Ach so, das hatte ich überlesen. Geht so: (wegen b):
\( 0=\operatorname{det}\left(\begin{array}{ll}a+c & b+d \\ a+c & b+d\end{array}\right) \) wegen a)
\(= \operatorname{det}\left(\begin{array}{ll}a & b \\ a+c & b+d\end{array}\right) + \operatorname{det}\left(\begin{array}{ll}c & d \\ a+c & b+d\end{array}\right) \)
Bei beiden nochmal a) in der 2. Zeile anwenden:
\(= \operatorname{det}\left(\begin{array}{ll}a & b \\ a & b\end{array}\right) + \operatorname{det}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) +\operatorname{det}\left(\begin{array}{ll}c & d \\ a & b\end{array}\right) +\operatorname{det}\left(\begin{array}{ll}c & d \\ c & d\end{array}\right) \)
Wegen b)
\(= \operatorname{det}\left(\begin{array}{ll}a & b \\ a & b\end{array}\right) + 0 +\operatorname{det}\left(\begin{array}{ll}c & d \\ a & b\end{array}\right) +0 \)
\(= \operatorname{det}\left(\begin{array}{ll}a & b \\ a & b\end{array}\right) +\operatorname{det}\left(\begin{array}{ll}c & d \\ a & b\end{array}\right) \)
Und wegen \(0 = \operatorname{det}\left(\begin{array}{ll}a & b \\ a & b\end{array}\right) +\operatorname{det}\left(\begin{array}{ll}c & d \\ a & b\end{array}\right) \)
gilt also auch \( \operatorname{det}\left(\begin{array}{ll}a & b \\ a & b\end{array}\right) = -\operatorname{det}\left(\begin{array}{ll}c & d \\ a & b\end{array}\right) \)