0 xyz x-z y-x 0 y-zz-x z - y 0 is equal to
(z−x)(y−z)(y−x+xyz)
(x−z)(z−y)(y−x+xyz)
(z+x)(y+z)(y+x−xyz)
(z+x)(y−z)(y+x−xyz)
We have ,
0 xyz x-z y-x 0 y-zz-x z - y 0=z-x xyz x-z z-x 0 y-zz-x z - y 0 ∵ C1→C1- C3
=(z-x)1 xyz x-z 1 0 y-z1 z - y 0 taking(z-x)common from column 1
(Expanding along R1)
=(z-x)[1.{-(y-z)(z-y)-xyz(z-y)+(x-z)(z-y)]=(z-x)(z-y)(-y+z-xyz+x-z)=(z-x)(z-y)(x-y-xyz)=(z-x)(y-z)(y-x+xyz)