Пример нахождения обратной матрицы
Пример. Найти обратную матрицу для матрицы \(A = \left( {\begin{array}{*{20}{c}} {{\text{ }}1}&0&2 \\ { - 1}&3&2 \\ {{\text{ }}1}&4&0 \end{array}} \right)\).
Решение. Вычислим определитель матрицы \(A\) $$\left| {A{\text{ }}} \right| = \left| {\begin{array}{*{20}{c}} {{\text{ }}1}&0&2 \\ { - 1}&3&2 \\ {{\text{ }}1}&4&0 \end{array}} \right| = 1 \cdot 3 \cdot 0 + 0 \cdot 2 \cdot 1 + ( - 1) \cdot 4 \cdot 2 - 2 \cdot 3 \cdot 1 - 1 \cdot 2 \cdot 4 - 0 \cdot ( - 1) \cdot 0 = - 22.$$ Так как \(\left| {A{\text{ }}} \right| \ne 0\), то обратная матрица существует $$A{{\text{ }}^{ - 1}} = - \frac{1}{{22}}\left( {\begin{array}{*{20}{c}} {A{{\text{ }}_{11}}}&{A{{\text{ }}_{21}}}&{A{{\text{ }}_{31}}} \\ {A{{\text{ }}_{12}}}&{A{{\text{ }}_{22}}}&{A{{\text{ }}_{32}}} \\ {A{{\text{ }}_{13}}}&{A{{\text{ }}_{23}}}&{A{{\text{ }}_{33}}} \end{array}} \right)$$ Найдём алгебраические дополнения \({A_{ij}}\) элементов матрицы: $$\begin{gathered} A{{\text{ }}_{11}} = \left| {\begin{array}{*{20}{c}} 3&2 \\ 4&0 \end{array}} \right| = - 8,{\text{ }}A{{\text{ }}_{12}} = - \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ {{\text{ }}1}&0 \end{array}} \right| = 2,{\text{ }}A{{\text{ }}_{13}} = \left| {\begin{array}{*{20}{c}} { - 1}&3 \\ {{\text{ }}1}&4 \end{array}} \right| = - 7, \\ A{{\text{ }}_{21}} = - \left| {\begin{array}{*{20}{c}} 0&2 \\ 4&0 \end{array}} \right| = 8,{\text{ }}A{{\text{ }}_{22}} = \left| {\begin{array}{*{20}{c}} 1&2 \\ 1&0 \end{array}} \right| = - 2,{\text{ }}A{{\text{ }}_{23}} = - \left| {\begin{array}{*{20}{c}} 1&0 \\ 1&4 \end{array}} \right| = - 4, \\ \end{gathered} $$ $$A{{\text{ }}_{31}} = \left| {\begin{array}{*{20}{c}} 0&2 \\ 3&2 \end{array}} \right| = - 6,{\text{ }}A{{\text{ }}_{32}} = - \left| {\begin{array}{*{20}{c}} {{\text{ }}1}&2 \\ { - 1}&2 \end{array}} \right| = - 4,{\text{ }}A{{\text{ }}_{33}} = \left| {\begin{array}{*{20}{c}} {{\text{ }}1}&0 \\ { - 1}&3 \end{array}} \right| = 3.$$ Таким образом, $$A{{\text{ }}^{ - 1}} = - \frac{1}{{22}}\left( {\begin{array}{*{20}{c}} { - 8}&{{\text{ }}8}&{ - 6} \\ {{\text{ }}2}&{ - 2}&{ - 4} \\ { - 7}&{ - 4}&{{\text{ }}3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{\text{ }}\frac{4}{{11}}}&{ - \frac{4}{{11}}}&{{\text{ }}\frac{3}{{11}}} \\ { - \frac{1}{{11}}}&{{\text{ }}\frac{1}{{11}}}&{{\text{ }}\frac{2}{{11}}} \\ {{\text{ }}\frac{7}{{22}}}&{{\text{ }}\frac{2}{{11}}}&{ - \frac{3}{{22}}} \end{array}} \right)$$
