Determinantes por Sarrus
Calcula aplicando la Regla de Sarrus el determinante de la siguiente matriz:
SOLUCIÓN
Determinante 3×3 — Regla de Sarrus
Diagonales principales ↘ (productos positivos):
![\left|\begin{array}{ccc}{\color[RGB]{0,0,0}{1}} & {\color[RGB]{30,100,220}{-1}} & {\color[RGB]{200,30,30}{1}} \\ {\color[RGB]{200,30,30}{2}} & {\color[RGB]{0,0,0}{0}} & {\color[RGB]{30,100,220}{1}} \\ {\color[RGB]{30,100,220}{1}} & {\color[RGB]{200,30,30}{2}} & {\color[RGB]{0,0,0}{-1}}\end{array}\right|](local/cache-TeX/dbe0cf367bf5cf9f43100a0113f878cf.png "\left|\begin{array}{ccc}{\color[RGB]{0,0,0}{1}} & {\color[RGB]{30,100,220}{-1}} & {\color[RGB]{200,30,30}{1}} \\ {\color[RGB]{200,30,30}{2}} & {\color[RGB]{0,0,0}{0}} & {\color[RGB]{30,100,220}{1}} \\ {\color[RGB]{30,100,220}{1}} & {\color[RGB]{200,30,30}{2}} & {\color[RGB]{0,0,0}{-1}}\end{array}\right|")
![{\color[RGB]{0,0,0}{1}}{\cdot}{\color[RGB]{0,0,0}{0}}{\cdot}{\color[RGB]{0,0,0}{(-1)}}+{\color[RGB]{30,100,220}{(-1)}}{\cdot}{\color[RGB]{30,100,220}{1}}{\cdot}{\color[RGB]{30,100,220}{1}}+{\color[RGB]{200,30,30}{1}}{\cdot}{\color[RGB]{200,30,30}{2}}{\cdot}{\color[RGB]{200,30,30}{2}}={\color[RGB]{0,0,0}{0}}+{\color[RGB]{30,100,220}{(-1)}}+{\color[RGB]{200,30,30}{4}}=3](local/cache-TeX/0c2be350ad0df862de054883e6328aff.png "{\color[RGB]{0,0,0}{1}}{\cdot}{\color[RGB]{0,0,0}{0}}{\cdot}{\color[RGB]{0,0,0}{(-1)}}+{\color[RGB]{30,100,220}{(-1)}}{\cdot}{\color[RGB]{30,100,220}{1}}{\cdot}{\color[RGB]{30,100,220}{1}}+{\color[RGB]{200,30,30}{1}}{\cdot}{\color[RGB]{200,30,30}{2}}{\cdot}{\color[RGB]{200,30,30}{2}}={\color[RGB]{0,0,0}{0}}+{\color[RGB]{30,100,220}{(-1)}}+{\color[RGB]{200,30,30}{4}}=3"){kind=small}
Diagonales secundarias ↗ (productos negativos):
![\left|\begin{array}{ccc}{\color[RGB]{0,155,50}{1}} & {\color[RGB]{110,0,200}{-1}} & {\color[RGB]{0,0,0}{1}} \\ {\color[RGB]{110,0,200}{2}} & {\color[RGB]{0,0,0}{0}} & {\color[RGB]{0,155,50}{1}} \\ {\color[RGB]{0,0,0}{1}} & {\color[RGB]{0,155,50}{2}} & {\color[RGB]{110,0,200}{-1}}\end{array}\right|](local/cache-TeX/fcde5afd73ffee3b5d20c6eb17554d7d.png "\left|\begin{array}{ccc}{\color[RGB]{0,155,50}{1}} & {\color[RGB]{110,0,200}{-1}} & {\color[RGB]{0,0,0}{1}} \\ {\color[RGB]{110,0,200}{2}} & {\color[RGB]{0,0,0}{0}} & {\color[RGB]{0,155,50}{1}} \\ {\color[RGB]{0,0,0}{1}} & {\color[RGB]{0,155,50}{2}} & {\color[RGB]{110,0,200}{-1}}\end{array}\right|")
![{\color[RGB]{0,0,0}{1}}{\cdot}{\color[RGB]{0,0,0}{0}}{\cdot}{\color[RGB]{0,0,0}{1}}+{\color[RGB]{0,155,50}{1}}{\cdot}{\color[RGB]{0,155,50}{1}}{\cdot}{\color[RGB]{0,155,50}{2}}+{\color[RGB]{110,0,200}{(-1)}}{\cdot}{\color[RGB]{110,0,200}{2}}{\cdot}{\color[RGB]{110,0,200}{(-1)}}={\color[RGB]{0,0,0}{0}}+{\color[RGB]{0,155,50}{2}}+{\color[RGB]{110,0,200}{2}}=4](local/cache-TeX/7ab425d482b7b3977c253d43301bab28.png "{\color[RGB]{0,0,0}{1}}{\cdot}{\color[RGB]{0,0,0}{0}}{\cdot}{\color[RGB]{0,0,0}{1}}+{\color[RGB]{0,155,50}{1}}{\cdot}{\color[RGB]{0,155,50}{1}}{\cdot}{\color[RGB]{0,155,50}{2}}+{\color[RGB]{110,0,200}{(-1)}}{\cdot}{\color[RGB]{110,0,200}{2}}{\cdot}{\color[RGB]{110,0,200}{(-1)}}={\color[RGB]{0,0,0}{0}}+{\color[RGB]{0,155,50}{2}}+{\color[RGB]{110,0,200}{2}}=4"){kind=small}
Calcula aplicando la Regla de Sarrus el determinante de la siguiente matriz:
Determinante 3×3 — Regla de Sarrus
Diagonales principales ↘ (productos positivos):
Diagonales secundarias ↗ (productos negativos):
Mensajes
23 de junio de 2017, 19:11, por vicente alvarez aulestia
calculo de matrices
20 de junio de 2018, 20:33, por Alison
La respuesta no seria -1?
21 de junio de 2018, 11:20, por dani
Efectivamente, el resultado es -1