Question

Решение системы 3-х линейных уравнений с тремя неизвестными. Метод Крамера.Метод подстановки. Метод Гаусса.

3х+4у+2z=8
2x-y-3z=-1
x+5y+z=0

Answer 1

$\left\{\begin{array}{c}3x+4y+2z=8,\\2x-y-3z=-1,\\x+5y+z=0;\end{array}\right.$

Метод Крамера.
$\Delta= \left|\begin{array}{ccc}3&4&2\\2&-1&-3\\1&5&1\end{array}\right|=44, \\ \Delta_1= \left|\begin{array}{ccc}8&4&2\\-1&-1&-3\\0&5&1\end{array}\right|=106, \\ \Delta_2= \left|\begin{array}{ccc}3&8&2\\2&-1&-3\\1&0&1\end{array}\right|=-41, \\ \Delta_3= \left|\begin{array}{ccc}3&4&8\\2&-1&-1\\1&5&0\end{array}\right|=99, \\ x=\frac{\Delta_1}{\Delta}=\frac{53}{22}=2\frac{9}{22}, \\ y=\frac{\Delta_2}{\Delta}=-\frac{41}{44}, \\ z=\frac{\Delta_3}{\Delta}=\frac{9}{4}=2\frac{1}{4};$

Метод подстановки.
$\left\{\begin{array}{c}x=\frac{8}{3}-\frac{4}{3}y-\frac{2}{3}z,\\2x-y-3z=-1,\\x+5y+z=0;\end{array}\right. \left\{\begin{array}{c}x=\frac{8}{3}-\frac{4}{3}y-\frac{2}{3}z,\\2(\frac{8}{3}-\frac{4}{3}y-\frac{2}{3}z)-y-3z=-1,\\\frac{8}{3}-\frac{4}{3}y-\frac{2}{3}z+5y+z=0;\end{array}\right.$
$\left\{\begin{array}{c}x=\frac{8}{3}-\frac{4}{3}y-\frac{2}{3}z,\\-\frac{11}{3}y-\frac{13}{3} z=-\frac{19}{3},\\\frac{11}{3}y+\frac{1}{3}z=-\frac{8}{3};\end{array}\right. \left\{\begin{array}{c}x=\frac{8}{3}-\frac{4}{3}y-\frac{2}{3}z,\\y=\frac{19}{11}-\frac{13}{11}z,\\\frac{11}{3}y+\frac{1}{3}z=-\frac{8}{3};\end{array}\right.$
$\left\{\begin{array}{c}x=\frac{8}{3}-\frac{4}{3}(\frac{19}{11}-\frac{13}{11}z)-\frac{2}{3}z,\\y=\frac{19}{11}-\frac{13}{11}z,\\\frac{11}{3}(\frac{19}{11}-\frac{13}{11}z)+\frac{1}{3}z=-\frac{8}{3};\end{array}\right. \left\{\begin{array}{c}x=\frac{4}{11}+\frac{10}{11}z,\\y=\frac{19}{11}-\frac{13}{11}z,\\-4z=-9;\end{array}\right.$
$\left\{\begin{array}{c}x=\frac{53}{22},\\y=-\frac{41}{44},\\z=\frac{9}{4}.\end{array}\right.$

Метод Гаусса.
$\left(\begin{array}{ccc|c}3&4&2&8\\2&-1&-3&-1\\1&5&1&0\end{array}\right)=\left(\begin{array}{ccc|c}1&\frac{4}{3}&\frac{2}{3}&\frac{8}{3}\\2&-1&-3&-1\\1&5&1&0\end{array}\right)=$
$=\left(\begin{array}{ccc|c}1&\frac{4}{3}&\frac{2}{3}&\frac{8}{3}\\0&-\frac{11}{3}&-\frac{13}{3}&-\frac{19}{3}\\0&\frac{11}{3}&\frac{1}{3}&-\frac{8}{3}\end{array}\right)=\left(\begin{array}{ccc|c}1&\frac{4}{3}&\frac{2}{3}&\frac{8}{3}\\0&1&\frac{13}{11}&\frac{19}{11}\\0&\frac{11}{3}&\frac{1}{3}&-\frac{8}{3}\end{array}\right)=$
$=\left(\begin{array}{ccc|c}1&0&-\frac{10}{11}&\frac{4}{11}\\0&1&\frac{13}{11}&\frac{19}{11}\\0&0&-4&-9\end{array}\right)=\left(\begin{array}{ccc|c}1&0&-\frac{10}{11}&\frac{4}{11}\\0&1&\frac{13}{11}&\frac{19}{11}\\0&0&1& \frac{9}{4}\end{array}\right)=$
$=\left(\begin{array}{ccc|c}1&0&0&\frac{53}{22}\\0&1&0&-\frac{41}{44}\\0&0&1&\frac{9}{4}\end{array}\right). \\ x=2\frac{9}{22}, y=-\frac{41}{44}, z=2\frac{1}{4}.$