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The determinant $\begin{vmatrix} 1 & 1 &1 \\ x & y & z \\ x^2 & y^2 &z^2 \\ \end{vmatrix}$ can be factored to the form $(x-y)(y-z)(z-x)$


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) $


Is there a geometric reason for this?

The determinant of this matrix is the volume of a parallelopiped with sides as vectors whose tail is at the origin and head at x,y,z coordinates being equal to the columns(or rows) of the matrix.$^{[1]}$

So is the volume of this parallelopiped equals $(x-y)(y-z)(z-x)$ in any obvious geometric way?


References

[1] Nykamp DQ, “The relationship between determinants and area or volume.” From Math Insight. http://mathinsight.org/relationship_determinants_area_volume

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  • $\begingroup$ Consider the determinant $D$ as a quadratic in $X$ : Let $D= Ax^2+Bx +C$ where $A,B,C$ depend on $y,z$ but not on $x.$ If $y\ne z$ then the zeroes of this quadratic are $y,z,$ as can be seen by looking at $D$. So $D=A(x-y)(x-z)$ whenever $y\ne z$. And the co-efficient $A$ of $ x^2$ in $D$ is $z-y.$ $\endgroup$ Jun 20 at 5:34
  • $\begingroup$ The question in the title and in the body are not quite the same. Maybe you could try to clarify? $\endgroup$ Jun 20 at 7:22
  • $\begingroup$ @HansLundmark I don't quite understand what you mean...Maybe you suggest a better title? $\endgroup$ Jun 20 at 7:28
  • $\begingroup$ It's clear why the determinant can be factored like that – you've even given a proof yourself! So it seems like you have already answered your own question (as it's formulated in the title). You might want to come up with a title that makes clear that you are looking for a geometric rather than an algebraic explanation, maybe involving the words "geometric" and/or "volume". $\endgroup$ Jun 20 at 7:31
  • $\begingroup$ @HansLundmark Ok,thanks... $\endgroup$ Jun 20 at 8:11
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Since $x=y, y=z, z=x$ give thre determinant $D$ as the determinant as zero, and it being homogeneous cubic (see the product of diagonal element), D needs to be $D=A(x-y)(y-z)(z-x)$. Further, set $z=0, x=1,y=2$ to get $A=1$.

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Subtracting a multiple of another column (or row) to an existing column (or row) does not change the determinant. $$ \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \\ \end{vmatrix} $$ $$= \begin{vmatrix} 1 & 0 &0 \\ x & y-x & z-x \\ x^2 & y^2-x^2 &z^2-x^2 \\ \end{vmatrix}$$ $$= \begin{vmatrix} 1 & 0 &0 \\ 0 & y-x & z-x \\ 0 & y^2-x^2 &z^2-x^2 \\ \end{vmatrix}$$ $$={(y-x)(z-x) \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & y+x &z+x \\ \end{vmatrix} } $$ $$=(y-x)(z-x) \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & y+x & z-y \\ \end{vmatrix} $$ $$=(y-x)(z-x)(z-y)$$

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  • $\begingroup$ Even though this is not an answer to my question, it is helpful.So thanks... $\endgroup$ Jun 20 at 6:23
  • $\begingroup$ I subtracted suitable multiples of row 1 from rows 2 and 3 to get the zeros. A geometric reason why the row & column operations don't change volume would answer your question. $\endgroup$
    – David
    Jun 20 at 6:33

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