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Prove that there exists exactly one solution

  1. Nov 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that the linear system of the form:

    [tex]
    a_{1,1}x + a_{1,2}y = b_{1}\\
    a_{2,1}x + a_{2,2}y = b_{2}
    [/tex]

    has exactly one solution in x and y if [itex]a_{1,1}a_{2,2} \neq a_{1,2}a_{2,1}[/itex]



    3. The attempt at a solution

    I proved the contrapositive:

    Assume that the given system does not have exactly one solution in x and y, then it either has no solution or it has infinitely many solutions.

    Case 1: Assume that there is no solution to this system. Since it is a system of two equations, I can refer to a geometric explanation. Two equations do not intersect if they are parallel, i.e. their slopes are equal and b1 != b2. For their slopes to be equal it has to be the case that [itex]a_{1,1}a_{2,2} = a_{1,2}a_{2,1}[/itex].

    Case 2: Assume there are infinitely many solutions, in that case the equations are the same and their slopes are equal, therefore again fulfilling [itex]a_{1,1}a_{2,2} = a_{1,2}a_{2,1}[/itex].

    Now I think the proof is correct, but I don't like to rely on geometric intuition, if it would have been a system of 3 equations I wouldn't be able to prove this type of problem. I tried applying the Gauss algorithm to this system, but it did not yield anything useful.
    Could anybody give me a hint for another approach please?
     
  2. jcsd
  3. Nov 2, 2013 #2

    UltrafastPED

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    From matrix theory: your original condition is equivalent to "the determinant is non-zero"; this means that the coefficient matrix can be inverted. So AX=B can be rewritten as X=A^-1*B ... which is the only solution.

    If det(A)=0 then A^-1 does not exist ... and then there is not a unique solution.
     
  4. Nov 2, 2013 #3
    Thanks for your quick reply. We haven't covered matrices and determinants yet. Is there another way, without using determinants to prove this?
     
  5. Nov 2, 2013 #4

    UltrafastPED

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    In that case just draw the possible lines, using the given condition to control the slopes. Note that if you put both equations into y=mx+b form you get the slopes ... and from the condition given the slopes are not equal.

    Hence the lines are not parallel. Now you have to remember a famous theorem from Euclid's geometry ...
     
  6. Nov 2, 2013 #5

    Ray Vickson

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    Just start solving the equations by hand. Since ##a_{11}a_{22} \neq a_{12} a_{21}## we cannot have both ##a_{11}## and ##a_{12}## equal to zero, so at least one of them is nonzero. Say, for example, that ##a_{11} \neq 0##. That means that you can solve for ##x## in terms of ##y## from the first equation (since dividing by ##a_{11}## is OK). If you plug that expression for ##x## into the second equation, you will finally get an equation involving ##y## alone. This will have the form ##A y = B##. When does that equation have a unique solution? When does it have no solutions? When does it have infinitely many solutions?
     
  7. Nov 3, 2013 #6
    Ok thanks a lot. I did as you said and when solving for y I got a term with a11a22 - a12a21 in the denominator. So if a11a22 - a12a21 != 0, then the term is defined and there are unique solutions for y and x. If a11a22 - a12a21 = 0, the system would have no solutions, and I'm not sure about the case of infinitely many solutions. For that to happen I would need A=0 and B=0 in the equation Ay=B, so when solving for y both the term in the numerator and denominator would have to be 0, I guess.
     
  8. Nov 3, 2013 #7

    UltrafastPED

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    If your expression is zero the lines are parallel ... how many lines satisfy this condition?
     
  9. Nov 3, 2013 #8

    Ray Vickson

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    You are confusing the issue by premature "division". You obtain and equation of the form
    [tex] A y = B, \: A = a_{11} a_{22} - a_{21} a_{12}, \: B = a_{11} b_2 - a_{21}b_1.[/tex]
    If ##A \neq 0## you get a unique solution because at that point you know you are allowed to divide by ##A##. Next: if ##A = 0## you get infinitely many solutions if ##B = 0## (that is, any y value will satisfy 0*y=0) and there are no solutions at all if ##B \neq 0## (that is, '0*y = nonzero' is impossible).
     
    Last edited: Nov 3, 2013
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