Question

|yуг9.zx =(x-y) (y-z) (z-x) (xy + yz + zr)

Answer 1

The given left hand side determinant is,

Δ=| x x 2 yz y y 2 zx z z 2 xy |

Apply row operation R 1 → R 1 − R 2 ,

Δ=| x−y x 2 − y 2 yz−zx y y 2 zx z z 2 xy | =( x−y )| 1 x+y −z y y 2 zx z z 2 xy |

Apply row operation R 2 → R 2 − R 3 ,

Δ=( x−y )| 1 x+y −z y−z y 2 − z 2 zx−xy z z 2 xy | =( x−y )( y−z )| 1 x+y −z 1 y+z −x z z 2 xy |

Apply row operation R 1 → R 1 − R 2 ,

Δ=( x−y )( y−z )| 1−1 x+y−y−z −z+x 1 y+z −x z z 2 xy | =( x−y )( y−z )| 0 x−z −z+x 1 y+z −x z z 2 xy | =( x−y )( y−z )( z−x )| 0 −1 −1 1 y+z −x z z 2 xy |

Apply column operation C 2 → C 2 − C 3 ,

Δ=( x−y )( y−z )( z−x )| 0 −1+1 −1 1 y+z+x −x z z 2 −xy xy | =( x−y )( y−z )( z−x )| 0 0 −1 1 y+z+x −x z z 2 −xy xy |

Expand along R 1 ,

Δ=( x−y )( y−z )( z−x )[ 0−0− z 2 +xy+zx+zy+ z 2 ] =( x−y )( y−z )( z−x )( xy+zx+zy )

Hence, the left hand side of the determinant is equal to the right hand side.