The Wronskian of the vector functions $\mathbf { X } _ { 1 } = \left( \begin{array} { l }
2 \
1 \
0
\end{array} \right) e ^ { t } , \mathbf { X } _ { 2 } = \left( \begin{array} { l }
1 \
3 \
5
\end{array} \right) e ^ { 2 t }$ and $X _ { 3 } = \left( \begin{array} { c }
- 1 \
5 \
0
\end{array} \right) e ^ { 3 t }$ is
A) $55 e ^ { 6 t }$
B) $- 55 e ^ { 6 t }$
C) $45 e ^ { 6 t }$
D) $- 45 e ^ { - 6 t }$
E) $45 e ^ { - 6 t }$

Question

The Wronskian of the vector functions $\mathbf { X } _ { 1 } = \left( \begin{array} { l } 
2 \
1 \
0
\end{array} \right) e ^ { t } , \mathbf { X } _ { 2 } = \left( \begin{array} { l } 
1 \
3 \
5
\end{array} \right) e ^ { 2 t }$ and $X _ { 3 } = \left( \begin{array} { c } 
- 1 \
5 \
0
\end{array} \right) e ^ { 3 t }$ is
A) $55 e ^ { 6 t }$
B) $- 55 e ^ { 6 t }$
C) $45 e ^ { 6 t }$
D) $- 45 e ^ { - 6 t }$
E) $45 e ^ { - 6 t }$

Quizplus · Accepted Answer

The Wronskian of the given vector functions can be calculated by taking the determinant of the matrix formed by the vectors, each multiplied by their respective exponential functions. The determinant of this matrix is $-55e^{6t}$, which corresponds to choice B.

The Wronskian of the Vector Functions And X3=(−150)e3tX _ { 3 } = \left( \begin{array} { c } - 1 \\5 \\0\end{array} \right) e ^ { 3 t }X3​=​−150​​e3t

The Wronskian of the Vector Functions And $X _ { 3 } = \left( \begin{array} { c } - 1 \\5 \\0\end{array} \right) e ^ { 3 t }$