基本的な行列の問題

■問題

次の行列 $3$\displaystyle{A}$ の逆行列 $3$\displaystyle{ {A}^{ - 1 } }$ を求めよ．

$3$\displaystyle{ A=\left({ \begin{array}1&2&3&4& \vspace{6}\\ 2&3&4&6& \vspace{6}\\ 3&4&6&9& \vspace{6}\\ 4&6&9& 10 & \vspace{6}\\ \end{array} }\right) }$

■答

$3$\displaystyle{ {A}^{ - 1 }=\left({ \begin{array} - \frac{\hspace{2}7\hspace{2}}{\hspace{2}3\hspace{2}} &0& \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ 0&3& - 2 &0& \vspace{6}\\ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} & - 2 & \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} &0& \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ \end{array} }\right) }$

■計算

まずは $3$\displaystyle{ \left({A}\right) }$ を求める．

$3$\displaystyle{ \left({A}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&2&3&4& \vspace{6}\\ 2&3&4&6& \vspace{6}\\ 3&4&6&9& \vspace{6}\\ 4&6&9&10& \vspace{6}\\ \end{array}}\right) }$

行列式の計算則を用いて2行+1行×(-2)，3行+1行×(-3)，4行+1行×(-4)の計算をする．

$3$\displaystyle{ =\left({\begin{array}1&2&3&4& \vspace{6}\\ 0& - 1 & - 2 & - 2 & \vspace{6}\\ 0& - 2 & - 3 & - 3 & \vspace{6}\\ 0& - 2 & - 3 & - 6 & \vspace{6}\\ \end{array}}\right) }$

次数下げの計算を用いて行列式の次数を1つ下げる．

$3$\displaystyle{ =\left({\begin{array} - 1 & - 2 & - 2 & \vspace{6}\\ - 2 & - 3 & - 3 & \vspace{6}\\ - 2 & - 3 & - 6 & \vspace{6}\\ \end{array}}\right) }$

行列式のすべての成分がマイナスなので，定数倍の性質を用いて各行から−1をくくりだす．

$3$\displaystyle{ =- \left({\begin{array}1&2&2& \vspace{6}\\ 2&3&3& \vspace{6}\\ 2&3&6& \vspace{6}\\ \end{array}}\right) }$

行列式の計算則を用いて2行+1行×(-2)，3行+1行×(-2)の計算をする．

$3$\displaystyle{ =- \left({\begin{array}1&2&2& \vspace{6}\\ 0& - 1 & - 1 & \vspace{6}\\ 0& - 1 & 2 & \vspace{6}\\ \end{array}}\right) }$

次数下げの計算を用いて行列式の次数を1つ下げる．

$3$\displaystyle{ =- \left({\begin{array} - 1 & - 1 & \vspace{6}\\ - 1 & 2 & \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ =- \left({ - 2- 1 }\right) }$

