I have a set of 6 equations containing 6 unknowns.  When I create the matrices to solve them using Cramer's Rule, the determinants, all of them, evaluate out as a zero.

In the following example, the determinant of the first matrix is the numerator and the second the denominator when solving for X1.

20 2 0 1 0 3

20 0 2 0 0 3

0 2 -2 1 0 0

0 2 0 0 5 0

0 0 -2 1 -5 0

20 2 2 0 5 3





0 2 0 1 0 3

4 0 2 0 0 3

-4 2 -2 1 0 0

-4 2 0 0 5 0

0 0 -2 1 -5 0

0 2 2 0 5 3 

I am using EXCEL to evaluate the determinants.  Doing a 6X6 determinant manually is, well, tedious.

Is there some special relationship or message which I should know (but don't) when the determinants are all zero?


ETA - is there no way to post a decent table in these forums?

What is the RHS of the equations?

Determinant=0 means that the matrix is singular, which tells you (usually) that the equations are not linearly independent.  This means that your solution for X will be implicit.

Thank you for the help.  

What is RHS?

I believe the equations are linear and independent.  They are loop voltage equations for a 6 resistor network.  The network is shown in the figure below.

As you can see, I already have an algebraic solution and any valid solution is adequate but that is not the point.  I am trying to learn something and to do more but "bumped my head" on this anomaly.



I reviewed the equations and the matrices for typo errors but found none.

Knowing R5 carries no current, I took that out of the matrices but they still evaluate to zero.

I tried swapping the last column with the first but they still evaluate to zero.


I must have made a mistake, perhaps a polarity or other mistake.  <scurries off to review for errors, again>

Check the image I posted above.  It is a schematic of the circuit.  Was there something else you wanted?

The algebraic solution is quite easy to do, bordering on trivial once you recognize it is a balanced bridge.  

As I said, though, that's not the point.  I'm wondering why I can't use Cramer's Rule - what aspect of this disallows its use, causes the dependence or otherwise causes the method to fail.  

Is it the "balanced bridge" that's causing the failure - R5 is actually irrelevant to the circuit causing me to over-specify the order of the matrix?  I expected I5 to calculate out as zero but not to have the whole thing bomb.



Loop Equations

1)  E1 - I2R2 - I4R4 - IsRs = 0
2)  E1 - I1R1 - I3R3 - IsRs = 0
3)  I1R1 - I2R2 + I3R3 - I4R4 = 0
4)  I1R1 - I2R2 - I5R5 = 0
5)  I3R3 -I4R4 + I5R5 = 0
6)  E1 - I2R2 - I3R3 - I5R5 - IsRs = 0

The coefficients in the matrices are the values for Ri with polarity.  Values for Ri are given in the figure, above.  



You asked, why six equations?

In its general form, before you know it is actually a balanced bridge, there are six unknown currents (I1, I2, ...I6).  Six unknowns requires six equations.  I actually have 7 loop equations but need and used only 6.

I must be doing something fundamentally wrong, or misunderstanding how to use Cramer's Rule.

Could it be the professor selected a circuit and its resistor values knowing it was determinant (solvable using algebra) but incompatible with Cramer's Rule because the determinants all equal zero?  If so, that would be wicked! and evil of him.  


This is from Wikipedia:

Incompatible and indeterminate cases

A system of equations is said to be incompatible or inconsistent when there are no solutions and it is called indeterminate when there is more than one solution. For linear equations, an indeterminate system will have infinitely many solutions (if it is over an infinite field), since the solutions can be expressed in terms of one or more parameters that can take arbitrary values.

Cramer's rule applies to the case where the coefficient determinant is nonzero. In the 2 × 2 case, if the coefficient determinant is zero, then the system is incompatible if the numerator determinants are nonzero, or indeterminate if the numerator determinants are zero.

For 3x3 or higher systems, the only thing one can say when the coefficient determinant equals zero is that if any of the numerator determinants are nonzero, then the system must be incompatible. However, having all determinants zero does not imply that the system is indeterminate. A simple example where all determinants vanish (equal zero) but the system is still incompatible is the 3x3 system x+y+z=1, x+y+z=2, x+y+z=3.

I found a dependence problem with the loop equations.  For example:

equation 6 = equation 1 - equation 5
equation 6 = equation 2 + equation 4

This dependence would explain the issue.

I need six independent equations.  I thought I had that but I see my method for getting them was flawed.

Perhaps I can replace 3 KVL loops with 3 current nodes, such as: Is - I1 - I2 = 0.

I went ahead with 3 voltage loops and 3 current nodes and the determinants are no longer zero.  That's good.

The problem is, I am getting answers which do not make sense and which do not match my algebraic solution.  That's bad.


Still, I am back in the hunt!

It's done!

The three voltage loops and three current nodes worked, I just had to find and fix some sloppy typing (or was it sloppy thinking?).

I learned a few things along the way, so that's good.  What I haven't figured out is why this particular problem was so difficult.  I've used Cramer's Rule before, once just a few weeks ago.  Hmmmm,.....???

It was a dependence problem.  

In the future, I will be sure to use a good mix of current nodes and voltage loops.  I believe I figured out how to prevent dependence by looking at the topology of the loops but am not 100% certain of that yet.  It worked for this example but are my "new rules" generally true?  We will see.

I built a small spreadsheet in Excel to do the matrices.  There it so much repetition in the inputs, copy and paste made it a breeze.  Excel has a determinant function, MDETERM(array), that solves the determinants.  Inputs to the matrices are so easy in Excel, too.  Heck, they even look like matrices in the worksheet I built.

Thanks for the help and guidance.


Math & Science are fun!

Over specified.

You only have three loops.

Two loops if you remove R5.

Agreed, if you don't know R5 carries no current, it is possible there are three independent loops.  The trick is, until you can prove it, you don't know the bridge is balanced and can ignore/remove R5.  

As OP noted it is a simple enough circuit to analyze almost by inspection.

Throw in some reactive elements and plot a frequency response (Bode plot) and a matrix solution might be worth the effort.

Many years ago I had a prof that had a painfully complex circuit on a final exam.
If you looked carefully you realized it was a balanced bridge circuit and the answer was 0V (AC and DC).

And of course it was worth 20 points or so (out of 100).

Have you attempted to use a delta to wye conversion or Thevenin's yet?


or are you not to that section in circuit analysis?





Breaking it up is easier to solve and verify your mesh analysis, but if you aren't that far, don't confuse yourself further by trying it.

Problem can be solved by inspection.  Since V3 = V4, I5 = O.  If I5 is 0 then I4 = I2 and I3 = I1.

RHS is 'Right Hand Side.'

When I taught classes problems that would not solve correctly in SPICE occurred in many homework problems.

I picked them out carefully.

A PLL class had whatever loops had been used in my last SIGINT designs.

It showed the difference between the 'plug and chug' students and the ones that really understood.