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Math Help - Solutions to systems

  1. #1
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    Solutions to systems

    For which values of "a" will the following system have no solutions?
    exactly one solution?
    infinitely many solutions?
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  2. #2
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    Quote Originally Posted by Kbotz View Post
    For which values of "a" will the following system have no solutions?
    exactly one solution?
    infinitely many solutions?
    Set up the system in matrix form: AX = B where A is the coefficient matrix.

    There's a unique solution if det(A) \neq 0.

    If det(A) = 0 then there is either no solution or an infinite number of solutions. You need to examine this situation closely to decide.

    Alternatively, you can set up the augmented matrix and do row operations. Then note that a row looking like 0 0 0 | 0 means infinite solutions and a row looking like 0 0 0 | * where * is not zero means no solutions. Otherwise there is a unique solution.

    Details of the theory briefly outlined above are certain to be in your classnotes and textbook. If you need more help please show all that you have tried and say where you get stuck.
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    Quote Originally Posted by Kbotz View Post
    For which values of "a" will the following system have no solutions?
    exactly one solution?
    infinitely many solutions?
    \begin{bmatrix}<br />
1 & 2 & -3 & :4\\ <br />
3 & -1 & 5 & :2\\ <br />
4 & 1 & a^2-14 &  :a+2<br />
\end{bmatrix}

    Solve this augment matrix until you arrive at the given forms that Mr Fantastic mentioned.
    Last edited by dwsmith; May 21st 2010 at 09:23 PM.
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  4. #4
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    sorry it was ment to be positive 5 in the 2nd row.

    this is what i got from the elimination.
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  5. #5
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    Quote Originally Posted by Kbotz View Post
    sorry it was ment to be positive 5 in the 2nd row.

    this is what i got from the elimination.
    a_{24}=-10\neq10

    Multiply row 2 by \frac{-1}{7}

    a^2-16=(a-4)(a+4) so divide row 3 by (a-4)
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  6. #6
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    sorry my bad it is -10.

    so the exact solutions to this sytem are -4 and 3?
    i substituted this back into the original equation and formed a matrix, when a=-4, reduced form is not possible and when a=3, reduced form is possible
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  7. #7
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    Quote Originally Posted by Kbotz View Post
    sorry my bad it is -10.

    so the exact solutions to this sytem are -4 and 3?
    i substituted this back into the original equation and formed a matrix, when a=-4, reduced form is not possible and when a=3, reduced form is possible
    If a=-4, then the system is inconsistent. Therefore, when a\neq -4, the system is consistent.
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  8. #8
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    cool,
    conclusion the system has no solutions when a =-4 and exactly one solution when a=3.

    and if i want to get infinite solutions, i will have to get a row full of zeros.
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  9. #9
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    Quote Originally Posted by Kbotz View Post
    cool,
    conclusion the system has no solutions when a =-4 and exactly one solution when a=3.

    and if i want to get infinite solutions, i will have to get a row full of zeros.
    It will have one solution solution for this set { \forall a\in\mathbb{R}| a\neq -4}
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