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[Linear Algebra] solution to A^TCAx=f

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data
    With conductances [tex]c_{1}=1, c_{2}=c_{3}=2[/tex], multiply matrices to find
    [tex] A^TCAx = f [/tex].
    For [tex] f = (1,0,-1) [/tex] find a solution to [tex] A^TCAx = f [/tex].
    Write the potentials [tex] x [/tex] and currents [tex] y = -CAx [/tex] on the triangle graph, when the current source [tex] f [/tex] goes into node 1 and out from node 3.

    2. Relevant equations

    3. The attempt at a solution

    A^TCA =
    \left[ \begin{array}{ccc} 3 & -1 & -2 \\ -1 & 3 & -2 \\ -2 & -2 & 4 \\ \end{array} \right]


    After elimination I get:

    A^TCA =
    \left[ \begin{array}{cccc} 3 & -1 & -2 & 1 \\ 0 & 8 & -8 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right]


    Which to me means that the loop is broken, and that [tex] x_{3} [/tex] is a free variable.

    The book chooses [tex] x_{3}=7/8 [/tex]. Is there a good reason for that value?
    Maybe I am missing some important point here?

  2. jcsd
  3. Oct 15, 2009 #2


    Staff: Mentor

    How does ATCA go from a 3x3 matrix to a 3x4 matrix?
  4. Oct 15, 2009 #3


    User Avatar
    Science Advisor

    Mark44, what Dafe wrote as his second matrix is NOT "ATCA". It is the row echelon form of the augmented matrix with [1 0 -1] as the fourth column.

    Dafe, the only good reason I can see is that makes [itex]x_2= x_3+ 1/8= 1[/itex] but then [itex]x_1[/itex] is not an integer. Are you sure they "chose [itex]x_3= 7/8[/itex]". I would suspect that they solved for [itex]x_1[/itex] and [itex]x_3[/itex] in terms of [itex]x_2[/itex] and then chose [itex]x_2= 1[/itex].
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