0

I have the following problem:

Find a general solution of the system $x'(t)=Ax(t)$ for the given matrix $A$.

$A=\begin{bmatrix}9 & -6\\15 & -9\end{bmatrix}$

When I calculate the eigenvectors by hand I get $\pm\sqrt{171}$, but by Matlab I get these complex values of ridiculously small, complex fractions.

I can't figure out the disconnect.

My calculations are: $$(9-r)(-9-r)-6(15)=0$$ $$-81-9r+9r+r^2=0$$ $$r^2-171=0$$ $$r=\sqrt{171}$$

I'm not sure where I'm going wrong, since this section is supposed to be about complex solutions, but when I calculate it I don't get one and Matlab's doesn't seem right.

max_zorn
  • 4,875
TobyTobyo
  • 377
  • I don't have Matlab but I suspect nothing is wrong. The roots are found numerically and if you get a tiny imaginary part then chances are it can be real. If you set the imaginary part to $0$, does it match your answer? – max_zorn Feb 16 '18 at 04:04
  • 2
    Shouldn't it be $(9 - r) (-9 - r) + 6 * 15 = 0$, so $r^2 + 9 = 0$? – Nicolas Pelletier Feb 16 '18 at 04:22

1 Answers1

1

Your characteristic polynomial is $$\left|\begin{matrix} 9-r & -6\\ 15 & -(9+r)\\ \end{matrix}\right|=-(9-r)(9+r)-(-6)\cdot 15=-81+r^2+90=r^2+9,$$ so the eigenvalues (not eigenvectors) of $A$ are $r_1=3i$ and $r_2=-3i$.

Alejandro Nasif Salum
  • 5,703