I would like to show that $\textsf f(e_1) \wedge \textsf f(e_2) I^{-1} = \det (\textsf f) \bar{\textsf f}^{-1} (e_3)$
in geometric algebra in 3 dimensions, as in Geometric Algebra for Physicists (Doran & Lasenby) pg. 222.
Attempt (focusing on the wedge product factor at first):
\begin{align*} \textsf f(e_1) \wedge \textsf f(e_2) &= \textsf f(e_1) \wedge \textsf f(e_2) \wedge e_3^2 \\ &= \textsf f(e_1) \wedge \textsf f(e_2) \wedge (\textsf f^{-1} \textsf f (e_3) \cdot e_3) \\ &= \textsf f(e_1) \wedge \textsf f(e_2) \wedge (\textsf f (e_3) \cdot \bar{\textsf f}^{-1}(e_3)) \\ &= \textsf f(e_1) \wedge \textsf f(e_2) \wedge (\textsf f (e_3) \bar{\textsf f}^{-1}(e_3) - \textsf f (e_3) \wedge \bar{\textsf f}^{-1}(e_3) ) \\ &= \textsf f(e_1) \wedge \textsf f(e_2) \wedge (\textsf f (e_3) \bar{\textsf f}^{-1}(e_3)) - I \wedge \bar{\textsf f}^{-1}(e_3) \det \textsf f \\ &= \textsf f(e_1) \wedge \textsf f(e_2) \wedge (\textsf f (e_3) \bar{\textsf f}^{-1}(e_3)) \end{align*}
$ \newcommand\lcontr{\mathbin\rfloor} \newcommand\adj\bar \newcommand\l\mathsf $Doran & Lasenby's equations (4.4.2) are the key. Another way to state them is $$ \l f(\adj{\l f}(X)\lcontr Y) = X\lcontr\l f(Y). $$ Now recalling that $XJ = X\lcontr J$ for any pseudoscalar $J$, apply $\adj{\l f}^{-1}$ to the equation $$ e_3 = e_1\wedge e_2\,I^{-1}. $$