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Please help me with this math problem?

Express the general solution of x'=(2, 9/5, -5/2, -1)x in terms of real-valued functions. (this is 2x2 matrix, 2 and 9/5 on the left, -5/2 and -1 on the right. I know that the roots are 1/2+3/2i and 1/2-3/2i, complex numbers. But how do I find a and b for the first root?)

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1 Answer

The first eigenvalue is λ=1/2+3/2i. To find the corresponding eigenvector u=[a,b], substitute this into the eigenvalue equation, (A-λI)u=0. You get the following two equations:
(3/2-3/2i)a-5/2b=0
9/5a-(3/2+3/2i)b=0
 
You can now choose either a or b to be 1, and see what the other component has to be. If a=1, the first equation tells us
(3/2-3/2i)-5/2b=0, or b=3/5-3/5i.
We don't get any new information from the second equation. So the eigenvector is
u1=[1,3/5-3/5i]
 
Similarly, the second eigenvalue λ=1/2-3/2i gives the eigenvector
u2=[1,3/5+3/5i].
 
Let me know if you know how to proceed from here.

Comments

For the a and b, I got a=1, b=(3/5)+(3/5)i for the first root. And for the second root, I got a=1, b=(3/5)-(3/5)i.

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