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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 - Mathematics

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Sum

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 R1R1+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 C2C2C1 and C3C3C1 , 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

Concept: Properties of Determinants
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