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# ejercicio regla de cramer

Module by: Danna Amézquita. E-mail the author

Summary: Un ejercicio en el cual se usa la regla de cramer para solucionar el sistema de ecuaciones

greyrgb0.75,0.75,0.75 orangergb1.0,0.5,0.5 brownrgb0.5,0.25,0.0 pinkrgb1.0,0.5,0.5

Utilizar la regla de cramer, cuando sea aplicable, para resolver el sistema

x - 2 y - 3 z = - 1 x - 2 y - 3 z = - 1
(1)
2 x + y + z = 6 2 x + y + z = 6
(2)
x + 3 y - 2 z = 13 x + 3 y - 2 z = 13
(3)

redsage] blue B=matrix (QQ,[[1,-2,-3],[2,1,1],[1,3,-2]])

redsage] blueB

1-2-321113-21-2-321113-2 = A


redsage] blueB.determinant()

-30-30 = |A|


redsage] blue C = matrix ( QQ,[[-1,-2,-3],[6,1,1],[13,3,-2]])

redsage] blueC

-1-2-3611133-2-1-2-3611133-2 = A x


redsage] blueC.determinant()

-60-60 = | A x |


redsage] blue D = matrix (QQ,[[1,-1,-3],[2,6,1],[1,13,-2]])

redsage] blueD

1-1-3261113-21-1-3261113-2 = A y


redsage] blueD.determinant()

-90-90 = |Ay|


redsage] blue

redsage] blue E=matrix (QQ,[[1,-2,-1],[2,1,6],[1,3,13]])

redsage] blueC

-1-2-3611133-2-1-2-3611133-2 = A z


redsage] blueC.determinant ()

-60-60 =  |A z |


redsage] blue

x = | A x | | A | = - 60 - 30 = 2 x = | A x | | A | = - 60 - 30 = 2
(4)
y = | A y | | A | = - 90 - 30 = 3 y = | A y | | A | = - 90 - 30 = 3
(5)
z = | A z | | A | = 30 - 30 = - 1 z = | A z | | A | = 30 - 30 = - 1
(6)

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