układ Cramera A*w=b
: 9 cze 2011, o 11:02
Witam Proszę o sprawdzenie zadania, układu Cramera Zadanie muszę rozwiązać z tego wzoru: W=\(\displaystyle{ A^{-1}}\)*b
A= \(\displaystyle{ \begin{bmatrix} 1&1&2\\2&3&3\\3&4&6\end{bmatrix}}\) b= \(\displaystyle{ \begin{bmatrix} 2\\3\\0\end{bmatrix}}\)
A det = \(\displaystyle{ \begin{bmatrix} 1&1&2\\2&3&3\\3&4&6\end{bmatrix}}\)=18+16+9-18-12-12=1
\(\displaystyle{ d_{11}}\) =\(\displaystyle{ (-1)^{2}}\)\(\displaystyle{ \begin{bmatrix} 3&3\\4&6\end{bmatrix}}\)=18-12=6
\(\displaystyle{ d_{12}}\) =\(\displaystyle{ (-1)^{3}}\)\(\displaystyle{ \begin{bmatrix} 2&3\\3&6\end{bmatrix}}\)=-1(12-9)=-3
\(\displaystyle{ d_{13}}\) =\(\displaystyle{ (-1)^{4}}\)\(\displaystyle{ \begin{bmatrix} 2&3\\3&4\end{bmatrix}}\)=8-9=-1
\(\displaystyle{ d_{21}}\) =\(\displaystyle{ (-1)^{3}}\)\(\displaystyle{ \begin{bmatrix} 1&2\\4&6\end{bmatrix}}\)=(-1)6-8=2
\(\displaystyle{ d_{22}}\) =\(\displaystyle{ (-1)^{4}}\)\(\displaystyle{ \begin{bmatrix} 1&2\\3&6\end{bmatrix}}\)=6-6=0
\(\displaystyle{ d_{23}}\) =\(\displaystyle{ (-1)^{5}}\)\(\displaystyle{ \begin{bmatrix} 1&1\\3&4\end{bmatrix}}\)=-1(4-3)=-1
\(\displaystyle{ d_{31}}\) =\(\displaystyle{ (-1)^{6}}\)\(\displaystyle{ \begin{bmatrix} 1&1\\2&3\end{bmatrix}}\)=1
\(\displaystyle{ d_{32}}\) =\(\displaystyle{ (-1)^{5}}\)\(\displaystyle{ \begin{bmatrix} 1&2\\2&3\end{bmatrix}}\)=(-1)*(-1)=1
\(\displaystyle{ d_{33}}\) =\(\displaystyle{ (-1)^{6}}\)\(\displaystyle{ \begin{bmatrix} 1&1\\2&3\end{bmatrix}}\)=1
D=\(\displaystyle{ \begin{bmatrix} 6&-3&-1\\2&0&-1\\1&1&1\end{bmatrix}}\)
\(\displaystyle{ D^{T}}\)=\(\displaystyle{ \begin{bmatrix} 6&2&1\\-3&0&1\\-1&-1&1\end{bmatrix}}\)
\(\displaystyle{ A^{-1}}\)=\(\displaystyle{ \frac{1}{det(A)}}\)\(\displaystyle{ D^{T}}\)
\(\displaystyle{ A^{-1}}\)=\(\displaystyle{ \frac{1}{1}}\)\(\displaystyle{ \begin{bmatrix} 6&2&1\-32&0&1\\-1&-1&1\end{bmatrix}}\)=-3-2+6-6=-5
\(\displaystyle{ \begin{bmatrix} 6&2&1\\-3&0&1\\-1&-1&1\end{bmatrix}}\) * \(\displaystyle{ \begin{bmatrix} 2\\3\\0\end{bmatrix}}\)= \(\displaystyle{ \begin{bmatrix} 18\\-6\\-5\end{bmatrix}}\)
w=\(\displaystyle{ \begin{bmatrix} 18\\-6\\-5\end{bmatrix}}\)
A= \(\displaystyle{ \begin{bmatrix} 1&1&2\\2&3&3\\3&4&6\end{bmatrix}}\) b= \(\displaystyle{ \begin{bmatrix} 2\\3\\0\end{bmatrix}}\)
A det = \(\displaystyle{ \begin{bmatrix} 1&1&2\\2&3&3\\3&4&6\end{bmatrix}}\)=18+16+9-18-12-12=1
\(\displaystyle{ d_{11}}\) =\(\displaystyle{ (-1)^{2}}\)\(\displaystyle{ \begin{bmatrix} 3&3\\4&6\end{bmatrix}}\)=18-12=6
\(\displaystyle{ d_{12}}\) =\(\displaystyle{ (-1)^{3}}\)\(\displaystyle{ \begin{bmatrix} 2&3\\3&6\end{bmatrix}}\)=-1(12-9)=-3
\(\displaystyle{ d_{13}}\) =\(\displaystyle{ (-1)^{4}}\)\(\displaystyle{ \begin{bmatrix} 2&3\\3&4\end{bmatrix}}\)=8-9=-1
\(\displaystyle{ d_{21}}\) =\(\displaystyle{ (-1)^{3}}\)\(\displaystyle{ \begin{bmatrix} 1&2\\4&6\end{bmatrix}}\)=(-1)6-8=2
\(\displaystyle{ d_{22}}\) =\(\displaystyle{ (-1)^{4}}\)\(\displaystyle{ \begin{bmatrix} 1&2\\3&6\end{bmatrix}}\)=6-6=0
\(\displaystyle{ d_{23}}\) =\(\displaystyle{ (-1)^{5}}\)\(\displaystyle{ \begin{bmatrix} 1&1\\3&4\end{bmatrix}}\)=-1(4-3)=-1
\(\displaystyle{ d_{31}}\) =\(\displaystyle{ (-1)^{6}}\)\(\displaystyle{ \begin{bmatrix} 1&1\\2&3\end{bmatrix}}\)=1
\(\displaystyle{ d_{32}}\) =\(\displaystyle{ (-1)^{5}}\)\(\displaystyle{ \begin{bmatrix} 1&2\\2&3\end{bmatrix}}\)=(-1)*(-1)=1
\(\displaystyle{ d_{33}}\) =\(\displaystyle{ (-1)^{6}}\)\(\displaystyle{ \begin{bmatrix} 1&1\\2&3\end{bmatrix}}\)=1
D=\(\displaystyle{ \begin{bmatrix} 6&-3&-1\\2&0&-1\\1&1&1\end{bmatrix}}\)
\(\displaystyle{ D^{T}}\)=\(\displaystyle{ \begin{bmatrix} 6&2&1\\-3&0&1\\-1&-1&1\end{bmatrix}}\)
\(\displaystyle{ A^{-1}}\)=\(\displaystyle{ \frac{1}{det(A)}}\)\(\displaystyle{ D^{T}}\)
\(\displaystyle{ A^{-1}}\)=\(\displaystyle{ \frac{1}{1}}\)\(\displaystyle{ \begin{bmatrix} 6&2&1\-32&0&1\\-1&-1&1\end{bmatrix}}\)=-3-2+6-6=-5
\(\displaystyle{ \begin{bmatrix} 6&2&1\\-3&0&1\\-1&-1&1\end{bmatrix}}\) * \(\displaystyle{ \begin{bmatrix} 2\\3\\0\end{bmatrix}}\)= \(\displaystyle{ \begin{bmatrix} 18\\-6\\-5\end{bmatrix}}\)
w=\(\displaystyle{ \begin{bmatrix} 18\\-6\\-5\end{bmatrix}}\)