일차변환과 벡터 각

$M =$$\displaystyle \begin{pmatrix} {1} & {- 1} \\ {3} & {- 2} \end{pmatrix}$ $M ^{2}$$\displaystyle = \begin{pmatrix} {1} & {- 1} \\ {3} & {- 2} \end{pmatrix}$$\displaystyle \begin{pmatrix} {1} & {- 1} \\ {3} & {- 2} \end{pmatrix}$$\displaystyle = \begin{pmatrix} {- 2} & {1} \\ {- 3} & {1} \end{pmatrix}$ $M ^{3}$$\displaystyle = \begin{pmatrix} {- 2} & {1} \\ {- 3} & {1} \end{pmatrix}$$\displaystyle \begin{pmatrix} {1} & {- 1} \\ {3} & {- 2} \end{pmatrix}$$\displaystyle = \begin{pmatrix} {1} & {0} \\ {0} & {1} \end{pmatrix}$ $M$$\displaystyle \begin{pmatrix} {1} \\ {2} \end{pmatrix}$$=$$\displaystyle \begin{pmatrix} {1} & {- 1} \\ {3} & {- 2} \end{pmatrix}$$\displaystyle \begin{pmatrix} {1} \\ {2} \end{pmatrix}$$\displaystyle = \begin{pmatrix} {- 1} \\ {- 1} \end{pmatrix}$이므로 $\mathrm{B} ( - 1 , - 1 )$ $M ^{3}$$\displaystyle \begin{pmatrix} {2} \\ {0} \end{pmatrix}$$=$$\displaystyle \begin{pmatrix} {1} & {0} \\ {0} & {1} \end{pmatrix}$$\displaystyle \begin{pmatrix} {2} \\ {0} \end{pmatrix}$$=$$\displaystyle \begin{pmatrix} {2} \\ {0} \end{pmatrix}$이므로 $\mathrm{D} ( 2 , 0 )$ ∴ $\vec{\mathrm{OB}} = ( - 1 , - 1 )$, $\vec{\mathrm{BD}}$$= \mathrm{\vec{OD}} - \vec{OB}$$= ( 3 , 1 )$ ∴ $\cos \theta$$\displaystyle = \frac{\mathrm{\vec{OB} \cdot \vec{BD}}}{\left| \vec{OB} \right| \left| \vec{BD} \right|}$ $\displaystyle = \frac{( - 1 , - 1 ) \cdot ( 3 , 1 )}{\sqrt{( - 1 ) ^{2} + ( - 1 ) ^{2}} \cdot \sqrt{3 ^{2} + 1 ^{2}}}$ $\displaystyle = \frac{- 4}{\sqrt{2} \cdot \sqrt{10}}$ $\displaystyle = - \frac{2 \sqrt{5}}{5}$