$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$, As $A^{-1} \times A = I$, $\begin{bmatrix} x_{11} & x_{12 }&x_{13} \\ x_{21} & x_{22 }&x_{23} \\ x_{31} & x_{32 }&x_{33} \end{bmatrix}\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
From this we can see that:
$ax_{11}+dx_{12}+gx_{13}=1$, $ax_{21}+dx_{22}+gx_{23}=0$,
$ax_{31}+dx_{32}+gx_{33}=0$
$bx_{11}+ex_{12}+hx_{13}=0$, $bx_{21}+ex_{22}+hx_{23}=1$,
$bx_{31}+ex_{32}+hx_{33}=0$
$cx_{11}+fx_{12}+ix_{13}=0$, $cx_{21}+fx_{22}+ix_{23}=0$,
$cx_{31}+fx_{32}+ix_{33}=1$
$\det A = a(ei-fh)-b(di-fg)+c(dh-eg)$
Using the above, can we prove a formula matrix for inverting 3x3 matrices?
\begin{eqnarray*} A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \end{eqnarray*} Assuming $aei-afh+bfg-bdi+cdh-ceg \neq 0$ \begin{eqnarray*} A^{-1} =\frac{1}{aei-afh+bfg-bdi+cdh-ceg} \begin{bmatrix} ei-fh & ch-bi & bf-ce \\ fg-di & ai-cg & cd-af \\ dh-eg & bg-ah & ae-bd \end{bmatrix}. \end{eqnarray*}