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Introductory Linear Algebra por Bernard Kolman 8/E 2005pgn209ejc25

Module by: Julian Daza. E-mail the author

Summary: Ejercicio 32.25 pgn209 Algebra linear Bernard Kolman 2005

kol2005pgn209ejc21.tm por Julian Camilo Daza

21. solve the ginven linear system by Cramer's rule

x + y + z - 2w = -4

2y + z + 3w = 4

2x + y - z + 2w = 5

x - y + w = 4

 sage: M = matrix(QQ,[[1,1,1,-2],[0,2,1,3],[2,1,-1,2],[1,-1,0,1]])

 sage: M

       [ 1 1 1 -2]

       [ 0 2 1 3]

       [ 2 1 -1 2]

       [ 1 -1 0 1]

 sage: m = matrix(QQ,[[-4,1,1,-2],[4,2,1,3],[5,1,-1,2],[4,-1,0,1]])

 sage: m

      [-4 1 1 -2]

      [ 4 2 1 3]

      [ 5 1 -1 2]

      [ 4 -1 0 1]

 sage: n = matrix(QQ,[[1,-4,1,-2],[0,4,1,3],[2,5,-1,2],[1,4,0,1]])

 sage: n

      [ 1 -4 1 -2]

      [ 0 4 1 3]

      [ 2 5 -1 2]

      [ 1 4 0 1]

 sage: o = matrix(QQ,[[1,1,-4,-2],[0,2,4,3],[2,1,5,2],[1,-1,4,1]])

 sage: o

      [ 1 1 -4 -2]

      [ 0 2 4 3]

      [ 2 1 5 2]

      [ 1 -1 4 1]

 sage: p = matrix(QQ,[[1,1,1,-4],[0,2,1,4],[2,1,-1,5],[1,-1,0,4]])

 sage: p

      [ 1 1 1 -4]

      [ 0 2 1 4]

      [ 2 1 -1 5]

      [ 1 -1 0 4]

 

Ahora para el valor de x:

 sage: (m.determinant())/(M.determinant())

      1

 

x=1

Ahora para el valor de y:

 sage: (n.determinant())/(M.determinant())

      -1

 

y=-1

Ahora para el valor de z:

 sage: (o.determinant())/(M.determinant())

      0

 

z=0

Ahora para el valor de w:

 sage: (p.determinant())/(M.determinant())

      2

 

w=2

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