The solution of $$X ^ { \prime } = \left( \begin{array} { c c c }
4 & 0 & 0 \
0 & 3 & 1 \
0 & - 1 & 1
\end{array} \right) X$$ are
A) $$X = c _ { 1 } \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
B) $$X = c _ { 1 } \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
C) $$X = c _ { 1 } \left( \begin{array} { l }
1 \
0 \
0
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c }
0 \
1 \
- 1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
D) $$X = c _ { 1 } \left( \begin{array} { l }
1 \
0 \
0
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { l }
0 \
1 \
1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l }
0 \
1 \
1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
E) $$X = c _ { 1 } \left( \begin{array} { l }
1 \
0 \
0
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { c }
1 \
- 1 \
0
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c }
1 \
- 1 \
0
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l }
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$

Question

The solution of $$X ^ { \prime } = \left( \begin{array} { c c c } 
4 & 0 & 0 \
0 & 3 & 1 \
0 & - 1 & 1
\end{array} \right) X$$ are
A) $$X = c _ { 1 } \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
B) $$X = c _ { 1 } \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
C) $$X = c _ { 1 } \left( \begin{array} { l } 
1 \
0 \
0
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 
0 \
1 \
- 1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
D) $$X = c _ { 1 } \left( \begin{array} { l } 
1 \
0 \
0
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { l } 
0 \
1 \
1
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l } 
0 \
1 \
1
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$
E) $$X = c _ { 1 } \left( \begin{array} { l } 
1 \
0 \
0
\end{array} \right) e ^ { 4 t } + c _ { 2 } \left( \begin{array} { c } 
1 \
- 1 \
0
\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 
1 \
- 1 \
0
\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 
0 \
1 \
0
\end{array} \right) e ^ { 2 t } \right]$$

Quizplus · Accepted Answer

The given matrix is diagonalizable, and the eigenvalues can be directly read from the diagonal: 4, 2 (with algebraic multiplicity 2). The eigenvector corresponding to the eigenvalue 4 is $\left( \begin{array}{l} 1 \ 0 \ 0 \end{array} \right)$, indicating exponential growth at a rate of $e^{4t}$ for that component. The eigenvalue 2, having algebraic multiplicity 2, suggests a generalized eigenvector is needed due to the repeated eigenvalue, leading to the solution structure in choice C.

The Solution Of X′=(4000310−11)XX ^ { \prime } = \left( \begin{array} { c c c } 4 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) XX′=​400​03−1​011​​X

The Solution Of $X ^ { \prime } = \left( \begin{array} { c c c } 4 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) X$