Witam,
Mam ogromną prośbę: Czy mógłby mi ktoś sprawdzić te zadanie i poprawić błędy? Bardzo mi zależy na tych zadaniach.
zadanie 1
Wyznaczyć proste niezmiennicze przekształcenia afinicznego \(\displaystyle{ \begin{cases} y^{1} = x^{1} - 5 x^{2}-3 \\ y^{2} = -5 x^{1} + x^{2}-3 \end{cases}}\).
Moje rozwiązanie:
\(\displaystyle{ f:\begin{cases} y^{1} = x^{1} - 5 x^{2}-3 \\ y^{2} = -5 x^{1} + x^{2}-3 \end{cases}}\)
\(\displaystyle{ p: A x^{1} + B x^{2} + C=0}\) \(\displaystyle{ \xrightarrow f}\) \(\displaystyle{ f\left( p\right): A y^{1} + B y^{2} + C=0}\)
\(\displaystyle{ f^{-1} \left( f \left( p\right) \right) : A \left( x^{1} -5 x^{2} -3\right) + B\left( -5 x^{1} + x^{2} -3\right) +C= 0}\)
\(\displaystyle{ f^{-1}\left( f\left( p\right) \right) = \lambda p}\)
\(\displaystyle{ \left( A- 5B\right)x^{1} + \left( -5A+B\right)x^{2}+ \left( -3A - 3B +C \right)= \lambda A x^{1}+ \lambda B x^{2} + \lambda C}\)
\(\displaystyle{ \left\{\begin{array}{l} A- 5B = \lambda A\\-5A + B = \lambda B \\-3A - 3B + C = \lambda C\end{array}}\)
\(\displaystyle{ \begin{cases} \left(1- \lambda\right)A - 5B = 0 \\-5A + \left( 1- \lambda\right)B = 0 \end{cases}}\)
\(\displaystyle{ 0= \begin{bmatrix} 1 - \lambda & 5 \\ -5 & 1- \lambda\end{bmatrix}= \left( 1- \lambda\right)^{2} - 25 = \left( 1-\lambda-5\right) \left( 1-\lambda +5\right) = \left( -4-\lambda\right) \left(6-\lambda\right)=0}\)
\(\displaystyle{ \lambda_{1} = -4}\)
\(\displaystyle{ \lambda_{2} = 6}\)
\(\displaystyle{ \begin{cases} 5A - 5B = 0 \\-5A + 5B = 0\end{cases}}\)
\(\displaystyle{ A=B}\)
\(\displaystyle{ -3A-3B+C=-4C}\)
\(\displaystyle{ -3B-3B =-5C}\)
\(\displaystyle{ -6B = -5C}\)
\(\displaystyle{ C= \frac{6}{5} B}\)
\(\displaystyle{ \lambda_{1} = 4}\)
\(\displaystyle{ p: B x^{1} + B x^ {2} + \frac{6}{5}B = 0}\)
\(\displaystyle{ p: 4 x^{1} + 5 x^{2} + 6= 0}\)
\(\displaystyle{ \lambda_{2}=6}\)
\(\displaystyle{ \begin{cases} -5A - 5B = 0\\-5A-5B=0\end{cases}}\)
\(\displaystyle{ B = -A}\)
\(\displaystyle{ -3A-3B+C=6C}\)
\(\displaystyle{ -3A+3A+C = 6C}\)
\(\displaystyle{ C=0}\)
\(\displaystyle{ p: A x^{1} - A x^{2} = 0}\)
\(\displaystyle{ p: x^{1} - x^{2} = 0}\)
zadanie 2
Wyzanczyć równania wszystkich przekształceń afinicznych pzrestrzeni \(\displaystyle{ E^{2}}\), przeprowadzających prostą \(\displaystyle{ k: x^{1}- x^{2} = 0}\) na prostą \(\displaystyle{ l: x^{1} + x^{2} - 2 = 0}\), zaś prosta \(\displaystyle{ l}\) na prostą \(\displaystyle{ k}\). Znaleźć i zapisać te z otrzymanych pzrekształceń, które są izometriami.
