# Lineare Abbildung/Auf Basis/(0,1,2) nach (3,-2) und (1,4,1) nach (1,0) und (2,1,3) nach (7,2)/(3,-5,4)/Aufgabe/Lösung

Wir lösen zuerst das lineare Gleichungssystem

${\displaystyle a{\begin{pmatrix}0\\1\\2\end{pmatrix}}+b{\begin{pmatrix}1\\4\\1\end{pmatrix}}+c{\begin{pmatrix}2\\1\\3\end{pmatrix}}={\begin{pmatrix}3\\-5\\4\end{pmatrix}}.}$

Die Zeilenoperation ${\displaystyle {}IV=2II-III}$ führt auf

${\displaystyle (IV)\,\,\,7b-c=-14}$

und ${\displaystyle {}V=I+2IV}$ führt auf

${\displaystyle (V)\,\,\,15b=-25.}$

Damit ist

${\displaystyle {}b={\frac {-25}{15}}=-{\frac {5}{3}}\,}$

und

${\displaystyle {}2c=3-b=3+{\frac {5}{3}}={\frac {14}{3}}\,,}$

also

${\displaystyle {}c={\frac {7}{3}}\,,}$

und

${\displaystyle {}a=-5-4b-c=-5-4{\left({\frac {-5}{3}}\right)}-{\frac {7}{3}}={\frac {-15}{3}}+{\frac {20}{3}}-{\frac {7}{3}}=-{\frac {2}{3}}\,.}$

Also ist

{\displaystyle {}{\begin{aligned}\varphi {\begin{pmatrix}3\\-5\\4\end{pmatrix}}&=\varphi {\left(-{\frac {2}{3}}{\begin{pmatrix}0\\1\\2\end{pmatrix}}-{\frac {5}{3}}{\begin{pmatrix}1\\4\\1\end{pmatrix}}+{\frac {7}{3}}{\begin{pmatrix}2\\1\\3\end{pmatrix}}\right)}\\&=-{\frac {2}{3}}\cdot \varphi {\begin{pmatrix}0\\1\\2\end{pmatrix}}-{\frac {5}{3}}\cdot \varphi {\begin{pmatrix}1\\4\\1\end{pmatrix}}+{\frac {7}{3}}\cdot \varphi {\begin{pmatrix}2\\1\\3\end{pmatrix}}\\&=-{\frac {2}{3}}\cdot {\begin{pmatrix}3\\-2\end{pmatrix}}-{\frac {5}{3}}\cdot {\begin{pmatrix}1\\0\end{pmatrix}}+{\frac {7}{3}}\cdot {\begin{pmatrix}7\\2\end{pmatrix}}\\&={\begin{pmatrix}-2-{\frac {5}{3}}+{\frac {49}{3}}\\{\frac {4}{3}}+{\frac {14}{3}}\end{pmatrix}}\\&={\begin{pmatrix}{\frac {38}{3}}\\{\frac {18}{3}}\end{pmatrix}}\\&={\begin{pmatrix}{\frac {38}{3}}\\6\end{pmatrix}}.\end{aligned}}}