anonymous
  • anonymous
Find the general solution of x'=(2, -1, 2+i, -1-i)x. (this is a matrix, 2 and -1 on the left, 2+i, -1-i on the right, and I know that the 2 roots are -i, 1. but I don't know how to find a, b for the first root.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Sorry about the wait. \[x'=\begin{pmatrix}2&2+i\\-1&-1-i\end{pmatrix}x\] This is your equation, right?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
\[\begin{vmatrix}2-\lambda&2+i\\-1&-1-i-\lambda\end{vmatrix}=\lambda^2-(1-i)\lambda-i=0~~\Rightarrow~~\lambda_1=1,\lambda_2=-i\] And you're having trouble with the eigenvectors, right?

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anonymous
  • anonymous
Yes, continue.
anonymous
  • anonymous
For \(\lambda_1=1\), you have the following equation: \[\begin{pmatrix}1&2+i\\-1&-2-i\end{pmatrix}\begin{pmatrix}a_1\\a_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\] where \(\vec{a}=\begin{pmatrix}a_1\\a_2\end{pmatrix}\) is the eigenvectors for \(\lambda_1\). The second row is (-1) times the first, so you're left with the following equation with two unknowns: \[a_1+(2+i)a_2=0\] Letting \(a_2=1\), you get \(a_1=-2-i\), so \[\vec{a}=\begin{pmatrix}-2-i\\1\end{pmatrix}\]
anonymous
  • anonymous
For \(\lambda_2=-i\), you have the following equation: \[\begin{pmatrix}2+i&2+i\\-1&-1\end{pmatrix}\begin{pmatrix}b_1\\b_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\] where \(\vec{b}=\begin{pmatrix}b_1\\b_2\end{pmatrix}\) is the eigenvectors for \(\lambda_2\). The two rows are the same, so you have \[-b_1-b_2=0\] Letting \(b_2=1\), you get \(b_1=-1\), so \[\vec{b}=\begin{pmatrix}-1\\1\end{pmatrix}\]
anonymous
  • anonymous
But the first vector is 2+i, -1. How do I get that? How do I find a, b for the first one?
anonymous
  • anonymous
Are you saying that's what your answer key says? This usually comes up as an issue, but it's really not. If you let \(a_2=-1\), you'll get \(a_1=2+i\). It all depends on the values you choose. They're really the same eigenvector.
anonymous
  • anonymous
But how do you get that? a+b(2+i)=0, and -a-b(2-i)=0, how do you solve for a and b?
anonymous
  • anonymous
You fix one of the values. Pick an easy one to work with.
anonymous
  • anonymous
I get it, thanks.
anonymous
  • anonymous
Your welcome. Check out Paul's online notes for this topic. It helped me out when I first learned this stuff.

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