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Math Help - Two lin alg question

  1. #1
    Newbie Db0ne's Avatar
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    Two lin alg question

    1. Use gauss-jordan elimination to solve the following systems
    x - 2y + z = -6
    2x - 3y = 7
    -x + 3y - 3z =11
    I got a finite solution for this one which was wrong so I'm redoing it. Is it suppose to be infinite?

    Ok. so I got x-3z=4
    y-2z=5
    and 0=0
    Now im stuck.

    2. Determine the polynomial function whose graph passes through the points: (1,4), (2,0), (3,12)
    This one i'm stuck on. Curve fitting?
    Last edited by Db0ne; June 12th 2009 at 07:10 PM.
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  2. #2
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    Quote Originally Posted by Db0ne View Post
    1. Use gauss-jordan elimination to solve the following systems
    x - 2y + z = -6
    2x - 3y = 7
    -x + 3y - 3z =11
    I got a finite solution for this one which was wrong so I'm redoing it. Is it suppose to be infinite?

    Ok. so I got x-3z=4
    y-2z=5
    and 0=0
    Now im stuck.

    2. Determine the polynomial function whose graph passes through the points: (1,4), (2,0), (3,12)
    This one i'm stuck on. Curve fitting?
    1. Since the determinant of the coefficient matrix is equal to zero, there is either no solution or an infinite number of solutions. I haven't checked your work but assuming that x - 3z = 4 and y - 2z = 5 are correct, the solution can be written as

    x = 3t + 4, y = 2t + 5, z = t

    where t is any real number. That is, an infinite number of solutions.

    2. Assuming that a polynomial of minimum degree is required, you should use the model y = ax^2 + bx + c and substitute the given data to get three equations involving a, b and c. Solve these equations simultaneously.
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  3. #3
    MHF Contributor

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    Quote Originally Posted by Db0ne View Post
    1. Use gauss-jordan elimination to solve the following systems
    x - 2y + z = -6
    2x - 3y = 7
    -x + 3y - 3z =11
    I got a finite solution for this one which was wrong so I'm redoing it. Is it suppose to be infinite?

    Ok. so I got x-3z=4
    Then you've added wrong! Or, perhaps, subtracted where you should have added. Adding the last two equations cancels the "y" term : 2x-x- 3z= x- 3z= 7+ 11= 18, not 4.

    y-2z=5
    and 0=0
    Now im stuck.
    Since x- 3z= 18, x= 3z+ 18 and the first equation becomes 3z+ 18- 2y+ z= -6 so 4z- 2y= -24. Dividing by 2, 2z- y= -12 or y= 2z+ 12.

    Putting x= 3z+ 18 and y= 2z+ 12 into the last equation, -(3z+ 18)+ 3(2z+ 12)- 3z= 11 or -3z+ 6z- 3z- 18+ 36= 0z+ 18= 11 which is never true. The is no x, y, z that satisfy these three equations.

    2. Determine the polynomial function whose graph passes through the points: (1,4), (2,0), (3,12)
    This one i'm stuck on. Curve fitting?
    A quadratic function will fit any three points (in general, a function of degree n-1 will fit n points). Let y= ax^2+ bx+ c and put in those three points:
    at (1, 4): 4= a(1^2)+ b(1)+ c so a+ b+ c= 4
    at (2, 0): 0= a(2^2)+ b(2)+ c so 4a+ 2b+ c= 0
    at (3 12): 12= a(3^2)+ b(3)+ c so 9a+ 3b+ c= 12.

    Solve those three equations for a, b, and c.
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