evaluate: `|(3x,-x+y,-x+z),(x-y,3y,z-y),(x-z,y-z,3z)|`

Options

A. `3(x+y+z)(yz-yx-xz)`
B. `3(x+2y+z)(yz+yx+xz)`
C. `(x+y+z)(yz+yx+xz)`
D. `3(x+y+z)(yz+yx+xz)`

Answer

D

Explanation

We have `|(3x,-x+y,-x+z),(x-y,3y,z-y),(x-z,y-z,3z)|`
Applyig `C_(1)toC_(1)+C_(2)+C_(3)`
`=|(x+y+z,-x+y,-x+z),(x+y+z,3y,z-y),(x+y+z,y-z,3z)|`
`=(x+y+z)|(1,-x+y,-x+z),(1,3y,z-y),(1,y-z,3z)|`
[taking `(x+y+z)` common from column `C_(1)`]
`=x(x+y+z)|(1,-x+y,-x+z),(0,2y+x,x-y),(0,x-z,2z+x)|`
`[ :' R_(2)toR_(2)-R_(1) "and" R_(3)toR_(1)]`
Now expanding along first column we get
`(x+y+z).1[(2y+x)(2z+x)-(x-y)(x-z)]`
`=(x+y+z)(4yz+2yx+xy+x^(2)-x^(2)+xz+yx-yz)`
`(x+y+z)(3yz+3yx+3xz)`
`=3(x+y+z)(yz+yx+xz)`