So I was given the following system$$x'(t)=r_2y-(r_1x+r_3x),$$$$y'(t)=r_3x-r_2y$$and told to solve for its eigenvalues. Well, this is where I got:$$\begin{bmatrix}
x\
y
\end{bmatrix}'=\begin{bmatrix}
-(r_1+r_3) & r_2\
r_3 & -r_2
\end{bmatrix}\begin{bmatrix}
x\
y
\end{bmatrix},$$$$\begin{vmatrix}
-(r_1+r_3)-\lambda & r_2\
r_3 & -r_2-\lambda
\end{vmatrix}=(\lambda+r_1)(\lambda+r_2)+\lambda r_3=0.$$I have no idea how to solve for the eigenvalues, and this is the best I could simplify the expression. :( Help?

Question

So I was given the following system$$x'(t)=r_2y-(r_1x+r_3x),$$$$y'(t)=r_3x-r_2y$$and told to solve for its eigenvalues. Well, this is where I got:$$\begin{bmatrix}
x\ 
y
\end{bmatrix}'=\begin{bmatrix}
-(r_1+r_3) & r_2\ 
r_3 & -r_2
\end{bmatrix}\begin{bmatrix}
x\ 
y
\end{bmatrix},$$$$\begin{vmatrix}
-(r_1+r_3)-\lambda & r_2\ 
r_3 & -r_2-\lambda
\end{vmatrix}=(\lambda+r_1)(\lambda+r_2)+\lambda r_3=0.$$I have no idea how to solve for the eigenvalues, and this is the best I could simplify the expression. :( Help?

anonymous · Answer

I feel so silly right now...

anonymous · Answer

sorry this is slightly off topic but how did you format the question so beautifully?  does the web site accept latex?

anonymous · Answer

i just joined the site 60 seconds ago and am learning my way around

anonymous · Answer

It does. You have to put it in between these brackets: \[ \.] (without the dot on the second one).

anonymous · Answer

thanks!  unfortunately i don't know the answer to your question though.

anonymous · Answer

That's fine :) and welcome!

anonymous · Answer

You can deal with it as a quadratic equation in $\lambda$.