Generalisation of formula for area of triangle in determinant form? It is well known that the area of triangle in the Euclidean plane is given by the formula
$$A = \dfrac 1 2 {\left| \begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{vmatrix} \right|},$$
where $(x_i, y_i)$ are the coordinates of the three vertices of the triangle.
I was wondering if this admits a generalisation to higher dimensions, since the standard proof of this formula (something along the lines of this) seems to result in a determinant almost accidentally.
For example, might the volume of a tetrahedron be given by the following?
$$A = \dfrac 1 2 {\left| \begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
x_4 & y_4 & z_4 & 1 \\
\end{vmatrix} \right|}.$$
I suspect this is too naive a generalisation, but I'd be curious how you generalise this determinant formula anyway, if possible.
 A: Good attempt on generalization but the volume of tetrahedron is
$$A = \dfrac 1 6 {\left| \begin{vmatrix}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
x_4 & y_4 & z_4 & 1 \\
\end{vmatrix} \right|}.$$
as you can see the formula extends as such
$$V = \frac{1}{n!} {\left| \begin{vmatrix}
x_1 & . & . & N_1&1 \\
. & . &  .& .&1 \\
. & . & .& .&1 \\
x_{n+1} & . & . &N_{n+1} &1 \\
\end{vmatrix} \right|}$$
where n $=$ number of dimensions and N denotes the $n^{th}$ dimension.
but this formula is only when a triangle sided shape extends its dimensions like you can see for triangle (2-D), tetrahedron (3-D), pentagonal tetrahedron and so on..., all of them have triangles as their sides.
Not all shapes of higher dimensions follow this trend as you can see with shapes like parallelopiped and so on...,
If the parallelopiped has the sides with direction cosines $x_1 \widehat i+y_1 \widehat j+z_1\widehat k$, $x_2 \widehat i+y_2 \widehat j+z_2\widehat k$, $x_3 \widehat i+y_3 \widehat j+z_3\widehat k$
$$V={\left| \begin{matrix}
x_1&x_2&x_3\\
y_1&y_2&y_3\\
z_1&z_2&z_3\\
\end{matrix}\right|}\tag{volume of parallelopiped}$$
A: According to wikipedia, the volume of an n-dimensional triangle ('simplex') determined by vertices $v_0, v_1, \ldots, v_n$ is given by
$$
\left| \frac{1}{n!}
\text{det}
\begin{pmatrix}
v_0 & v_1 & \cdots  & v_n \newline
1 & 1 & \cdots & 1
\end{pmatrix} \right|
$$
which uses the transpose of your suggested matrix with the resulting determinant scaled by $\frac{1}{n!}$.
For example, in four dimensions, you would have
$$
\begin{align*}
\left| \frac{1}{n!}
\text{det}
\begin{pmatrix}
v_0 & v_1 & v_2 & v_3 & v_4 \newline
1 & 1 & 1 & 1 & 1
\end{pmatrix} \right|
&=\left| \frac{1}{4!}
\text{det}
\begin{pmatrix}
x_0 & x_1 & x_2 & x_3 & x_4 \newline
y_0 & y_1 & y_2 & y_3 & y_4 \newline
z_0 & z_1 & z_2 & z_3 & z_4 \newline
w_0 & w_1 & w_2 & w_3 & w_4 \newline
1 & 1 & 1 & 1 & 1
\end{pmatrix} \right| \\
&=\left| \frac{1}{24}
\text{det}
\begin{pmatrix}
x_0 & y_0 & z_0 & w_0 & 1 \newline
x_1 & y_1 & z_1 & w_1 & 1 \newline
x_2 & y_2 & z_2 & w_2 & 1 \newline
x_3 & y_3 & z_3 & w_3 & 1 \newline
x_4 & y_4 & z_4 & w_4 & 1
\end{pmatrix} \right|
\end{align*}
$$
The proof involves using induction by using n = 2 as the base case and then deriving the n-dimensional formula from the (n-1)-dimensional formula for the general case with n > 2. The full proof is in P. Stein, A Note on the Volume of a Simplex, which is available at jstor.org/stable/2315353.
