# 基本的な1次変換の問題

## ■問題

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

## ■答

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

## ■解き方

$R\left(\frac{\pi }{3}\right)=\left(\begin{array}{cc}\mathrm{cos}\frac{\pi }{3}& -\mathrm{sin}\frac{\pi }{3}\\ \mathrm{sin}\frac{\pi }{3}& \mathrm{cos}\frac{\pi }{3}\end{array}\right)$$=\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$$=\frac{1}{2}\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)$

$\left(3,3\right)$ を原点を中心に$\frac{\pi }{3}$ 回転すると，

$R\left(\frac{\pi }{3}\right)\left(\begin{array}{c}3\\ 3\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)\left(\begin{array}{c}3\\ 3\end{array}\right)$$=\frac{1}{2}\left(\begin{array}{c}3-3\sqrt{3}\\ 3\sqrt{3}+3\end{array}\right)$$=\left(\begin{array}{c}\frac{3}{2}-\frac{3}{2}\sqrt{3}\\ \frac{3}{2}\sqrt{3}+\frac{3}{2}\end{array}\right)$

となり，点$\left(\frac{3}{2}-\frac{3}{2}\sqrt{3},\frac{3}{2}\sqrt{3}+\frac{3}{2}\right)$ に移る．

(2)

$R\left(\frac{\pi }{2}\right)=\left(\begin{array}{cc}\mathrm{cos}\frac{\pi }{2}& -\mathrm{sin}\frac{\pi }{2}\\ \mathrm{sin}\frac{\pi }{2}& \mathrm{cos}\frac{\pi }{2}\end{array}\right)$$=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$

$\left(3,3\right)$ を原点を中心に$\frac{\pi }{2}$ 回転すると，

$R\left(\frac{\pi }{2}\right)\left(\begin{array}{c}3\\ 3\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}3\\ 3\end{array}\right)$$=\left(\begin{array}{c}-3\\ 3\end{array}\right)$

となり，点$\left(-3,3\right)$に移る．

(3)

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

$\left(3,3\right)$ を原点を中心に$\frac{\pi }{6}$ 回転すると，

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

となり，点$\left(\frac{3}{2}\sqrt{3}-\frac{3}{2},\frac{3}{2}+\frac{3}{2}\sqrt{3}\right)$ に移る．

(4)

$R\left(\frac{\pi }{4}\right)=\left(\begin{array}{cc}\mathrm{cos}\frac{\pi }{4}& -\mathrm{sin}\frac{\pi }{4}\\ \mathrm{sin}\frac{\pi }{4}& \mathrm{cos}\frac{\pi }{4}\end{array}\right)$$=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)$$=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$

$\left(3,3\right)$ を原点を中心に$\frac{\pi }{4}$ 回転すると，

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

となり，点$\left(0,3\sqrt{2}\right)$ 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