Rozwiązanie:
Zadanie [tex]5.[/tex]
[tex]\left\{\begin{array}{ccc}2x-4y+2z=8\\-2x-y+z=-3\\-3x+y-2z=-7\end{array}\right[/tex]
Rozwiązujemy:
[tex]W=\left|\begin{array}{ccc}2&-4&2\\-2&-1&1\\-3&1&-2\end{array}\right|[/tex]
Korzystając z metody Sarrusa:
[tex]W=\left|\begin{array}{ccc}2&-4&2\\-2&-1&1\\-3&1&-2\end{array}\right| \left\begin{array}{cc}2&-4\\-2&-1\\-3&1\end{array}\right =(4+12-4)-(6+2-16)=12+8=20\neq 0[/tex]
Dalej mamy:
[tex]W_{x}=\left|\begin{array}{ccc}8&-4&2\\-3&-1&1\\-7&1&-2\end{array}\right|[/tex]
[tex]W_{y}=\left|\begin{array}{ccc}2&8&2\\-2&-3&1\\-3&-7&-2\end{array}\right|[/tex]
[tex]W_{z}=\left|\begin{array}{ccc}2&-4&8\\-2&-1&-3\\-3&1&-7\end{array}\right|[/tex]
Korzystając z metody Sarrusa:
[tex]W_{x}=\left|\begin{array}{ccc}8&-4&2\\-3&-1&1\\-7&1&-2\end{array}\right|\left\begin{array}{cc}8&-4\\-3&-1\\-7&1\end{array}\right=(16+28-6)-(14+8-24)=38+2=40[/tex]
[tex]W_{y}=\left|\begin{array}{ccc}2&8&2\\-2&-3&1\\-3&-7&-2\end{array}\right\left|\begin{array}{cc}2&8\\-2&-3\\-3&-7\end{array}\right=(12-24+28)-(18-14+32)=16-36=-20[/tex]
[tex]W_{z}=\left|\begin{array}{ccc}2&-4&8\\-2&-1&-3\\-3&1&-7\end{array}\right\left|\begin{array}{cc}2&-4\\-2&-1\\-3&1\end{array}\right=(14-36-16)-(24-6-56)=-38+38=0[/tex]
Zatem:
[tex]x=\frac{W_{x}}{W}=\frac{40}{20}=2\\y=\frac{W_{y}}{W}=\frac{-20}{20}=-1\\z=\frac{W_{z}}{W}=\frac{0}{20}=0[/tex]
