Question Number 210971 by black_mamba234 last updated on 25/Aug/24
Answered by A5T last updated on 25/Aug/24
$$\left.\mathrm{2}\right)\:\mathrm{2}{x}_{\mathrm{1}} +\mathrm{4}{y}_{\mathrm{1}} −\mathrm{2}{z}_{\mathrm{1}} =\mathrm{2}\left({x}_{\mathrm{1}} +{k}\right)+\mathrm{4}{y}_{\mathrm{1}} −\mathrm{2}\left({z}_{\mathrm{1}} +{k}\right) \\ $$$${f}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right)={f}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} ,{z}_{\mathrm{2}} \right)\nRightarrow\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right)=\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} ,{z}_{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{f}\:{is}\:{not}\:{injective}. \\ $$