A rational map $\varphi:X\dashrightarrow Y$ (written with a dashed arrow because it is not defined at every point) between two (irreducible) algebraic varieties over $\mathbb{C}$, is said to be dominant if $\varphi(X)$ is dense in $Y$ for the Zariski's topology.
For an integer $n\geqslant 2$ identify the space of all $n\times n$ matrices $A=(a_{i,j})$ with $\mathbb{C}^{n^2}$. Denote by $\Delta_{j,k}$ the determinant of the $(n-1)\times(n-1)$-matrix obtained from $A$ deleting its $j$-th row and $k$-th column. I'm trying to prove (or disprove) the following proposition:
For all $k=1,\ldots,n$ the association $A=(a_{i,j})\mapsto \Big(a_{k,i}\frac{(-1)^{k+j}\Delta_{j,k}}{\det A}\Big)_{i,j}$ defines $\pi_k:\mathbb{C}^{n^2}\dashrightarrow \mathbb{C}^{n^2}$, which is a dominant rational map.
The obvious strategy of equating the coordinates of $\pi_k$ to an element $(\alpha_{i,j})$ of $\mathbb{C}^{n^2}$ and eliminating $\det\, A$ leads in too many relations ($n^4-n^3$, I'd say) involving the $a's$, $\Delta's$ and $\alpha's$ that cannot be obviously reduced. Hence it is not clear to me how to identify the open set (if any) on the target at which $\pi_k$ surjects.
Any idea? Thank you.
I just realized that the above maps are not dominant. In the case $n=2$ already, say for $k=1$: $$\pi_1\Big(\begin{equation*} \begin{pmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ \end{pmatrix} \end{equation*}\Big)= \frac{1}{\det A}\begin{pmatrix} a_{1,1}\Delta_{1,1} & -a_{1,1}\Delta_{2,1} \\ a_{1,2}\Delta_{1,1} & -a_{1,2}\Delta_{2,1} \\ \end{pmatrix} = \frac{a_{1,1}a_{1,2}}{\det A}\begin{pmatrix} \Delta_{1,1} & -\Delta_{2,1} \\ \Delta_{1,1} & -\Delta_{2,1} \\ \end{pmatrix}. $$
Hence the image of $\pi_1$ is contained in the hypersurface of $\mathbb{C}^{2^2}=\mathbb{C}^{4}$ determined by the equation $\det=0$. It follows that the image is not dense for the Zariski topology.
Similar reasoning shows that for higher $n$ and all $k=1,\ldots,n$ that the $\pi_k$ are not dominant.