Differential Equations Systems problem (3X3 matrix) Please help

[VinnyCee]

Here is the problem:

X'\,=\,\left(\begin{array}{ccc}3 & -1 & -1\\1 & 1 & -1\\1 & -1 & 1\end{array}\right)\,X

Here is what i have so far:

det(A\,-\,rI)\,=\,0

det\left[\left(\begin{array}{ccc}3 & -1 & -1\\1 & 1 & -1\\1 & -1 & 1\end{array}\right)\,-\,\left(\begin{array}{ccc}r & 0 & 0\\0 & r & 0\\0 & 0 & r\end{array}\right)\right]\,=\,\left(\begin{array}{ccc}3\,-\,r & -1 & -1\\1 & 1\,-\,r & -1\\1 & -1 & 1\,-\,r\end{array}\right)

-(r\,-\,1)\,(r^2\,-\,4r\,+4)\,=\,0

r_1\,=\,1,\,\,\,\,r_2\,=\,2,\,\,\,\,r_3\,=\,2

r\,=\,2 is repeated once.

(A\,-\,r_1I)\,\xi\,=\,0

\left[\left(\begin{array}{ccc}3 & -1 & -1\\1 & 1 & -1\\1 & -1 & 1\end{array}\right)\,-\,\left(\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right)\right]\,\left(\begin{array}{ccc} \xi_1 \\ \xi_2 \\ \xi_3 \end{array}\right)=\,\left(\begin{array}{ccc}2 & -1 & -1\\1 & 0 & -1\\1 & -1 & 0\end{array}\right)\,\left(\begin{array}{ccc} \xi_1 \\ \xi_2 \\ \xi_3 \end{array}\right)\,=\,0

Now, multiplying out the right hand side of the equation above, and rref'ing, I get:

\left(\begin{array}{cccc} 1 & 0 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)

Which means that,

\xi_1\,-\,\xi_3\,=\,0
\xi_2\,-\,\xi_3\,=\,0

So then,

\xi_1\,=\,\xi_2\,=\,\xi_3?

Here is where I am stuck! What am I supposed to do now to solve this system?

 

[PhilG]

Just pick a number, like \xi_1 = \xi_2 = \xi_3 = 1 or something. That's just saying that any vector with all three components the same is an eigenvector corresponding to eigenvalue r = 1. Is that what you are asking?

 

[VinnyCee]

Ok, \xi_1=\xi_2=\xi_3=1.

What do I do with this information?

Do I just solve for the r_2=r_3=2 part and that is the answer?

I am still not understnding exactly what it is that I am supposed to be doing with the \xi_n values.

 

[geosonel]

(note: unfortunately, there's a little difference between VinnyCee's notation and what follows. VinnyCee uses ξ₁, ξ₂, and ξ₃ to represent eigenvector components. In what follows, ξ₁, ξ₂, and ξ₃ are the actual eigenvectors themselves. sorry.)

from your results (& standard textbook): for single unique eigenvalue r₁=1 and its eigenvector ξ₁=<1, 1, 1>, a specific solution is:

X_{1}(t) \ = \ C_{1} \cdot \exp(r_1 t) \cdot \xi_{1} \ = \ C_1 \cdot \exp(t) \cdot \left ( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right )

then for repeated eigenvalue r₂₃=2 and its eigenvectors ξ₂=<2, 1, 1> and ξ₃=<1.33, -1, 2.33>, a specific solution is:

X_{23}(t) \ = \ C_{2} \cdot \exp(r_{23} t) \cdot \xi_{2} \ + \ C_{3} \cdot \exp(r_{23} t) \cdot ( \xi_{2} t \ + \ \xi_{3} ) \ = \ C_{2} \cdot \exp(2 t) \cdot \left ( \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right ) \ + \ C_3 \cdot \exp(2t) \cdot \left ( \left ( \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right ) \cdot t \ + \ \left ( \begin{array}{c} 1.33 \\ -1 \\ 2.33 \end{array} \right ) \right )

\mbox{General Solution} \ = \ \textbf{X(t)} \ = \ X_{1}(t) \ + \ X_{23}(t)

general solution is (X₁(t) + X₂₃(t)), and constants C₁, C₂, and C₃ obtained from initial cond.

 

[VinnyCee]

The general solution is...

Is this the correct general solution?:

X\,=\,\ C_1 \cdot e^t \cdot \left ( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right ) \,+\,\ C_2 \cdot e^{2t} \cdot \left ( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right ) \,+\,\ C_3 \cdot e^{2t} \cdot \left [ \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right )\cdot t\,+\,\left ( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right ) \right]

This is using different eigenvectors than you calculated. I got:

\xi^{(2)}\,=\,\left ( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right ) and \xi^{(2)}\,=\,\left ( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right ).

I think that they are just multipules of your solutions.