The author of a book draft I am reading pointed out if a vector transforms as
\Bv \rightarrow M \Bv,

then cross products must transform as
\Ba \cross \Bb \rightarrow \lr{ \textrm{adj}\, M }^\T \lr{ \Ba \cross \Bb }.

Bivectors clearly must transform in the same fashion. We also noticed that the adjoint is related to the reciprocal frame vectors of the columns of $$M$$, but didn’t examine the reciprocal frame formulation of the adjoint in any detail.

Before we do that, let’s consider a slightly simpler case, the transformation of a pseudoscalar. That is
\begin{aligned}
M(\Ba) \wedge M(\Bb) \wedge M(\Bc)
&\rightarrow
\sum_{ijk}
\lr{ \Bm_i a_i } \wedge
\lr{ \Bm_j a_j } \wedge
\lr{ \Bm_k a_k } \\
&=
\sum_{ijk}
\lr{ \Bm_i \wedge \Bm_j \wedge \Bm_k } a_i b_j c_k \\
&=
\sum_{ijk}
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 } \epsilon_{ijk} a_i b_j c_k \\
&=
\Abs{M}
\sum_{ijk} \epsilon_{ijk} a_i b_j c_k \\
&=
\Abs{M} \lr{ \Ba \wedge \Bb \wedge \Bc }.
\end{aligned}

This is a well known geometric algebra result (called an outermorphism transformation.)

It’s somewhat amusing that an outermorphism transformation with two wedged vectors is a bit more complicated to express than the same for three. Let’s see if we can find a coordinate free form for such a transformation.
\begin{aligned}
M(\Ba) \wedge M(\Bb)
&=
\sum_{ij} \lr{ \Bm_i a_i } \wedge \lr{ \Bm_j b_j } \\
&=
\sum_{ij} \lr{ \Bm_i \wedge \Bm_j } a_i b_j \\
&=
\sum_{i < j} \lr{ \Bm_i \wedge \Bm_j }
\begin{vmatrix}
a_i & a_j \\
b_i & b_j
\end{vmatrix} \\
&=
\sum_{i < j} \lr{ \Bm_i \wedge \Bm_j } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_j \wedge \Be_i } }.
\end{aligned}

Recall that the reciprocal frame with respect to the basis $$\setlr{ \Bm_1, \Bm_2, \Bm_3 }$$, assuming this is a non-degenerate basis, has elements of the form
\begin{aligned}
\Bm^1 &= \lr{ \Bm_2 \wedge \Bm_3 } \inv{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 } \\
\Bm^2 &= \lr{ \Bm_3 \wedge \Bm_1 } \inv{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 } \\
\Bm^3 &= \lr{ \Bm_1 \wedge \Bm_2 } \inv{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }.
\end{aligned}

This can be flipped around as
\begin{aligned}
\Bm_2 \wedge \Bm_3 &= \Bm^1 \Abs{M} I \\
\Bm_3 \wedge \Bm_1 &= \Bm^2 \Abs{M} I \\
\Bm_1 \wedge \Bm_2 &= \Bm^3 \Abs{M} I \\
\end{aligned}

\begin{aligned}
M&(\Ba) \wedge M(\Bb) \\
&=
\lr{ \Bm_1 \wedge \Bm_2 } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } }
+
\lr{ \Bm_2 \wedge \Bm_3 } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_3 \wedge \Be_2 } }
+
\lr{ \Bm_3 \wedge \Bm_1 } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_1 \wedge \Be_3 } } \\
&=
I \Abs{M} \lr{
\Bm^3 \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } }
+
\Bm^1 \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_3 \wedge \Be_2 } }
+
\Bm^2 \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_1 \wedge \Be_3 } }
}
\end{aligned}

Let’s see if we can simplify one of these double index quantities
\begin{aligned}
I \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } }
&=
\gpgradethree{ I \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } } } \\
&=
\gpgradethree{ I \lr{ \Ba \wedge \Bb } \lr{ \Be_2 \wedge \Be_1 } } \\
&=
\gpgradethree{ \lr{ \Ba \wedge \Bb } \Be_{12321} } \\
&=
\gpgradethree{ \lr{ \Ba \wedge \Bb } \Be_{3} } \\
&=
\Ba \wedge \Bb \wedge \Be_3.
\end{aligned}

We have
M(\Ba) \wedge M(\Bb) = \Abs{M} \lr{
\lr{ \Ba \wedge \Bb \wedge \Be_1 } \Bm^1
+
\lr{ \Ba \wedge \Bb \wedge \Be_2 } \Bm^2
+
\lr{ \Ba \wedge \Bb \wedge \Be_3 } \Bm^3
}.

Using summation convention, we can now express the transformation of a bivector $$B$$ as just
B \rightarrow \Abs{M} \lr{ B \wedge \Be_i } \Bm^i.

If we are interested in the transformation of a pseudovector $$\Bv$$ defined implicitly as the dual of a bivector $$B = I \Bv$$, where
As intuited, the determinant weighted reciprocal frame vectors for the columns of the transformation $$M$$, are the components of the adjoint. That is