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Need help on matrices using cramer's rule

  1. Jan 21, 2005 #1

    qdv

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    I am learning how to solve a matrice using cramer's rule, and not sure if this is the correct answer.

    Solve the following systems of equations
    x - y + 3z = 8
    3x + y - 2z = -2
    2x + 4y + z = 0
    so I figured out the solution is x = 1, y = -1, z = 2

    but is this equation consider a
    dependent equation that all solutions that satisfy x - y + 3z = 8 ??

    thanks
     
  2. jcsd
  3. Jan 21, 2005 #2

    dextercioby

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    Homework Helper

    It's okay.The solution satisfies all equations and it's unique,therefore...Congratulations!! :smile:

    Daniel.
     
  4. Mar 18, 2008 #3
    Online calculator

    I solve it using a online calculator and I got

    Cramer rule's solver step by step
    Coeficients Matrix
    1 -1 3 8
    3 1 -2 -2
    2 4 1 0
    Δ = determinant1 -1 3
    3 1 -2
    2 4 1
    Δ sub x = det8 -1 3
    -2 1 -2
    0 4 1
    Δ sub y = det1 8 3
    3 -2 -2
    2 0 1
    Δ sub z = det1 -1 8
    3 1 -2
    2 4 0
    Δ = det1 -1 3
    3 1 -2
    2 4 1
    1 -1 3
    3 1 -2

    [(1) (1) (1) + (3) (4) (3) + (2) (-1) (-2)] - [(3) (-1) (1) + (1) (4) (-2) + (2) (1) (3)]
    (1) + (36) + (4)- (-3) + (-8) + (6)
    ( 41) - ( -5)
    Δ = 46

    Δx = det8 -1 3
    -2 1 -2
    0 4 1
    8 -1 3
    -2 1 -2

    [(8) (1) (1) + (-2) (4) (3) + (0) (-1) (-2)] - [(-2) (-1) (1) + (8) (4) (-2) + (0) (1) (3)]
    (8) + (-24) + (0)- (2) + (-64) + (0)
    ( -16) - ( -62)
    Δx = 46

    Δy = det1 8 3
    3 -2 -2
    2 0 1
    1 8 3
    3 -2 -2

    [(1) (-2) (1) + (3) (0) (3) + (2) (8) (-2)] - [(3) (8) (1) + (1) (0) (-2) + (2) (-2) (3)]
    (-2) + (0) + (-32)- (24) + (0) + (-12)
    ( -34) - ( 12)
    Δy = -46

    Δz = det1 -1 8
    3 1 -2
    2 4 0
    1 -1 8
    3 1 -2

    [(1) (1) (0) + (3) (4) (8) + (2) (-1) (-2)] - [(3) (-1) (0) + (1) (4) (-2) + (2) (1) (8)]
    (0) + (96) + (4)- (0) + (-8) + (16)
    ( 100) - ( 8)
    Δz = 92

    x =46/46

    y =-46/46

    z =-46/46

    x =1

    y =-1

    z =2

    -------
    www.algebrasolver.totalh.com
     
    Last edited: Mar 18, 2008
  5. Mar 20, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What do you mean by "this equation"? It's not at all clear what your question is. Yes, as dextercioby said, and you could easily have checked, x= 1, y= -1, z= 2 satisfies the three equations and, since the determinant of coefficients is not 0, is the only solution to that system of equations.
     
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