# Linear Algebra Problem concerning a circuit

1. Jul 7, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

http://img216.imageshack.us/img216/302/problem8310en9.jpg [Broken]

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.

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

2. Relevant equations

Linear algebra, matrices, etc.

3. 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 $i_2\,=\,i_3\,=\,\frac{E}{3R}$?

Last edited by a moderator: May 3, 2017
2. Jul 7, 2007

### Dick

Look fine to me.

3. Jul 8, 2007

### huyen_vyvy

yeah, that's right but in the initial equations, it should be:
I2R2 + I1R1=E
I3R3 + I1R1 = E

4. Jul 8, 2007

### Dick

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