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Posted by on Friday, December 4, 2009 at 9:03pm.

I previously posted this question under the wrong topic.

Which of the following choices is a solution for:

(-3-i)x1 + (1-2i)x2 = 0
(-7-9i)x1 + (7-4i)x2 = 0

for any value of the variable t

[x1] = ?
[x2]

A: [5]
[1+7i] * t

B: [5]
[1-7i] * t

C: [5]
[-1-7i] * t

D: [5]
[-1+7i] * t

  • Linear Algebra - , Friday, December 4, 2009 at 10:00pm

    (-3-i)x1 + (1-2i)x2 = 0.....(1)
    (-7-9i)x1 + (7-4i)x2 = 0 ....(2)
    or
    Ax=0
    The system is a homogeneous equation. A trivial solution is x1=0 and x2=0.

    For the system to have non-trivial solutions, the determinant of A must vanish, or |A|=0.
    Calculate
    |A|
    =(-3-i)(7-4i) - (-7-9i)(1-2i)
    = -21-4+5i - (-25+5i)
    =0

    So non-trivial solutions exist, because the two equations are linearly dependent.
    Prove that the two equations are linearly dependent by applying Gauss elimination which results in two identical equations. Eliminate the second equation.

    Take equation (1),
    (-3-i)x1 + (1-2i)x2 = 0
    We multiply by the conjugate of (1-2i) to get
    (1+2i)(-3-i)x1 + (1+2i)(1-2i)x2 = 0
    -(1+7i)x1 + 5x2 = 0
    Let x1=5t, where t is a variable parameter, then
    -(1+7i)5t + 5x2 = 0
    Solving, x2=1+7i
    Therefore
    x1=5t, x2=1+7i

    Make your pick for the answer.

  • Linear Algebra - , Friday, December 4, 2009 at 10:39pm

    Thanks a lot!

  • Linear Algebra - , Friday, December 4, 2009 at 11:42pm

    You're welcome!

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