If we have a $3\times3$ matrix $A=[A^i_j]$, show that
\begin{equation}
\epsilon^{ijk}\epsilon_{pqr}\text{det}A=
\begin{vmatrix}
A^i_p & A^i_q & A^i_r\\
A^j_p & A^j_q & A^j_r\\
A^k_p & A^k_q & A^k_r\\
\end{vmatrix}
\end{equation}
In the equation, $\epsilon^{ijk}$ and $\epsilon_{pqr}$ are the components of the permutation tensor P.
$\epsilon^{ijk}=(\boldsymbol{g^i} \times \boldsymbol{g^j}) \cdot \boldsymbol{g^k}$
$\epsilon_{pqr}=(\boldsymbol{g_p} \times \boldsymbol{g_q}) \cdot \boldsymbol{g_r}$
where {$\boldsymbol{g^{i,j,k}}$} and {$\boldsymbol{g_{p,q,r}}$} are the general reciprocal base and base vectors in space.
Already known and tries:
$\epsilon^{ijk}\epsilon_{pqr}=
\begin{vmatrix}
\delta^i_p & \delta^i_q & \delta^i_r\\
\delta^j_p & \delta^j_q & \delta^j_r\\
\delta^k_p & \delta^k_q & \delta^k_r\\
\end{vmatrix}$, where $\delta$ is the Kronecker delta.
And I know that
$(\boldsymbol{g^i} \times \boldsymbol{g^j}) \cdot \boldsymbol{g^k} =
\begin{vmatrix}
g^{ix} & g^{iy} & g^{iz}\\
g^{jx} & g^{jy} & g^{jz}\\
g^{kx} & g^{ky} & g^{kz}\\
\end{vmatrix}$
The result is intuitively direct to me, but I don't know how to prove it in mathematical language. Thank you!