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Math Help - Linear equations with a parameter

  1. #1
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    Linear equations with a parameter



    How do I find for which K parameter we have zero solutions, single solution and infinite number of solutions? I can't seem to do it with the Gaussian Elimination.
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  2. #2
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    Hello, GIPC!

    Here's a start . . .



    \begin{array}{ccccccc}<br />
x & + & (k^2-6)y &+& (4k+4)z &=& 5k+3 \\<br />
\text{-}x &+& (2k^2-6)y &+& (4k+4)z &=& 6k+3 \\<br />
& +&(3k^2-12)y &+& (6k+6)z &=& 9k+6 \end{array}

    Find for which parameter k we have zero solutions, single solution and infinite number of solutions.
    Consider the determinant of the system: . D \;=\;\begin{vmatrix}1 & k^2-6 & 4(k+1) \\ \text{-}1 & 2(k^2-3) & 4(k+1) \\ 0 & 3(k^2-4) & 6(k+1) \end{vmatrix}

    Factor 2(k+1) from column 3: . D \;=\;2(k+1)\begin{vmatrix}1 & k^2-6 & 2 \\ \text{-}1 & 2(k^2-3) & 2 \\ 0 & 3(k^2-4) & 3 \end{vmatrix}

    Evaluate: . D \;=\;6(k+1)\bigg[\left[2(k^2-3) - 2(k^2-4)\right] + \left[(k^2-6) - 2(k^2-4)\right] + 0\,\bigg]

    And we have: . D \;=\;6(k+1)(-2k^2+8) \;=\;-12(k+1)(k^2-4) \;=\;-12(k+1)(k-2)(k+2)


    . . Hence: . D \:=\: 0\,\text{ for }\,k \:=\:\text{-}1,\: 2,\:\text{-}2


    Can you take it from here?

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  3. #3
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    I have another question, if the system changes to:

    x + (k^2 -6)y + (4k+4)z = 5k+3
    (3k^2 -3)y +(6k+6)z = 9k+6

    for which k do i have a single solution, no solution and infinite solotions? How do I solve it?
    Last edited by GIPC; March 27th 2010 at 04:18 PM.
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  4. #4
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    bump
    anyone?

    this problem is giving me fits.
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  5. #5
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    Are you not trying anything yourself? For example, it shouldn't take long to see that you can reduce the second equation to (k^2- 1)y+ 2(k+1)z= 3k+ 2. And, as long as k\ne \pm 1, we must have y= \frac{3k+ 2- 2(k+1)z}{k^2- 1} for any value of z. And then x= 5k+3- (4k+4)z- (k^2- 6)y.
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