The given function-
T(β£β‘βxyzββ¦β€β)=β£β‘βx+yβz2xyx+z+1ββ¦β€β
LetΞ±=β£β‘βΞ±1βΞ±2βΞ±3βββ¦β€β,Ξ²=β£β‘βΞ²1βΞ²2βΞ²3βββ¦β€ββR3
T(Ξ±+Ξ²)=T(β£β‘βΞ±1βΞ±2βΞ±3βββ¦β€β+β£β‘βΞ²1βΞ²2βΞ²3βββ¦β€β)=T(β£β‘βΞ±1β+Ξ²1βΞ±2β+Ξ²2βΞ±3β+Ξ²3βββ¦β€β)=β£β‘βΞ±1β+Ξ²1β+Ξ±2β+Ξ²2ββ(Ξ±3β+Ξ²3β)2(Ξ±1β+Ξ²1β)(Ξ±2β+Ξ²2β)Ξ±1β+Ξ²1β+Ξ±3β+Ξ²3β+1ββ¦β€β=β£β‘βΞ±1β+Ξ±2ββΞ±3β2Ξ±1βΞ±2βΞ±1β+Ξ±3β+1ββ¦β€β+β£β‘βΞ²1β+Ξ²2ββΞ²3β2Ξ²1βΞ²2β+2(Ξ±1βΞ²2β+Ξ²1βΞ±2β)Ξ²1β+Ξ²3βββ¦β€βξ =T(Ξ±)+T(Ξ²)
Since, T(Ξ²)=β£β‘βΞ±1β+Ξ±2ββΞ±3β2Ξ±1βalpha2βΞ±1β+alpha3β+1ββ¦β€βξ =β£β‘βΞ²1β+Ξ²2ββΞ²3β2Ξ²1βΞ²2β+2(Ξ±1βΞ²2β+Ξ²1βΞ±2β)Ξ²1β+Ξ²3βββ¦β€β
Therefore T is not liner transformation between vector space.
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