The solution of $$X ^ { \prime } = \left( \begin{array} { c c c }
3 & 0 & 0 \
0 & 3 & 1 \
0 & - 1 & 1
\end{array} \right) \mathrm { X }$$ is
A) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
B) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
C) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { l }
1 \
0 \
0
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
D) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { l }
1 \
0 \
0
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { l }
0 \
1 \
1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l }
0 \
1 \
1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
E) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { l }
1 \
0 \
0
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c }
1 \
- 1 \
0
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c }
1 \
- 1 \
0
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$

Question

The solution of $$X ^ { \prime } = \left( \begin{array} { c c c } 
3 & 0 & 0 \
0 & 3 & 1 \
0 & - 1 & 1
\end{array} \right) \mathrm { X }$$ is
A) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
B) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
C) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 
1 \
0 \
0
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
D) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 
1 \
0 \
0
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { l } 
0 \
1 \
1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l } 
0 \
1 \
1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
E) $$\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 
1 \
0 \
0
\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 
1 \
- 1 \
0
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 
1 \
- 1 \
0
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$

Quizplus · Accepted Answer

The given matrix has an eigenvalue of 3 with an eigenvector $\left( \begin{array} { l } 1 \ 0 \ 0 \end{array} \right)$, and an eigenvalue of 2 with a generalized eigenvector structure due to the 2x2 block in the bottom right. The correct solution must reflect these eigenvalues and the corresponding (generalized) eigenvectors. Option C correctly identifies the eigenvalue 3 with its eigenvector and the eigenvalue 2 with its generalized eigenvector and associated structures.

The Solution Of X′=(3000310−11)XX ^ { \prime } = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) \mathrm { X }X′=​300​03−1​011​​X

The Solution Of $X ^ { \prime } = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) \mathrm { X }$