I previously posted this question under the wrong topic.

Question

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

Answer 1

(-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.

Answer 2

Thanks a lot!

Answer 3

You're welcome!