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

■答

$3$\displaystyle{ \left({\begin{array} - \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} &\frac{\hspace{2}1\hspace{2}}{\hspace{2}28\hspace{2}}&\frac{\hspace{2}11\hspace{2}}{\hspace{2}28\hspace{2}}& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}& - \frac{\hspace{2}5\hspace{2}}{\hspace{2}28\hspace{2}} &\frac{\hspace{2}1\hspace{2}}{\hspace{2}28\hspace{2}}& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}&\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}& - \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} & \vspace{6}\\ \end{array}}\right) }$

■計算

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

$3$\displaystyle{ \left({A}\right) }$ $3$\displaystyle{ =\left({\begin{array}1&3&2& \vspace{6}\\ 2& - 1 &3& \vspace{6}\\ 3&2&1& \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)の計算をする．

$3$\displaystyle{ =\left({\begin{array}1&3&2& \vspace{6}\\ 0& - 7 & - 1 & \vspace{6}\\ 0& - 7 & - 5 & \vspace{6}\\ \end{array}}\right) }$

次数下げの計算を用いて行列式の次数を $3$\displaystyle{1}$ つ下げる．

$3$\displaystyle{ =\left({\begin{array} - 7 & - 1 & \vspace{6}\\ - 7 & - 5 & \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ =28 }$ $3$\displaystyle{ \neq 0 }$

各成分の余因子を求める．

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{11} ={\left({ - 1 }\right)}^{ 1+1 }\left({\begin{array} - 1 &3& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array}}\right) =- 7 }$ ， $3$\displaystyle{ \hspace{1}{\tilde{a}}_{12} ={\left({ - 1 }\right)}^{ 1+2 }\left({\begin{array}2&3& \vspace{6}\\ 3&1& \vspace{6}\\ \end{array}}\right) =7 }$ ， $3$\displaystyle{ \hspace{1}{\tilde{a}}_{13} ={\left({ - 1 }\right)}^{ 1+3 }\left({\begin{array}2& - 1 & \vspace{6}\\ 3&2& \vspace{6}\\ \end{array}}\right) =7 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{21} ={\left({ - 1 }\right)}^{ 2+1 }\left({\begin{array}3&2& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array}}\right) =1 }$ ， $3$\displaystyle{ \hspace{1}{\tilde{a}}_{22} ={\left({ - 1 }\right)}^{ 2+2 }\left({\begin{array}1&2& \vspace{6}\\ 3&1& \vspace{6}\\ \end{array}}\right) =- 5 }$ ， $3$\displaystyle{ \hspace{1}{\tilde{a}}_{23} ={\left({ - 1 }\right)}^{ 2+3 }\left({\begin{array}1&3& \vspace{6}\\ 3&2& \vspace{6}\\ \end{array}}\right) =7 }$

$3$\displaystyle{ \hspace{1}{\tilde{a}}_{31} ={\left({ - 1 }\right)}^{ 3+1 }\left({\begin{array}3&2& \vspace{6}\\ - 1 &3& \vspace{6}\\ \end{array}}\right) =11 }$ ， $3$\displaystyle{ \hspace{1}{\tilde{a}}_{32} ={\left({ - 1 }\right)}^{ 3+2 }\left({\begin{array}1&2& \vspace{6}\\ 2&3& \vspace{6}\\ \end{array}}\right) =1 }$ ， $3$\displaystyle{ \hspace{1}{\tilde{a}}_{33} ={\left({ - 1 }\right)}^{ 3+3 }\left({\begin{array}1&3& \vspace{6}\\ 2& - 1 & \vspace{6}\\ \end{array}}\right) =- 7 }$

求めた数値を逆行列の式

$3$\displaystyle{ {A}^{ - 1 }=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \mid A\mid \hspace{2}} $\begin{array}\hspace{1}{\tilde{a}}_{11}&\hspace{1}{\tilde{a}}_{21}&\cdots &\hspace{1}{\tilde{a}}_{ n1 }& \vspace{6}\\ \hspace{1}{\tilde{a}}_{12}&\hspace{1}{\tilde{a}}_{22}&\cdots &\hspace{1}{\tilde{a}}_{ n2 }& \vspace{6}\\ \vdots &\vdots & &\vdots & \vspace{6}\\ \hspace{1}{\tilde{a}}_{ 1n }&\hspace{1}{\tilde{a}}_{ 2n }&\cdots &\hspace{1}{\tilde{a}}_{ nn }& \vspace{6}\\ \end{array}$ }$

に代入する.この問題では $3$\displaystyle{ n=3 }$ である．

$3$\displaystyle{ {A}^{ - 1 } }$ $3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}28\hspace{2}}\left({\begin{array} - 7 &1&11& \vspace{6}\\ 7& - 5 &1& \vspace{6}\\ 7&7& - 7 & \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ =\left({\begin{array} - \frac{\hspace{2}7\hspace{2}}{\hspace{2}28\hspace{2}} &\frac{\hspace{2}1\hspace{2}}{\hspace{2}28\hspace{2}}&\frac{\hspace{2}11\hspace{2}}{\hspace{2}28\hspace{2}}& \vspace{6}\\ \frac{\hspace{2}7\hspace{2}}{\hspace{2}28\hspace{2}}& - \frac{\hspace{2}5\hspace{2}}{\hspace{2}28\hspace{2}} &\frac{\hspace{2}1\hspace{2}}{\hspace{2}28\hspace{2}}& \vspace{6}\\ \frac{\hspace{2}7\hspace{2}}{\hspace{2}28\hspace{2}}&\frac{\hspace{2}7\hspace{2}}{\hspace{2}28\hspace{2}}& - \frac{\hspace{2}7\hspace{2}}{\hspace{2}28\hspace{2}} & \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{\displaystyle{=\left({\begin{array} - \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} &\frac{\hspace{2}1\hspace{2}}{\hspace{2}28\hspace{2}}&\frac{\hspace{2}11\hspace{2}}{\hspace{2}28\hspace{2}}& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}& - \frac{\hspace{2}5\hspace{2}}{\hspace{2}28\hspace{2}} &\frac{\hspace{2}1\hspace{2}}{\hspace{2}28\hspace{2}}& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}&\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}& - \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} & \vspace{6}\\ \end{array}}\right)}}$

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最終更新日： 2022年6月14日