Rozwiązanie:
\(\displaystyle{ k: x^{1}- x^{2} = 0}\) \(\displaystyle{ \xrightarrow f}\) \(\displaystyle{ l: x^{1} + x^{2} - 2 = 0}\)
\(\displaystyle{ l: x^{1} + x^{2} - 2 = 0}\) \(\displaystyle{ \xrightarrow f}\) \(\displaystyle{ k: x^{1}- x^{2} = 0}\)
\(\displaystyle{ \begin{cases} \alpha k\left( x\right) = l\left( y \right) \\ \beta l\left( x\right)= k\left( y\right) \end{cases}}\)
\(\displaystyle{ \begin{cases} y^{1} + y^{2}-2 = \alpha x^{1} - \alpha x^{2} \\y^{1} - y^{2} = \beta x^{1} + \beta x^{2} - 2 \beta \end{cases}}\)
\(\displaystyle{ 2y^{1}- 2 = \left( \alpha + \beta \right)x^{1} + \left( - \alpha + \beta \right)x^{2} - 2 \beta}\)
\(\displaystyle{ y^{1} = \frac{ \alpha + \beta }{2} x^{1} + \frac{- \alpha + \beta }{2} x^{2} - \beta +1}\)
\(\displaystyle{ 2y^{1} - 2 = \left( \alpha - \beta \right)x^{1} + \left( - \alpha - \beta \right)X^{2} + 2 \beta}\)
\(\displaystyle{ y^{2} = \frac{ \alpha - \beta }{2} x^{1} + \frac{- \alpha - \beta }{2} x^{2} + \beta +1}\)
\(\displaystyle{ f: \begin{cases} y^{1} = \frac{ \alpha + \beta }{2} x^{1} + \frac{- \alpha + \beta }{2} x^{2} - \beta +1 \\y^{2} = \frac{ \alpha - \beta }{2} x^{1} + \frac{- \alpha - \beta }{2} x^{2} + \beta +1\end{cases}}\)
\(\displaystyle{ A= \left[\begin{array}{ccc} \frac{ \alpha + \beta }{2} & \frac{- \alpha - \beta }{2} \\ \frac{ \alpha - \beta }{2} & \frac{- \alpha + \beta }{2} \end{array}\right]}\)
\(\displaystyle{ f}\) jest izometrią, gdy \(\displaystyle{ A \cdot A^{T} = I i A^{T} \cdot A= I}\)
\(\displaystyle{ \left[\begin{array}{ccc} \frac{ \alpha + \beta }{2} & \frac{- \alpha - \beta }{2} \\ \frac{ \alpha - \beta }{2} & \frac{- \alpha + \beta }{2} \end{array}\right]}\) \(\displaystyle{ \cdot}\) \(\displaystyle{ \left[\begin{array}{ccc} \frac{ \alpha + \beta }{2} & \frac{\alpha - \beta }{2} \\ \frac{ -\alpha - \beta }{2} & \frac{- \alpha + \beta }{2} \end{array}\right]}\) = \(\displaystyle{ \left[\begin{array}{ccc} \frac{ \alpha ^{2}+ \beta ^{2}}{2} & \frac{ \alpha^{2} - \beta ^{2}}{2} \\ \frac{ \alpha ^{2}- \beta ^{2} }{2} & \frac{ \alpha ^{2}+ \beta {2}}{2} \end{array}\right]}\) = \(\displaystyle{ \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]}\)
\(\displaystyle{ \left[\begin{array}{ccc} \frac{ \alpha + \beta }{2} & \frac{\alpha - \beta }{2} \\ \frac{ -\alpha - \beta }{2} & \frac{- \alpha + \beta }{2} \end{array}\right]}\) \(\displaystyle{ \cdot}\) \(\displaystyle{ \left[\begin{array}{ccc} \frac{ \alpha + \beta }{2} & \frac{-\alpha - \beta }{2} \\ \frac{ \alpha - \beta }{2} & \frac{- \alpha + \beta }{2} \end{array}\right]}\) = \(\displaystyle{ \left[\begin{array}{ccc} \frac{ \alpha ^{2}+ \beta ^{2}}{2} & \frac{ \beta ^{2} - \alpha ^{2}}{2} \\ \frac{ \beta ^{2}- \alpha ^{2} }{2} & \frac{ \alpha ^{2}+ \beta {2}}{2} \end{array}\right]}\) = \(\displaystyle{ \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]}\)
