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Solving equation

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Express the determinant as a product of 4 linear factors

    [itex]
    \left(
    \begin{array}{ccc}
    a & \text{bc} & b+c \\
    b & \text{ac} & a+c \\
    c & \text{ab} & a+b
    \end{array}
    \right)[/itex]

    b hence or otherwise find the values of a for which the sumultaneous equations.

    [itex]
    ax+2t+3z=0
    [/itex]

    [itex]
    2x+ay+(1+a)z=0
    [/itex]

    [itex]
    x+2ay+(2+a)z=0
    [/itex]

    have a solution other than x=y=z=0

    ii) solve the equations when a=-3


    3. The attempt at a solution
    [itex]
    \text{Det}\left[\left(
    \begin{array}{ccc}
    a & \text{bc} & b+c \\
    b & \text{ac} & a+c \\
    c & \text{ab} & a+b
    \end{array}
    \right)\right] = (a-b)(c-b)(a-c)(a+b+c)
    [/itex]


    I get the right answers for part 2 of 1,2,-3. I don't know why that determinant of 0 implies that solution.

    for the last part i get

    [itex] x=\frac{5\lambda}{3}[/itex]
    [itex] y=\lambda[/itex]
    [itex] z=\lambda[/itex]

    [itex] r=\lambda\left(
    \begin{array}{c}
    \frac{5}{3} \\
    1 \\
    1
    \end{array}
    \right)[/itex]

    The answer is

    [itex] r = \lambda\left(
    \begin{array}{c}
    1 \\
    0 \\
    1
    \end{array}
    \right)[/itex] ???
     
  2. jcsd
  3. Jun 16, 2009 #2
    [itex]3x-18y-3z=0[/itex]

    [itex]-3x+2y+3z=0[/itex]

    [itex]y=0[/itex]

    [itex]x=z[/itex]

    [itex]z=\lambda[/itex]

    [itex]r=\lambda\left(
    \begin{array}{c}
    1 \\
    0 \\
    1
    \end{array}
    \right)[/itex]

    I wonder why it messes up when I use the other equations
     
  4. Jun 16, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I presume you mean "y" rather than "t" here.

    Determinant 0 does not imply any particular solution. It implies that there is NOT a unique solution. Since the given matrix equation has (0 0 0) as a solution, that means that there must be other solutions- in fact an infinite number of solutions, forming a subspace of R3. Since you have shown that the determinant is (a- b)(c- b)(a- c)(a+ b+ c), it will be 0 when any one of those factors is 0. Further, the determinant in (b) is the same as in (a) with b= 2 and c= 1. So its determinant is (a-2)(a-1)(a+ 3)= 0. The determinant will be 0 when a= 2, a= 1 or a= -3, making those factors 0.

    With a= 3, the equations become 3x+ 2y+ 3z= 0, 2x+ 3y+ 4z= 0, and x+ 6y+ 5z= 0. [itex]x= \lambda[/itex], y= 0, [itex]z= \lambda[/itex] makes the first equation [itex]3\lambda+ 3\lambda= 6\lambda= 0[/itex] which is true only for [itex]\lambda[/itex] equals 0 and that is the "trivial" solution.
    [itex] r = \lambda\left(
    \begin{array}{c}
    1 \\
    0 \\
    1
    \end{array}
    \right)[/itex]
    is definitely NOT a solution.
     
  5. Jun 16, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Where did you get these three equations? They are NOT the equation from yyour first post.
     
  6. Jun 16, 2009 #5
    (3) x-6y-z=0
    3x-18y-3z=0
    from (1) -3x+2y+3z=0

    (1)+(3) = -16y=0
    y=0

    x=z
    z=x
    etc?
     
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