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# By using properties of determinants show that determinant x x^2 yz y y^2 zx z z^2 xy =(x-y)(y-z)(z-x)(xy++yz+zx) Ex 4.2 Q : 9

Q : 9        By using properties of determinants, show that:

$\dpi{100} \begin{vmatrix} x & x^2 & yz\\ y & y^2 &zx \\ z & z^2 & xy \end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$

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We have the determinant:

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

Applying the row transformations $R_{1} \rightarrow R_{1}- R_{3}$ and then  $R_{2} \rightarrow R_{2}- R_{3}$, we have;

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

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

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

Now, applying $R_{1} \rightarrow R_{1} - R_{2}$; we have

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

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

Now, expanding the remaining determinant;

$= (x-z)(y-z)(x-y) \left [ (xy+zx) + (z^2 - zy-z^2) \right]$

$= -(x-z)(y-z)(x-y) \left [ xy+zx + zy \right]$

$= (x-y)(y-z)(z-x) \left [ xy+zx + zy \right]$

Hence proved.

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