회전·확대 변환 타원

$\displaystyle f \circ g : \begin{pmatrix} {\frac{1}{2}} & {- \frac{\sqrt{3}}{2}} \\ {\frac{\sqrt{3}}{2}} & {\frac{1}{2}} \end{pmatrix} \begin{pmatrix} {1} & {0} \\ {0} & {2} \end{pmatrix} = \begin{pmatrix} {\frac{1}{2}} & {- \sqrt{3}} \\ {\frac{\sqrt{3}}{2}} & {1} \end{pmatrix}$ $\displaystyle \begin{pmatrix} {x'} & \\ {y'} \end{pmatrix} = \begin{pmatrix} {\frac{1}{2}} & {- \sqrt{3}} \\ {\frac{\sqrt{3}}{2}} & {1} \end{pmatrix} \begin{pmatrix} {x} & \\ {y} \end{pmatrix}$$\displaystyle \begin{pmatrix} {x} & \\ {y} \end{pmatrix} = \frac{1}{2} \begin{pmatrix} {1} & {\sqrt{3}} \\ {- \frac{\sqrt{3}}{2}} & {\frac{1}{2}} \end{pmatrix} \begin{pmatrix} {x'} & \\ {y'} \end{pmatrix}$ $\displaystyle x = \frac{1}{2} ( x' + \sqrt{3} y' ) , y = \frac{1}{2} ( - \frac{\sqrt{3}}{2} x' + \frac{1}{2} y' )$을 $\displaystyle \frac{x ^{2}}{4} + y ^{2} = 1$에 대입하여 정리하면 $x^{\prime 2} + y^{\prime 2} = 4$ 둘레의 길이는 $4 \pi$