Решение систем линейных уравнений - примеры

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Метод Крамера

Задание:


Решить систему уравнений

$$\left\{\begin{array}{l}{{{z+2 x-y}={2}}}\\{{{-\frac{1}{4} z-x-\frac{1}{2} y}={5}}}\\{{{-5 x+y}={3}}}\end{array}\right.$$

Решение:


$$\left\{\begin{array}{l}{{{z+2 x-y}={2}}}\\{{{-\frac{1}{4} z-x-\frac{1}{2} y}={5}}}\\{{{-5 x+y}={3}}}\end{array}\right.$$

Представим систему в матричном виде:

$${{{{\left(\begin{array}{ccc}2&-1&1\\-1&-\frac{1}{2}&-\frac{1}{4}\\-5&1&0\end{array}\right)}\cdot {\left(\begin{array}{c}x\\y\\z\end{array}\right)}}}={\left(\begin{array}{c}2\\5\\3\end{array}\right)}}$$

$${{x}={{\frac {{\Delta}_{{0} }}{\Delta}}}}$$ ;      $${{y}={{\frac {{\Delta}_{{1} }}{\Delta}}}}$$ ;      $${{z}={{\frac {{\Delta}_{{2} }}{\Delta}}}}$$ ;

$$\Delta$$ = $${\begin{vmatrix}{2}&{-1}&{1}\\{-1}&{-\frac{1}{2}}&{-\frac{1}{4}}\\{-5}&{1}&{0}\end{vmatrix}}$$ = $${\begin{vmatrix}{{{2}-{{{1}\cdot {{\frac {2}{1}}}}}}}&{{{-1}-{{{1}\cdot {{\frac {-1}{1}}}}}}}&{1}\\{{{-1}-\left({{{-\frac{1}{4}}\cdot {{\frac {2}{1}}}}}\right)}}&{{{-\frac{1}{2}}-\left({{{-\frac{1}{4}}\cdot {{\frac {-1}{1}}}}}\right)}}&{-\frac{1}{4}}\\{{{-5}-{{{0}\cdot {{\frac {2}{1}}}}}}}&{{{1}-{{{0}\cdot {{\frac {-1}{1}}}}}}}&{0}\end{vmatrix}}$$ = $${\begin{vmatrix}{0}&{0}&{1}\\{-\frac{1}{2}}&{-\frac{3}{4}}&{-\frac{1}{4}}\\{-5}&{1}&{0}\end{vmatrix}}$$ = $${{1}\cdot {{\begin{vmatrix}{-\frac{1}{2}}&{-\frac{3}{4}}\\{-5}&{1}\end{vmatrix}}}}$$ = $${{1}\cdot \left({{{{{-\frac{1}{2}}\cdot {1}}}-\left({{{-\frac{3}{4}}\cdot \left({-5}\right)}}\right)}}\right)}$$ = $$-\frac{17}{4}$$

$${\Delta}_{{0} }$$ = $${\begin{vmatrix}{2}&{-1}&{1}\\{5}&{-\frac{1}{2}}&{-\frac{1}{4}}\\{3}&{1}&{0}\end{vmatrix}}$$ = $${\begin{vmatrix}{{{2}-{{{1}\cdot {{\frac {2}{1}}}}}}}&{{{-1}-{{{1}\cdot {{\frac {-1}{1}}}}}}}&{1}\\{{{5}-\left({{{-\frac{1}{4}}\cdot {{\frac {2}{1}}}}}\right)}}&{{{-\frac{1}{2}}-\left({{{-\frac{1}{4}}\cdot {{\frac {-1}{1}}}}}\right)}}&{-\frac{1}{4}}\\{{{3}-{{{0}\cdot {{\frac {2}{1}}}}}}}&{{{1}-{{{0}\cdot {{\frac {-1}{1}}}}}}}&{0}\end{vmatrix}}$$ = $${\begin{vmatrix}{0}&{0}&{1}\\{\frac{11}{2}}&{-\frac{3}{4}}&{-\frac{1}{4}}\\{3}&{1}&{0}\end{vmatrix}}$$ = $${{1}\cdot {{\begin{vmatrix}{\frac{11}{2}}&{-\frac{3}{4}}\\{3}&{1}\end{vmatrix}}}}$$ = $${{1}\cdot \left({{{{{\frac{11}{2}}\cdot {1}}}-\left({{{-\frac{3}{4}}\cdot {3}}}\right)}}\right)}$$ = $$\frac{31}{4}$$

