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Solving System of Equations

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Well, I am in the middle of solving a (quite difficult) differential equation using the Method of Undetermined Coefficients, and have finally come across a system of equations and I need to solve for A(sub 1), or (A1), (A2), (B1), and (B2), but I cannot figure out how to (or if it is possible at all, seeing as how I could have done the rest of the problem wrong before it). The system of equations I now have is:

    (1) -7(A1) + 3(A2) + 6(B1) - 21(B2) = 2
    (2) 3(A1) + 12(A2) - 7(B1) + 3(B2) = -34
    (3) 21(A1) + 5(B1) = 150

    2. Relevant equations
    ---

    3. The attempt at a solution

    Using equation (3), I can determine that
    (A1) = (50/7) - (5/21)(B1)
    and that
    (B1) = 30 - (21/5)(A1)

    Using these and plugging them into equation (2), I get a 4th equation:

    (4) 1029(A1) + 420(A2) - 25(B1) + 105(B2) = 5410

    Now I have a new system of equations:

    (1) -7(A1) + 3(A2) + 6(B1) - 21(B2) = 2
    (2) 3(A1) + 12(A2) - 7(B1) + 3(B2) = -34
    (3) 21(A1) + 5(B1) = 150
    (4) 1029(A1) + 420(A2) - 25(B1) + 105(B2) = 5410

    Now, I tried using my calculator to solve this system, but it only gives me an error. Is there another way to solve for the coefficients?? Any help would be great!
     
  2. jcsd
  3. Mar 30, 2008 #2
    Hi,
    I'm not the best at linear algebra, but it looks to me like you have 3 equations and 4 unknowns. That can't be solved. You do need one more equation, but it seems to me like if you solve the last equation for A1 and A2 and then plug that back into equation 2, you have a dependent system. If you solve number 3 for A1 and then plug that in to 1 and 2, then you have 2 equations and 3 unknowns...still can't solve it.

    I may be wrong here...I'm wrong a lot,
    But that's what I remember.
    CC
     
  4. Mar 30, 2008 #3
    Well I started with only 3 equations, but under "attempt at the solution" I did what you said and got a 4th equation. So now I have 4 equations with 4 variables which should work, but when I use matrices in my calculator, it won't solve it.
     
  5. Mar 30, 2008 #4
    You have to allow one of the variables to be arbitrary, and then solve for the rest of them in terms of the arbitrary variable. This will give you a set of solutions, which are known as the "solution space"
     
  6. Mar 31, 2008 #5

    rock.freak667

    User Avatar
    Homework Helper

    What you did to obtain that 4th equation was putting A1 and B1 from (3) into another equation. While that may appear to give you a fourth equation, if you actually try to solve it via row-reduction, you should see that you will get a row of zero's. Meaning that you will have an infinite number of solutions

    You could post the differential equation you tried to solve and then post your attempt at it and we can attempt to see if you went wrong somewhere in the process.
     
    Last edited: Mar 31, 2008
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