Expressing the determinant of a restricted map in terms of its cofactor matrix $\newcommand{\Cof}{\operatorname{Cof}}$
Consider an invertible linear map $A:\mathbb{R}^d \to \mathbb{R}^d$. 
Let $V \subseteq \mathbb{R}^d$ be a $(d-1)$-dimensional subspace, and set $ W = A(V) \subseteq \mathbb{R}^d$. Suppose we also have a preferred orientation on $ V$ (this amount to choosing a direction of the normal vector, since $ V$ is of codimension $1$).
Consider the restriction $A|_V:V \to W$. This is a linear map between two oriented inner product spaces (I endow $W$ with the orientation which makes the $A|_V$ orientation-preserving*), so it has a well-defined notion of determinant (the ratio between the pullback volume form on $W$, and the volume form on $V$).
I am trying to express $\det (A|_V)$ in terms of the cofactor** of $A$.
Is the following statement true?

Conjecture: Let $V \subseteq \mathbb{R}^d$ be a $(d-1)$-dimensional subspace. Then $\det(A|_V)= \|\Cof A(v^{\perp})\|_{\mathbb{R}^d}$, where $v^{\perp}$ is a unit vector which spans $V^\perp$.

If this is false, is there any other way to express $\det (A|_V)$ in terms of $\Cof A$?
Here is a proof of the two dimensional case:
Set $A= \begin{pmatrix} a & b \\\ c & d \end{pmatrix}$, then
$\operatorname{Cof}A= \begin{pmatrix} d & -c \\\ -b & a \end{pmatrix}$.
Suppose $V=\operatorname{span}\{v\}$ where $v=(x,y)$ is a unit vector in $\mathbb{R}^2$. Since the domain and codomain of $A|_V$ are one-dimensional, $$\det(A|_V)=\frac{\|Av\|}{\|v\|}=\frac{\|(ax+by,cx+dy)\|}{\| (x,y)\|}=\|(ax+by,cx+dy)\|. \tag{1}$$
Let $v^{\perp}=(-y,x) \in V^{\perp}$. Then $$ \|\Cof A(v^{\perp})\|=\|(-cx-dy,ax+by)\|  .\tag{2}$$
Combine equations $(1),(2)$.

*Alternatively, we don't have to select an orientation on $W$, and ask what is $|\det(A|_V)|$ which is defined regardless any orientation on $W$. (In that case we do not need $A$ to be invertible).
**The cofactor of $A$ is defined as follows:
$$ (Cof A)_{ij}=(-1)^{i+j}M_{ij},$$ where $M_{ij}$ is the $i-j$ minor  of $A$.
 A: I think it's probably true, here is a sketch of how one might prove it.  Suppose the standard orthonormal basis $ e_1, \cdots, e_d $, so that the volume form (multivector) on $ \mathbb{R}^d $ is $ \text{vol} = e_1 \wedge \cdots \wedge e_d $. For a codimension one subspace $ V \subset \mathbb{R}^d $, with normal vector $ v^{\perp} $, the volume form on $ V $ can be expressed as
$$
\text{vol}_V = \sum_i \pm v^{\perp}_i \; e_1 \wedge \cdots \widehat{e_i} \cdots \wedge e_d
$$
(since $ v^{\perp} \wedge \text{vol}_V  = \text{vol} $).
Now, you can check that for a $(d-1)$-multivector, the action of $ A $ is precisely given by minors
$$
A_* ( e_1 \wedge \cdots \widehat{e_i} \cdots \wedge e_d ) = \sum_j \pm M_{ij} ( e_1 \wedge \cdots \widehat{e_j} \cdots \wedge e_d ) 
$$
So that action of $ A $ on $ \text{vol}_V$ is
$$
A_* \text{vol}_V = \sum_j \sum_i Cof(A)_{ij} v^{\perp}_i ( e_1 \wedge \cdots \widehat{e_j} \cdots \wedge e_d )\\ 
= \sum \left( Cof A(v^{\perp}) \right)_j \; ( e_1 \wedge \cdots \widehat{e_j} \cdots \wedge e_d )
$$
Now since the image of $ V $ is $ W $, $A_* \text{vol}_W $ must be proportional to the volume form on $ W $, $$ \sum \left( Cof A(v^{\perp}) \right)_j \; ( e_1 \wedge \cdots \widehat{e_j} \cdots \wedge e_d ) = A_* \text{vol}_V = \det(A |_V) \text{vol}_W = \\
 \det(A |_V) \sum_i \pm w^{\perp}_i \; e_1 \wedge \cdots \widehat{e_i} \cdots \wedge e_d. $$ Taking the norm gives the proportionality and so the determinant.
