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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`
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Solution
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
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