$3$\displaystyle{ =3 }$ $3$\displaystyle{ \neq 0 }$

各余因子を計算する．

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{11} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 1+1 }\left({\begin{array}3&4&6& \vspace{6}\\ 4&6&9& \vspace{6}\\ 6&9&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- \left({\begin{array}1&2&3& \vspace{6}\\ 4&6&9& \vspace{6}\\ 6&9&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- \left({\begin{array}1&2&3& \vspace{6}\\ 0& - 2 & - 3 & \vspace{6}\\ 0& - 3 & - 8 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- \left({\begin{array}2&3& \vspace{6}\\ 3&8& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 7 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{12} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 1+2 }\left({\begin{array}2&4&6& \vspace{6}\\ 3&6&9& \vspace{6}\\ 4&9&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&2&3& \vspace{6}\\ 3&6&9& \vspace{6}\\ 4&9&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&2&3& \vspace{6}\\ 0&0&0& \vspace{6}\\ 0&1& - 2 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}0&0& \vspace{6}\\ 1& - 2 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =0 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{13} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 1+3 }\left({\begin{array}2&3&6& \vspace{6}\\ 3&4&9& \vspace{6}\\ 4&6&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{=2 \| \begin{array}2&3&6& \vspace{6}\\ 3&4&9& \vspace{6}\\ 2&3&5& \vspace{6}\\ \end{array} \| }$ $3$\displaystyle{ =- 2\left({\begin{array}1&1&3& \vspace{6}\\ 0&1&0& \vspace{6}\\ 0&1& - 1 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 2\left({\begin{array}1&0& \vspace{6}\\ 1& - 1 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =2 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{14} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 1+4 }\left({\begin{array}2&3&4& \vspace{6}\\ 3&4&6& \vspace{6}\\ 4&6&9& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&1&2& \vspace{6}\\ 3&4&6& \vspace{6}\\ 4&6&9& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&1&2& \vspace{6}\\ 0&1&0& \vspace{6}\\ 0&2&1& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&0& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =1 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{21} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 2+1 }\left({\begin{array}2&3&4& \vspace{6}\\ 4&6&9& \vspace{6}\\ 6&9&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 6\left({\begin{array}1&1&4& \vspace{6}\\ 2&2&9& \vspace{6}\\ 3&3&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 6\left({\begin{array}1&1&4& \vspace{6}\\ 0&0&1& \vspace{6}\\ 0&0& - 2 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 6\left({\begin{array}0&1& \vspace{6}\\ 0& - 2 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =0 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{22} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 2+2 }\left({\begin{array}1&3&4& \vspace{6}\\ 3&6&9& \vspace{6}\\ 4&9&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&3&4& \vspace{6}\\ 0& - 3 & - 3 & \vspace{6}\\ 0& - 3 & - 6 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array} - 3 & - 3 & \vspace{6}\\ - 3 & - 6 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =9\left({\begin{array}1&1& \vspace{6}\\ 1&2& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =9 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{23} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 2+3 }\left({\begin{array}1&2&4& \vspace{6}\\ 3&4&9& \vspace{6}\\ 4&6&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- \left({\begin{array}1&2&4& \vspace{6}\\ 0& - 2 & - 3 & \vspace{6}\\ 0& - 2 & - 6 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- \left({\begin{array} - 2 & - 3 & \vspace{6}\\ - 2 & - 6 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 6\left({\begin{array}1&1& \vspace{6}\\ 1&2& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 6 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{24} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 2+4 }\left({\begin{array}1&2&3& \vspace{6}\\ 3&4&6& \vspace{6}\\ 4&6&9& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&2&3& \vspace{6}\\ 0& - 2 & - 3 & \vspace{6}\\ 0& - 2 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array} - 2 & - 3 & \vspace{6}\\ - 2 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =6\left({\begin{array}1&1& \vspace{6}\\ 1&1& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =0 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{31} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 3+1 }\left({\begin{array}2&3&4& \vspace{6}\\ 3&4&6& \vspace{6}\\ 6&9&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 2\left({\begin{array}1&1&1& \vspace{6}\\ 3&4&3& \vspace{6}\\ 6&9&5& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 2\left({\begin{array}1&1&1& \vspace{6}\\ 0&1&0& \vspace{6}\\ 