Let a Be the Matrix of the Previous Two Problems $\mathbf { X } ^ { \prime } = A \mathbf { X }$

Multiple Choice

Let A be the matrix of the previous two problems. The solution of $\mathbf { X } ^ { \prime } = A \mathbf { X }$ is

A) $c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]$
B) $c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { - 3 t } + c _ { 2 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { - 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) t e ^ { - 2 t } + \left( \begin{array} { c } 0 \\- 1 \\0\end{array} \right) e ^ { - 2 t } \right]$
C) $c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { - 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) t e ^ { - 2 t } + \left( \begin{array} { c } 0 \\- 1 \\0\end{array} \right) e ^ { - 2 t } \right]$
D) $c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { - 3 t } + c _ { 2 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { c } 0 \\- 1 \\0\end{array} \right) e ^ { 2 t } \right]$
E) $c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { c } 0 \\- 1 \\0\end{array} \right) e ^ { 2 t } \right]$