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    I just need help with this.

    Consider the following equations
    x + 2y + 2z = 1
    2x + y + 3z = 3
    4x + 5y + 7z = \lambda
    (i) Find the value of \lambda for which these equations are consistent.
    (ii) Find the general solution corresponding to this value of \lambda.
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    (Original post by StephenP91)
    I just need help with this.

    Consider the following equations
    x + 2y + 2z = 1
    2x + y + 3z = 3
    4x + 5y + 7z = \lambda
    (i) Find the value of \lambda for which these equations are consistent.
    (ii) Find the general solution corresponding to this value of \lambda.
    2*equation 1 + equation 2 =
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    (Original post by Get me off the £\?%!^@ computer)
    2*equation 1 + equation 2 =
    Equation 3, so \lambda would = 5 if they are consistent, but why? Ignore this. I understand. Simple enough.

    I am more concerned about the general solution question anyway.
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    (Original post by StephenP91)
    Equation 3, so \lambda would = 5 if they are consistent, but why? Ignore this. I understand. Simple enough.

    I am more concerned about the general solution question anyway.
    Well what have you done?
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    (Original post by Get me off the £\?%!^@ computer)
    Well what have you done?
    It's not the question that I find hard, it's simply that I don't know the method. Though I think I've guessed at it now.

    You row reduce to echelon. You get Z to be some value and then you substitute from then, unless you get all 0s on the bottom, then you simply say z=z and go from there.

    Though I am not entirely sure, I think it could be something like that.
 
 
 
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