Constant-Coefficient Systems. Find a real general solution of the following system. Find a real general solution of the following system. (Show the
details.)
$y'_1 = 9 y_1 + 13.5 y_2 \;\;\;\;\;\;\; y′_2 = 1.5 y_1 + 9 y_2$
$y' = \begin{pmatrix} 9& 13.5\\1.5& 9\end{pmatrix}$y
$\det$ (A - $\lambda$I$)= \begin{pmatrix} 9-\lambda& 13.5\\1.5& 9-\lambda\end{pmatrix}$ = $\lambda^2 - 18\lambda+60.75$
$\lambda \Rightarrow \lambda_1 = 13.5\;\;\;\lambda_2 = 4.5\;\;(eigen- values) $
Eigen vectors
$For \;\lambda = 13.5$
I get 
$\begin{pmatrix} -4.5& 13.5\\1.5& -4.5\end{pmatrix}$$\begin{pmatrix} 0\\0 \end{pmatrix}$ 
Using Gaussian elimination I get:
$\begin{pmatrix} -1& 3\\0& 0\end{pmatrix}$$\begin{pmatrix} 0\\0 \end{pmatrix}$
Which implies that:
$-x_1 + 3x_2 =0 $
$Let\;\; x_2 = t$
$\therefore\;\;x_1 = 3t$
Hence 
$\begin{pmatrix} x_1\\x_2 \end{pmatrix}$$\begin{pmatrix} 3\\1 \end{pmatrix}$t
Thus the eigen vectors are:
$x^{(1)}$=$\begin{pmatrix} 3\\1 \end{pmatrix}$
$For \;\lambda = 4.5$
$\begin{pmatrix} 4.5& 13.5\\1.5& 4.5\end{pmatrix}$$\begin{pmatrix} 0\\0 \end{pmatrix}$ 
Using Gaussian elimination I get:
$\begin{pmatrix} 1& 3\\0& 0\end{pmatrix}$$\begin{pmatrix} 0\\0 \end{pmatrix}$
Which implies that:
$x_1 + 3x_2 =0 $
$Let\;\; x_2 = t$
$\therefore\;\;x_1 = -3t$
Hence 
$\begin{pmatrix} x_1\\x_2 \end{pmatrix}$$\begin{pmatrix} -3\\1 \end{pmatrix}$t
Thus the eigen vectors are:
$x^{(2)}$=$\begin{pmatrix} -3\\1 \end{pmatrix}$
My instructors solution for $\;for \;\lambda = 4.5$
$x^{(2)}$=$\begin{pmatrix} 3\\-1 \end{pmatrix}$
I just want to know if my working is correct or there is another simple and better 
way. Just confused with instructors answers. Get confused when calculating eigen vectors because my solutions does not match instructors solution for other same type of questions. 
 A: We are given:
$$A = \begin{bmatrix}9 & 27/2\\3/2 & 9\end{bmatrix}$$
The characteristic equation and eigenvalues are found by solving $[A-\lambda I] = 0$, which gives:
$$1/4 (2 \lambda-27) (2 \lambda-9) = 0 \rightarrow \lambda_1 = \dfrac{27}{2}~,\lambda_2 = \dfrac{9}{2}$$
We now find the eigenvectors for each distinct eigenvalue by solving $[A-\lambda_i I]v_i = 0$.
For $\lambda_1 = 27/2$, we have the RREF of:
$$\begin{bmatrix} 1 & -3 \\ 0 & 0 \end{bmatrix}v_1 = 0$$
This leads to:
$a - 3b = 0 \rightarrow a = 3b, ~\mbox{so let}~ b = 1 \rightarrow a = 3$, so our eigenvector is:
$$v_1 = \begin{bmatrix}3\\ 1 \end{bmatrix}$$
Repeating this process for $\lambda_1 = 9/2$, we have the RREF of:
$$\begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}v_2 = 0$$
This leads to:
$a + 3b = 0 \rightarrow a = -3b, ~\mbox{so let}~ b = 1 \rightarrow a = -3$, so our eigenvector is:
$$v_2 = \begin{bmatrix} -3 \\ 1 \end{bmatrix}$$
Here is one approach to finding the exponential matrix using the eigenvalues / eigenvectors.
We can write the solution to this system as:
$$x(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = c_1 e^{27t/2 }\begin{bmatrix} 3 \\ 1 \end{bmatrix}+ c_2 e^{9t/2 }\begin{bmatrix} -3 \\ 1 \end{bmatrix}$$
This gives us a fundamental matrix of:
$$\phi(t) = \begin{bmatrix} 3e^{27t/2} & -3e^{9t/2} \\ e^{27t/2} & e^{9t/2} \end{bmatrix}$$
From this, we can find the matrix exponential using:
$$e^{A t} = \phi(t)(\phi(0))^{-1} = \begin{bmatrix}1/2(e^{9t/2} + e^{27t/2}) & 3/2 (-e^{9t/2} + e^{27t/2})\\ ~1/6(-e^{9t/2} + e^{27t/2}) & 1/2(e^{9t/2} + e^{27t/2}) \end{bmatrix}$$
