[tex]\left\{\begin{array}{l}x+3y+3z=16\\-3x+y-z=0\\2x+y+2z=8\end{array} \right[/tex]
Skorzystajmy z elementarnych przekształceń wierszowych.
[tex]\left[\begin{array}{ccc|c}1&3&3&16\\-3&1&-1& 0\\2&1&2&8\end{array} \right]\begin{array}{l} &w_2+3w_1&w_3-2w_1\end{array}\longrightarrow\left[\begin{array}{ccc|c}1&3&3&16\\0&10&8& 48\\0&-5&-4&-24\end{array} \right]\begin{array}{l} &w_2:2& &\end{array}\longrightarrow[/tex]
[tex]\longrightarrow\left[\begin{array}{ccc|c}1&3&3&16\\0&5&4& 24\\0&-5&-4&-24\end{array} \right]\begin{array}{l} &&w_3+w_2\end{array}\longrightarrow\left[\begin{array}{ccc|c}1&3&3&16\\0&5&4& 24\\0&0&0&0\end{array} \right]\longrightarrow\left[\begin{array}{ccc|c}1&3&3&16\\0&5&4& 24\end{array} \right][/tex]Wróćmy do układu równań i wyraźmy zmienne x i y w zależności od zmiennej z.
[tex]\left \{ {{x+3y+3z=16} \atop {5y+4z=24}} \right. \longrightarrow\left \{ {{x+3y+3z=16} \atop {5y=-4z+24\ |:5}} \right. \longrightarrow\left \{ {{x+3y+3z=16} \atop {y=-\frac{4}{5}z+4\frac{4}{5}}} \right. \longrightarrow\left \{ {{x+3(-\frac{4}{5}z+4\frac{4}{5})+3z=16} \atop {y=-\frac{4}{5}z+4\frac{4}{5}}} \right. \longrightarrow[/tex][tex]\longrightarrow\left \{ {{x-2\frac{2}{5}z+14\frac{2}{5}+3z=16} \atop {y=-\frac{4}{5}z+4\frac{4}{5}}} \right. \longrightarrow\left \{ {{x=-\frac{3}{5}z+1\frac{3}{5}} \atop {y=-\frac{4}{5}z+4\frac{4}{5}}} \right.[/tex]
lub parametrycznie
[tex]\left\{\begin{array}{l}x=-\frac{3}{5}t+1\frac{3}{5}\\y=-\frac{4}{5}t+4\frac{4}{5}\\z=t\end{array} \right[/tex]
