회전변환과 대칭변환의 합성

회전변환 $f$의 행렬을 $A$, 대칭변환 $g$의 행렬을 $B$라 하면 $\displaystyle A = {\begin{pmatrix} \cos \frac{\pi}{3} & & - \sin \frac{\pi}{3} \\ \sin \frac{\pi}{3} & & \cos \frac{\pi}{3} \end{pmatrix}} = {\begin{pmatrix} \frac{1}{2} & & - \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & & \frac{1}{2} \end{pmatrix}}$, $\displaystyle B = {\begin{pmatrix} 0 & & 1 \\ 1 & & 0 \end{pmatrix}}$이고 합성변환 $g ^{- 1} \circ f \circ g$의 행렬은 $B ^{- 1} AB$이다. $\displaystyle B ^{- 1} AB = {\begin{pmatrix} 0 & & 1 \\ 1 & & 0 \end{pmatrix}} {\begin{pmatrix} \frac{1}{2} & & - \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & & \frac{1}{2} \end{pmatrix}} {\begin{pmatrix} 0 & & 1 \\ 1 & & 0 \end{pmatrix}}$ $\displaystyle = {\begin{pmatrix} \frac{\sqrt{3}}{2} & & \frac{1}{2} \\ \frac{1}{2} & & - \frac{\sqrt{3}}{2} \end{pmatrix}} {\begin{pmatrix} 0 & & 1 \\ 1 & & 0 \end{pmatrix}} = {\begin{pmatrix} \frac{1}{2} & & \frac{\sqrt{3}}{2} \\ - \frac{\sqrt{3}}{2} & & \frac{1}{2} \end{pmatrix}}$ $\displaystyle {\begin{pmatrix} x' \\ y' \end{pmatrix}} = {\begin{pmatrix} \frac{1}{2} & & \frac{\sqrt{3}}{2} \\ - \frac{\sqrt{3}}{2} & & \frac{1}{2} \end{pmatrix}} {\begin{pmatrix} x \\ y \end{pmatrix}}$에서 $\displaystyle x' = \frac{1}{2} x + \frac{\sqrt{3}}{2} y , y' = - \frac{\sqrt{3}}{2} x + \frac{1}{2} y$ $x + 2 y + 5 = 0$이 $ax' + by' + 5 = 0$으로 옮겨지므로 $\displaystyle a \left( \frac{1}{2} x + \frac{\sqrt{3}}{2} y \right) + b \left( - \frac{\sqrt{3}}{2} x + \frac{1}{2} y \right) + 5 = 0$ $\displaystyle \left( \frac{1}{2} a - \frac{\sqrt{3}}{2} b \right) x + \left( \frac{\sqrt{3}}{2} a + \frac{1}{2} b \right) y + 5 = 0 \cdots \cdots \text{㉠}$ ㉠이 $x + 2 y + 5 = 0$와 같아야 하므로 따라서 $\displaystyle \frac{1}{2} a - \frac{\sqrt{3}}{2} b = 1 , \frac{\sqrt{3}}{2} a + \frac{1}{2} b = 2$ 연립하여 풀면 $\displaystyle a = \frac{1}{2} + \sqrt{3} , b = - \frac{\sqrt{3}}{2} + 1$ $\displaystyle \therefore a + 2 b = \frac{5}{2}$