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Linear Algebra (Conditions for solutions).

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the following system:
    x + by = -1
    ax + 2y = 5
    Find the conditions on a and b such that the system has no solution, one solution or infinitely many solutions.


    2. Relevant equations
    General Algebra really.


    3. The attempt at a solution
    Previously we had been using augmented matrices to solve these sort of problems but using one here doesn't really help.
    Rearranging the equations you get x = (5b + 2)/(ab - 2) and y =(5 + a) / (2 - ab)

    The answer is:
    If ab≠2 then the unique solution is x = (5b + 2)/(ab - 2) and y =(5 + a) / (2 - ab).
    If ab = 2 and ab ≠ -5 then there are no solutions.
    If ab =2 and a = -5 then x = 2t/5 -1 and y = t.

    I understand why ab cannot be 2 for the first part of the answer but why is the fact that a can or cannot equal 5 when ab = 2 relevant for the second and third parts of the answer? Thanks.
     
  2. jcsd
  3. Apr 17, 2013 #2

    Mark44

    Staff: Mentor

    I used an augmented matrix.
    The above should be - If ab = 2 and a ≠ -5, then there are no solutions.

    The reason for this can be seen in a reduced augmented matrix.
    $$ \begin{bmatrix} 1 & b & | & -1 \\ 0 & 2 - ab & | & 5 + a\end{bmatrix}$$

    If ab = 2, the bottom row is 0 0 | 5 + a.

    If a = -5, the bottom row is 0 0 | 0, which implies that there are an infinite number of solutions. Geometrically, the system represents two lines that coincide.

    If a ≠ -5, the bottom row is 0 0 <nonzero>, which makes the system inconsistent. Geometrically, the system represents two parallel lines that do not coincide, hence don't intersect.
    If a
     
  4. Apr 17, 2013 #3
    Brilliant. Thanks a million for that!
     
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