We are now ready to return to our original system \({\mathbf x}' = A {\mathbf x}\text{,}\) where
\begin{equation*}
A
=
\begin{pmatrix}
-17 & -46 & -7 \\
8 & 21 & 3 \\
-7 & -15 & -1
\end{pmatrix}.
\end{equation*}
This matrix has a single eigenvalue \(\lambda = 1\text{.}\) It is easy to show that the only nonzero powers of \(A - \lambda I = A - I\) are
\begin{align*}
(A + I) & =
\begin{pmatrix}
-18 & -46 & -7 \\
8 & 20 & 3 \\
-7 & -15 & -2
\end{pmatrix}\\
(A + I)^2 & =
\begin{pmatrix}
5 & 13 & 2 \\
-5 & -13 & -2 \\
20 & 52 & 8
\end{pmatrix}.
\end{align*}
Therefore,
\begin{align*}
e^{tA} & =
e^{t} \left(
I + t (A+I ) + \frac{t^2}{2!} (A + I )^2
\right)\\
& =
e^{t}
\begin{pmatrix}
1 - 18t + 5t^2/2 \amp -46t +13t^2/2 \amp - 7t + t^2 \\
8t - 5t^2/2 \amp 1 + 20t -13t^2/2 \amp 3t - t^2 \\
-7t + 10t^2 \amp -15t + 26t^2 \amp 1 - 2t + 4t^2
\end{pmatrix}.
\end{align*}
Now, to compute three linearly independent solutions for \({\mathbf x}' = A {\mathbf x}\text{,}\) we simply compute \(e^{t A} {\mathbf v}\) for three linearly independent vectors. We will use the standard basis vectors
\begin{equation*}
{\mathbf e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},
{\mathbf e}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},
{\mathbf e}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.
\end{equation*}
Thus, the general solution to our system is
\begin{align*}
{\mathbf x}(t)
& =
c_1
e^{-t}
\begin{pmatrix}
1 - 18t + 5t^2/2 \\
8t - 5t^2/2 \\
-7t + 10t^2
\end{pmatrix}
+
c_2
e^{-t}
\begin{pmatrix}
-46t +13t^2/2 \\
1 + 20t -13t^2/2 \\
-15t + 26t^2
\end{pmatrix}\\
& +
c_3
e^{-t}
\begin{pmatrix}
- 7t + t^2 \\
3t - t^2 \\
1 - 2t + 4t^2
\end{pmatrix}.
\end{align*}