$${\Delta}_{{1} }$$ = $${\begin{vmatrix}{2}&{2}&{1}\\{-1}&{5}&{-\frac{1}{4}}\\{-5}&{3}&{0}\end{vmatrix}}$$ = $${\begin{vmatrix}{{{2}-{{{1}\cdot {{\frac {2}{1}}}}}}}&{{{2}-{{{1}\cdot {{\frac {2}{1}}}}}}}&{1}\\{{{-1}-\left({{{-\frac{1}{4}}\cdot {{\frac {2}{1}}}}}\right)}}&{{{5}-\left({{{-\frac{1}{4}}\cdot {{\frac {2}{1}}}}}\right)}}&{-\frac{1}{4}}\\{{{-5}-{{{0}\cdot {{\frac {2}{1}}}}}}}&{{{3}-{{{0}\cdot {{\frac {2}{1}}}}}}}&{0}\end{vmatrix}}$$ = $${\begin{vmatrix}{0}&{0}&{1}\\{-\frac{1}{2}}&{\frac{11}{2}}&{-\frac{1}{4}}\\{-5}&{3}&{0}\end{vmatrix}}$$ = $${{1}\cdot {{\begin{vmatrix}{-\frac{1}{2}}&{\frac{11}{2}}\\{-5}&{3}\end{vmatrix}}}}$$ = $${{1}\cdot \left({{{{{-\frac{1}{2}}\cdot {3}}}-{{{\frac{11}{2}}\cdot \left({-5}\right)}}}}\right)}$$ = $$26$$

$${\Delta}_{{2} }$$ = $${\begin{vmatrix}{2}&{-1}&{2}\\{-1}&{-\frac{1}{2}}&{5}\\{-5}&{1}&{3}\end{vmatrix}}$$ = $${\begin{vmatrix}{{{2}-\left({{{-1}\cdot {{\frac {-5}{1}}}}}\right)}}&{-1}&{{{2}-\left({{{-1}\cdot {{\frac {3}{1}}}}}\right)}}\\{{{-1}-\left({{{-\frac{1}{2}}\cdot {{\frac {-5}{1}}}}}\right)}}&{-\frac{1}{2}}&{{{5}-\left({{{-\frac{1}{2}}\cdot {{\frac {3}{1}}}}}\right)}}\\{{{-5}-{{{1}\cdot {{\frac {-5}{1}}}}}}}&{1}&{{{3}-{{{1}\cdot {{\frac {3}{1}}}}}}}\end{vmatrix}}$$ = $${\begin{vmatrix}{-3}&{-1}&{5}\\{-\frac{7}{2}}&{-\frac{1}{2}}&{\frac{13}{2}}\\{0}&{1}&{0}\end{vmatrix}}$$ = $${-{{{1}\cdot {{\begin{vmatrix}{-3}&{5}\\{-\frac{7}{2}}&{\frac{13}{2}}\end{vmatrix}}}}}}$$ = $${-{{{1}\cdot \left({{{{{-3}\cdot {\frac{13}{2}}}}-{{{5}\cdot \left({-\frac{7}{2}}\right)}}}}\right)}}}$$ = $$2$$

$${{x}={-\frac{31}{17}}}$$ ;      $${{y}={-\frac{104}{17}}}$$ ;      $${{z}={-\frac{8}{17}}}$$ ;

Проверка

$$\left\{\begin{array}{l}{{{2}={2}}}\\{{{5}={5}}}\\{{{3}={3}}}\end{array}\right.$$

Ответ:


$${{x}={-\frac{31}{17}}}$$ ;      $${{y}={-\frac{104}{17}}}$$ ;      $${{z}={-\frac{8}{17}}}$$ ;



