1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Solutions in matrices
  1. Sets of matrices... (Replies: 8)

Loading...