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This is my first time posting here so I'm exciting to join the community and gain as much knowledge as possible. My algebra is quite lacking and I'm a first year Physicist. Please could you help me solve the following three equations to find $I_3$ in terms of only $R$s and $\xi$ (i.e. no $I_1$ or $I_2$ terms). Thank you!

$$\begin{eqnarray} \xi_1 - I_1 R_1 - (I_1 - I_2) R_4 &=& 0\\ - I_2 R_2 - (I_2 - I_3) R_5 - (I_2 - I_1) R_4 &=& 0\\ - I_3 R_3- I_3 R_6 - (I_3 - I_2) R_5 &=& 0 \end{eqnarray}$$

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  • $\begingroup$ this System should be solved for $$I_1,I_2,I_3$$? $\endgroup$ – Dr. Sonnhard Graubner Dec 27 '17 at 22:37
  • $\begingroup$ Why don't you write out the linear system in $I_k$ and solve it? $\endgroup$ – copper.hat Dec 27 '17 at 22:38
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    $\begingroup$ you really should negate everything, put the constant $\xi_1$ on the other side of the equals, and carefully write out the three by three linear system, where the "variables" are $I_1, I_2, I_3.$ Unless nonmaximal rank, there should be just one solution $\endgroup$ – Will Jagy Dec 27 '17 at 22:39
  • $\begingroup$ $\xi$ is called Xi, not epsilon. $\endgroup$ – miracle173 Dec 27 '17 at 22:41
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    $\begingroup$ The first equation gives you $I_2$ in terms of $I_1$. Replace in the other equations, and you get a system of just two equations with two unknowns $I_1, I_3\,$. Repeat once more to eliminate $I_3$, and you are down to one equation with the single unknown $I_1$. Solve that and you are done. $\endgroup$ – dxiv Dec 27 '17 at 22:43
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So far, I get "augmented matrix" $$ \left( \begin{array}{ccc|c} R_1 + R_4 & - R_4 & 0 & \xi_1 \\ -R_4 & R_2 + R_4 + R_5 & - R_5 & 0 \\ 0 & - R_5 & R_3 + R_5 + R_6 & 0 \end{array} \right) $$

OOH, symmetric square part

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I calculated the solution with Maxima

(%i1) list_of_equations:[xi1-I1*R1-(I1-I2)*R4=0, -I2*R2-(I2-I3)*R5-(I2-I1)*R4=0,-I3*R3-I3*R6-(I3-I2)*R5=0];
(%o1) [(I2 - I1) R4 - I1 R1 + xi1 = 0, 
 - (I2 - I3) R5 + (I1 - I2) R4 - I2 R2 = 0, - I3 R6 + (I2 - I3) R5 - I3 R3 = 0]
(%i2) list_of_unknowns:[I1,I2,I3];
(%o2)                            [I1, I2, I3]
(%i3) list_of_solutions:solve(list_of_equations,list_of_unknowns);
(%o3) [[I1 = (xi1 (R5 (R6 + R3 + R2) + R2 (R6 + R3)) + xi1 R4 (R6 + R5 + R3))
/(R4 (R1 (R6 + R5 + R3) + R5 (R6 + R3 + R2) + R2 (R6 + R3))
 + R1 (R5 (R6 + R3 + R2) + R2 (R6 + R3))), 
I2 = (xi1 R4 (R6 + R5 + R3))/(R4 (R1 (R6 + R5 + R3) + R5 (R6 + R3 + R2)
 + R2 (R6 + R3)) + R1 (R5 (R6 + R3 + R2) + R2 (R6 + R3))), 
I3 = (xi1 R4 R5)/(R4 (R1 (R6 + R5 + R3) + R5 (R6 + R3 + R2) + R2 (R6 + R3))
 + R1 (R5 (R6 + R3 + R2) + R2 (R6 + R3)))]]
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  • $\begingroup$ I am sorry but you don't help the OP by doing that. The OP needs guiding and method. You should at least make him the remark that the denominators are the same... $\endgroup$ – Jean Marie Dec 27 '17 at 23:04
  • $\begingroup$ @JeanMarie I was only interested to check if a solution exists by using a cas. And decided to communicate this finding to the OP. To do this in a comment would be more appropriate but I see no way to format a comment usefully. $\endgroup$ – miracle173 Dec 28 '17 at 8:31

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