[Linear Algebra] solution to A^TCAx=f

[Dafe]

Homework Statement

With conductances $$c_{1}=1, c_{2}=c_{3}=2$$, multiply matrices to find
$$A^TCAx = f$$.
For $$f = (1,0,-1)$$ find a solution to $$A^TCAx = f$$.
Write the potentials $$x$$ and currents $$y = -CAx$$ on the triangle graph, when the current source $$f$$ goes into node 1 and out from node 3.

Homework Equations

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 $$x_{3}$$ is a free variable.

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

Thanks.

 

[Mark44]

Homework Statement

With conductances $$c_{1}=1, c_{2}=c_{3}=2$$, multiply matrices to find
$$A^TCAx = f$$.
For $$f = (1,0,-1)$$ find a solution to $$A^TCAx = f$$.
Write the potentials $$x$$ and currents $$y = -CAx$$ on the triangle graph, when the current source $$f$$ goes into node 1 and out from node 3.

Homework Equations

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 $$x_{3}$$ is a free variable.

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

Thanks.

How does A^TCA go from a 3x3 matrix to a 3x4 matrix?

 

[HallsofIvy]

Mark44, what Dafe wrote as his second matrix is NOT "A^TCA". 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 $x_2= x_3+ 1/8= 1$ but then $x_1$ is not an integer. Are you sure they "chose $x_3= 7/8$". I would suspect that they solved for $x_1$ and $x_3$ in terms of $x_2$ and then chose $x_2= 1$.