Follow the link to the circuit that I need to solve for all the resistor voltage drops.
I have to make 5 linearly independent equations in order to solve for all 5 unknown voltages.
Using the loops as drawn in the image following the link I get the equations:
1. iR1*4000 + iR2*2000 = 8
2. -iR2*2000 + iR3*1000 = 4
3. iR4*6000 + iR5*1000 = -4
4. iR1*4000 + iR3*1000 = 12
5. -iR2*2000 + iR3*1000 + iR4*6000 - iR5*1000 = 0
And the Matrix:
4000 2000 0 0 0 8
0 -2000 1000 0 0 4
0 0 0 6000 1000 -4
4000 0 1000 0 0 12
0 -2000 1000 6000 -1000 0
When I row reduce this matrix, I get a row of all zeros which indicates that there isn't one unique solution to this matrix. I cant figure out what I am doing wrong as this is impossible. I could probably use different loops but I still need to understand why this configuration does not work. I have the understanding that all of my equations are linearly independent.
$$\begin{cases}V_1+R_1i_1+R_2(i_1-i_3)=0,\\R_2(i_3-i_1)+R_3i_3+V_2=0,\\-V_2+R_4i_4+R_5i_4=0.\end{cases}$$
The matrix of the system is
$$\begin{pmatrix}R_1+R_2&-R_2&0\\-R_2&R_2+R_3&0\\0&0&R_4+R_5\end{pmatrix}.$$
In fact, as there is no resistor in the edge $V_2$, the loop of $R_4$ is stand-alone (as shown by the matrix structure). And the determinant is obviously positive.
By superposition:
Set $V_2=0$ (short-circuit), and $V_1$ sees an equivalent resistance
$$R_1+\frac1{\dfrac1{R_2}+\dfrac1{R_3}}.$$
Then set $V_1=0$, and $V_2$ sees a resistance
$$\frac1{\dfrac1{R_1}+\dfrac1{R_2}}+R_3$$ on one side and $$R_4+R_5$$ on the other. With that you can find all currents.