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Math Help - Linear solutions

  1. #1
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    Linear solutions

    Not sure how to go about this?

    For which values of a will the following system have no solutions? one solution? or infinite solutions?

    System
    x + 2y - 3z = 4
    3x - y + 5z = 2
    4x + y + (a^2 - 14)z = a + 2
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  2. #2
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    Quote Originally Posted by chadlyter View Post
    Not sure how to go about this?

    For which values of a will the following system have no solutions? one solution? or infinite solutions?

    System
    x + 2y - 3z = 4
    3x - y + 5z = 2
    4x + y + (a^2 - 14)z = a + 2
    Multiply equation 1 by -3 and add it to equation 2 to get

    -7y+14z=-10

    Multiply eqaution 1 by -4 and add it to equation 3 to get

    -7y+(a^2-2)z=a-14

    Now multiply the first of these equations by -1 and add it to the other to get

    (a^2-16)z=a-4

    Now we factor the left hand side to get

    (a-4)(a+4)z=a-4

    Now lets check some values of a

    Case I: a=4

    we get 0=0 so the system is consistant dependant (has an infinite number of solutions.)

    Case II: a=-4

    We get 0=-8 So the system in inconsistant(has no solutions)

    Case III: a \ne 4 \mbox{ or } a \ne -4

    we get

    (a-4)(a+4)z=a-4 \iff z =\frac{1}{a+4}

    The system has exactly one solution and it is given above.
    So the system is consistant independant.

    I hope this clear it up.

    Note this could have been done with a matrix and putting the matrix into eschelon form.
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