Answer
See below
Work Step by Step
Given:
$\begin{bmatrix}
-17& 0 & -42\\
-7 & 4 & -14\\
7& 0 & 18
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-17-\lambda & 0 & -42\\
-7 & 4-\lambda & -14\\
7 & 0 & 18-\lambda
\end{vmatrix}=(\lambda-4)^2(\lambda-3)$
so that A has eigenvalues $\lambda_1=4\\
\lambda_2=4\\
\lambda_3=3$
Eigenvalue $λ_1 =4$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=0\\
v_2=1\\
v_3=0$
The solution is $v = r(0,1,0)$. Therefore,
$v_1= e^{4t}\begin{bmatrix}
0\\
1 \\
0\end{bmatrix}$
Do the same for $v_2$ and $v_3$,
$v_2= e^{4t}\begin{bmatrix}
-2\\
0 \\
1
\end{bmatrix}$
$v_3= e^{3t}\begin{bmatrix}
-3\\
-1\\
1
\end{bmatrix}$
Hence, the general solution is:
$x(t)=c_1e^{4t}\begin{bmatrix}
0\\
1 \\
0\end{bmatrix}+c_2e^{4t}\begin{bmatrix}
-2\\
0\\
1
\end{bmatrix}+c_3e^{3t}\begin{bmatrix}
-3\\
-1\\
0 \end{bmatrix}=\begin{bmatrix}
-2c_2e^{4t} -3c_3e^{3t}\\
c_1e^{4t}-c_3e^{3t}\\
c_2e^{4t}+c_3e^{3t}
\end{bmatrix}$
