基本的な1次変換の問題

■問題

次の回転行列 $3$\displaystyle{ R\left({{\theta}}\right) }$ を求め，点 $3$\displaystyle{ $ 3,3 $ }$ が回転で移る座標を求めよ．

(1) $3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}3\hspace{2}} }\right)$ (2) $3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }\right)$ (3) $3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}6\hspace{2}} }\right)$ (4) $3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} }\right)$

■答

(1) $3$\displaystyle{ $ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3},\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3}+\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} $ }$ (2) $3$\displaystyle{ $ - 3,3 $ }$ (3) $3$\displaystyle{ $ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3}- \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}},\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}+\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3} $ }$ (4) $3$\displaystyle{ $ 0,3\sqrt{2} $ }$

■解き方

$3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}3\hspace{2}} }\right) \displaystyle{= $ \begin{array} \cos \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}3\hspace{2}} & - \sin \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ \sin \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}3\hspace{2}} & \cos \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{= $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} & - \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \begin{array}1& - \sqrt{3} & \vspace{6}\\ \sqrt{3} &1& \vspace{6}\\ \end{array} $ }$

点 $3$\displaystyle{ $ 3,3 $ }$ を原点を中心に $3$\displaystyle{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}3\hspace{2}} }$ 回転すると，

$3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}3\hspace{2}} }\right) \displaystyle{ $ \begin{array}3& \vspace{6}\\ 3& \vspace{6}\\ \end{array} $ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \begin{array}1& - \sqrt{3} & \vspace{6}\\ \sqrt{3} &1& \vspace{6}\\ \end{array} $ $ \begin{array}3& \vspace{6}\\ 3& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \begin{array} 3- 3\sqrt{3} & \vspace{6}\\ 3\sqrt{3}+3 & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{= $ \begin{array} \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3} & \vspace{6}\\ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3}+\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \end{array} $ }$

となり，点 $3$\displaystyle{ $ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3},\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3}+\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} $ }$ に移る．

(2)

$3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }\right) \displaystyle{= $ \begin{array} \cos \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} & - \sin \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \sin \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} & \cos \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{= $ \begin{array}0& - 1 & \vspace{6}\\ 1&0& \vspace{6}\\ \end{array} $ }$

点 $3$\displaystyle{ $ 3,3 $ }$ を原点を中心に $3$\displaystyle{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }$ 回転すると，

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

となり，点 $3$\displaystyle{ $ - 3,3 $ }$ に移る．

(3)

$3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}6\hspace{2}} }\right) \displaystyle{= $ \begin{array} \cos \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}6\hspace{2}} & - \sin \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}6\hspace{2}} & \vspace{6}\\ \sin \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}6\hspace{2}} & \cos \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}6\hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{= $ \begin{array} \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} & \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \begin{array} \sqrt{3} & - 1 & \vspace{6}\\ 1& \sqrt{3} & \vspace{6}\\ \end{array} $ }$

点 $3$\displaystyle{ $ 3,3 $ }$ を原点を中心に $3$\displaystyle{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}6\hspace{2}} }$ 回転すると，

$3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}6\hspace{2}} }\right) \displaystyle{ $ \begin{array}3& \vspace{6}\\ 3& \vspace{6}\\ \end{array} $ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \begin{array} \sqrt{3} & - 1 & \vspace{6}\\ 1& \sqrt{3} & \vspace{6}\\ \end{array} $ $ \begin{array}3& \vspace{6}\\ 3& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \begin{array} 3\sqrt{3}- 3 & \vspace{6}\\ 3+3\sqrt{3} & \vspace{6}\\ \end{array} $ = $ \begin{array} \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3}- \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}+\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3} & \vspace{6}\\ \end{array} $ }$

となり，点 $3$\displaystyle{ $ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3}- \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}},\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}+\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\sqrt{3} $ }$ に移る．

(4)

$3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} }\right) \displaystyle{= $ \begin{array} \cos \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} & - \sin \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} & \vspace{6}\\ \sin \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} & \cos \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{= $ \begin{array} \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} $ \begin{array}1& - 1 & \vspace{6}\\ 1&1& \vspace{6}\\ \end{array} $ }$

点 $3$\displaystyle{ $ 3,3 $ }$ を原点を中心に $3$\displaystyle{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} }$ 回転すると，

$3$ R\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}6\hspace{2}} }\right) \displaystyle{ $ \begin{array}3& \vspace{6}\\ 3& \vspace{6}\\ \end{array} $ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} $ \begin{array}1& - 1 & \vspace{6}\\ 1&1& \vspace{6}\\ \end{array} $ $ \begin{array}3& \vspace{6}\\ 3& \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} $ \begin{array} 3- 3 & \vspace{6}\\ 3+3 & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{= $ \begin{array}0& \vspace{6}\\ \frac{\hspace{2}6\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{= $ \begin{array}0& \vspace{6}\\ 3\sqrt{2} & \vspace{6}\\ \end{array} $ }$

となり，点 $3$\displaystyle{ $ 0,3\sqrt{2} $ }$ に移る．

ホーム>>カテゴリー分類>>行列>>線形代数>>線形代数に関する問題>>線形写像・1次変換>>基本的な1次変換の問題

作成：学生スタッフ

最終更新日：2023年2月10日