Question

Допоможіть будь ласка з прикладом СЛАР методом Гаусса

Answer 1

Ответ:

$(\frac1{16};\frac18;-\frac3{16})$

Пошаговое объяснение:

$\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left\{\begin{array}{ccc}5x-y+z=0\\3x+y+2z=-1\\x+y+z=0\end{array}\right\\\\\left(\begin{array}{ccc}5&-1&1\\3&1&2\\1&1&1\end{array}\right|\left\begin{array}{ccc}0\\-1\\0\end{array}\right)\asymp\left(\begin{array}{ccc}5&-1&1\\3&1&2\\6&0&2\end{array}\right|\left\begin{array}{ccc}0\\-1\\0\end{array}\right)\asymp\left(\begin{array}{ccc}5&-1&1\\8&0&3\\6&0&2\end{array}\right|\left\begin{array}{ccc}0\\-1\\0\end{array}\right)\asymp$

$\displaystyle \asymp\left(\begin{array}{ccc}5&-1&1\\6&0&\frac94\\6&0&2\end{array}\right|\left\begin{array}{ccc}0\\-\frac34\\0\end{array}\right)\asymp\left(\begin{array}{ccc}5&-1&1\\6&0&\frac94\\0&0&\frac14\end{array}\right|\left\begin{array}{ccc}0\\-\frac34\\-\frac34\end{array}\right)\asymp\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right|\left\begin{array}{ccc}\frac1{16}\\\frac18\\-\frac3{16}\end{array}\right)$

Тогда

$\left\{\begin{array}{ccc}5x-y+z=0\\3x+y+2z=-1\\x+y+z=0\end{array}\right \Leftrightarrow \left\{\begin{array}{ccc}x=\frac1{16}\\y=\frac18\\z=-\frac3{16}\end{array}\right$