Aloha :)
In Matrix-Darstellung lautet die Funktion:$$f(x,y,z)=\binom{x+y}{2z-x}=\binom{1}{-1}x+\binom{1}{0}y+\binom{0}{2}z=\left(\begin{array}{c}1 & 1 & 0\\-1 & 0 & 2\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)$$Wir wenden die Funktion auf die Vektoren der Basis \(B=(\vec b_1,\vec b_2,\vec b_3)\) an und drücken das Ergebnis in Vektoren der Basis \(C=(\vec c_1,\vec c_2)\) aus:
$$f(\vec b_1)=\left(\begin{array}{c}1 & 1 & 0\\-1 & 0 & 2\end{array}\right)\left(\begin{array}{c}1\\0\\-1\end{array}\right)=\binom{1}{-3}=\binom{0}{1}\cdot(-3)+\binom{1}{0}\cdot1=\binom{-3}{1}_C$$$$f(\vec b_2)=\left(\begin{array}{c}1 & 1 & 0\\-1 & 0 & 2\end{array}\right)\left(\begin{array}{c}1\\1\\1\end{array}\right)=\binom{2}{1}=\binom{0}{1}\cdot1+\binom{1}{0}\cdot2=\binom{1}{2}_C$$$$f(\vec b_2)=\left(\begin{array}{c}1 & 1 & 0\\-1 & 0 & 2\end{array}\right)\left(\begin{array}{c}1\\0\\0\end{array}\right)=\binom{1}{-1}=\binom{0}{1}\cdot(-1)+\binom{1}{0}\cdot1=\binom{-1}{1}_C$$$$\Rightarrow\quad{_C[f]_B}=\left(\begin{array}{c}-3 & 1 & -1\\1 & 2 & 1\end{array}\right)$$Jetzt sollst du die Basen wechseln:
$${_{C\,'}}[f]_{B\,'}={_{C\,'}}id_{\,C}\cdot {_C[f]_B}\cdot{_B\,}id_{\,B\,'}$$$$={_{C\,'}}id_{E}\cdot{_E}id_{\,C}\cdot {_C[f]_B}\cdot{_B\,}id_{\,E}\cdot{_E\,}id_{\,B\,'}$$$$=\left({_E}id_{C\,'}\right)^{-1}\cdot{_E}id_{\,C}\cdot {_C[f]_B}\cdot\left({_E\,}id_{\,B}\right)^{-1}\cdot{_E\,}id_{\,B\,'}$$$$=\left(\begin{array}{c}1 & 1\\1 & -1\end{array}\right)^{-1}\left(\begin{array}{c}0 & 1\\1 & 0\end{array}\right)\left(\begin{array}{c}-3 & 1 & -1\\1 & 2 & 1\end{array}\right)\left(\begin{array}{c}1 & 1 & 1\\0 & 1 & 0\\-1 & 1 & 0\end{array}\right)^{-1}\left(\begin{array}{c}1 & 1 & 1\\1 & 0 & 0\\0 & 0 & -1\end{array}\right)$$$$=\left(\begin{array}{c}\frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & -\frac{1}{2}\end{array}\right)\left(\begin{array}{c}0 & 1\\1 & 0\end{array}\right)\left(\begin{array}{c}-3 & 1 & -1\\1 & 2 & 1\end{array}\right)\left(\begin{array}{c}0 & 1 & -1\\0 & 1 & 0\\1 & -2 & 1\end{array}\right)\left(\begin{array}{c}1 & 1 & 1\\1 & 0 & 0\\0 & 0 & -1\end{array}\right)$$$$=\left(\begin{array}{c}\frac{1}{2} & \frac{1}{2}\\-\frac{1}{2} & \frac{1}{2}\end{array}\right)\left(\begin{array}{c}-3 & 1 & -1\\1 & 2 & 1\end{array}\right)\left(\begin{array}{c}1 & 0 & 1\\1 & 0 & 0\\-1 & 1 & 0\end{array}\right)$$$$=\left(\begin{array}{c}\frac{1}{2} & \frac{1}{2}\\-\frac{1}{2} & \frac{1}{2}\end{array}\right)\left(\begin{array}{c}-1 & -1 & -3\\2 & 1 & 1\end{array}\right)$$$$=\left(\begin{array}{c}\frac{1}{2} & 0 & -1\\\frac{3}{2} & 1 & 2\end{array}\right)$$