\(\displaystyle{ \begin{cases} \frac{ \alpha ^{2}+ \beta^{2}}{2} =1 \\ \alpha ^{2} - \beta ^{2} = 0\end{cases}}\)
\(\displaystyle{ \alpha ^{2} = \beta ^{2}}\)
\(\displaystyle{ \frac{2 \beta ^{2}}{2}= 1}\)
\(\displaystyle{ \beta ^{2} =1}\)
\(\displaystyle{ \beta = 1}\) \(\displaystyle{ \vee}\) \(\displaystyle{ \beta =-1}\)
\(\displaystyle{ \beta = 1}\)
\(\displaystyle{ \alpha^{2} = 1}\)
\(\displaystyle{ \alpha =1}\) \(\displaystyle{ \vee}\) \(\displaystyle{ \alpha =-1}\)
\(\displaystyle{ \left( \alpha , \beta \right) = \left( 1,1\right)}\)
\(\displaystyle{ \left( \alpha , \beta \right) = \left( -1,1\right)}\)
\(\displaystyle{ \beta = -1}\) \(\displaystyle{ \alpha ^ {2} = 1}\)
\(\displaystyle{ \alpha =1}\) \(\displaystyle{ \vee}\) \(\displaystyle{ \alpha =-1}\)
\(\displaystyle{ \left( \alpha , \beta \right) = \left( 1,1\right)}\)
\(\displaystyle{ \left( \alpha , \beta \right) = \left( -1,1\right)}\)
\(\displaystyle{ \begin{cases} y^{2} = \frac{ \alpha + \beta }{2} x^{1} + \frac{- \alpha - \beta }{2} x^{2} - \beta +1 \\ y^{2} = \frac{ \alpha - \beta }{2} x^{1} + \frac{- \alpha - \beta }{2} x^{2} + \beta + 1 \end{cases}}\)
\(\displaystyle{ 1.}\) \(\displaystyle{ \left( 1,1\right)}\)
\(\displaystyle{ \begin{cases} y^{1} = x^{1}\\y^{2} = -x^{2}+2\end{cases}}\)
\(\displaystyle{ 2.}\) \(\displaystyle{ \left( -1,1\right)}\)
\(\displaystyle{ \begin{cases} y^{1} = x^{2}\\y^{2} = -x^{1}+2\end{cases}}\)
\(\displaystyle{ 3.}\) \(\displaystyle{ \left( 1,-1\right)}\)
\(\displaystyle{ \begin{cases} y^{1} = -x^{2}+2 \\y^{2} = x^{1}\end{cases}}\)
\(\displaystyle{ 4.}\) \(\displaystyle{ \left( -1,-1\right)}\)
\(\displaystyle{ \begin{cases} y^{1} = -x^{1}+2\\y^{2} = x^{2}\end{cases}}\)
zadanie 3 - z tym mam problem i nie potrafię go zrobić w całości jakby ktoś mógł mi rozwiązać krok po kroku tak, abym mogła zrozumieć bylabym bardzo wdzięczna.
a) Zapisać we współrzędnych ortonormalnych izometrię \(\displaystyle{ f}\) przestrzeni \(\displaystyle{ E^{2}}\), ktora jest złożeniem obrotu dookoła początku układu o kąt \(\displaystyle{ \frac{ \pi }{2}}\) i translacji o wektor \(\displaystyle{ \overline{v} \left(2,2 \right)}\).
\(\displaystyle{ f\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=\left[\begin{array}{cc}\cos\frac\pi2&-\sin\frac\pi2\\\sin\frac\pi2&\cos\frac\pi2\end{array}\right]\cdot\left[\begin{array}{c}x\\y\end{array}\right]+\left[\begin{array}{c}-2\\2\end{array}\right]}\)
b) Przedstawić tę izometrię jako złożenie dwóch symetrii osiowych \(\displaystyle{ s_{\left( 2\right) } \circ s_{\left( 1\right) }}\).
c) Sprawdzić rachunkiem, czy rzeczywiście \(\displaystyle{ f =s_{\left( 2\right) } \circ s_{\left( 1\right) }}\).
d) Podać nazwę izometrii \(\displaystyle{ f}\).