Question

Using properties of determinants, prove that `|[2y,y-z-x,2y],[2z,2z,z-x-y],[x-y-z,2x,2x]|=(x+y+z)^3`

Answer 1

We need to prove that  `|[2y,y-z-x,2y],[2z,2z,z-x-y],[x-y-z,2x,2x]|=(x+y+z)^3`

`|[2y,y-z-x,2y],[2z,2z,z-x-y],[x-y-z,2x,2x]|`

On applying R₁→R₁+R₂+R₃, we get

`=|[x+y+z,x+y+z,x+y+z],[2z,2z,z-x-y],[x-y-z,2x,2x]|`

Taking x+y+z common from the first row, we get

`=(x+y+z)|[1,1,1],[2z,2z,z-x-y],[x-y-z,2x,2x]|`

Now, applying C₂→C₂−C₁ and C₃→C₃−C₁ , we get: 

`=(x+y+z)|[1,0,0],[2z,0, (z-x-y)-(2z)],[x-y-z, 2x-(x-y-z), 2x-(x-y-z)]|`

`= (xy+z) = |(1,0,0),(2z,0,-(z+x+y)),(x-y-z,(x+y+z) , x+y+z)|`

`= (x+y+z)^3 |(1,0,0),(2z,0,-1),(x-y-z , 1 , 1)|`

`= (x+y+z)^3 [1|(0,-1),(1,1)| - 0 |+0|]`

`= (x+y+z)^3 [|(0,-1),(1,1)|]`

`= (x+y+z)^3` = RHS