Polynomring/Einsetzung/Matrix/X^2+(1+4i)X+3-i/2-i 1+3i 5 -3+4i/Aufgabe/Lösung

Es ist

${}{\begin{aligned}{\begin{pmatrix}2-{\mathrm {i} }&1+3{\mathrm {i} }\\5&-3+4{\mathrm {i} }\end{pmatrix}}^{2}&={\begin{pmatrix}2-{\mathrm {i} }&1+3{\mathrm {i} }\\5&-3+4{\mathrm {i} }\end{pmatrix}}\circ {\begin{pmatrix}2-{\mathrm {i} }&1+3{\mathrm {i} }\\5&-3+4{\mathrm {i} }\end{pmatrix}}\\&={\begin{pmatrix}(2-{\mathrm {i} })^{2}+5(1+3{\mathrm {i} })&(1+3{\mathrm {i} })(2-{\mathrm {i} }-3+4{\mathrm {i} })\\5(2-{\mathrm {i} }-3+4{\mathrm {i} })&5(1+3{\mathrm {i} })+(-3+4{\mathrm {i} })^{2}\end{pmatrix}}\\&={\begin{pmatrix}4-1-4{\mathrm {i} }+5+15{\mathrm {i} }&(1+3{\mathrm {i} })(-1+3{\mathrm {i} })\\5(-1+3{\mathrm {i} })&5+15{\mathrm {i} }+9-16-24{\mathrm {i} }\end{pmatrix}}\\&={\begin{pmatrix}8+11{\mathrm {i} }&-10\\-5+15{\mathrm {i} }&-2-9{\mathrm {i} }\end{pmatrix}}.\end{aligned}}$

Somit ist

${}{\begin{aligned}{\begin{pmatrix}2-{\mathrm {i} }&1+3{\mathrm {i} }\\5&-3+4{\mathrm {i} }\end{pmatrix}}^{2}+(1+4{\mathrm {i} }){\begin{pmatrix}2-{\mathrm {i} }&1+3{\mathrm {i} }\\5&-3+4{\mathrm {i} }\end{pmatrix}}+(3+{\mathrm {i} }){\begin{pmatrix}1&0\\0&1\end{pmatrix}}&={\begin{pmatrix}8+11{\mathrm {i} }&-10\\-5+15{\mathrm {i} }&-2-9{\mathrm {i} }\end{pmatrix}}+{\begin{pmatrix}6+7{\mathrm {i} }&-11+7{\mathrm {i} }\\5+20{\mathrm {i} }&-19-8{\mathrm {i} }\end{pmatrix}}+{\begin{pmatrix}3+{\mathrm {i} }&0\\0&3+{\mathrm {i} }\end{pmatrix}}\\&={\begin{pmatrix}17+19{\mathrm {i} }&-21+7{\mathrm {i} }\\35{\mathrm {i} }&-18-16{\mathrm {i} }\end{pmatrix}}.\,\end{aligned}}$

Zur gelösten Aufgabe