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Math Help - [SOLVED] Need help on matricies

  1. #1
    SPOIDUR
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    [SOLVED] Need help on matricies

    I'm doing homework and there is one question that I am suck on, the example in the text does not explain it enough to really get, for me so I figured if someone could go through it and show / explain me the steps, it would be much appreciated.

    Anyways the question itself is

    In each of the following, find conditions on a and b such that the system has no solution, one solution, and infinitely many solutions

    x + by = -1
    ax + 2y = 5








    The Solution in the book says
    If ab =/= 2, unique solution x = (-2-5b)/(2-ab) y= (a+5)/(2-ab)
    If ab=2: no solution if a =/= -5 ; if a = -5, the solutions are x=-1+2/5t, y=t.


    Could someone please go through this problem explaining how I find this answer?
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  2. #2
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    Quote Originally Posted by SPOIDUR View Post
    I'm doing homework and there is one question that I am suck on, the example in the text does not explain it enough to really get, for me so I figured if someone could go through it and show / explain me the steps, it would be much appreciated.

    Anyways the question itself is

    In each of the following, find conditions on a and b such that the system has no solution, one solution, and infinitely many solutions

    x + by = -1
    ax + 2y = 5


    The Solution in the book says
    If ab =/= 2, unique solution x = (-2-5b)/(2-ab) y= (a+5)/(2-ab)
    If ab=2: no solution if a =/= -5 ; if a = -5, the solutions are x=-1+2/5t, y=t.


    Could someone please go through this problem explaining how I find this answer?
    I'll assume you know:
    1. How to express your linear equations in the matrix form AX = B.
    2. You know how to find the determinant of A when A is a 2x2 matrix.
    3. You understand that the equations have no unique solution when the determinant of A, det(A), is equal to zero.

    In your problem, det(A) = 2 - ab.

    det(A) = 0 => 2 - ab = 0 => ab = 2.

    So there will be NO unique solution when ab = 2. Therefore there WILL be a unique solution when ab \neq 2.

    When there's no unique solution, that means that either there is an infinite number of solutions or no solution at all. To test which of these happens when ab = 2, you might do something like the following:

    ab = 2 \Rightarrow b = \frac{2}{a}. Substitute this into you equations and do some re-arranging:

    Equation (1): x +  by  = -1 \Rightarrow x + \frac{2}{a} y = -1 \Rightarrow ax + 2y = -a.

    Equation (2): ax + 2y  = 5.

    Now compare equations (1) and (2). You can see that when -a = 5 \Rightarrow a = -5 the equations are the same. One equation and two unknowns => infinite number of solutions.

    So when ab = 2 AND a = -5, there an infinite number of solutions.

    It follows that there will be no solution at all when ab = 2 AND a \neq -5.

    When a = -5 and ab = 2, b = -\frac{2}{5}. So the equations boil down to -5x + 2y = 5. The form of the infinite number of solutions can be expressed in several ways. To see where the books answer came from, note that if you let y = t, where t can be any real number:

    -5x + 2t = 5 \Rightarrow -5x = 5 - 2t \Rightarrow 5x = 2t - 5 \Rightarrow x = \frac{2}{5}t - 1.

    As to where the unique solution comes from when ab \neq 2:

    x + by = -1 .... (1)

    ax + 2y = 5 .... (2)

    Solve using the elimination method:

    Multiply equation (1) by a, subtract it from equation (2) and then solve for y:

    (2 - ab)y = 5 + a => ......

    Then sub y into either equation and solve for x.
    Last edited by mr fantastic; January 31st 2008 at 04:58 PM.
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  3. #3
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    Alternatively, if you're NOT familiar with matrices, here's an edited version of my earlier reply that doesn't use matrices:

    Quote Originally Posted by mr fantastic View Post

    Unique solution:

    x + by = -1 .... (1)

    ax + 2y = 5 .... (2)

    Solve using the elimination method:

    Multiply equation (1) by a, subtract it from equation (2) and then solve for y:

    (2 - ab)y = 5 + a \Rightarrow y = \frac{5 + a}{2 - ab}.

    Note that this solution is OK provided the denominator is NOT equal to zero, that is, 2 - ab \neq 0 \Rightarrow ab \ne 2.

    Now sub y = \frac{5 + a}{2 - ab} into either equation and solve for x. Using equation (1):

    x +  by  = -1 \Rightarrow x = -1 - by = -1 - b \left( \frac{5 + a}{2 - ab} \right) = \frac{-(2 - ab)}{2 - ab} - \frac{b(5 + a)}{2 - ab} = \frac{-2 + ab - 5b - ab}{2 - ab} = \frac{-2 - 5b}{2 - ab}.

    So there's a unique solution, given by the above, when ab \neq 2.

    --------------------------------------------------------------------------

    No unique solution: ab = 2.


    When there's no unique solution, that means that either there is an infinite number of solutions or no solution at all. To test which of these happens when ab = 2, you might do something like the following:

    ab = 2 \Rightarrow b = \frac{2}{a}. Substitute this into you equations and do some re-arranging:

    Equation (1): x +  by  = -1 \Rightarrow x + \frac{2}{a} y = -1 \Rightarrow ax + 2y = -a.

    Equation (2): ax + 2y  = 5.

    Now compare equations (1) and (2). You can see that when -a = 5 \Rightarrow a = -5 the equations are the same. One equation and two unknowns => infinite number of solutions.

    So when ab = 2 AND a = -5, there an infinite number of solutions.

    It follows that there will be no solution at all when ab = 2 AND a \neq -5.

    When a = -5 and ab = 2, b = -\frac{2}{5}. So the equations boil down to -5x + 2y = 5. The form of the infinite number of solutions can be expressed in several ways. To see where the books answer came from, note that if you let y = t, where t can be any real number:

    -5x + 2t = 5 \Rightarrow -5x = 5 - 2t \Rightarrow 5x = 2t - 5 \Rightarrow x = \frac{2}{5}t - 1.
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