Question #194951

Determine whether the function T: R

3 β†’R

3 by 𝑇 ([

π‘₯

𝑦

𝑧

]) = [

π‘₯ + 𝑦 βˆ’ 𝑧

2π‘₯𝑦

π‘₯ + 𝑧 + 1

] is a linear 

transformation between vector space.



1
Expert's answer
2021-05-19T17:17:35-0400

The given function-


T([xyz])=[x+yβˆ’z2xyx+z+1]T(\begin{bmatrix} x\\y\\z \end{bmatrix})=\begin{bmatrix}x+y-z\\2xy\\x+z+1\end{bmatrix}


    LetΞ±=[Ξ±1Ξ±2Ξ±3],Ξ²=[Ξ²1Ξ²2Ξ²3]∈R3\alpha=\begin{bmatrix} \alpha_1\\\alpha_2\\\alpha_3 \end{bmatrix},\beta=\begin{bmatrix} \beta_1\\\beta_2\\\beta_3 \end{bmatrix}\in R^3


T(Ξ±+Ξ²)=T([Ξ±1Ξ±2Ξ±3]+[Ξ²1Ξ²2Ξ²3])=T([Ξ±1+Ξ²1Ξ±2+Ξ²2Ξ±3+Ξ²3])=[Ξ±1+Ξ²1+Ξ±2+Ξ²2βˆ’(Ξ±3+Ξ²3)2(Ξ±1+Ξ²1)(Ξ±2+Ξ²2)Ξ±1+Ξ²1+Ξ±3+Ξ²3+1]=[Ξ±1+Ξ±2βˆ’Ξ±32Ξ±1Ξ±2Ξ±1+Ξ±3+1]+[Ξ²1+Ξ²2βˆ’Ξ²32Ξ²1Ξ²2+2(Ξ±1Ξ²2+Ξ²1Ξ±2)Ξ²1+Ξ²3]β‰ T(Ξ±)+T(Ξ²)T(\alpha+\beta)=T(\begin{bmatrix} \alpha_1\\\alpha_2\\\alpha_3 \end{bmatrix}+\begin{bmatrix} \beta_1\\\beta_2\\\beta_3 \end{bmatrix}) \\[9pt] =T(\begin{bmatrix} \alpha_1+\beta_1\\\alpha_2+\beta_2\\\alpha_3+\beta_3 \end{bmatrix}) \\[9pt] =\begin{bmatrix} \alpha_1+\beta_1+\alpha_2+\beta_2-(\alpha_3+\beta_3)\\2(\alpha_1+\beta_1)(\alpha_2+\beta_2)\\\alpha_1+\beta_1+\alpha_3+\beta_3+1 \end{bmatrix}\\[9pt] =\begin{bmatrix} \alpha_1+\alpha_2-\alpha_3\\2\alpha_1 \alpha_2\\\alpha_1+\alpha_3+1 \end{bmatrix}+\begin{bmatrix} \beta_1+\beta_2-\beta_3\\2\beta_1 \beta_2+2(\alpha_1 \beta_2+\beta_1 \alpha_2)\\\beta_1+\beta_3 \end{bmatrix}\\[9pt]\neq T(\alpha)+T(\beta)




Since, T(Ξ²)=[Ξ±1+Ξ±2βˆ’Ξ±32Ξ±1alpha2Ξ±1+alpha3+1]β‰ [Ξ²1+Ξ²2βˆ’Ξ²32Ξ²1Ξ²2+2(Ξ±1Ξ²2+Ξ²1Ξ±2)Ξ²1+Ξ²3]T(\beta)=\begin{bmatrix} \alpha_1+\alpha_2-\alpha_3\\2\alpha_1 alpha_2\\\alpha_1+alpha_3+1 \end{bmatrix}\neq\begin{bmatrix} \beta_1+\beta_2-\beta_3\\2\beta_1 \beta_2+2(\alpha_1 \beta_2+\beta_1 \alpha_2)\\\beta_1+\beta_3 \end{bmatrix}


Therefore T is not liner transformation between vector space.

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