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Homework Help: Solutions in matrices

  1. Nov 13, 2006 #1
    The augmented matrix of a system of linear equations in the unknowns x, y, z, u and v, has the form
    [tex]\left(\begin{array}{ccccc|c}
    1 & -3 & 1 & -1 & 0 & -1\\
    0 & 0 & 1 & 1 & 2 & 1\\
    0 & 0 & 0 & 0 & 1 & -1\\
    0 & 0 & 0 & 0 & 0 & 0
    \end{array}
    \right)[/tex]
    Write down all solutions of the system.

    The answer to this is (x, y, z, u, v) = (-4 + 3s + 2t, s, 3 - t, t, -1)


    I have:
    v = -1
    u = 1 - 2v - z = 3 - z
    z = 1 - 2v - u = 3 - u
    -3y = -4 + 2t - x
    x = -4 + 2t + 3y

    Now, I can see that they just went ahead and made u and y arbitrary. It is perfectly fine to do that?

    I'm thinking that we can allow u = t because at that point we do not know what z is; and when we solve for y we do not know what x is. Am I right in thinking that?
     
  2. jcsd
  3. Nov 13, 2006 #2
    Yeah you can do that . The basic variables are [tex] x, z , v [/tex]. The free variables are [tex] y, u [/tex]. If we let [tex] y = s [/tex] and [tex] u = t [/tex] then the solution is:

    [tex]\left(\begin{array}{c}
    -4+3s+2t\\
    s\\
    3-t\\
    t\\
    -1\\
    \end{array}
    \right) = \left(\begin{array}{c}
    -4\\
    0\\
    3\\
    0\\
    -1\\
    \end{array}
    \right) + s\left(\begin{array}{c}
    3\\
    1\\
    0\\
    0\\
    0\\
    \end{array}
    \right) + t \left(\begin{array}{c}
    2\\
    0\\
    -1\\
    1\\
    0\\
    \end{array}
    \right)[/tex]
     
    Last edited: Nov 13, 2006
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