdots \\
\pa{x_{nn}}{x_{11}} & \pa{x_{nn}}{x_{12}} & \cdots\cdots & \pa{x_{nn}}{x_{nn}}
\end{pmatrix}
$$
 A: For convenience, define the variables
$$\alpha = \det(A),\quad a={\rm vec}(A),\quad c={\rm vec}(C)$$
From your previous result
$$\frac{\partial\alpha}{\partial A} = \alpha A^{-T} = C$$
Continue on to find the differential and gradient of $C$.
$$\eqalign{
dC
 &= A^{-T}d\alpha + \alpha\,dA^{-T} \cr
 &= C\alpha^{-1}\,d\alpha + \alpha\,dA^{-T} \cr
 &= C\alpha^{-1}(C:dA) - \alpha\,A^{-T}\,dA^T\,A^{-T} \cr
}$$
Vectorize, i.e. flatten all the $n\times n$ matrices into $n^2\times 1$ vectors.
$$\eqalign{
dc &= \alpha^{-1}c(c^Tda) - \alpha\,(A^{-1}\otimes A^{-T})K\,da \cr
\frac{\partial c}{\partial a}
 &= \alpha^{-1}cc^T - \alpha\,(A^{-1}\otimes A^{-T})K \cr
 &= \alpha^{-1}\Big(cc^T - \big(C^T\otimes C\big)K\Big) \cr
}$$
This result is the desired $n^2\times n^2$ matrix.
And $K$ is the commutation matrix associated with the vec-operation.
Update
The above result assumes that $\alpha\ne 0$, but it can be rearranged into a form valid for all $\alpha$, and with a more symmetric appearance.
$$\eqalign{
\alpha\,\Big(\frac{\partial c}{\partial a}\Big)
 &= \Big(c^T\otimes c \,-\, \big(C^T\otimes C\big)K\Big) 
 \, = \alpha\,\bigg(\frac{\partial^2 \alpha}{\partial a\,\partial a^T}\bigg) \cr
}$$
