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Matrix algebra - Gauss 0=0, determinant = 0

  1. Dec 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Mesh Analysis, find current z:

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

    b= [tex]\left(\begin{array}{c}+10\\-10\\0\end{array}\right)[/tex]

    2. Relevant equations

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

    Gauss elimination

    Cramer's method

    3. 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?
     
    Last edited: Dec 12, 2012
  2. jcsd
  3. Dec 12, 2012 #2

    HallsofIvy

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    If the equation "Ax= B" reduces to "0= 0" then the original equation is true for all x.
     
  4. Dec 12, 2012 #3
    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?
     
  5. Dec 12, 2012 #4

    SteamKing

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    Try posting the original problem. There may be some error in the matrix formulation.
     
  6. Dec 12, 2012 #5
    TttHQ.jpg

    All R's are 15Ω And all U's are 10V

    [tex]\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}[/tex]

    Sorting into:

    [tex]\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}[/tex]

    A=[tex]\left(\begin{array}{ccc}30&-15&-15\\-15&30&-15\\-15&-15&30\end{array}\right)[/tex]

    x=[tex]\left(\begin{array}{c}x\\y\\z\end{array}\right)[/tex]

    b=[tex]\left(\begin{array}{c}10\\-10\\0\end{array}\right)[/tex]

    A*x=b

    Find current z. Answer is given as 0A.
     
    Last edited: Dec 12, 2012
  7. Dec 12, 2012 #6

    Ray Vickson

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    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
    [tex] 30x - 15y - 15z = 10\\
    -15x + 30y - 15z = -10\\
    -15x - 15y + 30z = 0. [/tex]
    So, solving the third equation for x, say, gives x = 2z-y, and substituting that into the first two equations gives
    [tex] -45y + 45z = 10\\
    45y - 45z = -10.[/tex]
    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.
     
  8. Dec 12, 2012 #7
    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.
     
  9. Dec 12, 2012 #8

    Ray Vickson

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    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
    [tex]R_1(x−z)+R_3(x-y)=U_1\\
    R_2(y-z)+R_3(y-x) = -U_2\\
    R_1(z-x)+R_2(z-y)+R_4z=0[/tex]
    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!
     
  10. Dec 12, 2012 #9
    ... 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!
     
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