Matrix algebra - Gauss 0=0, determinant = 0

[DrOnline]

Homework Statement

Mesh Analysis, find current z:

A= $$\left(\begin{array}{ccc}+30x&-15y&-15z\\-15x&+30y&-15z\\-15x&-15y&+30z\end{array}\right)$$

b= $$\left(\begin{array}{c}+10\\-10\\0\end{array}\right)$$

Homework Equations

A*x=b
A= resistance
x= currents
b= voltage sources

Gauss elimination

Cramer's method

The Attempt at a Solution

I first tried Cramer's method. This requires the determinant of A. Unfortunately, that determinant = zero, make the method impossible as I cannot divide by zero.

Then I try Gauss elimination, and end up with zero = zero.

The correct answer for current z should be 0.

What should I do? None of the methods work. What does this prove? I have checked, and rechecked my matrices...

So how can I calculate the current z?

 

[HallsofIvy]

If the equation "Ax= B" reduces to "0= 0" then the original equation is true for all x.

 

[DrOnline]

I wonder how that actually translate to the electrical circuit...

It's so strange. With just batteries and resistors, surely there always be currents != zero, and the answers must be concrete.

Can you shed some more light on this?

 

[SteamKing]

Try posting the original problem. There may be some error in the matrix formulation.

 

[DrOnline]

All R's are 15Ω And all U's are 10V

$$\begin{array}{ccc}R1(x-z)&0&R3(x-y)&0&=U1\\0&R2(y-z)&R3(y-x)&0&=-U2\\R1(z-x)&R2(z-y)&0&R4z&=0\end{array}$$

Sorting into:

$$\begin{array}{ccc}(R1+R3)*x&-R3*y&-R1*z\\-R3*x&(R2+R3)*y&-R2*z\\-R1*x&-R2*y&(R1+R2)*z\end{array}$$

A=$$\left(\begin{array}{ccc}30&-15&-15\\-15&30&-15\\-15&-15&30\end{array}\right)$$

x=$$\left(\begin{array}{c}x\\y\\z\end{array}\right)$$

b=$$\left(\begin{array}{c}10\\-10\\0\end{array}\right)$$

A*x=b

Find current z. Answer is given as 0A.

 

[Ray Vickson]

Homework Statement

Mesh Analysis, find current z:

A= $$\left(\begin{array}{ccc}+30x&-15y&-15z\\-15x&+30y&-15z\\-15x&-15y&+30z\end{array}\right)$$

b= $$\left(\begin{array}{c}+10\\-10\\0\end{array}\right)$$

Homework Equations

A*x=b
A= resistance
x= currents
b= voltage sources

Gauss elimination

Cramer's method

The Attempt at a Solution

I first tried Cramer's method. This requires the determinant of A. Unfortunately, that determinant = zero, make the method impossible as I cannot divide by zero.

Then I try Gauss elimination, and end up with zero = zero.

The correct answer for current z should be 0.

What should I do? None of the methods work. What does this prove? I have checked, and rechecked my matrices...

So how can I calculate the current z?

The matrix A should not have x,y and z it it; with A as you have written it, the system Ax = b will be a set of coupled quadratic equations!

So, you have the system of equations
$$30x - 15y - 15z = 10\\<br /> -15x + 30y - 15z = -10\\<br /> -15x - 15y + 30z = 0.$$
So, solving the third equation for x, say, gives x = 2z-y, and substituting that into the first two equations gives
$$-45y + 45z = 10\\<br /> 45y - 45z = -10.$$
One of these is redundant, so all we end up with is 45y = 45z - 10, or y = z - (2/9).
This gives x = z + (2/9). So that is the form of solution: x = z + (2/9), y = z - (2/9), z = anything.

That is always what happens when you do Gaussian elimination on a system with determinant = 0: you either get no solution at all, or else you get an infinite number of solutions (as in this case). Just saying 0 = 0 misses a whole lot of important information.

 

[DrOnline]

Wow, thanks, but if z has an infinite number of solutions, how can the answer be stated as z=0?

I know the currents should not be in the A matrix, that was a lazy mistake on my part because it takes a while to write all the matrices in tex I kinda mashed two together for Gauss. See my new post that clearly outlines my method. I don't think I made any mistakes creating the matrices.

 

[Ray Vickson]

Wow, thanks, but if z has an infinite number of solutions, how can the answer be stated as z=0?

I know the currents should not be in the A matrix, that was a lazy mistake on my part because it takes a while to write all the matrices in tex I kinda mashed two together for Gauss. See my new post that clearly outlines my method. I don't think I made any mistakes creating the matrices.

Well, in answer to your question: z = 0 is one of the possible solutions, but there are infinitely many others as well. However, tht does not solve your physical problem, which arises because your matrix is wrong. Your original equations are
$$R_1(x−z)+R_3(x-y)=U_1\\<br /> R_2(y-z)+R_3(y-x) = -U_2\\<br /> R_1(z-x)+R_2(z-y)+R_4z=0$$
so the last equation is -R1 x - R2 y + (R1 + R2 + R4)z = 0, or -15x -15y + 45z = 0. Therefore, A should have '45' (not '30') in the bottom right corner.

Whenever you see an 'unsolvable' problem in a formulation of a simple physical question, you (should) know that you must have done something wrong!

 

[DrOnline]

... sorry for wasting your time. Like I wrote, I checked and rechecked and spent a while on it, and all the time it was a stupid mistake like that.

Thanks!