Answer
See below
Work Step by Step
Given:
$\begin{bmatrix}
2 & -4 & 3\\
-9 & -3 & -9\\
4 & 4 &3
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
2-\lambda & -4 & 3\\
-9 & -3-\lambda & -9\\
4 & 4 & 3-\lambda
\end{vmatrix}=(\lambda-6)(\lambda+3)(\lambda+1)$
so that A has eigenvalues $\lambda_1=6\\
\lambda_2=-3\\
\lambda_3=-1$
Eigenvalue $λ_1 =6$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=-1\\
v_2=1\\
v_3=0$
The solution is $v = r(-1,1,0)$. Therefore,
$v_1= e^{6t}\begin{bmatrix}
-1\\
1 \\
0\end{bmatrix}$
Do the same for $v_2$ and $v_3$,
$v_2= e^{-3t}\begin{bmatrix}
-2\\
-1 \\
2
\end{bmatrix}$
$v_3= e^{-t}\begin{bmatrix}
-1\\
0 \\
1
\end{bmatrix}$
Hence, the general solution is:
$x(t)=c_1e^{6t}\begin{bmatrix}
-1\\
1 \\
0\end{bmatrix}+c_2e^{-3t}\begin{bmatrix}
-2\\
-1 \\
2
\end{bmatrix}+c_3e^{-t}\begin{bmatrix}
-1\\
0 \\
1\end{bmatrix}=\begin{bmatrix}
-c_1e^{6t}-2c_2e^{-3t} -c_3e^{-t}\\
c_1e^{6t}-c_2e^{-3t}\\
2c_2e^{-3t}+c_3e^{-t}
\end{bmatrix}$