Метод Гаусса

Задание:


Решить систему уравнений

$$\left\{\begin{array}{l}{{{z+2 x-y}={2}}}\\{{{-\frac{1}{4} z-x-\frac{1}{2} y}={5}}}\\{{{-5 x+y}={3}}}\end{array}\right.$$

Решение:


$$\left\{\begin{array}{l}{{{z+2 x-y}={2}}}\\{{{-\frac{1}{4} z-x-\frac{1}{2} y}={5}}}\\{{{-5 x+y}={3}}}\end{array}\right.$$

$${\left(\begin{array}{ccc|c}{2}&{-1}&{1}&{2}\\{-1}&{-\frac{1}{2}}&{-\frac{1}{4}}&{5}\\{-5}&{1}&{0}&{3}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{-\frac{1}{2}}&{\frac{1}{2}}&{1}\\{-1}&{-\frac{1}{2}}&{-\frac{1}{4}}&{5}\\{-5}&{1}&{0}&{3}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{-\frac{1}{2}}&{\frac{1}{2}}&{1}\\{0}&{-1}&{\frac{1}{4}}&{6}\\{-5}&{1}&{0}&{3}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{-\frac{1}{2}}&{\frac{1}{2}}&{1}\\{0}&{-1}&{\frac{1}{4}}&{6}\\{0}&{-\frac{3}{2}}&{\frac{5}{2}}&{8}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{-\frac{1}{2}}&{\frac{1}{2}}&{1}\\{0}&{1}&{-\frac{1}{4}}&{-6}\\{0}&{-\frac{3}{2}}&{\frac{5}{2}}&{8}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{\frac{3}{8}}&{-2}\\{0}&{1}&{-\frac{1}{4}}&{-6}\\{0}&{-\frac{3}{2}}&{\frac{5}{2}}&{8}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{\frac{3}{8}}&{-2}\\{0}&{1}&{-\frac{1}{4}}&{-6}\\{0}&{0}&{\frac{17}{8}}&{-1}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{\frac{3}{8}}&{-2}\\{0}&{1}&{-\frac{1}{4}}&{-6}\\{0}&{0}&{1}&{-\frac{8}{17}}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{0}&{-\frac{31}{17}}\\{0}&{1}&{-\frac{1}{4}}&{-6}\\{0}&{0}&{1}&{-\frac{8}{17}}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{0}&{-\frac{31}{17}}\\{0}&{1}&{0}&{-\frac{104}{17}}\\{0}&{0}&{1}&{-\frac{8}{17}}\end{array}\right)}$$

$$\left\{\begin{array}{l}{{{x}={-\frac{31}{17}}}}\\{{{y}={-\frac{104}{17}}}}\\{{{z}={-\frac{8}{17}}}}\end{array}\right.$$

Проверка:

$$\left\{\begin{array}{l}{{{2}={2}}}\\{{{5}={5}}}\\{{{3}={3}}}\end{array}\right.$$

Ответ:

$$\left\{\begin{array}{l}{{{x}={-\frac{31}{17}}}}\\{{{y}={-\frac{104}{17}}}}\\{{{z}={-\frac{8}{17}}}}\end{array}\right.$$



Метод Обратных матриц

Задание:


Решить систему уравнений

$$\left\{\begin{array}{l}{{{z+2 x-y}={2}}}\\{{{-\frac{1}{4} z-x-\frac{1}{2} y}={5}}}\\{{{-5 x+y}={3}}}\end{array}\right.$$

Решение:


$$\left\{\begin{array}{l}{{{z+2 x-y}={2}}}\\{{{-\frac{1}{4} z-x-\frac{1}{2} y}={5}}}\\{{{-5 x+y}={3}}}\end{array}\right.$$

