Understanding why $\det(A) = \det(A^T)$ via the 3D Paralleliped

Question

I am trying to understand why, geometrically, we have that

$$ \det(A) = \det(A^T). $$

To build intuition, I am thinking in 3 dimensions. So let $A$ be a $3 \times 3$ matrix of real numbers.

First, I know that if

$$ \det(A) = \left| \begin{array} xx_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array} \right| $$

then $|\det(A)|$ can be thought of as the area of a parallelepiped formed by $(x_1, x_2, x_3)$, $(y_1, y_2, y_3)$, and $(z_1, z_2, z_3)$.

Moreover, the fact that $\det(A) = \det(A^T)$ implies that $|\det(A)|$ can also be thought of as the area of the parallelepiped formed by

$(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$.

Question: Does this also mean that the area of a parallelepiped formed by any permutation of the $x_i$, $y_i$, and $z_i$ is equal to $|\det(A)|$? For example, does the area of a parallelepiped formed by the coordinates $(x_1, y_2, z_3)$, $(x_2, y_1, z_2)$, and $(x_3, y_3, z_1)$ equal $|\det(A)$|?

Answer 1

Short answer is no.

For example suppose that $A$ is the identity matrix. The vectors are then $(x_1, y_1, z_1)=(1,0,0)$, $(x_2, y_2, z_2)=(0,1,0)$, and $(x_3,y_3,z_3)=(0,0,1)$. And $\text{det}(A)=1$. However, you could make a permutation $(x_1, y_2, z_3)=(1, 1, 1)$ and $(x_2, y_1, z_1)=(0, 0, 0)$ and $(x_3, y_3, z_2)=(0, 0, 0)$. The volume enclosed by those vectors is certainly not 1--in fact it is $0$.

Answer 2

Here is a 2D example.

$$\begin{pmatrix} 2&3\\ 1&4\\ \end{pmatrix} \mapsto_{transpose} \begin{pmatrix} 2&1\\ 3&4\\ \end{pmatrix}$$

I'm still thinking of a satisfying (geometric) explanation as to why this always works, but pictures are always good.