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Question about a System of Equations

  1. Aug 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve the following system using Gauss-Jordan Elimination
    x - y + 2z - w = -1
    2x + y - 2z - 2w = -2
    -x + 2y - 4z + w = 1
    3x + 0y + 0z - 3w = -3

    3. The attempt at a solution
    I first wanted to ask what the difference between Gauss Jordan Elimination and Gaussian Elimination was.

    none of the equations are homogeneous and it has an equal number of unknowns and equations so I can rule out that it may have infinitely many solutions.

    The answer was written in parametric form as the following:

    x = t - 1
    y = 2s
    z = s
    w= t

    Which I also don't understand, what's the point of writing the answers in parametric form if it doesn't have infinitely many solutions, perhaps it does? Rather, in general, what's the point of writing the solutions of a system of equations in parametric form? I remember from Calculus that it gave a sort of orientation or direction of movement to functions. But in this sense is it the same or even similar?
    I am very rusty with systems as trivial as they may be at first, and i'm only a week into my Linear Algebra course, so it would be helpful to get the ball rolling on some of these things.

    Thanks, Icesalmon
     
  2. jcsd
  3. Aug 25, 2011 #2

    Mark44

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    No you can't. If one of the equations is a linear combination of the others, the system is dependent and will have an infinite number of solutions.
    I haven't solved the system, but the answer suggests that the system is dependent (has multiple solutions).
    Since there are two parameters in the answers you show, the solutions form a two-dimensional plane. All of the solution vectors are in this plane, in the sense that their tails are at the origin and their heads define points in the plane.

    The solution set could also be written like this:
    <x;y;z;w> = s<0; 2; 1; 0> + t<1; 0; 0; 1> + <-1; 0; 0; 0>
    (All vectors above should be considered to be column vectors.)
     
  4. Aug 25, 2011 #3

    vela

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    I'm pretty sure they are two different names for the same method.
    This isn't correct. You can't rule out that possibility until you verify that the equations are independent.
    This system indeed has an infinite number of solutions.
    Just curious, but how else do you propose to express the solution to the system?

    You can write those equations in the form x = t(1,0,0,1)+s(0,2,1,0)+(-1,0,0,0), which you will find useful later on in the course.
     
  5. Aug 25, 2011 #4
    How do I show this?

    I wasn't sure there were infinitely many solutions so I thought the solutions would be unique, but don't there have to be equations that are all constant multiples of one another to give infinitely many solutions?
     
  6. Aug 25, 2011 #5

    vela

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    The most straightforward way is simply to solve the system.
    No, you just need the equations not to be independent. If one equation is a multiple of another, it's not independent, but that's not the most general case. As Mark mentioned above, if an equation is a linear combination of the others, it's not independent. For example, take the system
    \begin{align*}
    x + y + z &= 1 \\
    x - y - z &= 0 \\
    3x + y + z &= 2
    \end{align*}
    You can make the third equation by multiplying the first equation by 2 and adding it to the second equation, so it's not independent of the other two. This fact would show up during Gaussian elimination as a row of all zeros.
     
  7. Aug 25, 2011 #6
    Okay, so:

    Eq. #4: 3x + 0y + 0z - 3w = -3 Is a linear combination of
    Eq. #1: x - y + 2z - w = -1
    And
    Eq. #2: 2x + y - 2z - 2w = -2

    Because if I add #1 and #2 together I get #4
     
    Last edited: Aug 25, 2011
  8. Aug 25, 2011 #7

    vela

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    Right, and the fact you end up with two parameters in the solution means you should find yet another equation that's dependent.
     
  9. Aug 25, 2011 #8
    Okay, I found my answers and got my parameters, but I can't find the other dependent Equation. But my question about this final dependent equation, is it the reason I can generalize my answers because right now I have x1 = w1 - 1 and y2 = 2z2 where w1 = t and z2 = s, I need just w = t and z = s, and i'm thinking that finding the second dependent equation would give me the ability to say that.
     
  10. Aug 25, 2011 #9

    vela

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    From the reduced matrix, you end up with the equations

    x = w - 1
    y = 2z

    There are no restrictions on what w and z are equal to, so you can say z=s and w=t, where t and s are your parameters. Then x = t-1 and y = 2s.
     
  11. Aug 26, 2011 #10

    Ray Vickson

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    Gauss vs. Gauss-Jordan: in Gaussian elimination we produce a lower-triangular system, so we get a system of equations like
    a11*x1 + a12*x2 + a13*x3 = r1
    a22*x2 + a23*x3 = r2
    a33*x3 = r3
    We can, of course, solve this by back-substitution: get x3 from eq (3), then put that into eq (2) and get x2, etc. In Gauss-Jordan elimination we also eliminate "upwards", subtracting a multiple of the above second equation from the first, etc., to end up with a diagonal system like c11*x1 = d1, c22*x2 = d2, c33*x3 = d3. Basically, one method performs "LU decomposition" while the other essentially computes the matrix inverse.

    RGV
     
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