Представим систему в матричном виде:

$${{{{\left(\begin{array}{ccc}2&-1&1\\-1&-\frac{1}{2}&-\frac{1}{4}\\-5&1&0\end{array}\right)}\cdot {\left(\begin{array}{c}x\\y\\z\end{array}\right)}}}={\left(\begin{array}{c}2\\5\\3\end{array}\right)}}$$

$${{A}={\left(\begin{array}{ccc}2&-1&1\\-1&-\frac{1}{2}&-\frac{1}{4}\\-5&1&0\end{array}\right)}}$$ ;      $${{X}={\left(\begin{array}{c}x\\y\\z\end{array}\right)}}$$ ;      $${{B}={\left(\begin{array}{c}2\\5\\3\end{array}\right)}}$$ ;

$${{X}={{{{{A}^{-1}}}\cdot {B}}}}$$

$${{A}={\left(\begin{array}{ccc}2&-1&1\\-1&-\frac{1}{2}&-\frac{1}{4}\\-5&1&0\end{array}\right)}}$$ ;

$${{A}^{-1}}$$ = $${\frac {\left(\begin{array}{ccc}{A}_{{1} {1} }&{A}_{{2} {1} }&{A}_{{3} {1} }\\{A}_{{1} {2} }&{A}_{{2} {2} }&{A}_{{3} {2} }\\{A}_{{1} {3} }&{A}_{{2} {3} }&{A}_{{3} {3} }\end{array}\right)}{|{A}|}}$$

$$|{A}|$$ = $${\begin{vmatrix}{2}&{-1}&{1}\\{-1}&{-\frac{1}{2}}&{-\frac{1}{4}}\\{-5}&{1}&{0}\end{vmatrix}}$$ = $${\begin{vmatrix}{{{2}-{{{1}\cdot {{\frac {2}{1}}}}}}}&{{{-1}-{{{1}\cdot {{\frac {-1}{1}}}}}}}&{1}\\{{{-1}-\left({{{-\frac{1}{4}}\cdot {{\frac {2}{1}}}}}\right)}}&{{{-\frac{1}{2}}-\left({{{-\frac{1}{4}}\cdot {{\frac {-1}{1}}}}}\right)}}&{-\frac{1}{4}}\\{{{-5}-{{{0}\cdot {{\frac {2}{1}}}}}}}&{{{1}-{{{0}\cdot {{\frac {-1}{1}}}}}}}&{0}\end{vmatrix}}$$ = $${\begin{vmatrix}{0}&{0}&{1}\\{-\frac{1}{2}}&{-\frac{3}{4}}&{-\frac{1}{4}}\\{-5}&{1}&{0}\end{vmatrix}}$$ = $${{1}\cdot {{\begin{vmatrix}{-\frac{1}{2}}&{-\frac{3}{4}}\\{-5}&{1}\end{vmatrix}}}}$$ = $${{1}\cdot \left({{{{{-\frac{1}{2}}\cdot {1}}}-\left({{{-\frac{3}{4}}\cdot \left({-5}\right)}}\right)}}\right)}$$ = $$-\frac{17}{4}$$

$${A}_{{1} {1} }$$ = $${\begin{vmatrix}{-\frac{1}{2}}&{-\frac{1}{4}}\\{1}&{0}\end{vmatrix}}$$ = $${{{{-\frac{1}{2}}\cdot {0}}}-\left({{{-\frac{1}{4}}\cdot {1}}}\right)}$$ = $$\frac{1}{4}$$

$$- {A}_{{2} {1} }$$ = $${\begin{vmatrix}{-1}&{1}\\{1}&{0}\end{vmatrix}}$$ = $${{{{-1}\cdot {0}}}-{{{1}\cdot {1}}}}$$ = $$-1$$

$${A}_{{3} {1} }$$ = $${\begin{vmatrix}{-1}&{1}\\{-\frac{1}{2}}&{-\frac{1}{4}}\end{vmatrix}}$$ = $${{{{-1}\cdot \left({-\frac{1}{4}}\right)}}-{{{1}\cdot \left({-\frac{1}{2}}\right)}}}$$ = $$\frac{3}{4}$$

