Answer
See below
Work Step by Step
Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
6-\lambda & -2 & -1\\
8 & -2-\lambda & -2 \\
4 & -2 & 1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\ v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\ 0
\end{bmatrix}$
$\begin{bmatrix}
6-\lambda & -2 & -1\\
8 & -2-\lambda & -2 \\
4 & -2 & 1-\lambda
\end{bmatrix}=0$
$-(\lambda-2)^2(\lambda-1)=0$
$\lambda_1=\lambda_2=2,\lambda_3=1$
The eigenvalues of $A$ are $\lambda_1 = 2$ 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 $\lambda_3$:
$v_1=(2,1,0)\\
v_2=(1,0,1)\\
v_3=(\frac{3}{5},2,1)$
It follows from Theorem 7.4.3 that if we set $S =\begin{bmatrix}
2 & 1 & \frac{3}{5}\\
1 & 0 & 2\\
0 & 1 & 1
\end{bmatrix}$ and $D =diag(2,2,1)$
then $e^{At} = Se^{Dt}S^{−1}$
It is easily shown that
$S^{-1}=\begin{bmatrix}
\frac{5}{11} & \frac{1}{11} & -\frac{5}{11}\\
-\frac{5}{22} & -\frac{5}{11} & \frac{7}{22}\\
-\frac{5}{22} & \frac{5}{11} & \frac{5}{22}
\end{bmatrix}$
Substituting:
$e^{At}=\begin{bmatrix}
2 & 1 & \frac{3}{5}\\
1 & 0 & 2\\
0 & 1 & 1
\end{bmatrix}\begin{bmatrix}
e^{2t} & 0 & 0\\
0 & e^{2t} & 0\\
0 & 0 & e^{t}
\end{bmatrix}\begin{bmatrix}
\frac{5}{11} & \frac{1}{11} & -\frac{5}{11}\\
-\frac{5}{22} & -\frac{5}{11} & \frac{7}{22}\\
-\frac{5}{22} & \frac{5}{11} & \frac{5}{22}
\end{bmatrix}=\begin{bmatrix}
2e^{2t} & e^{2t} & \frac{3}{5}e^{t}\\
e^{2t} &0 & 2e^{t}\\
0 & e^{2t} & e^{t}
\end{bmatrix}\begin{bmatrix}
\frac{5}{11} & \frac{1}{11} & -\frac{5}{11}\\
-\frac{5}{22} & -\frac{5}{11} & \frac{7}{22}\\
-\frac{5}{22} & \frac{5}{11} & \frac{5}{22}
\end{bmatrix}=\begin{bmatrix}
-\frac{3}{22}e^t+\frac{25}{22}e^{2t} & \frac{3}{11}e^t-\frac{3}{11}e^{2t} & \frac{3}{22}e^t-\frac{3}{22}e^{2t}\\
-\frac{5}{11}e^t+\frac{5}{11}e^{2t}& \frac{10}{11}e^t+\frac{1}{11}e^{2t} & \frac{5}{11}e^t-\frac{5}{11}e^{2t}\\
-\frac{5}{22}e^t+\frac{5}{22}e^{2t} & \frac{5}{11}e^t-\frac{5}{11}e^{2t} & \frac{5}{22}e^{t}+\frac{17}{22}e^{2t}
\end{bmatrix}$
