高専の数学解説

第15章 15.8

L=(cos⁡60∘−sin⁡60∘sin⁡60∘cos⁡60∘)=(12−323212)M=(cos⁡135∘−sin⁡135∘sin⁡135∘cos⁡135∘)=(−22−2222−22)L = \begin{pmatrix} \cos 60^\circ & -\sin 60^\circ \\ \sin 60^\circ & \cos 60^\circ \end{pmatrix} = \begin{pmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix} \qquad M = \begin{pmatrix} \cos 135^\circ & -\sin 135^\circ \\ \sin 135^\circ & \cos 135^\circ \end{pmatrix} = \begin{pmatrix} -\dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2} \\ \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2} \end{pmatrix}

(1)

Ⓛ

(12−323212)(2−4)=(1+233−2)\begin{pmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix} \begin{pmatrix} 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 1+2\sqrt{3} \\ \sqrt{3}-2 \end{pmatrix}

Ⓜ

(−22−2222−22)(2−4)=(−2+222+22)=(232)\begin{pmatrix} -\dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2} \\ \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2} \end{pmatrix} \begin{pmatrix} 2 \\ -4 \end{pmatrix} = \begin{pmatrix} -\sqrt{2}+2\sqrt{2} \\ \sqrt{2}+2\sqrt{2} \end{pmatrix} = \begin{pmatrix} \sqrt{2} \\ 3\sqrt{2} \end{pmatrix}

(2)

Ⓛ

(12−323212)(32)=(32−332+1)=(−3252)\begin{pmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix} \begin{pmatrix} \sqrt{3} \\ 2 \end{pmatrix} = \begin{pmatrix} \dfrac{\sqrt{3}}{2}-\sqrt{3} \\ \dfrac{3}{2}+1 \end{pmatrix} = \begin{pmatrix} -\dfrac{\sqrt{3}}{2} \\ \dfrac{5}{2} \end{pmatrix}

Ⓜ

(−22−2222−22)(32)=(−62−262−2)\begin{pmatrix} -\dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2} \\ \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2} \end{pmatrix} \begin{pmatrix} \sqrt{3} \\ 2 \end{pmatrix} = \begin{pmatrix} -\dfrac{\sqrt{6}}{2}-\sqrt{2} \\ \dfrac{\sqrt{6}}{2}-\sqrt{2} \end{pmatrix}

解説: r31bn1z