Eigenpairs Differential equation Consider the linear system $Y'=AY$ given by $$Y'=\begin{bmatrix}3 & 2 \\ -9 & -6\end{bmatrix}Y$$

A. Compute the eigenpairs for the coefficient matrix $A$

So far I have $\lambda = 0\text{ and }-3$ and

B. For each eigenpair in (A), form a solution of $Y'=AY$.

Which when I used $0$ for my lambda and I substitute $x_1=1$ and then once I compute for my $x_2=-\dfrac{3}{2}$

Then I I used $x_1=1$ and then my $x_2=3$ resulting to forming my $y(t)=c_1[1, -3/2]+ c_2e^{3t} [-1,3]$
C. Is the set of solutions found a fundamental set of solutions?Justify.

I'm not sure how to justify this.
 A: Given:
$\mathbf A = \begin{bmatrix}~~3&~~2\\-9&-6\end{bmatrix}$
Part a: You should get an Eigensystem as follows:
$\lambda_1 = -3, v_1 = (-1, 3)$
$\lambda_2 = 0, v_2 = (-2, 3)$
Lets add the details for how we arrived at this first eigenvector.
$A - \lambda_1 I = 0$, yields: $\begin{bmatrix}~~6&~~2\\-9&-3\end{bmatrix}.v_1 = \begin{bmatrix}0\\0\end{bmatrix} $
The Row-Reduced-Echelon-Form (RREF) for this is:
$$\begin{bmatrix}~~1&~\frac{1}{3}\\0&0\end{bmatrix}.v_1 = \begin{bmatrix}0\\0\end{bmatrix} $$
So, we have: $\displaystyle a + \frac{1}{3} b = 0, ~~\text{choose}~~ b = 3 \rightarrow a = -1$.
Part b: To write the general solution, we have distinct eigenvalues and one of them is zero, so we get:
$$y(t) = c_1 e^{\lambda_1 t} v_1 + c_2 v_2 = c_1 e^{-3t}\begin{bmatrix}-1\\3\end{bmatrix} + c_2 \begin{bmatrix}-2\\-3\end{bmatrix}.$$
For Part c, you need to show that $y_1(t)$ and $y_2(t)$ are two solutions to the DEQ and that the Wronskian$(y_1,y_2)(t) \ne 0$.  Then the two solutions are called a fundamental set of solutions.
You should get $W(y_1(t), y_2(t)) = -9 c_1 c_2 e^{-3 t} \ne 0$ for any $t$.
You can work the rest out and look up the theory since you probably discussed in class.
