# Finding general solution of $3{\times}3$ system

I am given the following: $$x'= \begin{bmatrix} 2 &0 &0 \\ -7&9 &7 \\ 0&0 &2 \end{bmatrix} x$$

Solving $$\det(A-\lambda I)$$, I get $$\lambda = 2,2,9$$. Solving $$\det(A-2\lambda)$$, I get $$\begin{bmatrix} 0&0 &0 \\ -7&7 &7 \\ 0&0 &0 \end{bmatrix}$$ So we have geom multi. = alg. multi, so our matrix is complete.

Take $$v_1 = \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$$

Similarly, solve $$\det(A-9\lambda)$$ to get $$\begin{bmatrix} -7&0 &0 \\ -7&0 &7 \\ 0&0 &-7 \end{bmatrix}$$ So take $$v_2 = \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$$

So the general solution should be $$x(t) = C_1e^{2t}\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} +C_2e^{2t} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} + C_3e^{9t}\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$$

However, according to the back of the book solution, this is incorrect. What am I missing here? Thank you.

You are right that $$v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$$ is an eigenvector of $$A$$ with eigenvalue $$2$$; i.e $$v_1 \in \ker(A-2I)$$. But like you mentioned, the dimension of this kernel is $$2$$, so you need to find another linearly independent eigenvector. It is easy to verify that $$\xi = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$$ is such a vector. In other words, $$\{v_1, \xi\}$$ form a basis for $$\ker(A-2I)$$. Hence, your general solution will be $$$$x(t) = C_1 e^{2t} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + C_2 e^{2t} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + C_3 e^{9t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$$$

Your eigenvectors are not right for $$\lambda=2.$$ You have the right system, but you should get two solutions: $$v_1=(1,1,0)$$ and $$v_2=(1,0,1).$$

The system $$x'= \begin{bmatrix} 2 &0 &0 \\ -7&9 &7 \\ 0&0 &2 \end{bmatrix}$$ is simply $$x_1'=2x_1$$

$$x_2'=-7x_1+9x_2+7x_3$$

$$x_3'=2x_3$$

Therefore $$x_1=c_1e^{2t}$$ and $$x_3=c_2 e^{2t}$$

Solving for $$x_2'=9x_2+7(c_2-c_1)e^{2t}$$

we get $$x_2=c_3e^{9t}+(c_1-c_2)e^{2t}$$

Thus we have $$x(t) = c_1 e^{2t} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} 0\\-1 \\ 1 \end{pmatrix} + c_3 e^{9t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$

For $$\lambda=2$$, $$(A-2I)v=0$$ gives $$-x+y+z=0$$.

Letting $$y=k_1$$ and $$z=k_2$$ you get $$x=k_1+k_2$$.

Then $$v=\begin{pmatrix}k_1+k_2\\k_1\\k_2\end{pmatrix}=k_1\begin{pmatrix}1\\1\\0\end{pmatrix}=k_2\begin{pmatrix}1\\0\\1\end{pmatrix}$$.

$$\implies v_1=\begin{pmatrix}1\\1\\0\end{pmatrix}$$ and $$v_2=\begin{pmatrix}1\\0\\1\end{pmatrix}$$ are two corresponding eigenvectors.