基本的な行列の問題

■問題

$3$\displaystyle{A=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} $ \begin{array}9&1& \vspace{6}\\ - 6 & - 1 & \vspace{6}\\ \end{array} $ }$ ， $3$\displaystyle{B= $ \begin{array}3&1& \vspace{6}\\ - 5 & - 3 & \vspace{6}\\ \end{array} $ }$ のとき， $3$\displaystyle{A{B}^{ 5 }{A}^{ - 1 }}$ を求めよ．

■答え

$3$\displaystyle{A{B}^{5}{A}^{ - 1 }=\frac{\hspace{2} 16 \hspace{2}}{\hspace{2}3\hspace{2}}\left({ \begin{array} - 14 & - 32 & \vspace{6}\\ 5& 14 & \vspace{6}\\ \end{array} }\right)}$

■解説

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

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

$3$\displaystyle{AB{A}^{ - 1 }}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ \begin{array}9&1& \vspace{6}\\ - 6 & - 1 & \vspace{6}\\ \end{array} }\right)\left({ \begin{array}3&1& \vspace{6}\\ - 5 & - 3 & \vspace{6}\\ \end{array} }\right)\left({ \begin{array}1&1& \vspace{6}\\ - 6 & - 9 & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ \begin{array}9&1& \vspace{6}\\ - 6 & - 1 & \vspace{6}\\ \end{array} }\right)\left({ \begin{array} - 3 & - 6 & \vspace{6}\\ 13 & 22 & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ \begin{array} - 14 & - 32 & \vspace{6}\\ 5& 14 & \vspace{6}\\ \end{array} }\right)}$

よって

$3$\displaystyle{{\left({ AB{A}^{ - 1 } }\right)}^{2}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ \begin{array} - 14 & - 32 & \vspace{6}\\ 5& 14 & \vspace{6}\\ \end{array} }\right)\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ \begin{array} - 14 & - 32 & \vspace{6}\\ 5& 14 & \vspace{6}\\ \end{array} }\right)}$ $3$\displaystyle{=4\left({ \begin{array}1&0& \vspace{6}\\ 0&1& \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{n}$ を自然数とすると

$3$\displaystyle{{ $ AB{A}^{ - 1 } $ }^{n}=A{B}^{n}{A}^{ - 1 }}$ となるから， $3$\displaystyle{A{B}^{ 5 }{A}^{ - 1 }={ $ AB{A}^{ - 1 } $ }^{ 5 }}$ と変換できる．

$3$\displaystyle{{\left({ AB{A}^{ - 1 } }\right)}^{5}}$ $3$\displaystyle{={\left({ AB{A}^{ - 1 } }\right)}^{2}{\left({ AB{A}^{ - 1 } }\right)}^{2}\left({ AB{A}^{ - 1 } }\right)}$

$3$\displaystyle{=4\left({ \begin{array}1&0& \vspace{6}\\ 0&1& \vspace{6}\\ \end{array} }\right)4\left({ \begin{array}1&0& \vspace{6}\\ 0&1& \vspace{6}\\ \end{array} }\right)\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left({ \begin{array} - 14 & - 32 & \vspace{6}\\ 5& 14 & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{=\frac{\hspace{2} 16 \hspace{2}}{\hspace{2}3\hspace{2}}\left({ \begin{array} - 14 & - 32 & \vspace{6}\\ 5& 14 & \vspace{6}\\ \end{array} }\right)}$

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最終更新日： 2023年10月10日