A function that looks like determinant Let $A$ be the $n\times n$ matrix $(a_{ij})$. By Laplace formula, the cofactor expansion along the $j$th row is
$$\det(A)=\sum_{j=1}^n (-1)^{i+j}a_{ij}M_{ij}.$$
I'm studying the function
$$f(A)=\sum_{j=1}^n a_{ij}M_{ij}.$$
It's easy to show that $f(A)$ does not depend on the row of the expansion.
Is this function well known? Is there an alternative way to compute it? (I know this question is too general, but any known fact about $f$ can be helpful)
I've shown that $f(A)$ is the coefficient of the term $x^{[(n+1)^n-1]/n}$ of the polynomial $\prod_i p_i(x)$, where $p_i(x)=\prod_j a_{ij}x^{(n+1)^{j-1}}$. This is interesting but not so helpful to compute $f(A)$.
 A: (Edit: The OP has modified the definition of $f$. The following answer no longer applies.)

"It's easy to show that $f(A)$ does not depend on the row of the expansion."

Are you sure? Consider $A=\pmatrix{1&1\\ 1&-1}$. Expand along the first row, we get $f(A)=2$. Expand along the second row, we get $f(A)=0$. Your definition of $f$ does depend on the row, and I think the defining formula itself provides the easiest way to evaluate $f$.
A: This type of idea reminds of something I stumbled on quite some time ago.  I will outline a bit of the connection that I see here, but it might not be exactly what you are looking for.  
There is an old construction from Tensor analysis that in some sense generalizes the determinant on the level of tensors.  It is a monstrous little creature of sorts but it can be used to contract tensors invariantly and a number of fundamental results in classical invariant theory depend on it quite heavily.  Since your focus is on matrices, I will focus on that, but it is really a tensor operation so you need to be a little bit careful here
Let $A = \left(a_{ij}\right)$ be an $n \times n$ matrix.  One way of defining the determinant of $A$ is as follows:
$$
\det (A) = \sum\limits_{\sigma \in S(n)} \operatorname{sgn}\left(\sigma\right) a_{\sigma(1) 1} a_{\sigma(2) 2} \cdots a_{\sigma(n) n},
$$
where $S(n)$ is the permutation group on $\left\{1 ,2,\ldots, n\right\}$ and $\operatorname{sgn}\left(\sigma\right)$ is the sign of the permutation $\sigma$.
What isn't altogether clear when one first encounters this definition is whether or not the choice of row or column matters in the expansion and one might be inclined to try something along the following lines:
$$
\sum\limits_{\left(\sigma_1 , \sigma_2\right) \in S(n) \times S(n)} \operatorname{sgn}\left(\sigma_{1}\right)\cdot \operatorname{sgn}\left(\sigma_{2}\right) a_{\sigma_{1}(1) \sigma_{2}(1)} a_{\sigma_{1}(2) \sigma_{2}(2)} \cdots a_{\sigma_{1}(n) \sigma_{2}(n)},
$$
and, if I am not mistaken, you will obtain $n! \det(A)$.  However, what you quickly realize in this example is that you have even more choices than you might have realized at first glance.  Namely, each term on the summation contains $2n$ indices and you can select where each of the permutations $\sigma_1$ and $\sigma_2$ will actually operate. 
Now, due to the growth of the permutation group as $n$ increases, these summations get out of hand in a hurry, but if you focus on $2 \times 2$ matrices and sums over $S_2 \times S_2$, you can get a nice feel for what is happening.
Consider the following examples (where $A = \left(a_{ij}\right)$ is a two by two matrix (you technically should to think of the $a_{ij}$ as forming the components of a covariant tensor as opposed to a matrix if you are thinking about what happens with a change of basis):
\begin{align*}
\sum\limits_{\left(\sigma_1 , \sigma_2\right) \in S(2) \times S(2)} \operatorname{sgn}\left(\sigma_{1}\right)\cdot \operatorname{sgn}\left(\sigma_{2}\right) a_{\sigma_{1}(1) \sigma_{2}(1)} a_{\sigma_{1}(2) \sigma_{2}(2)} &=\\
&= a_{11}a_{22} - a_{12}a_{21} - a_{21}a_{12} + a_{22}a_{11}\\
&= 2 \det (A),\\
\end{align*}
and
\begin{align*}
\sum\limits_{\left(\sigma_1 , \sigma_2\right) \in S(2) \times S(2)} \operatorname{sgn}\left(\sigma_{1}\right)\cdot \operatorname{sgn}\left(\sigma_{2}\right) a_{\sigma_{1}(1) \sigma_{1}(2)} a_{\sigma_{2}(1) \sigma_{2}(2)} &=\\
&= a_{12}a_{12} - a_{12}a_{21} - a_{21}a_{12} + a_{21}a_{21}\\
&= a_{12}^2 - 2a_{12}a_{21} + a_{21}^2,\\
\end{align*}
where the last equality holds under the assumption the ground field is commutative.
This type of operation shows up sometimes in geometry and physics and the physicists make quite a bit of use out of it using what is called the Levi-Civita Symbol (http://en.wikipedia.org/wiki/Levi-Civita_symbol).  This makes it clear it clear that it is really an operation on tensors.
To the best of my knowledge, there is no agreed upon generalization of the determinant to higher dimensional arrays (and I would be happy to hear otherwise).  I believe that Gelfand et. al. have one version, but I am not sure how well it is established Gelfand et. al.
