Linearkombination/R/3/Aufgabe/Lösung

Es geht darum, das lineare Gleichungssystem

${\displaystyle {\begin{matrix}2x+y+z&=&0\\\,\,\,\,-y+2z&=&1\\2x+3y+3z&=&2\,\end{matrix}}}$

zu lösen. Die Differenz

${\displaystyle {}IV=III-I\,}$

führt auf

${\displaystyle {}2y+2z=2\,.}$

Die Differenz

${\displaystyle {}V=IV-II\,}$

führt auf

${\displaystyle {}3y=1\,.}$

Also ist

${\displaystyle {}y={\frac {1}{3}}\,,}$

${\displaystyle {}z=1-y={\frac {2}{3}}\,}$

und

${\displaystyle {}x=-{\frac {1}{2}}y-{\frac {1}{2}}z=-{\frac {1}{2}}\cdot {\frac {1}{3}}-{\frac {1}{2}}\cdot {\frac {2}{3}}=-{\frac {1}{2}}\,.}$

Also ist

${\displaystyle {}{\begin{pmatrix}0\\1\\2\end{pmatrix}}=-{\frac {1}{2}}{\begin{pmatrix}2\\0\\2\end{pmatrix}}+{\frac {1}{3}}{\begin{pmatrix}1\\-1\\3\end{pmatrix}}+{\frac {2}{3}}{\begin{pmatrix}1\\2\\3\end{pmatrix}}\,.}$