Вычислить обратную матрицу - пример

Подождите несколько секунд до окончания загрузки формул! (Подробнее)

Задание:

Найти матрицу, обратную матрице $${{A}={\left(\begin{array}{ccc}5&1&2\\0&7&-4\\2&6&3\end{array}\right)}}$$

Решение:


$${{A}={\left(\begin{array}{ccc}5&1&2\\0&7&-4\\2&6&3\end{array}\right)}}$$ ;

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

$$|{A}|$$ = $${\begin{vmatrix}{5}&{1}&{2}\\{0}&{7}&{-4}\\{2}&{6}&{3}\end{vmatrix}}$$ = $${\begin{vmatrix}{{{5}-{{{1}\cdot {{\frac {5}{1}}}}}}}&{1}&{{{2}-{{{1}\cdot {{\frac {2}{1}}}}}}}\\{{{0}-{{{7}\cdot {{\frac {5}{1}}}}}}}&{7}&{{{-4}-{{{7}\cdot {{\frac {2}{1}}}}}}}\\{{{2}-{{{6}\cdot {{\frac {5}{1}}}}}}}&{6}&{{{3}-{{{6}\cdot {{\frac {2}{1}}}}}}}\end{vmatrix}}$$ = $${\begin{vmatrix}{0}&{1}&{0}\\{-35}&{7}&{-18}\\{-28}&{6}&{-9}\end{vmatrix}}$$ = $${-{{{1}\cdot {{\begin{vmatrix}{-35}&{-18}\\{-28}&{-9}\end{vmatrix}}}}}}$$ = $${-{{{1}\cdot \left({{{{{-35}\cdot \left({-9}\right)}}-\left({{{-18}\cdot \left({-28}\right)}}\right)}}\right)}}}$$ = $$189$$

$${A}_{{1} {1} }$$ = $${\begin{vmatrix}{7}&{-4}\\{6}&{3}\end{vmatrix}}$$ = $${{{{7}\cdot {3}}}-\left({{{-4}\cdot {6}}}\right)}$$ = $$45$$

$$- {A}_{{2} {1} }$$ = $${\begin{vmatrix}{1}&{2}\\{6}&{3}\end{vmatrix}}$$ = $${{{{1}\cdot {3}}}-{{{2}\cdot {6}}}}$$ = $$-9$$

$${A}_{{3} {1} }$$ = $${\begin{vmatrix}{1}&{2}\\{7}&{-4}\end{vmatrix}}$$ = $${{{{1}\cdot \left({-4}\right)}}-{{{2}\cdot {7}}}}$$ = $$-18$$

$$- {A}_{{1} {2} }$$ = $${\begin{vmatrix}{0}&{-4}\\{2}&{3}\end{vmatrix}}$$ = $${{{{0}\cdot {3}}}-\left({{{-4}\cdot {2}}}\right)}$$ = $$8$$

$${A}_{{2} {2} }$$ = $${\begin{vmatrix}{5}&{2}\\{2}&{3}\end{vmatrix}}$$ = $${{{{5}\cdot {3}}}-{{{2}\cdot {2}}}}$$ = $$11$$

$$- {A}_{{3} {2} }$$ = $${\begin{vmatrix}{5}&{2}\\{0}&{-4}\end{vmatrix}}$$ = $${{{{5}\cdot \left({-4}\right)}}-{{{2}\cdot {0}}}}$$ = $$-20$$

$${A}_{{1} {3} }$$ = $${\begin{vmatrix}{0}&{7}\\{2}&{6}\end{vmatrix}}$$ = $${{{{0}\cdot {6}}}-{{{7}\cdot {2}}}}$$ = $$-14$$

$$- {A}_{{2} {3} }$$ = $${\begin{vmatrix}{5}&{1}\\{2}&{6}\end{vmatrix}}$$ = $${{{{5}\cdot {6}}}-{{{1}\cdot {2}}}}$$ = $$28$$

$${A}_{{3} {3} }$$ = $${\begin{vmatrix}{5}&{1}\\{0}&{7}\end{vmatrix}}$$ = $${{{{5}\cdot {7}}}-{{{1}\cdot {0}}}}$$ = $$35$$

$${{A}^{-1}}$$ = $${\frac {\left(\begin{array}{ccc}45&9&-18\\-8&11&20\\-14&-28&35\end{array}\right)}{189}}$$ = $$\left(\begin{array}{ccc}\frac{5}{21}&\frac{1}{21}&-\frac{2}{21}\\-\frac{8}{189}&\frac{11}{189}&\frac{20}{189}\\-\frac{2}{27}&-\frac{4}{27}&\frac{5}{27}\end{array}\right)$$

Проверка

$${{A}\cdot {{{A}^{-1}}}}$$ = $${{\left(\begin{array}{ccc}5&1&2\\0&7&-4\\2&6&3\end{array}\right)}\cdot {\left(\begin{array}{ccc}\frac{5}{21}&\frac{1}{21}&-\frac{2}{21}\\-\frac{8}{189}&\frac{11}{189}&\frac{20}{189}\\-\frac{2}{27}&-\frac{4}{27}&\frac{5}{27}\end{array}\right)}}$$ = $$\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$$

Результат верный!

Ответ:

$${{{{A}^{-1}}}={\left(\begin{array}{ccc}\frac{5}{21}&\frac{1}{21}&-\frac{2}{21}\\-\frac{8}{189}&\frac{11}{189}&\frac{20}{189}\\-\frac{2}{27}&-\frac{4}{27}&\frac{5}{27}\end{array}\right)}}$$