Answer
See below
Work Step by Step
Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
3-\lambda & -2 & -2\\
1 & -\lambda & -2 \\
0 & & 3-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\ v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\ 0
\end{bmatrix}$
$\begin{bmatrix}
3-\lambda & -2 & -2\\
1 & -\lambda & -2 \\
0 & & 3-\lambda
\end{bmatrix}=0$
$(3-\lambda)(\lambda^2-3\lambda+2)=0$
$(3-\lambda)(\lambda-2)(\lambda-1)$
$\lambda_1=3, \lambda_2=2,\lambda_3=1$
The eigenvalues of $A$ are $\lambda_1 = 3$ and $\lambda_2=2$, $\lambda_3=1$, therefore A is defective.
A straightforward computation yields the following eigenvectors, which correspond respectively to $\lambda_1,\lambda_2$ and $\lambda_3$:
$v_1=(3,1,0)\\
v_2=(4,1,0)\\
v_3=(5,1,-1)$
It follows from Theorem 7.4.3 that if we set $S =\begin{bmatrix}
3 & 4 & 5\\
1 & 1 & 1\\
0 & 0 & -1
\end{bmatrix}$ and $D = \begin{bmatrix}
1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{bmatrix}$
then $e^{At} = Se^{Dt}S^{−1}$
It is easily shown that
$S^{-1}=\begin{bmatrix}
-1 & 4 & -1\\
1 & -3 & 2\\
0 & 0 & -1
\end{bmatrix}$
Substituting:
$e^{At}=\begin{bmatrix}
3 & 4 & 5\\
1 & 1 & 1\\
0 & 0 & -1
\end{bmatrix}\begin{bmatrix}
e^{t} & 0 & 0\\
0 & e^{2t} & 0\\
0 & 0 & 3^{3t}
\end{bmatrix}\begin{bmatrix}
-1 & 4 & -1\\
1 & -3 & 2\\
0 & 0 & -1
\end{bmatrix}=\begin{bmatrix}
3e^{t} & 4e^{2t} &5e^{3t}\\
e^{t} & e^{2t} & e^{3t}\\
0 & & e^{-3t}
\end{bmatrix}\begin{bmatrix}
-1 & 4 & -1\\
1 & -3 & 2\\
0 & 0 &-1
\end{bmatrix}=\begin{bmatrix}
-3e^t+4e^{2t} & 12e^t-12e^{2t} & -3e^t+8e^{2t}-5e^{3t}\\
-e^t+e^{2t}&4e^t-3e^{2t} & -e^t+2e^{2t}-e^{3t}\\
0 & 0 & e^{3t}
\end{bmatrix}$
