R^3/(2,3,2) \wedge (4,-1,5)/Standardbasis/Aufgabe/Lösung

Es ist

${}{\begin{aligned}{\begin{pmatrix}2\\3\\2\end{pmatrix}}\wedge {\begin{pmatrix}4\\-1\\5\end{pmatrix}}&=(2e_{1}+3e_{2}+2e_{3})\wedge (4e_{1}-e_{2}+5e_{3})\\&=-2e_{1}\wedge e_{2}+10e_{1}\wedge e_{3}+12e_{2}\wedge e_{1}+15e_{2}\wedge e_{3}+8e_{3}\wedge e_{1}-2e_{3}\wedge e_{2}\\&=-2e_{1}\wedge e_{2}+10e_{1}\wedge e_{3}-12e_{1}\wedge e_{2}+15e_{2}\wedge e_{3}-8e_{1}\wedge e_{3}+2e_{2}\wedge e_{3}\\&=-14e_{1}\wedge e_{2}+2e_{1}\wedge e_{3}+17e_{2}\wedge e_{3}.\end{aligned}}$

Zur gelösten Aufgabe