Rozwiązanie:
Zadanie [tex]4[/tex].
Najpierw obliczmy:
[tex]A^{T}=\left|\begin{array}{cc}-2&-1\\4&2\\1&4\end{array}\right|[/tex]
[tex]3C=3\left|\begin{array}{cc}1&0\\0&1\\-2&0\end{array}\right|=\left|\begin{array}{cc}3&0\\0&3\\-6&0\end{array}\right|[/tex]
Zatem:
[tex]A^{T}B-3C=\left|\begin{array}{cc}-2&-1\\4&2\\1&4\end{array}\right| \cdot \left|\begin{array}{cc}2&-4\\-1&2\end{array}\right| -\left|\begin{array}{cc}3&0\\0&3\\-6&0\end{array}\right|=\left|\begin{array}{cc}-3&6\\6&-12\\-2&4\end{array}\right|-\left|\begin{array}{cc}3&0\\0&3\\-6&0\end{array}\right|=\left|\begin{array}{cc}-6&6\\6&-15\\4&4\end{array}\right|[/tex]
