Determinant as the volume of a box in n-dimensions Below is a proof found in Gilbert Strang's book and here
Why is the determinant the volume of a parallelepiped in any dimensions?
that the determinant equals the volume of a box :

To find the volume of a box whose edges are given by a  set of vectors $\{v_{1},\ldots,v_{n}\}$ we apply Gram-Schmidt orthogonalization to $\{v_{1},\ldots,v_{n}\}$, so that
\begin{eqnarray*}
v_{1} & = & v_{1}\\
v_{2} & = & c_{12}v_{1}+v_{2}^{\perp}\\
v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp}\\
 & \vdots
\end{eqnarray*}
where $v_{2}^{\perp}$ is orthogonal to $v_{1}$; and $v_{3}^{\perp}$
is orthogonal to $span\left\{ v_{1},v_{2}\right\} $, etc.
Since determinant is multilinear, anti-symmetric, then
\begin{eqnarray*}
\det\left(v_{1},v_{2},v_{3},\ldots,v_{n}\right) & = & \det\left(v_{1},c_{12}v_{1}+v_{2}^{\perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp},\ldots\right)\\
 & = & \det\left(v_{1},v_{2}^{\perp},v_{3}^{\perp},\ldots,v_{n}^{\perp}\right)\\
 & =?& \mbox{signed volume}\left(v_{1},\ldots,v_{n}\right)
\end{eqnarray*}

Question: This only proves that the determinant of the original edges equals the volume of the new created box.
That is we don't show $\det(v_1,...,v_n)$ = signed volume $(v_1,...v_n)$.
What we do prove is $\det(v_1,...,v_n)$ = Volume of  the box with orthogonal edges.
 A: A sketch: Consider a parallelepided spanned by vectors $(v_1,v_2,...,v_n)$:
$$ K(v_1,v_2,...,v_n)= \{ \sum_{k=1}^n t_k v_k : 0 \leq t_k\leq 1, 1\leq k\leq n\}$$
Given $c\in [0,1]$ consider an elementary shear of this solid in the 1-2 direction:  $K'=K(v_1,v_2-cv_1,v_3,...,v_n)$. Note that the directions $v_3,...,v_n$ are left untouched. $K'$ may be cut into two pieces $K'=(K'\cap K) \cup (K'\setminus K)=: K_1\cup K_2$ (make a drawing). If you translate $K_2$ by the vector $-v_1$ you almost get $K\setminus K'$. True for the interior but not for the borders which, however, are of zero volume.
$K$ is in this way reconstructed from $K'$ by cutting the latter into two pieces and translating one of them (up to zero volume corrections).
Thus, the volume of $K$ and $K'$ are the same and as you see by inspection, the determinants as well. Repeating the operation and also using the inverse you see that the same holds true for all $c\in {\Bbb R}$.
The transformation described by Strang may be decomposed into $n(n-1)/2$ such operations.
