A=[2,3;3,2]
b=[1;-2]
gdy rozwiązuję metodą macierzy odwrotnej oraz metodą Cramera, to wynik to:
x=[-4;7/2]
natomiast metodą Gaussa wychodzi mi inny wynik!!!
o co chodzi?
Równanie Cramera, coś niezwykłego!!
-
agulka1987
- Użytkownik
- Posty: 3090
- Rejestracja: 24 paź 2008, o 15:23
- Płeć: Kobieta
- Lokalizacja: Opole
- Podziękował: 1 raz
- Pomógł: 879 razy
Równanie Cramera, coś niezwykłego!!
nie wim jak wychodzi CI Gaussem, ale to co odałeś jako rozwiazanie Cramerem i macierza odwrotną jest złe. przwidłowy wynik to:
\(\displaystyle{ x=\begin{bmatrix}- \frac{8}{5}\\ \frac{7}{5}\end{bmatrix}}\)
CRAMER
\(\displaystyle{ det A \begin{bmatrix}2 & 3\\3 & 2\end{bmatrix} = 4-9=-5}\)
\(\displaystyle{ det A_{1} \begin{bmatrix}1 & 3\\-2 & 2\end{bmatrix} = 2+6 = 8}\)
\(\displaystyle{ det A_{2} \begin{bmatrix}2 & 1\\3 & -2\end{bmatrix} = -4-3 = -7}\)
\(\displaystyle{ x_{1} = \frac{detA_{1}}{detA} = \frac{8}{-5} = - \frac{8}{5}}\)
\(\displaystyle{ x_{2} = \frac{detA_{2}}{detA} = \frac{-7}{-5} = \frac{7}{5}}\)
GAUSS
\(\displaystyle{ \begin{bmatrix}2 & 3 \left|1\\3 & 2 \left|-2\end{bmatrix}}\)
\(\displaystyle{ W_{2} - \frac{3}{2}W_{1} = \begin{bmatrix}2 & 3 \left|1\\0 & - \frac{5}{2} \left|- \frac{7}{2} \end{bmatrix}}\)
\(\displaystyle{ W_{2} \cdot (- \frac{2}{5}) = \begin{bmatrix}2 & 3 \left|1\\0 & 1 \left| \frac{7}{5} \end{bmatrix}}\)
\(\displaystyle{ W_{1}-3W_{2} = \begin{bmatrix}2 & 0 \left|- \frac{16}{5} \\0 & 1 \left| \frac{7}{5} \end{bmatrix}}\)
\(\displaystyle{ W_{1} \cdot \frac{1}{2} = \begin{bmatrix}1 & 0 \left|- \frac{8}{5} \\0 & 1 \left| \frac{7}{5} \end{bmatrix}}\)
MACIERZ ODWROTNA
\(\displaystyle{ A \cdot X = b}\)
\(\displaystyle{ X = A^{-1} \cdot b}\)
\(\displaystyle{ A^{-1} = \begin{bmatrix} - \frac{2}{5} & \frac{3}{5} \\ \frac{3}{5} & - \frac{2}{5} \end{bmatrix}}\)
\(\displaystyle{ X= \begin{bmatrix} - \frac{2}{5} & \frac{3}{5} \\ \frac{3}{5} & - \frac{2}{5} \end{bmatrix} \cdot \begin{bmatrix}1\\-2\end{bmatrix}}\)
\(\displaystyle{ X=\begin{bmatrix}- \frac{8}{5}\\ \frac{7}{5}\end{bmatrix}}\)
\(\displaystyle{ x=\begin{bmatrix}- \frac{8}{5}\\ \frac{7}{5}\end{bmatrix}}\)
CRAMER
\(\displaystyle{ det A \begin{bmatrix}2 & 3\\3 & 2\end{bmatrix} = 4-9=-5}\)
\(\displaystyle{ det A_{1} \begin{bmatrix}1 & 3\\-2 & 2\end{bmatrix} = 2+6 = 8}\)
\(\displaystyle{ det A_{2} \begin{bmatrix}2 & 1\\3 & -2\end{bmatrix} = -4-3 = -7}\)
\(\displaystyle{ x_{1} = \frac{detA_{1}}{detA} = \frac{8}{-5} = - \frac{8}{5}}\)
\(\displaystyle{ x_{2} = \frac{detA_{2}}{detA} = \frac{-7}{-5} = \frac{7}{5}}\)
GAUSS
\(\displaystyle{ \begin{bmatrix}2 & 3 \left|1\\3 & 2 \left|-2\end{bmatrix}}\)
\(\displaystyle{ W_{2} - \frac{3}{2}W_{1} = \begin{bmatrix}2 & 3 \left|1\\0 & - \frac{5}{2} \left|- \frac{7}{2} \end{bmatrix}}\)
\(\displaystyle{ W_{2} \cdot (- \frac{2}{5}) = \begin{bmatrix}2 & 3 \left|1\\0 & 1 \left| \frac{7}{5} \end{bmatrix}}\)
\(\displaystyle{ W_{1}-3W_{2} = \begin{bmatrix}2 & 0 \left|- \frac{16}{5} \\0 & 1 \left| \frac{7}{5} \end{bmatrix}}\)
\(\displaystyle{ W_{1} \cdot \frac{1}{2} = \begin{bmatrix}1 & 0 \left|- \frac{8}{5} \\0 & 1 \left| \frac{7}{5} \end{bmatrix}}\)
