Linear Algebra Problem concerning a circuit

VinnyCee
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Homework Statement



http://img216.imageshack.us/img216/302/problem8310en9.jpg

By applying Kirchoff's laws to the circuit above, we obtain the following equations:

i_1\,-\,i_2\,-\,i_3\,=\,0

i_1\,-\,i_2\,-\,i_3\,=\,0

R_2\,i_2\,-\,R_3\,i_3\,=\,0

R_1\,i_1\,-\,R_2\,i_2\,=\,E

R_1\,i_1\,-\,R_3\,i_3\,=\,E

Obtain the solution set of equations by Gauss elimination. If there is no solution, or if there is a non-unique solution, explain that result in physical terms.

R_1\,=\,R_2\,=\,R_3\,\equiv\,R

Homework Equations



Linear algebra, matrices, etc.

The Attempt at a Solution



First, I put the four non-identical equations into a matrix.

\left(\begin{array}{cccc}1&-1&-1&0\\0&R&-R&0\\R&R&0&E\\R&0&R&E\end{array}\right)

Now I reduce it down to R.R.E.F. using elementary row operations. (Note that one of the equations is redundant)

\left(\begin{array}{cccc}1&0&0&\frac{2E}{3R}\\0&1&0&\frac{E}{3R}\\0&0&1&\frac{E}{3R}\end{array}\right)

So then, i_1\,=\,\frac{2E}{3R} and i_2\,=\,i_3\,=\,\frac{E}{3R}?
 
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Look fine to me.
 
yeah, that's right but in the initial equations, it should be:
I2R2 + I1R1=E
I3R3 + I1R1 = E
 
huyen_vyvy said:
yeah, that's right but in the initial equations, it should be:
I2R2 + I1R1=E
I3R3 + I1R1 = E

True. But the signs were correct in the matrix. Took it as a typo.
 
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