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March 25, 2017

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For which values of 'a' will the following system have no solutions? Exactly one solution? Infinitely many solutions?

x + 2y - 3z = 4
3x - y +5z = 2
4x + y + (a^2-14)z = a+2

  • Linear Algebra - ,

    d =
    | 1 2 -3 |
    |3 -1 5 |
    | 4 1 (a^2-14)|

    d = -(a^2-14)-6(a^2-14)+40 - 9-12-5
    d = -7(a^2-14)+14
    d = -7a^2 + 8*14 = (-a^2 +16)7

    if d is zero, all solutions have 0 in the denominator
    so if a - 4 or -4, solution is undefined
    Now if a is such that equation 3 is a multiple of one of the first two, there is no unique solution (lines are parallel)
    now solutions are of form:
    P/d
    where P is the determinant of the matrix resulting from replacing row one with the right side for x, row 2 for the right for y, row 3 for the right for z
    so for z for example
    | +1 +2 +4 |
    | +3 -1 +2 |
    | +4 +1 (a+2)|
    --------------------
    (-a^2 +16)7

    [-(a+2)+16+12+16-2 -6(a+2) ] /(-a^2 +16)7

    [-7(a+2)+42]/(-a^2 +16)7

    [-7a + 28 ] /(-a^2 +16)7

    -7 [ a-4 ] / -7[a^2-16]
    1/(a+4)

    hmmm, z is undefined if a = -4
    etc

  • Linear Algebra - ,

    I do not see how the lines can be parallel and the only real problem I see so far is singularity when a = +/-4

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