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Gaussian Elemination

  1. Oct 24, 2005 #1
    use gaussian elimination to solve the following system of equation:

    2x + y - 3z = 3
    4x - y + 2z = 25
    -3x + 2y + 4z = -9

    ok so first i rearranged them so the inital numbers were in decending order (for the pivot)

    R1 -> 4x - y + 2z = 25
    R2 -> -3x + 2y + 4z = -9
    R3 -> 2x + y - 3z = 3

    pivot factor r = -3/4 for R2
    pivot factor r = 1/2 for R3

    R1 = 4x - y + 2z = 25
    R2 + (3/4)R1 = (-3x+(3/4)4x) + (2y + (-3/4y)) + (4z + (3/4)2z) = -9 + (3/4)25
    = 0x + 1.25y + 5.5z = 9.75

    R3 = (2x - 1/2x) + (y +1/2y) + (-3z-z) = 3 - 25/2
    = 0x + 1.5y - 4z = -9.5

    now solve for R2 and R3 to find Z

    pivoting factor r = 1.25/1.5 = 5/6

    R2 - 5/6R3 = (1.25y - 5/6(1.5y)) + (5.5z - 5/6(4z)) = 9.75 - 5/6(9.5)
    = 0y + ((33/6)z - (20/6)z) = 58.5/6 - 47.5/6
    = (13/6)z = 11/6
    z = 11/13

    and i know thats wrong...can anyone point out where i've went wrong?
     
  2. jcsd
  3. Oct 25, 2005 #2

    TD

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    Homework Helper

    You probably made some arithmetic mistakes, but I didn't check it all.
    Personally, when using Gaussian elimination, I'd use the extended matrix form and just work with the coëfficiënts.

    Try to put the following matrix in row-reduced form (if you've seen that of course) and 'translate' back into equations.

    [tex]\left( {\begin{array}{*{20}c}
    2 & 1 & { - 3} &\vline & 3 \\
    4 & { - 1} & 2 &\vline & {25} \\
    { - 3} & 2 & 4 &\vline & { - 9} \\

    \end{array} } \right)[/tex]
     
  4. Oct 25, 2005 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    presumably you meant here "R3- (1/2)R1" but the calculation is correct.

    Here's you're problem! Your signs are wrong on the last two terms.
    R3 is 1.5- 4x= -9.5 so R2-(5/6)R3 is
    (1.25y- (5/6)(1.5y))+ (5.5z-5/6(-4z))= 9.75- (5/6)(-9.5)
    (5.5+ 20/3)z= 9.75+ 95/12.

     
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