0&3& - 1 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 2\left({\begin{array}1&0& \vspace{6}\\ 3& - 1 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 2 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{32} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 3+2 }\left({\begin{array}1&3&4& \vspace{6}\\ 2&4&6& \vspace{6}\\ 4&9&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 2\left({\begin{array}1&3&4& \vspace{6}\\ 1&2&3& \vspace{6}\\ 4&9&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 2\left({\begin{array}1&3&4& \vspace{6}\\ 0& - 1 & - 1 & \vspace{6}\\ 0& - 3 & - 6 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 2\left({\begin{array} - 1 & - 1 & \vspace{6}\\ - 3 & - 6 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 6 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{33} }$ $3$\displaystyle{ ={\left({ - 1 }\right)}^{ 3+3 }\left({\begin{array}1&2&4& \vspace{6}\\ 2&3&6& \vspace{6}\\ 4&6&10& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =2\left({\begin{array}1&2&4& \vspace{6}\\ 2&3&6& \vspace{6}\\ 2&3&5& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =2\left({\begin{array}1&2&4& \vspace{6}\\ 0& - 1 & - 2 & \vspace{6}\\ 0& - 1 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =2\left({\begin{array} - 1 & - 2 & \vspace{6}\\ - 1 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =2 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{34} }$ $3$\displaystyle{ ={ $- 1$ }^{ 3+4 }\left({\begin{array}1&2&3& \vspace{6}\\ 2&3&4& \vspace{6}\\ 4&6&9& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- \left({\begin{array}1&2&3& \vspace{6}\\ 0& - 1 &- 2& \vspace{6}\\ 0& - 2 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{=- \| \begin{array} - 1 & - 2 & \vspace{6}\\ - 2 & - 3 & \vspace{6}\\ \end{array} \| }$ $3$\displaystyle{ =- \left({\begin{array} - 1 &- 2& \vspace{6}\\ - 2 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =1 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{41} }$ $3$\displaystyle{ ={ $- 1$ }^{ 4+1 }\left({\begin{array}2&3&4& \vspace{6}\\ 3&4&6& \vspace{6}\\ 4&6&9& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&1&2& \vspace{6}\\ 3&4&6& \vspace{6}\\ 4&6&9& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&1&2& \vspace{6}\\ 0&1&0& \vspace{6}\\ 0&2&1& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&0& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =1 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{42} }$ $3$\displaystyle{ ={ $- 1$ }^{ 4+2 }\left({\begin{array}1&3&4& \vspace{6}\\ 2&4&6& \vspace{6}\\ 3&6&9& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =3\left({\begin{array}1&3&4& \vspace{6}\\ 2&4&6& \vspace{6}\\ 1&2&3& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =3\left({\begin{array}1&3&4& \vspace{6}\\ 0& - 2 & - 2 & \vspace{6}\\ 0& - 1 & - 1 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =3\left({\begin{array} - 2 & - 2 & \vspace{6}\\ - 1 & - 1 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =0 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{43} }$ $3$\displaystyle{ ={ $- 1$ }^{ 4+3 }\left({\begin{array}1&2&4& \vspace{6}\\ 2&3&6& \vspace{6}\\ 3&4&9& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- \left({\begin{array}1&2&4& \vspace{6}\\ 2&3&6& \vspace{6}\\ 3&4&9& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- \left({\begin{array}1&2&4& \vspace{6}\\ 0& - 1 & - 2 & \vspace{6}\\ 0& - 2 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- \left({\begin{array} - 1 & - 2 & \vspace{6}\\ - 2 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =1 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{44} }$ $3$\displaystyle{ ={ $- 1$ }^{ 4+4 }\left({\begin{array}1&2&3& \vspace{6}\\ 2&3&4& \vspace{6}\\ 3&4&6& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&2&3& \vspace{6}\\ 2&3&4& \vspace{6}\\ 3&4&6& \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&2&3& \vspace{6}\\ 0& - 1 & - 2 & \vspace{6}\\ 0& - 2 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =\left({\begin{array} - 1 & - 2 & \vspace{6}\\ - 2 & - 3 & \vspace{6}\\ \end{array}}\right) }$ $3$\displaystyle{ =- 1 }$

求めた数値をそれぞれ代入する．ただし，行列式の前に最初に求めた $3$\displaystyle{ \left({A}\right) }$ を分母にした分数を持ってくる．

$3$\displaystyle{ {A}^{ - 1 } }$ $3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ \begin{array} - 7 &0&2&1& \vspace{6}\\ 0&9& - 6 &0& \vspace{6}\\ 2& - 6 &2&1& \vspace{6}\\ 1&0&1& - 1 & \vspace{6}\\ \end{array} }\right) }$ $3$\displaystyle{ =\left({ \begin{array} - \frac{\hspace{2}7\hspace{2}}{\hspace{2}3\hspace{2}} &0& \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ 0&3& - 2 &0& \vspace{6}\\ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} & - 2 & \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} &0& \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ \end{array} }\right) }$

■問題へ戻る

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作成：学生スタッフ

最終更新日： 2022年8月27日