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Ответ.

Метод Гаусса решения СЛАУ .

(обозначене: str - cтрока)

[tex]\left\{\begin{array}{ccc}2x-3y+z=14\\3x+2y+3z=3\\-x+2y+2z=-6\end{array}\right\\\\\\\left(\begin{array}{cccl}-1&2&2&|-6\\3&2&3&|\ 3\\2&-3&1&|\ 14\end{array}\right)\sim \ \ 1str\cdot 3+2str\ ;\ 1str\cdot 2+3str[/tex]

[tex]\sim \left(\begin{array}{cccl}-1&2&2&|-6\\0&8&9&|-15\\0&1&5&|\ \ 2\end{array}\right)2str\longleftrightarrow 3str\left(\begin{array}{cccl}-1&2&2&|-6\\0&1&5&|\ \ 2\\0&8&9&|-15\end{array}\right)\sim \\\\\\2str\cdot (-8)+3str\\\\\\\sim \left(\begin{array}{cccl}-1&2&2&|-6\\0&1&5&|\ \ 2\\0&0&-31&|-31\end{array}\right)[/tex]

[tex]\left\{\begin{array}{cccl}-x+2y+2z=-6\\\qquad \ \ y+5z=2\\\quad \qquad -31z=-31\end{array}\right\ \ \left\{\begin{array}{cccl}-x+2y+2=-6\\\qquad \ \ y+5=2\\\quad \qquad z=1\end{array}\right\ \ \left\{\begin{array}{cccl}-x-2\cdot 3=-8\\\qquad \ \ y=-3\\\quad \qquad z=1\end{array}\right\\\\\\\left\{\begin{array}{c}x=2\\\, y=-3\\z=1\end{array}\right\ \ \ \ \ Otvet:\ \ (\, 2\, ;-3\, ;\, 1\, )\ .[/tex]