Question

If one eigenvalue of a 2 by 2 matrix E2x2 is 1 = 7 and a corresponding eigenvector of is ( 2) , then the general solution of the system ODES: = (t) = E2x2 f(t) cost sint is ä(t) = C: + C2 sin cost can't be determined. Solving the system requires additional information sint COS is 8(t) + C2 pot cost sint cost sint is i(t) = C: + C2 sint COS

Answer 1

Now, a 2x2 matrix can have at most two distinct eigenvalues, and if one of the eigenvalues is a complex number, i, then the other eigenvalue must be it's conjugate -i.

Further, the eigenvectors corresponding to these eigenvalues are conjugate as well, that is, if we can write the eigenvector corresponding to the eigenvalue i as u+iv, then the eigenvector corresponding to the eigenvalue -i is u-iv, so the eigenvector corresponding to the eigenvalue -i is

$\small \begin{pmatrix} 1\\ -i \end{pmatrix}$

Now, we have

For two distinct complex eigenvalues 11 + 12, where 1 = a +iB, 12 = a - 13

and corresponding eigenvectors 11 + 12, where n1 = u + iv, 12 = u – iu

the general solution takes the form

T = Cieat (cos(8t) u – sin (8t) v) + c2eat (cos(8t) v + sin (8t) u)

Now, in our case

A=0+1=0 = 0;3=

and

$\small \eta=\begin{pmatrix} 1\\ \pm i \end{pmatrix}=\begin{pmatrix} 1\\ 0 \end{pmatrix}\pm i\begin{pmatrix} 0\\ 1 \end{pmatrix} \Rightarrow u=\begin{pmatrix} 1\\ 0 \end{pmatrix};\;v=\begin{pmatrix} 0\\ 1 \end{pmatrix}$

So, the general solution to the system is

$\small \vec x(t)=c_1e^{0t}\left (\cos(t)\begin{pmatrix} 1\\ 0 \end{pmatrix}-\sin(t)\begin{pmatrix} 0\\ 1 \end{pmatrix} \right )+c_2e^{0t}\left (\cos(t)\begin{pmatrix} 0\\ 1 \end{pmatrix}+\sin(t)\begin{pmatrix} 1\\ 0 \end{pmatrix} \right )$

Simplifying we get

$\small \vec x(t)=c_1\left (\cos(t)\begin{pmatrix} 1\\ 0 \end{pmatrix}-\sin(t)\begin{pmatrix} 0\\ 1 \end{pmatrix} \right )+c_2\left (\cos(t)\begin{pmatrix} 0\\ 1 \end{pmatrix}+\sin(t)\begin{pmatrix} 1\\ 0 \end{pmatrix} \right )$

Multiplying the trigonometric functions and adding the components we get

$\small \vec x(t)=c_1\cdot \begin{bmatrix} \cos t\\ -\sin t \end{bmatrix}+c_2\begin{bmatrix} \sin t\\ \cos t \end{bmatrix}$

which is our answer, that is, option 4.