Question

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

Answer 1

Let △=\(\begin{vmatrix} x & x^2 & yz\\ y & y^2 & zx\\z&z^2&xy \end{vmatrix}\)

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have: 

△=\(\begin{vmatrix} x & x^2 & yz\\ y-x & y^2-x^3 & zx-yz\\z-x&z^2-x^2&xy-yz \end{vmatrix}\)

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

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

 Applying R₃ → R₃ + R₂, we have: 

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

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

Expanding along R₃, we have: 

△=[(x-y)(z-x)(z-y)]\(\bigg[(-1)\)\(\begin{vmatrix} x &yz \\ -1 & z \end{vmatrix}\)+1\(\begin{vmatrix} x &x^2 \\ -1 & -x-y \end{vmatrix}\) \(\bigg]\)
=[(x-y)(z-x)(z-y)][(-xz-yz)+(-x²-xy+x²)] 
=(x-y)(y-z)(z-x)(xy+yz+zx) 
Hence, the given result is proved.