Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: General Solution Linear algebra

  1. Jan 21, 2010 #1
    The problem statement, all variables and given/known data
    Find the general solution to the system:

    [tex]ax+ by= 0[/tex]
    [tex]cx+ dy= 0[/tex]

    Consider the case when
    [tex]ad- bc\neq 0[/tex]

    The attempt at a solution
    I multiplied the first equation by "c" and the second equation by "a", and then I subtracted the two equations.

    I got the following matrix:
    [tex]0......ad-bc......0[/tex]
    [tex]ac......ad......0[/tex]

    Therefore, [tex]ad\neq bc[/tex]. I got the general solution to be [tex]cx+ dy= 0[/tex] from the second row of the matrix. Is it right?
     
  2. jcsd
  3. Jan 21, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Nope. The first line in your reduced matrix corresponds to the equation [tex](ad-bc)y=0[/tex]. Since [itex]ad-bc\ne 0[/itex], y must be zero. I'll let you figure out what x equals.
     
  4. Jan 21, 2010 #3
    Ok, since y= 0,

    [tex]acx+ ady= 0[/tex]
    [tex]acx+ ad(0)= 0[/tex]
    [tex]acx= 0[/tex]

    So [tex]x= 0[/tex] too?

    But that gives a general solution of [tex]0=0[/tex]. Does it make it infinetly many solutions?
     
  5. Jan 21, 2010 #4

    Mark44

    Staff: Mentor

    You're making this harder than it needs to be. After discovering that y = 0, substitute that value in either of your original equations to solve for x.

    There is only one solution to this system of equations.
     
  6. Jan 21, 2010 #5
    If I substitute 0 in for y into, say, the first equation, I still get x=0. So is that the general solution?
     
  7. Jan 21, 2010 #6

    Mark44

    Staff: Mentor

    The solution is x=0, y=0. That is the only solution.
     
  8. Jan 21, 2010 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, that's the general solution. But just plugging it into ax+by=0 doesn't prove that. Suppose a=0??
     
  9. Jan 21, 2010 #8
    Is there a way, within the scope of this question, to determine whether a=0? Otherwise, I stick with x=0?
     
  10. Jan 21, 2010 #9

    Mark44

    Staff: Mentor

    Substitute y = 0 into both of your original equations. What do you get? If a = 0, as Dick mentioned, how does the condition that ad - bc != 0 affect things?
     
  11. Jan 21, 2010 #10
    If I substitute y=0 into both equations, I get ax=0 and cx=0. If a and c are both 0, then it won't satisfy the condition I stated in the problem. So either one of them is zero or neither. But I still can't see where you are going with this.
     
  12. Jan 21, 2010 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You just got it. If a and c can't both be 0 then one of those equations tells you x=0.
     
  13. Jan 21, 2010 #12
    Ok, that makes sense. Thanks for the help all of you.
     
  14. Jan 21, 2010 #13

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The significance of this problem is that it tells you you don't have to actually solve the system of equations to see that x=0, y=0 is the only solution. You can just calculate ad-bc and see that it's non-zero.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook