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The determinant of a $3\times 3$ matrix $\begin{vmatrix} 1 & 1 &1 \\ x & y & z \\ x^2 & y^2 &z^2 \\ \end{vmatrix} $ is the volume of a parallelopiped with its three sides as the vectors whose tails rest on origin and heads at the coordinates $(1,x,x^2),(1,y,y^2)$ and $(1,z,z^2)$ $^{[1]} $.

The determinant of this matrix can be simplified to $(x-y)(y-z)(z-x)$.

Proof:

Subtracting column 1 from column 2, and putting that in column 2,

\begin{equation*} \begin{vmatrix} 1 & 1 &1 \\ x & y & z \\ x^2 & y^2 &z^2 \\ \end{vmatrix} = \begin{vmatrix} 1 & 0 &1 \\ x & y-x & z \\ x^2 & y^2-x^2 &z^2 \\ \end{vmatrix} \end{equation*}

$ = z^2(y-x)-z(y^2-x^2)+x(y^2-x^2)-x^2(y-x) $

rearranging the terms,

$ =z^2(y-x)-x^2(y-x)+x(y^2-x^2)-z(y^2-x^2) $

taking out the common terms $(y-x)$ and $(y^2-x^2)$,

$ =(y-x)(z^2-x^2)+(y^2-x^2)(x-z) $

expanding the terms $(z^2-x^2)$ and $(y^2-x^2)$

$ =(y-x)(z-x)(z+x)+(y-x)(y+x)(x-z) $

$ =(y-x)(z-x)(z+x)-(y-x)(z-x)(y+x) $

taking out the common term $(y-x)(z-x)$

$ =(y-x)(z-x) [z+x-y-x] $

$ =(y-x)(z-x)(z-y) $

$ =(x-y)(y-z)(z-x) $

As the $x$ coordinate of the heads of these three vectors is $1$, the head of these vectors lies in a plane perpendicular to the $x$-axis and a distance of $1$ unit away from the origin.

(If we connect these three points, we get a triangle.)

This plane will cut the parallelopiped into two equal triangular pyramids whose base lies in the plane.

The perpendicular distance from the base of the pyramid to the tip is $1$ unit.

The volume of the required parallelogram is the sum of the volume of the two triangular pyramids.

$\text{volume of a pyramid}=\frac{1}{3}bh$.

The height is $1$ units.

The area of a triangle is, by Shoelace formula, $$A = \frac{1}{2} \begin{vmatrix} 1 & 1 &1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ \end{vmatrix} $$ where the vertices of the triangle are $(x_1,y_1),(x_2.y_2),(x_3,y_3)$ $^{[2]}$

The vertices of the required traingle has the coordinates $(x,x^2),(y,y^2)$ and $(z,z^2)$.

So the area of the triangle,

$$A=\frac{1}{2}\begin{vmatrix} 1 & 1 &1 \\ x & y & z \\ x^2 & y^2 &z^2 \\ \end{vmatrix}$$

which, as shown above, can be simplified to $\frac{1}{2} (x-y)(y-z)(z-x)$

So, the volume is $$\frac{1}{3}bh=\frac{1}{3}\times\frac{1}{2}(x-y)(y-z)(z-x)\times 1$$ $$= \frac{1}{6}(x-y)(y-z)(z-x) $$ But, shouldn't the volume be equal to the determinant which is $(x-y)(y-z)(z-x)$ ?

References

[1]Youtube video by 3Blue1brown: https://youtu.be/Ip3X9LOh2dk?t=345

[2]Wikipedia article:https://en.wikipedia.org/wiki/Shoelace_formula

Sophile
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  • Indeed, the volume of the pyramid is $1/6$ times the determinant, see this post, why. – Dietrich Burde Jun 19 '21 at 11:10
  • "This plain will cut the parallelogram into two equal triangular pyramids whose base lies in the plane." - Assuming you mean "parallelepiped" here, doesn't a parallelepiped in 3 dimensions decompose into six triangular pyramids (aka tetrahedra, aka 3-simplices)? And in general, in $n$ dimensions, the factor is $n!$. –  Jun 19 '21 at 11:15
  • @DietrichBurde But I am not trying to find the volume of a pyramid from its vertices. I calculated the volume of a parallelopiped in two ways.1.with determinant method.2.By geometry.My question is - Why dont the two answers agree? – Sophile Jun 19 '21 at 11:17
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    @AbhinavPB But you've computed $\frac{1}{3}bh$, which as you say above is the volume of the pyramid, not the parallelepiped. –  Jun 19 '21 at 11:18
  • @Bungo I think i should add a picture.It is difficult to visualize a parallelopiped being cut into two equal triangular pyramids... – Sophile Jun 19 '21 at 11:20
  • It is cut into six equal triangular pyramids, not two. I agree that it's hard to visualize, but my geometric visualization skills are poor in general. –  Jun 19 '21 at 11:21
  • @Bungo Also, Thanks for pointing out the spelling mistake.I corrected the "parallelogram" to "parallelopiped" – Sophile Jun 19 '21 at 11:21
  • Although it's not important but the name of the channel is $\text{3Blue1Brown}$ – p_square Jun 19 '21 at 11:43
  • @Algebrology Oh thanks...i misspelled it – Sophile Jun 19 '21 at 11:44
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    @Bungo I think you are right... I think i made an error while saying that the plane can cut the parallelopiped to two equal pyramids. I realised it only when i tried to make a 3d model of it in blender,Turns out I am also not good at visualization of 3d geometry – Sophile Jun 19 '21 at 11:45
  • Realizing this error, what should i do? Should I delete the question? – Sophile Jun 19 '21 at 11:48
  • @AbhinavPB One way to see (without making a 3d model) that two triangular pyramids joined together can't be a parallelepiped is that the combined object would only have five vertices, whereas a parallelepiped has eight. I don't have a similar observation for why six triangular pyramids do form a parallelepiped, but this answer is reasonably clear. –  Jun 19 '21 at 21:44

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