Question about determinants and orthogonal complement. Prove that this vector belongs to $S^\perp$.

Question

This is one of my linear algebra questions:

Let $S=\{\textbf{v}_1,\textbf{v}_2,\cdots,\textbf{v}_{n-1}\}$ be a set of $n-1$ vectors in $\mathbb{R}^n$. Define an $n\times n$ matrix A such that the first column of A is $(\textbf{e}_1,\textbf{e}_2,\cdots,\textbf{e}_n)^T$, and the ith column of A is $\textbf{v}_{i-1}$ for $i=2,3,...,n$. We regard $\textbf{e}_1,\textbf{e}_2,\cdots,\textbf{e}_n$ as some variables and compute detA. The result obtained is a linear combination of $\textbf{e}_1,\textbf{e}_2,\cdots,\textbf{e}_n$, which can be interpreted as a vector in $\mathbb{R}^n$. Prove that this vector belongs to $S^\perp$.

I considered the case when n=3. Then it can be interpreted as the cross-product of $\textbf{v}_1,\textbf{v}_2$.
I also know that $S^\perp$ contains a nonzero vector. However, since I don't have explicit numbers for $\textbf{v}_1,\textbf{v}_2,\cdots,\textbf{v}_n$, I don't know how to proceed from here.

Can anyone help me? Thanks in advance!

[Edit]:

The notation here seemed confusing to me as well when I first saw this question. Actually, an explicit example is given.

(For example, if $n=3,\textbf{v}_1=(0,1,2)^T,\textbf{v}_2=(3,0,4)^T$, then we have $detA=\begin{vmatrix}\textbf{e}_1&0&3\\\textbf{e}_2&1&0\\\textbf{e}_3&2&4\end{vmatrix}=4\textbf{e}_1+6\textbf{e}_2-\textbf{e}_3=(4,6,-3)^T$)

Answer 1

Hint.

$$ \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} \cdot\, \det\begin{pmatrix} \mathbf{e}_1 & v_{11}~~ & \cdots & v_{1(n-1)}~~ \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{e}_n & v_{(n-1)1} & \cdots & v_{(n-1)(n-1)} \end{pmatrix} = \det \begin{pmatrix} a_1 & v_{11}~~ & \cdots & v_{1(n-1)}~~ \\ \vdots & \vdots & \ddots & \vdots \\ a_n & v_{(n-1)1} & \cdots & v_{(n-1)(n-1)} \end{pmatrix}. $$

Answer 2

Note: this answer is based on some interpretation of the question which is rather unclear. The main interpretation is that $\{\mathrm{e}_1, \dots,\mathrm{e}_n\}$ stands for the canonical basis in the question.

So let $V = \mathbb R^n$. The map

$$\begin{array}{l|rcl} \varphi : & V & \longrightarrow & \mathbb R \\ & x & \longmapsto & \det(x, v_1, \dots, v_{n-1}) \end{array} $$ is a linear form. We have $\operatorname{Span}\{v_1, \dots, v_{n-1}\} \subseteq \{\varphi\}^\perp$ as $\varphi(x)=0$ for $x \in \operatorname{Span}\{v_1, \dots, v_{n-1}\}$. Taking the annihilator of both sides of the above inclusion, we get

$$\mathbb R \varphi = \{\varphi\}^{\perp \perp} \subseteq \operatorname{Span}\{v_1, \dots, v_{n-1}\}^{\perp}$$ which is exactly meaning that the vector $\det((\mathrm{e}_1, \dots,\mathrm{e}_n)^T, v_1, \dots, v_{n-1})$ belongs to $S^\perp$ with the interpretation of the question that $\{\mathrm{e}_1, \dots,\mathrm{e}_n\}$ stands for the canonical basis.