Kreuzprodukt/Berechnung/C/3/Aufgabe/Lösung

Es ist

${}{\begin{aligned}{\begin{pmatrix}3-5{\mathrm {i} }\\4+8{\mathrm {i} }\\-{\mathrm {i} }\end{pmatrix}}\times {\begin{pmatrix}-2\\-6-7{\mathrm {i} }\\3-4{\mathrm {i} }\end{pmatrix}}&={\begin{pmatrix}{\left(4+8{\mathrm {i} }\right)}{\left(3-4{\mathrm {i} }\right)}-{\mathrm {i} }{\left(6+7{\mathrm {i} }\right)}\\-{\left(3-5{\mathrm {i} }\right)}{\left(3-4{\mathrm {i} }\right)}+2{\mathrm {i} }\\-{\left(3-5{\mathrm {i} }\right)}{\left(6+7{\mathrm {i} }\right)}+2{\left(4+8{\mathrm {i} }\right)}\end{pmatrix}}\\&={\begin{pmatrix}12+32-16{\mathrm {i} }+24{\mathrm {i} }-6{\mathrm {i} }+7\\-9+20+12{\mathrm {i} }+15{\mathrm {i} }+2{\mathrm {i} }\\-18-35+30{\mathrm {i} }-21{\mathrm {i} }+8+16{\mathrm {i} }\end{pmatrix}}\\&={\begin{pmatrix}51+2{\mathrm {i} }\\11+29{\mathrm {i} }\\-45+25{\mathrm {i} }\end{pmatrix}}.\end{aligned}}$

Zur gelösten Aufgabe