第14章 14.19

(1)

$A = aB_1B_2 + bC_1C_2 = a\begin{pmatrix}\cos\theta\\ \sin\theta\end{pmatrix}\begin{pmatrix}\cos\theta & \sin\theta\end{pmatrix} + b\begin{pmatrix}-\sin\theta\\ \cos\theta\end{pmatrix}\begin{pmatrix}-\sin\theta & \cos\theta\end{pmatrix}$

$= a\begin{pmatrix}\cos^2\theta & \cos\theta\sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{pmatrix} + b\begin{pmatrix}\sin^2\theta & -\cos\theta\sin\theta\\ -\cos\theta\sin\theta & \cos^2\theta\end{pmatrix} = \begin{pmatrix}a\cos^2\theta+b\sin^2\theta & (a-b)\cos\theta\sin\theta\\ (a-b)\cos\theta\sin\theta & a\sin^2\theta+b\cos^2\theta\end{pmatrix}$

$A^2 = (aB_1B_2+bC_1C_2)(aB_1B_2+bC_1C_2) = a^2B_1B_2B_1B_2 + b^2C_1C_2C_1C_2 + abB_1B_2C_1C_2 + abC_1C_2B_1B_2$

ここで

$B_1B_2C_1C_2 = \begin{pmatrix}\cos^2\theta & \cos\theta\sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{pmatrix}\begin{pmatrix}\sin^2\theta & -\cos\theta\sin\theta\\ -\cos\theta\sin\theta & \cos^2\theta\end{pmatrix} = \begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}$

$C_1C_2B_1B_2 = \begin{pmatrix}\sin^2\theta & -\cos\theta\sin\theta\\ -\cos\theta\sin\theta & \cos^2\theta\end{pmatrix}\begin{pmatrix}\cos^2\theta & \cos\theta\sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{pmatrix} = \begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}$

$B_1B_2B_1B_2 = \begin{pmatrix}\cos^2\theta & \cos\theta\sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{pmatrix}\begin{pmatrix}\cos^2\theta & \cos\theta\sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{pmatrix}$

$= \begin{pmatrix}\cos^4\theta+\cos^2\theta\sin^2\theta & \cos^3\theta\sin\theta+\cos\theta\sin^3\theta\\ \cos^3\theta\sin\theta+\cos\theta\sin^3\theta & \cos^2\theta\sin^2\theta+\sin^4\theta\end{pmatrix} = \begin{pmatrix}\cos^2\theta & \cos\theta\sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{pmatrix} = B_1B_2$

$C_1C_2C_1C_2 = \begin{pmatrix}\sin^2\theta & -\cos\theta\sin\theta\\ -\cos\theta\sin\theta & \cos^2\theta\end{pmatrix}\begin{pmatrix}\sin^2\theta & -\cos\theta\sin\theta\\ -\cos\theta\sin\theta & \cos^2\theta\end{pmatrix}$

$= \begin{pmatrix}\sin^4\theta+\cos^2\theta\sin^2\theta & -\cos\theta\sin^3\theta-\cos^3\theta\sin\theta\\ -\cos\theta\sin^3\theta-\cos^3\theta\sin\theta & \cos^2\theta\sin^2\theta+\cos^4\theta\end{pmatrix} = \begin{pmatrix}\sin^2\theta & -\cos\theta\sin\theta\\ -\cos\theta\sin\theta & \cos^2\theta\end{pmatrix} = C_1C_2$

よって

$A^2 = a^2B_1B_2 + b^2C_1C_2 \quad \text{となり}$

$A^n = a^nB_1B_2 + b^nC_1C_2 \quad \text{となる}$

(2)

$|A| = (a\cos^2\theta+b\sin^2\theta)(a\sin^2\theta+b\cos^2\theta) - \{(a-b)\cos\theta\sin\theta\}^2$

$= a^2\cos^2\theta\sin^2\theta + b^2\cos^2\theta\sin^2\theta + ab(\cos^4\theta+\sin^4\theta) - (a^2-2ab+b^2)\cos^2\theta\sin^2\theta$

$= ab(\cos^4\theta + 2\cos^2\theta\sin^2\theta + \sin^4\theta) = ab$

これから

$A^{-1} = \frac{1}{ab}\begin{pmatrix} a\sin^2\theta + b\cos^2\theta & -(a-b)\cos\theta\sin\theta \\ -(a-b)\cos\theta\sin\theta & a\cos^2\theta + b\sin^2\theta \end{pmatrix} = \begin{pmatrix} \dfrac{1}{a}\cos^2\theta + \dfrac{1}{b}\sin^2\theta & \left(\dfrac{1}{a}-\dfrac{1}{b}\right)\cos\theta\sin\theta \\ \left(\dfrac{1}{a}-\dfrac{1}{b}\right)\cos\theta\sin\theta & \dfrac{1}{a}\sin^2\theta + \dfrac{1}{b}\cos^2\theta \end{pmatrix}$

$= \frac{1}{a}B_1B_2 + \frac{1}{b}C_1C_2$