[Linear Algebra] solution to A^TCAx=f

  • Thread starter Dafe
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Homework Statement


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.


Homework Equations





The Attempt at a Solution



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

[/tex]

After elimination I get:

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

[/tex]

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?

Thanks.
 

Answers and Replies

  • #2
35,285
7,129

Homework Statement


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.


Homework Equations





The Attempt at a Solution



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

[/tex]

After elimination I get:

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

[/tex]

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?

Thanks.

How does ATCA go from a 3x3 matrix to a 3x4 matrix?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
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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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