The state transition matrix for the system $\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\ 1&1 \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 1\ 1 \end{array}} \right]u$ is

Question

The state transition matrix for the system $\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\ 1&1 \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 1\ 1 \end{array}} \right]u$ is

TuteeHUB · Accepted Answer

$\left[ {\begin{array}{*{20}{c}} {{e^t}}&{t{e^t}}\ 0&{{e^t}} \end{array}} \right]$

TuteeHUB · Answer

$\left[ {\begin{array}{*{20}{c}} {{e^t}}&0\ {{e^t}}&{{e^t}} \end{array}} \right]$

TuteeHUB · Answer

$\left[ {\begin{array}{*{20}{c}} {{e^t}}&0\ {{t^2}{e^t}}&{{e^t}} \end{array}} \right]$

TuteeHUB · Answer

$\left[ {\begin{array}{*{20}{c}} {{e^t}}&0\ {t{e^t}}&e^t \end{array}} \right]$

1.	The state transition matrix for the system \(\left[ {\begin{array}{{20}{c}} {{{\dot x}_1}}\\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} 1&0\\ 1&1 \end{array}} \right]\;\left[ {\begin{array}{{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{{20}{c}} 1\\ 1 \end{array}} \right]u\) is
A.	\(\left[ {\begin{array}{*{20}{c}} {{e^t}}&0\\ {{e^t}}&{{e^t}} \end{array}} \right]\)
B.	\(\left[ {\begin{array}{*{20}{c}} {{e^t}}&0\\ {{t^2}{e^t}}&{{e^t}} \end{array}} \right]\)
C.	\(\left[ {\begin{array}{*{20}{c}} {{e^t}}&0\\ {t{e^t}}&e^t \end{array}} \right]\)
D.	\(\left[ {\begin{array}{*{20}{c}} {{e^t}}&{t{e^t}}\\ 0&{{e^t}} \end{array}} \right]\)
Answer» D. \(\left[ {\begin{array}{*{20}{c}} {{e^t}}&{t{e^t}}\\ 0&{{e^t}} \end{array}} \right]\)

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