Results 1 to 3 of 3

Thread: [SOLVED] Need help on matricies

  1. #1
    SPOIDUR
    Guest

    [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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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 $\displaystyle ab \neq 2$.

    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:

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

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

    Equation (2): $\displaystyle ax + 2y = 5$.

    Now compare equations (1) and (2). You can see that when $\displaystyle -a = 5 \Rightarrow a = -5$ 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 $\displaystyle a \neq -5$.

    When a = -5 and ab = 2, $\displaystyle b = -\frac{2}{5}$. 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:

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

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

    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; Jan 31st 2008 at 04:58 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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:

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

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

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

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

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

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

    No unique solution: $\displaystyle ab = 2$.


    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:

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

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

    Equation (2): $\displaystyle ax + 2y = 5$.

    Now compare equations (1) and (2). You can see that when $\displaystyle -a = 5 \Rightarrow a = -5$ 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 $\displaystyle a \neq -5$.

    When a = -5 and ab = 2, $\displaystyle b = -\frac{2}{5}$. 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:

    $\displaystyle -5x + 2t = 5 \Rightarrow -5x = 5 - 2t \Rightarrow 5x = 2t - 5 \Rightarrow x = \frac{2}{5}t - 1$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Matricies, need help...
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Nov 26th 2010, 12:38 PM
  2. Matricies
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Aug 19th 2009, 04:25 AM
  3. [SOLVED] Solving systems of matricies and dets
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jul 13th 2009, 07:13 PM
  4. Matricies
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Dec 18th 2008, 08:46 AM
  5. Another set of matricies, but a different way of doing it.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jan 31st 2008, 07:25 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum