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Thread: When will this system have: one solution, infinity solutions. zero solutions?

  1. #1
    Super Member sakonpure6's Avatar
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    When will this system have: one solution, infinity solutions. zero solutions?

    Hello,

    I have the linear system of presumably 3 variables ( the book does not say):

    $\displaystyle x+ay=0$
    $\displaystyle y+bz=0$
    $\displaystyle z+cx=0$

    We are asked:
    Find (if possible) conditions a,b,c such that the system has infinitely many solutions, no solutions or one solution.
    In the answer, they claim that for $\displaystyle abc\neq-1$ there is a unique solution and so for $\displaystyle abc=-1$ there is are infinitely many solutions.

    My question is, how did they find the base case abc=-1?

    Thank you,
    -Sakon
    Last edited by sakonpure6; Mar 3rd 2015 at 04:58 PM.
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  2. #2
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    Re: When will this system have: one solution, infinity solutions. zero solutions?

    Hmm, try and solve z=-cx input Z in R2 => y-bcx=0 => y=bcx input Y in R1 => x+abcx= 0 => -1=abc I think.
    Last edited by Sodapopinski; Mar 3rd 2015 at 05:26 PM.
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  3. #3
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    Re: When will this system have: one solution, infinity solutions. zero solutions?

    Quote Originally Posted by sakonpure6 View Post
    Hello,

    I have the linear system of presumably 3 variables ( the book does not say):

    $\displaystyle x+ay=0$
    $\displaystyle y+bz=0$
    $\displaystyle z+cx=0$

    We are asked:


    In the answer, they claim that for $\displaystyle abc\neq-1$ there is a unique solution and so for $\displaystyle abc=-1$ there is are infinitely many solutions.

    My question is, how did they find the base case abc=-1?

    Thank you,
    -Sakon
    You have this system of equations:

    $\displaystyle \begin{align*} x + a\,y &= 0 \\ y + b\,z &= 0 \\ z + c\,x &= 0 \end{align*}$

    which can be written in matrix form as

    $\displaystyle \begin{align*} \left[ \begin{matrix} 1 & a & 0 \\ 0 & 1 & b \\ c & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \end{align*}$

    This will have a nontrivial solution for $\displaystyle \begin{align*} \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \end{align*}$ as long as $\displaystyle \begin{align*} \left| \begin{matrix} 1 & a & 0 \\ 0 & 1 & b \\ c & 0 & 1 \end{matrix} \right| \neq 0 \end{align*}$, so evaluate this determinant and see if you can find when it is not going to be 0...
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    Super Member sakonpure6's Avatar
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    Re: When will this system have: one solution, infinity solutions. zero solutions?

    Yes indeed, then if x(1+abc)=0 , one unique solution when abc is not -1 and infinitely many solutions when abc=-1 as x can be replaced by a parameter of any value?
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  5. #5
    Super Member sakonpure6's Avatar
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    Re: When will this system have: one solution, infinity solutions. zero solutions?

    Prove it, I am kind of lost. How would you go about solving that matrix in your last sentence?
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