Polynomring/Einsetzung/R nach Mat2R/2/Aufgabe/Lösung

Es ist

${}{\begin{pmatrix}3&-1\\6&7\end{pmatrix}}^{2}={\begin{pmatrix}3&-1\\6&7\end{pmatrix}}{\begin{pmatrix}3&-1\\6&7\end{pmatrix}}={\begin{pmatrix}3&-10\\60&43\end{pmatrix}}\,$

und

${}{\begin{pmatrix}3&-1\\6&7\end{pmatrix}}^{3}={\begin{pmatrix}3&-10\\60&43\end{pmatrix}}{\begin{pmatrix}3&-1\\6&7\end{pmatrix}}={\begin{pmatrix}-51&-73\\438&241\end{pmatrix}}\,.$

Daher ist

${}{\begin{aligned}\left(-X^{3}+4X^{2}+2X-5\right)\left({\begin{pmatrix}3&-1\\6&7\end{pmatrix}}\right)&=-{\begin{pmatrix}3&-1\\6&7\end{pmatrix}}^{3}+4{\begin{pmatrix}3&-1\\6&7\end{pmatrix}}^{2}+2{\begin{pmatrix}3&-1\\6&7\end{pmatrix}}-5{\begin{pmatrix}1&0\\0&1\end{pmatrix}}\\&=-{\begin{pmatrix}-51&-73\\438&241\end{pmatrix}}+4{\begin{pmatrix}3&-10\\60&43\end{pmatrix}}+2{\begin{pmatrix}3&-1\\6&7\end{pmatrix}}-5{\begin{pmatrix}1&0\\0&1\end{pmatrix}}\\&={\begin{pmatrix}64&31\\-186&-60\end{pmatrix}}.\,\end{aligned}}$

Zur gelösten Aufgabe