Решение. Вычислим определитель матрицы \(A\) $$\left| {A{\text{ }}} \right| = \left| {\begin{array}{*{20}{c}} {{\text{ }}1}&0&2 \\ { - 1}&3&2 \\ {{\text{ }}1}&4&0 \end{array}} \right| = 1 \cdot 3 \cdot 0 + 0 \cdot 2 \cdot 1 + ( - 1) \cdot 4 \cdot 2 - 2 \cdot 3 \cdot 1 - 1 \cdot 2 \cdot 4 - 0 \cdot ( - 1) \cdot 0 = - 22.$$ Так как \(\left| {A{\text{ }}} \right| \ne 0\), то обратная матрица существует $$A{{\text{ }}^{ - 1}} = - \frac{1}{{22}}\left( {\begin{array}{*{20}{c}} {A{{\text{ }}_{11}}}&{A{{\text{ }}_{21}}}&{A{{\text{ }}_{31}}} \\ {A{{\text{ }}_{12}}}&{A{{\text{ }}_{22}}}&{A{{\text{ }}_{32}}} \\ {A{{\text{ }}_{13}}}&{A{{\text{ }}_{23}}}&{A{{\text{ }}_{33}}} \end{array}} \right)$$ Найдём алгебраические дополнения \({A_{ij}}\) элементов матрицы: $$\begin{gathered} A{{\text{ }}_{11}} = \left| {\begin{array}{*{20}{c}} 3&2 \\ 4&0 \end{array}} \right| = - 8,{\text{ }}A{{\text{ }}_{12}} = - \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ {{\text{ }}1}&0 \end{array}} \right| = 2,{\text{ }}A{{\text{ }}_{13}} = \left| {\begin{array}{*{20}{c}} { - 1}&3 \\ {{\text{ }}1}&4 \end{array}} \right| = - 7, \\ A{{\text{ }}_{21}} = - \left| {\begin{array}{*{20}{c}} 0&2 \\ 4&0 \end{array}} \right| = 8,{\text{ }}A{{\text{ }}_{22}} = \left| {\begin{array}{*{20}{c}} 1&2 \\ 1&0 \end{array}} \right| = - 2,{\text{ }}A{{\text{ }}_{23}} = - \left| {\begin{array}{*{20}{c}} 1&0 \\ 1&4 \end{array}} \right| = - 4, \\ \end{gathered} $$ $$A{{\text{ }}_{31}} = \left| {\begin{array}{*{20}{c}} 0&2 \\ 3&2 \end{array}} \right| = - 6,{\text{ }}A{{\text{ }}_{32}} = - \left| {\begin{array}{*{20}{c}} {{\text{ }}1}&2 \\ { - 1}&2 \end{array}} \right| = - 4,{\text{ }}A{{\text{ }}_{33}} = \left| {\begin{array}{*{20}{c}} {{\text{ }}1}&0 \\ { - 1}&3 \end{array}} \right| = 3.$$ Таким образом, $$A{{\text{ }}^{ - 1}} = - \frac{1}{{22}}\left( {\begin{array}{*{20}{c}} { - 8}&{{\text{ }}8}&{ - 6} \\ {{\text{ }}2}&{ - 2}&{ - 4} \\ { - 7}&{ - 4}&{{\text{ }}3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{\text{ }}\frac{4}{{11}}}&{ - \frac{4}{{11}}}&{{\text{ }}\frac{3}{{11}}} \\ { - \frac{1}{{11}}}&{{\text{ }}\frac{1}{{11}}}&{{\text{ }}\frac{2}{{11}}} \\ {{\text{ }}\frac{7}{{22}}}&{{\text{ }}\frac{2}{{11}}}&{ - \frac{3}{{22}}} \end{array}} \right)$$
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