Factorize and prove that

$$ \Delta=\begin{vmatrix} yz-x^2&zx-y^2&xy-z^2\\ zx-y^2&xy-z^2&yz-x^2\\ xy-z^2&yz-x^2&zx-y^2 \end{vmatrix}\\=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2 $$ using factor theorem.

My Attempt:

$\Delta$ is a homogeneous symmetric polynomial of degree $6$.

When $(x-y)^2+(y-z)^2+(z-x)^2=0$, i.e. $x=y=z$ $$ \Delta=\begin{vmatrix} 0&0&0\\ 0&0&0\\ 0&0&0\\ \end{vmatrix}=0 $$ Thus, $(x-y)^2+(y-z)^2+(z-x)^2$ is a factor.

How do I extract the other $(x-y)^2+(y-z)^2+(z-x)^2$ from $\Delta$ $\color{red}{?}$

Does this have anything to do with all rows (or columns) being zero when $(x-y)^2+(y-z)^2+(z-x)^2=0$ $\color{red}{?}$

If I can extract that then i think I know how to proceed. The remaining factor must be a homogeneous quadratic symmetric polynomial, i.e. $p(x,y,z)=a(x^2+y^2+z^2)+b(xy+yz+zx)$ $$ \Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.a(x^2+y^2+z^2)+b(xy+yz+zx) $$ $$ \Delta(1,0,0)=\begin{vmatrix} -1&0&0\\ 0&0&-1\\ 0&-1&0\\ \end{vmatrix}=1=4.a\implies a=\frac{1}{4} $$ $$ \Delta(1,1,0)=\begin{vmatrix} -1&-1&1\\ -1&1&-1\\ 1&-1&-1\\ \end{vmatrix}=\begin{vmatrix} 0&0&1\\ -2&0&-1\\ 0&-2&-1\\ \end{vmatrix}\\ =\begin{vmatrix} -2&0\\ 0&-2\\ \end{vmatrix}=4=4.(2a+b)=4(1/2+b)=2+4b\\ \implies b=\frac{1}{2} $$ $$ \Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.\frac{1}{4}(x^2+y^2+z^2)+\frac{1}{2}(xy+yz+zx)\\ =\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.(x^2+y^2+z^2+2xy+2yz+2zx)\\ =\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2(x+y+z)^2 $$ Note:

I am trying to factorize the determinant using factor theorem given the fact that the determinant is a homogeneous symmetric polynomial of degree 6.

  • 2
    $\begingroup$ How do you infer that $(x-y)^2+(y-z)^2+(z-x)^2$ is a factor? Couldn't you use the same reasoning to determine that $(x-y)^2+(y-z)^4+(z-x)^6$ is a factor? $\endgroup$ – flawr Mar 4 '18 at 9:00
  • $\begingroup$ @flawr thnx. that seems true. then how do I extract $(x-y)^2+(y-z)^2+(z-x)^2$ in the first place ? $\endgroup$ – ss1729 Mar 4 '18 at 9:08
  • $\begingroup$ You could always expand the determinant and then factor it, which should not be difficult as you know the result. $\endgroup$ – flawr Mar 4 '18 at 9:10
  • $\begingroup$ @flawr Ofcause i knw how to do that. i'm trying to do it using factor theorem given the fact that the determinant is a symmetric polynomial of degree 6. $\endgroup$ – ss1729 Mar 4 '18 at 9:12
  • 1
    $\begingroup$ Does it help to swap rows 2 and 3, then note it's a circulant, and so the determinant is $-(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2 +c\omega)$ where $a=yz-x^2$ etc and $\omega$ is the cube root of unity? $\endgroup$ – ancientmathematician Mar 4 '18 at 9:27

Hint: Note that your matrix has the form

$$\pmatrix{a&b&c\cr b&c&a\cr c&a&b\cr }$$

Which has the determinant


which can again be factored into



By taking elementary operation:$$\Delta=\begin{vmatrix} yz-x^2&zx-y^2&xy-z^2\\ zx-y^2&xy-z^2&yz-x^2\\ xy-z^2&yz-x^2&zx-y^2 \end{vmatrix}=\begin{vmatrix} xy+yz+zx-x^2-y^2-z^2&zx-y^2&xy-z^2\\ xy+yz+zx-x^2-y^2-z^2&xy-z^2&yz-x^2\\ xy+yz+zx-x^2-y^2-z^2&yz-x^2&zx-y^2 \end{vmatrix}\\=-(x^2+y^2+z^2-xy-yz-zx)\begin{vmatrix} 1&zx-y^2&xy-z^2\\ 1&xy-z^2&yz-x^2\\ 1&yz-x^2&zx-y^2 \end{vmatrix}\\=-\frac{1}{2}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]\begin{vmatrix} 0&(x+y+z)(z-y)&(x+y+z)(x-z)\\ 0&(x+y+z)(x-z)&(x+y+z)(x-z)\\ 1&yz-x^2&zx-y^2 \end{vmatrix}\\=-\frac{1}{2}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]\begin{vmatrix} (z-y)&(x-z)\\ (x-z)&(x-z)\\ \end{vmatrix}=\frac{1}{2}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2, $$ as we know $x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]$.

  • $\begingroup$ pls read the post. i know this. i am not asking for solving it by usual matrix operations. $\endgroup$ – ss1729 Mar 4 '18 at 20:51

Note that $$ M:= \begin{bmatrix} yz-x^2&zx-y^2&xy-z^2\\ zx-y^2&xy-z^2&yz-x^2\\ xy-z^2&yz-x^2&zx-y^2 \end{bmatrix} $$ is the matrix of $2\times 2$ cofactors of the matrix: $$ N:= \begin{bmatrix} x & y & z\\ y & z & x\\ z& x & y \\ \end{bmatrix}. $$

Then as usual $MN=(\det N) I$, so that $\det M \det N =(\det N)^3$. As $\det N\not=0$ (as a polynomial) we have that $\Delta =\det M= (\det N)^2$ is the perfect square of a polynomial of degree $3$ -- which is what was asked.

But from this we can see everything: there is clearly a factor $(x+y+z)$ in $\det N$ and a factor $\frac{1}{2}( (x-y)^2 +(y-z)^2 +(z-x)^2))$ in $\det M$. Establishing the value of the constant is trivial.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.