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Thread: simultaneous equations/matrices

  1. #1
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    simultaneous equations/matrices

    Hello Everyone,

    a) Find the values of a for which the matrix ( a 3 ) is singular.
    2 a-1

    b) for these values of a solve the equation ax+3y=2a
    2x+(a-1)y= a square

    Thanks for your time!
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Oasis1993 View Post
    Hello Everyone,

    a) Find the values of a for which the matrix ( a 3 ) is singular.
    2 a-1

    b) for these values of a solve the equation ax+3y=2a
    2x+(a-1)y= a square

    Thanks for your time!
    $\displaystyle \displaystyle \begin{bmatrix}
    a& 3\\
    2 & a-1
    \end{bmatrix} $

    so to that matrix to be singular it's determinant must be equal to zero .....

    $\displaystyle D = \begin{vmatrix}
    a&3 \\
    2 & a-1
    \end{vmatrix} = 0$

    so you have

    $\displaystyle D = \begin{vmatrix}
    a&3 \\
    2 & a-1
    \end{vmatrix} = a\cdot (a-1) - 3\cdot 2 = a^2-a-6 =(a-3)(a+2) $




    now just solve system of equations

    $\displaystyle ax+3y=2a $
    $\displaystyle 2x +(a-1)y = a^2 $

    in any way that you know to do .... (do you know how to do that ?)
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  3. #3
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    thanks!
    I do know how to solve the equation however i can't get the answer correct..

    My book says that if we use 3 there's no solution but i can't seem to continue with -2 either...

    thanks
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Oasis1993 View Post
    thanks!
    I do know how to solve the equation however i can't get the answer correct..

    My book says that if we use 3 there's no solution but i can't seem to continue with -2 either...

    thanks
    $\displaystyle ax+3y=2a $
    $\displaystyle 2x +(a-1)y = a^2 $

    for a= -2
    $\displaystyle -2x+3y=-4 $
    $\displaystyle 2x -3y = 4 $

    Edit : here you have :

    two same equations :

    $\displaystyle 2x-3y= 4 $
    $\displaystyle 2x -3y = 4 $

    so you have

    $\displaystyle 4x-6y = 8 $

    so you have one equation with 2 unknowns....
    you have infinity of the solutions and equations are dependent (meaning for each new x or y you have another solution for that second unknown)

    for a = 3
    $\displaystyle 3x+3y=6 $
    $\displaystyle 2x +2y = 9 $

    Edit : here you have :

    $\displaystyle x+y = 2 $
    $\displaystyle x+y = 4.5 $

    now here it's easy to see there are no solutions because you have ones that x+y = 2 and another saying that x+y = 4.5 which cant be true for the both of the equations



    Edit: hm... this is moved to the university sub forum ( I think it was in pre-algebra at the start .... perhaps i'm wrong but ... ) , well if here you must show that based on determinants of that system you don't have / have solutions and if you have why and which , and if don't why ... and so on ... there are few cases by which you conclude this
    Last edited by yeKciM; Sep 18th 2010 at 01:52 PM.
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