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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.

Noldorin
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    I got here by having the same question as you. I'm using Larson's linear algebra book and it does give the formula $\pm \frac{1}{6}| \cdot |$ for the volume of a tetrahedron. Maybe it will generalize to $\pm \frac{1}{n!}| \cdot |$ for the n dimensional version? – BLUC Sep 30 '20 at 07:57
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    @BLUC yep, that sounds quite probable! – Noldorin Sep 30 '20 at 14:39
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    @BLUC Sorry, thought I already had. Done now. – Noldorin Oct 04 '20 at 00:58
  • @Jean Marie, I rolled back your edit, because the use of the double lines is quite deliberate — we want the absolute value of the determinant, not merely the determinant. – Noldorin May 23 '22 at 13:57
  • I have understood it afterwards. I have two commentaries 1) The first interpretation of these double bars is misleading because it evokes a "norm" or (as in old books, especially russian books) evokes a matrix 2) I would have preferred that the absolute values are abandonned by using the so important concept of signed area. – Jean Marie May 23 '22 at 19:59
  • @JeanMarie I agree it is slightly ambiguous, though it also quite conventional notation, I find. (Depends on the country, in part.) – Noldorin May 23 '22 at 20:08

2 Answers2

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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 $x_1 \widehat i+y_1 \widehat j+z_1\widehat k$, $x_2 \widehat i+y_2 \widehat j+z_2\widehat k$, $x_3 \widehat i+y_3 \widehat j+z_3\widehat k$ $$V={\left| \begin{matrix} x_1&x_2&x_3\\ y_1&y_2&y_3\\ z_1&z_2&z_3\\ \end{matrix}\right|}\tag{volume of parallelopiped}$$

rash
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  • Thank you, that's good to know. Do you have a good proof of the generalised formula, by any chance? – Noldorin Sep 30 '20 at 14:49
  • @Noldorin I searched up the generalisation online and wrote the answer as you can see the generalised formula here is quite intuitive. But the proof of this is beyond my knowledge and I am not able to understand it. You may have to search it up yourself. Sorry! :( – rash Sep 30 '20 at 15:29
  • yeah, I can’t find it online, but thanks anyway. – Noldorin Sep 30 '20 at 17:03
  • I accepted the other answer since he provided a link to the proof in the end, but I appreciate yours! Thanks. – Noldorin Sep 30 '20 at 21:03
  • P.S. You might want to clarify that in that formula for the volume of a parallelepiped, the column vectors are not position vectors (coordinates), but rather the lengths of three edges (stemming from a single vertex). – Noldorin Sep 30 '20 at 21:05
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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.

BLUC
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    Thanks! Now I'm curious about the proof. – Noldorin Sep 30 '20 at 14:50
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    The proof is in the article at https://www.jstor.org/stable/2315353. If you have trouble accessing it, give me an email or something similar and I'll send you a copy. – BLUC Sep 30 '20 at 20:20
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    Got it, thanks very much. Will accept your answer. (Feel free to edit in a link to the proof, by the way.) – Noldorin Sep 30 '20 at 21:02
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    If you use "det", you don't need to add bars... unless they mean absolute values but in this case it is very confusing... – Jean Marie May 23 '22 at 04:30