$$- {A}_{{1} {2} }$$ = $${\begin{vmatrix}{-1}&{-\frac{1}{4}}\\{-5}&{0}\end{vmatrix}}$$ = $${{{{-1}\cdot {0}}}-\left({{{-\frac{1}{4}}\cdot \left({-5}\right)}}\right)}$$ = $$-\frac{5}{4}$$

$${A}_{{2} {2} }$$ = $${\begin{vmatrix}{2}&{1}\\{-5}&{0}\end{vmatrix}}$$ = $${{{{2}\cdot {0}}}-{{{1}\cdot \left({-5}\right)}}}$$ = $$5$$

$$- {A}_{{3} {2} }$$ = $${\begin{vmatrix}{2}&{1}\\{-1}&{-\frac{1}{4}}\end{vmatrix}}$$ = $${{{{2}\cdot \left({-\frac{1}{4}}\right)}}-{{{1}\cdot \left({-1}\right)}}}$$ = $$\frac{1}{2}$$

$${A}_{{1} {3} }$$ = $${\begin{vmatrix}{-1}&{-\frac{1}{2}}\\{-5}&{1}\end{vmatrix}}$$ = $${{{{-1}\cdot {1}}}-\left({{{-\frac{1}{2}}\cdot \left({-5}\right)}}\right)}$$ = $$-\frac{7}{2}$$

$$- {A}_{{2} {3} }$$ = $${\begin{vmatrix}{2}&{-1}\\{-5}&{1}\end{vmatrix}}$$ = $${{{{2}\cdot {1}}}-\left({{{-1}\cdot \left({-5}\right)}}\right)}$$ = $$-3$$

$${A}_{{3} {3} }$$ = $${\begin{vmatrix}{2}&{-1}\\{-1}&{-\frac{1}{2}}\end{vmatrix}}$$ = $${{{{2}\cdot \left({-\frac{1}{2}}\right)}}-\left({{{-1}\cdot \left({-1}\right)}}\right)}$$ = $$-2$$

$${{A}^{-1}}$$ = $${\frac {\left(\begin{array}{ccc}\frac{1}{4}&1&\frac{3}{4}\\\frac{5}{4}&5&-\frac{1}{2}\\-\frac{7}{2}&3&-2\end{array}\right)}{-\frac{17}{4}}}$$ = $$\left(\begin{array}{ccc}-\frac{1}{17}&-\frac{4}{17}&-\frac{3}{17}\\-\frac{5}{17}&-\frac{20}{17}&\frac{2}{17}\\\frac{14}{17}&-\frac{12}{17}&\frac{8}{17}\end{array}\right)$$

$$X$$ = $${{{{A}^{-1}}}\cdot {B}}$$ = $$\left(\begin{array}{c}{{{{{{-\frac{1}{17}}\cdot {2}}}+\left({{{-\frac{4}{17}}\cdot {5}}}\right)}}+\left({{{-\frac{3}{17}}\cdot {3}}}\right)}\\{{{{{{-\frac{5}{17}}\cdot {2}}}+\left({{{-\frac{20}{17}}\cdot {5}}}\right)}}+{{{\frac{2}{17}}\cdot {3}}}}\\{{{{{{\frac{14}{17}}\cdot {2}}}+\left({{{-\frac{12}{17}}\cdot {5}}}\right)}}+{{{\frac{8}{17}}\cdot {3}}}}\end{array}\right)$$ = $$\left(\begin{array}{c}-\frac{31}{17}\\-\frac{104}{17}\\-\frac{8}{17}\end{array}\right)$$

$$\left\{\begin{array}{l}{{{x}={-\frac{31}{17}}}}\\{{{y}={-\frac{104}{17}}}}\\{{{z}={-\frac{8}{17}}}}\end{array}\right.$$

Проверка

$$\left\{\begin{array}{l}{{{2}={2}}}\\{{{5}={5}}}\\{{{3}={3}}}\end{array}\right.$$

Ответ:

$$\left\{\begin{array}{l}{{{x}={-\frac{31}{17}}}}\\{{{y}={-\frac{104}{17}}}}\\{{{z}={-\frac{8}{17}}}}\end{array}\right.$$