Answer
See below
Work Step by Step
Given: $\begin{bmatrix}
2 & 0 & 0\\
0 & 5 & -7\\
0 & 2 & -4
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
2-\lambda & 0 & 0\\
0 & 5-\lambda & -7\\
0 & 2 & -4-\lambda
\end{vmatrix}=(2-\lambda)(\lambda+2)(\lambda-3)$
so that A has eigenvalues $\lambda_1=2\\
\lambda_2=-2\\
\lambda_3=3$
Eigenvalue $λ_1 =2$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=1\\
\rightarrow v_2=v_3=0$
The solution is $v = r(1, 0, 0)$. Therefore,
$v_1=e^{2t} \begin{bmatrix}
1\\
0 \\
0
\end{bmatrix}$
Eigenvalue $\lambda_2 =-2$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=0\\
\rightarrow v_2=v_3=1$
The solution is $v = r(0, 0, 1)$. Therefore,
$v_2=e^{-2t} \begin{bmatrix}
0\\
0 \\
1
\end{bmatrix}$
Eigenvalue $\lambda_3 =3$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=0\\
v_3=2\\
v_2=\frac{7}{2}v_3=7$
The solution is $v = r(0,7,2)$. Therefore,
$v_3=e^{3t} \begin{bmatrix}
0 \\
7 \\
2
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1v_1+c_2v_2+c_3v_3=c_1e^{2t}\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}+c_2e^{-2t}\begin{bmatrix}
0\\
1\\
1
\end{bmatrix}+c_3e^{3t}\begin{bmatrix}
0\\
7\\
2
\end{bmatrix}=\begin{bmatrix}
c_1e^{2t}\\
c_2e^{-2t}+7c_3e^{3t}\\
c_2e^{-2t}+2c_3e^{3t}
\end{bmatrix}$
