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# By using the properties of determinants show that $\begin{vmatrix} x&x^2&yz \\ y&y^2&zx \\ z&z^2&xy \end{vmatrix} = (x-y)(y-z)(z-x)(xy+yz+zx)$

Toolbox:
• If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any one column of the determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\bigtriangleup=\begin{vmatrix}x & x^2 & yz\\y & y^2 & zx\\z &z^2 & xy\end{vmatrix}$

By applying $R_2\rightarrow R_2-R_1$ and $R_3\rightarrow R_3-R_1$ we get

$\bigtriangleup=\begin{vmatrix}x & x^2 & yz\\y-x & y^2-x^2 & zx-yz\\z-x &z^2-x^2 & xy-yz\end{vmatrix}$

But we know $y^2-x^2=(y-x)(y+x)$ and $z^2-x^2=(z+x)(z-x)$

We can write, zx-yz=z(x-y)and

xy-yz=y(x-z).

Hence $\bigtriangleup=\begin{vmatrix}x & x^2 & yz\\y-x & (y+x)(y-x) & z(x-y)\\z-x &(z+x)(z-x) & y(x-z)\end{vmatrix}$

Now taking (x-y) as the common factor from $R_2$ and (z-x) from $R_3$ we get

$\bigtriangleup=\begin{vmatrix}x & x^2 & yz\\-1& -(x+y) & z\\1 &x+z & -y\end{vmatrix}$

By applying $R_3\rightarrow R_3+R_2$,we have

$\bigtriangleup=(x-y)(z-x)\begin{vmatrix}x & x^2 & yz\\-1 & -(x+y) & z\\0 &z-y & z-y\end{vmatrix}$

now taking (z-y) as the common factor from $R_3$

$\bigtriangleup=(x-y)(z-x)(z-y)\begin{vmatrix}x & x^2 & yz\\-1 & -(x+y) & z\\0 &1 & 1\end{vmatrix}$

Now expanding along $R_3$ we have

$\bigtriangleup=(x-y)(z-x)(z-y)\begin{bmatrix}(-1)\begin{vmatrix}x &yz\\-1 &z\end{vmatrix}+1\begin{vmatrix}x & x^2\\-1 & -(x+y)\end{vmatrix}\end{bmatrix}$

$\qquad=(x-y)(z-x)(z-y)[(-xz-yz)+(-x^2-xy+x^2)]$

$\qquad=-(x-y)(z-x)(z-y)(xy+yz+zx)$

$\qquad=(x-y)(y-z)(z-x)(xy+yz+zx)$

Hence proved.

edited Feb 24, 2013