MCQOPTIONS
Saved Bookmarks
MCQOPTIONS
| 1. |
Consider the linear system \(\dot x = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 2}\end{array}} \right]x\), with initial condition \(x\left( 0 \right) = \left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\). The solution x(t) for this system is |
| A. | \(x\left( t \right) = \left[ {\begin{array}{*{20}{c}}{{e^{ - t}}}&{t{e^{ - 2t}}}\\0&{{e^{ - 2t}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\) |
| B. | \(x\left( t \right) = \left[ {\begin{array}{*{20}{c}}{{e^{ - t}}}&0\\0&{{e^{2t}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\) |
| C. | \(x\left( t \right) = \left[ {\begin{array}{*{20}{c}}{{e^{ - t}}}&{ - {t^2}{e^{ - 2t}}}\\0&{{e^{ - 2t}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\) |
| D. | \(x\left( t \right) = \left[ {\begin{array}{*{20}{c}}{{e^{ - t}}}&0\\0&{{e^{ - 2t}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\) |
| Answer» E. | |