What concise function $f$ satisfies $\prod_{i=1}^n \prod_{j=1}^n a_{i,j} = f \left( \det(\mathbf{A}) \operatorname{Adj}(\mathbf{A})^{-1} \right)$?

Question

Let $\mathbf{A} = [a_{i,j}]$ be an $n \times n$ matrix with determinant $\det (\mathbf{A})$. Consider the equality

$$\prod_{i=1}^n \prod_{j=1}^n a_{i,j} = f \left(\det(\mathbf{A}) \operatorname{Adj}(\mathbf{A})^{-1} \right)$$

where $f$ computes the product of the elements $\det(\mathbf{A}) \operatorname{Adj}(\mathbf{A})^{-1}$.

Since $\mathbf{A} = \det(\mathbf{A}) \operatorname{Adj}(\mathbf{A})^{-1}$, I tried to expand this expression. The product of the entries on the right would equal the product of the entries on the left... But I drowned in a huge puddle of series and products. I (unwisely) tried to use the Laplace expansion theorem.

Is there a concise expansion formula to express $f$?

Answer 1

Assume there is such a function $f$ and let $$ A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ~~~~~~~~~~~~ B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} $$

then we have $\det A = \det B = 1$ and the product of the entries is $0$ for $A$ and $2$ for $B$. We therefore get $$ 0 = f(\det A) = f(1) = f(\det B) = 2 $$ which is cotradictory.

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Comment: This was an answer to the first version of the question, which was whether there is a function which satisfies $\Pi_{i,j} a_{i,j} = f (\det A)$.