Proof that the determinant of a $3 \times 3$ matrix is the volume of the parallelepiped spanned by the columns I recently read about how the cross product formula $\boldsymbol{a} \times \boldsymbol{b} = \begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\a_1&a_2&a_3\\b_1&b_2&b_3  \end{vmatrix}$ comes from the following:
Consider the parallelepiped spanned by a vector $\boldsymbol{x}$ and the constant vectors  $\boldsymbol{r}$ and $\boldsymbol{q}$. Its signed volume is then  $vol(\boldsymbol{x})=\begin{vmatrix}x_1&x_2&x_3\\r_1&r_2&r_3\\q_1&q_2&q_3\end{vmatrix}$. I didn't think much of this as I've always thought of the determinant as being the signed area/volume of whatever is spanned by the column (or row) vectors in the matrix. This is easily seen in the two-dimensions:

Using the fact that the volume as a function of $\boldsymbol{x}$ is a linear transformation, we know there exists a matrix $P=\begin{pmatrix}p_1&p_2&p_3\end{pmatrix}$ such that $P\boldsymbol{x} =vol(\boldsymbol{x})$. This is the same as stating that there exists a vector $\boldsymbol{p}=\begin{pmatrix}p_1\\p_2\\p_3\end{pmatrix}$ such that $\boldsymbol{p}\cdot\boldsymbol{x}=vol(\boldsymbol{x})$. Now we also know that $(\boldsymbol{r} \times \boldsymbol{q})\cdot\boldsymbol{x}=vol(\boldsymbol{x})$, so if we can solve for $\boldsymbol{p}$ then we have also found the cross product $\boldsymbol{r} \times \boldsymbol{q}$. By abusing notation and saying $\boldsymbol{x} = \begin{pmatrix}\textbf{i}\\\textbf{j}\\\textbf{k}\end{pmatrix}$ we solve this equation.
My question is this: is there a geometric interpretation of the formula for the 3x3 determinant like there is for the 2x2? I know you can compute the scalar triple product of three vectors and show that it's equivalent to computing the determinant of the matrix with those three vectors as columns (or rows), but this doesn't explain the motivation behind the formula the same way the graphic above does for the 2x2 case.
 A: Addition / subtraction of one column with another does not change determinant so the determinant of matrix $A$ is equal to the determinant of diagonal matrix $B$, columns of which are linear combinations of matrix $A$'s column:
$$
B_{\cdot i}=\sum{c_{ij}A_{\cdot j}}\phantom{x}\to\phantom{x}\left|B\right|=\left|A\right|
$$
Shifting any "surface" of an object parallel to any of its other edges does not change its $n$-volume. So the two objects given by their respective set of vertices below have the same $n$-volume:
$$
\sum{p_{i}A_{\cdot i}}\phantom{x},\phantom{x}p_{i}\in\{0,1\}\\
\sum{q_{i}B_{\cdot i}}\phantom{x},\phantom{x}q_{i}\in\{0,1\}
$$
Since $B$ is a diagonal matrix it is quite obvious that its determinant gives the $n$-volume of the second object above. Therefore, determinant of $A$ equals determinant of $B$ which equals to $n$-volume of the second object which is equal to the $n$ volume of the first object.
A: Generally, we have that if $A \in GL(n, \mathbb{R}^n)$ and $S \subset \mathbb{R}^n$ is Borel (or Lebesgue) measurable, then $vol(A(S)) = |\det(A)|vol(S)$. A heuristic proof is not difficult (for a rigorous proof, by Caratheodory's theorem, it suffices to consider the case when $S$ is a product of intervals, then the rest of this heuristic proof will work). Since $\det(E_1E_2) = \det(E_1)\det(E_2)$, it suffices to prove the theorem when $A$ is a row operation. If $A$ is of the form $Ae_j = c_j$, $c_j \neq 0$, then it is geometrically obvious that $vol(A(S)) = |c_1|\dots|c_n|vol(S)$. If $A$ is of the form $Ae_j = e_{\sigma(j)}$, $\sigma \in S_n$, then it is geometrically obvious that $vol(A(S)) = vol(S)$. If $A$ is of the form $Ae_1 = e_1 + ce_2$, $Ae_j = e_j$ for $j \geq 2$, then by Fubini's theorem, $vol(A(S)) = vol(S)$.
