# Geometric reason why this determinant can be factored to (x-y)(y-z)(z-x)?

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

• 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 $z-y.$ Jun 20 at 5:34
• The question in the title and in the body are not quite the same. Maybe you could try to clarify? Jun 20 at 7:22
• @HansLundmark I don't quite understand what you mean...Maybe you suggest a better title? Jun 20 at 7:28
• 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". Jun 20 at 7:31
• @HansLundmark Ok,thanks... Jun 20 at 8:11

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