Systems of linear differential equation with linear coefficients $$ \frac{dx}{dt}=\left(\begin{matrix}1 & 9 & 9 \\0 & 19 & 18 \\ 0 & 9 & 10 \end{matrix} \right)\left(\begin{matrix}x_1\\x_2\\x_3 \end{matrix}\right)$$
My try:
We assume it has the solution of the form $x=\alpha e^{\lambda t}$
Setting the Determinant of the system to zeroes.
I have found three eigenvalues,
$$\lambda_1=1,\lambda_2=10, \lambda_3=19 $$
For $\lambda_1=1$
$$\left(\begin{matrix}\alpha_1 + 9\alpha_2 + 9\alpha_3 \\ 19\alpha_2 +18\alpha_3 \\ 9\alpha_2 + 10\alpha_3 \end{matrix} \right)=\left(\begin{matrix}\alpha_1\\ \alpha_2\\ \alpha_3 \end{matrix}\right)$$
Solving it we have,
$\alpha_1$ is independent and $\alpha_2=-\alpha_3$
For $\lambda_2=10$
$$\left(\begin{matrix}\alpha_1 + 9\alpha_2 + 9\alpha_3 \\ 19\alpha_2 +18\alpha_3 \\ 9\alpha_2 + 10\alpha_3 \end{matrix} \right)=\left(\begin{matrix}10\alpha_1\\ 10\alpha_2\\ 10\alpha_3 \end{matrix}\right)$$
$\alpha_1=\alpha_2+\alpha_3$ and $\alpha_2=-2\alpha_3$ and $\alpha_2=0$
For $\lambda_3=19$
I run into similar problems
I know that it is wrong, can someone help me in this matter?
 A: We have
$$X'= \left(\begin{matrix}x'_1\\x'_2\\x'_3 \end{matrix}\right) = \left(\begin{matrix}1 & 9 & 9 \\0 & 19 & 18 \\ 0 & 9 & 10 \end{matrix} \right)\left(\begin{matrix}x_1\\x_2\\x_3 \end{matrix}\right)$$
We find eigenvalues using the characteristic polynomial by solving $|A - \lambda I| = 0$ giving $-(\lambda -28) (\lambda -1)^2 = 0 $, so
$$\lambda_{1, 2} = 1, \lambda_3 = 28$$
Notice the repeated eigenvalue, so we may or may not have a deficient matrix. Are you familiar with algebraic and geometric multiplicities? 
To find the eigenvectors, for each eigenvalue we solve $[A - \lambda_i I]v_i = 0$. 
For $\lambda_{1,2} = 1$
$$RREF ~[A - I]v_{1,2} = \left(\begin{matrix}0 & 1 & 1 \\0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right)v_{1,2} = 0$$
It turns out that we can find two linearly independent eigenvectors for this eigenvalue as
$$v_1 = (1, 0, 0), v_2 = (0, -1, 1)$$
For $\lambda_3 = 28$
$$RREF ~[A - 28I]v_{3} = \left(\begin{matrix}1 & 0 & -1 \\0 & 1 & -2 \\ 0 & 0 & 0 \end{matrix} \right)v_{3} = 0$$
$$v_3 = (1, 2, 1)$$
Because we were able to find three linearly independent eigenvectors, we can write the solution as
$$X(t) = (c_1 v_1 + c_2 v_2)e^t + c_3 v_3 e^{28 t} = \left(c_1 \left(\begin{matrix}1\\ 0 \\ 0 \end{matrix}\right) + c_2 \left(\begin{matrix}0\\ -1 \\ 1 \end{matrix}\right)\right)e^t + c_3 \left(\begin{matrix}1\\ 2 \\ 1 \end{matrix}\right)e^{28t}$$
It is worth noting that there are many ways to solve such problems, for example see Systems of differential equations or Nineteen Dubious Ways to Compute the Exponential of a Matrix.
