Răspuns :
[tex] \hbox{Determinantul matricii A=}\left(\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right) \hbox{ este numarul: } \\\\ det(A)= \left|\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right|[/tex]
Regula lui Sarrus spune ca sub determinant se vor scrie primele 2 linii ale acestuia si se va face diferenta dintre suma produselor diagonalelor descendente si suma produselor diagonalelor ascendente care va fi egala cu 0.
[tex] \left|\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right| \\ \left\begin{array}{ccc}1&2&3\\4&5&6\ \end{array}\right \\\\\\ det(A)=1*5*9+4*8*3+7*2*6 -3*5*7-6*8*1-9*2*4 \\\\ det(A)=0 \\\\\\ det(A)=45+96+84-105-48-72= \\\\= 225-225= 0 \ \ \ 'A' [/tex]
Regula generala:
[tex] \left|\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right| = det(A) \\\\ .\ \ a_{11} \ \ \ a_{12} \ \ \ a_{13} \\ . \ \ a_{21} \ \ \ a_{22} \ \ \ a_{23} \\\\\\ a_{11}a_{22}a_{33}+a_{21}a_{32}a_{13}+a_{31}a_{12}a_{23}=a_{13}a_{22}a_{31}+a_{23}a_{32}a_{11}+a_{33}a_{12}a_{21}[/tex]
Regula lui Sarrus spune ca sub determinant se vor scrie primele 2 linii ale acestuia si se va face diferenta dintre suma produselor diagonalelor descendente si suma produselor diagonalelor ascendente care va fi egala cu 0.
[tex] \left|\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right| \\ \left\begin{array}{ccc}1&2&3\\4&5&6\ \end{array}\right \\\\\\ det(A)=1*5*9+4*8*3+7*2*6 -3*5*7-6*8*1-9*2*4 \\\\ det(A)=0 \\\\\\ det(A)=45+96+84-105-48-72= \\\\= 225-225= 0 \ \ \ 'A' [/tex]
Regula generala:
[tex] \left|\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right| = det(A) \\\\ .\ \ a_{11} \ \ \ a_{12} \ \ \ a_{13} \\ . \ \ a_{21} \ \ \ a_{22} \ \ \ a_{23} \\\\\\ a_{11}a_{22}a_{33}+a_{21}a_{32}a_{13}+a_{31}a_{12}a_{23}=a_{13}a_{22}a_{31}+a_{23}a_{32}a_{11}+a_{33}a_{12}a_{21}[/tex]