Ok... so I think I row reduced this matrix properly:
5x_1 + (3-i)x_2 = 0
(-3-i)x_1 - 2x_2 = 0
to
5x_1 + (3-i)x_2 = 0
0x_1 + 0x_2 = 0
I'm really having trouble finding an eigenvector for this matrix.... can anyone help me? The answer in the book says the eigenvector is -1-3i on first row and 5i on second row. I am in desperation for help please.... if anyone has any clue... please answer?
5x_1 + (3-i)x_2 = 0
(-3-i)x_1 - 2x_2 = 0
to
5x_1 + (3-i)x_2 = 0
0x_1 + 0x_2 = 0
I'm really having trouble finding an eigenvector for this matrix.... can anyone help me? The answer in the book says the eigenvector is -1-3i on first row and 5i on second row. I am in desperation for help please.... if anyone has any clue... please answer?
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The reduction to
5x₁ + (3-i) x₂ = 0
0x₁ + 0x₂ = 0 is correct.
So, x₁ = [(-3+i)/5] x₂, with x₂ as a free variable.
Letting x₂ = k, the general solution may be written as
[x₁]....[(-3+i)/5]
[x₂].=.[....1.....] * k for any scalar k.
In particular, if we rewrite k as k = 5i * c for some other scalar c, we have
[x₁]....[-1 - 3i]
[x₂].=.[..5i....] * c for any scalar c.
As long as your answer is a multiple of the book's answer, you will be fine.
I hope this helps!
5x₁ + (3-i) x₂ = 0
0x₁ + 0x₂ = 0 is correct.
So, x₁ = [(-3+i)/5] x₂, with x₂ as a free variable.
Letting x₂ = k, the general solution may be written as
[x₁]....[(-3+i)/5]
[x₂].=.[....1.....] * k for any scalar k.
In particular, if we rewrite k as k = 5i * c for some other scalar c, we have
[x₁]....[-1 - 3i]
[x₂].=.[..5i....] * c for any scalar c.
As long as your answer is a multiple of the book's answer, you will be fine.
I hope this helps!