[EDIT]
- For consistency, I switched the notations from $c(n)$ to $dc(n)$.
- It was asked by v s in the comments whether my answer generalize to higher dimensions. It does and gives an upper bound over any field:
$$dc(n)\le 2^n-1.$$
See my draft on this: An Upper Bound for the Permanent versus Determinant Problem.
[/EDIT]
[A side comment: I think you could edit your previous question instead of creating a new one.]
I have the following answer for you:
$$\operatorname{per}\begin{pmatrix} a&b&c\\d&e&f\\g&h&i\end{pmatrix}=
\det\begin{pmatrix}
0 & a & d & g & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & i & f & 0 \\
0 & 0 & 1 & 0 & 0 & c & i \\
0 & 0 & 0 & 1 & c & 0 & f \\
e & 0 & 0 & 0 & 1 & 0 & 0 \\
h & 0 & 0 & 0 & 0 & 1 & 0 \\
b & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}$$
Note that looking for such references about explicit examples, I could'nt find any and thus the example I give you is an example I built.
This question you are asking is commonly called the "Permanent vs Determinant problem". Suppose we are given an $(n\times n)$ matrix $A$, and we want the smallest matrix $B$ such that $\operatorname{per} A=\det B$. Let us denote by $dc(n)$ the dimensions of the smallest such $B$. Here are historical results:
- [Szegö 1913] $dc(n)\ge n+1$
- [von zur Gathen 1986] $dc(n)\ge n\sqrt 2-6\sqrt n$
- [Cai 1990] $dc(n)\ge n\sqrt 2$
- [Mignon & Ressayre 2004] $dc(n)\ge n^2/2$ in characteristic $0$
- [Cai, Chen & Li 2008] $dc(n)\ge n^2/2$ in characteristic $\neq 2$.
This shows that $5\le dc(3)\le 7$ (the upper bound is the matrix given above).
As I am lazy, I just give you one reference where you can find the other ones. It is the most recent paper I cited, by Cai, Chen and Li: A quadratic lower bound for the permanent and determinant problem over any characteristic $\neq 2$.
If you read French, you can also have a look to my slides on this subject: Permanent versus Déterminant.