MACIERZ ODWROTNA
\(\displaystyle{ A \cdot X = b}\)
\(\displaystyle{ X = A^{-1} \cdot b}\)
\(\displaystyle{ A^{-1} = \begin{bmatrix} - \frac{2}{5} & \frac{3}{5} \\ \frac{3}{5} & - \frac{2}{5} \end{bmatrix}}\)
\(\displaystyle{ X= \begin{bmatrix} - \frac{2}{5} & \frac{3}{5} \\ \frac{3}{5} & - \frac{2}{5} \end{bmatrix} \cdot \begin{bmatrix}1\\-2\end{bmatrix}}\)
\(\displaystyle{ X=\begin{bmatrix}- \frac{8}{5}\\ \frac{7}{5}\end{bmatrix}}\)
-
Mariusz M
- Użytkownik
- Posty: 6909
- Rejestracja: 25 wrz 2007, o 01:03
- Płeć: Mężczyzna
- Lokalizacja: 53°02'N 18°35'E
- Podziękował: 2 razy
- Pomógł: 1246 razy
Równanie Cramera, coś niezwykłego!!
ROZKŁAD LU
\(\displaystyle{ \left[ \begin{array}{cc} 2&3 \\ 3&2 \end{array} \right]= \left[ \begin{array}{cc} 1&0 \\ a_{21}&1 \end{array} \right] \cdot \left[ \begin{array}{cc} a_{11}&a_{12}\\0&a_{22} \end{array} \right]}\)
\(\displaystyle{ \left[ \begin{array}{cc} 2&3 \\ 3&2 \end{array} \right]= \left[ \begin{array}{cc} 1&0 \\ \frac{3}{2} &1 \end{array} \right] \cdot \left[ \begin{array} 2&3 \\ 0&- \frac{5}{2} \end{array} \right]}\)
\(\displaystyle{ \left[ \begin{array}{cc} 1&0 \\ \frac{3}{2} &1 \end{array} \right] \cdot \left[ \begin{array} {c}1&-2 \end{array} \right]}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ y_{2}= -\frac{7}{2} \end{cases}}\)
\(\displaystyle{ \left[ \begin{array}{cc} 2&3 \\ 0&- \frac{5}{2} \end{array} \right] \cdot \left[ \begin{array} {c}1&- \frac{7}{2} \end{array} \right]}\)
\(\displaystyle{ \begin{cases} x_{2}= \frac{7}{5} \\ 2x_{1}+ \frac{21}{5}= \frac{5}{5} \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{2}= \frac{7}{5} \\ 2x_{1}= -\frac{16}{5} \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{2}= \frac{7}{5} \\ x_{1}= -\frac{8}{5} \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{1}= -\frac{8}{5} \\ x_{1}= \frac{7}{5} \end{cases}}\)
\(\displaystyle{ \left[ \begin{array}{cc} 2&3 \\ 3&2 \end{array} \right]= \left[ \begin{array}{cc} 1&0 \\ a_{21}&1 \end{array} \right] \cdot \left[ \begin{array}{cc} a_{11}&a_{12}\\0&a_{22} \end{array} \right]}\)
\(\displaystyle{ \left[ \begin{array}{cc} 2&3 \\ 3&2 \end{array} \right]= \left[ \begin{array}{cc} 1&0 \\ \frac{3}{2} &1 \end{array} \right] \cdot \left[ \begin{array} 2&3 \\ 0&- \frac{5}{2} \end{array} \right]}\)
\(\displaystyle{ \left[ \begin{array}{cc} 1&0 \\ \frac{3}{2} &1 \end{array} \right] \cdot \left[ \begin{array} {c}1&-2 \end{array} \right]}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ y_{2}= -\frac{7}{2} \end{cases}}\)
\(\displaystyle{ \left[ \begin{array}{cc} 2&3 \\ 0&- \frac{5}{2} \end{array} \right] \cdot \left[ \begin{array} {c}1&- \frac{7}{2} \end{array} \right]}\)
\(\displaystyle{ \begin{cases} x_{2}= \frac{7}{5} \\ 2x_{1}+ \frac{21}{5}= \frac{5}{5} \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{2}= \frac{7}{5} \\ 2x_{1}= -\frac{16}{5} \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{2}= \frac{7}{5} \\ x_{1}= -\frac{8}{5} \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{1}= -\frac{8}{5} \\ x_{1}= \frac{7}{5} \end{cases}}\)