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

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