układ Cramera A*w=b

[Suey]

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=\(\displaystyle{ \begin{bmatrix} 18\\-6\\-5\end{bmatrix}}\)

[aalmond]

Wyznacznik.

[Suey]

To znaczy?

[sushi]

\(\displaystyle{ Det A= |..|}\) a nie \(\displaystyle{ Det A= [...]}\) --> to mial na mysli przedmówca


tak samo

\(\displaystyle{ d_{11}=(-1)^2 \cdot |...|}\)

[Suey]

Ach faktycznie Zapomniałam. A oprócz tego?

[sushi]

na koniec zawsze mozna sprawdzic czy

\(\displaystyle{ A \cdot W=B}\)