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The Solution Of X′=(1−214)X\mathbf { X } ^ { \prime } = \left( \begin{array} { c c } 1 & - 2 \\1 & 4\end{array} \right) \mathbf { X }

Question 19

Multiple Choice

The solution of X′=(1−214) X\mathbf { X } ^ { \prime } = \left( \begin{array} { c c } 1 & - 2 \\1 & 4\end{array} \right) \mathbf { X } is


A) X=c1(21) e−2t+c2(11) e−3t\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 2 \\1\end{array} \right) e ^ { - 2 t } + c _ { 2 } \left( \begin{array} { l } 1 \\1\end{array} \right) e ^ { - 3 t }
B) X=c1(21) e−2t+c2(1−1) e−3tX = c _ { 1 } \left( \begin{array} { l } 2 \\1\end{array} \right) e ^ { - 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { - 3 t }
C) X=c1(−2−1) e2t+c2(1−1) e3tX = c _ { 1 } \left( \begin{array} { c } - 2 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { 3 t }
D) X=c1(1−2) e2t+c2(1−1) e3tX = c _ { 1 } \left( \begin{array} { c } 1 \\- 2\end{array} \right) e ^ { 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { 3 t }
E) X=c1(2−1) e2t+c2(1−1) e3tX = c _ { 1 } \left( \begin{array} { c } 2 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { 3 t }

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