I've been looking for a way to express the solution of
$A_{pi}x_i=w_p$
using the identity
$\epsilon_{pqr}det(A)=\epsilon_{ijk}A_{pi}A_{qj}A_{rk}$
by using the fact that $det(A)=\epsilon_{ijk}A_{1i}A_{2j}A_{3k}$ and end up with
$x_{i}=\frac{w_{p}A_{qj}A_{rk}}{\epsilon_{pqr}A_{1i}A_{2j}A_{3k}}$
However, I've seen somewhere else the identity $S^{-1}_{ji}=\frac{1}{2det(A)}\epsilon_{ipq}\epsilon_{jkl}A_{pk}A_{ql}$ and have begun to think this might be the identity I need to use in order to get the right answer. Could anyone shed some light on the correctness of my approach and the identity listed above? I feel like I'm missing a key relation between the determinant and its inverse (we assume here that the inverse does exist since $det(A)\ne 0$.
There's a matrix called the adjugate of $A$, with components given by:
$$\mathrm{adj}(A)_{ji}=\frac 12\epsilon_{ipq}\epsilon_{jkl}A_{pk}A_{ql}.$$
The matrix and its adjugate obey the property that
$$\mathrm{adj}(A)A=\mathrm{det}(A)I=A\mathrm{adj}(A)$$
and so when we have $\mathrm{det}(A)\neq 0$ we can divide through by $\mathrm{det}(A)$ and get that
$$A^{-1}=\frac{\mathrm{adj}(A)}{\mathrm{det}(A)}.$$
So in components we have
$$A^{-1}_{\;\;\,ji}=\frac{\epsilon_{ipq}\epsilon_{jkl}A_{pk}A_{ql}}{2\epsilon_{abc}A_{a1}A_{b2}A_{c3}}$$
and since $x=A^{-1}w$ we have
$$x_j=\frac{\epsilon_{ipq}\epsilon_{jkl}A_{pk}A_{ql}w_i}{2\epsilon_{abc}A_{a1}A_{b2}A_{c3}}$$