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The matrix form of the linear system $$\frac{{{\text{dx}}}}{{{\text{dt}}}} = 3{\text{x}} - 5{\text{y}}$$ and $$\frac{{{\text{dy}}}}{{{\text{dt}}}} = 4{\text{x}} + 8{\text{y}}$$ is

A. \[\frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 3&{ - 5} \\ 4&8 \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\}\]

B. \[\frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 3&8 \\ 4&{ - 5} \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\}\]

C. \[\frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 4&{ - 5} \\ 3&8 \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\}\]

D. \[\frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 4&8 \\ 3&{ - 5} \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\}\]

Answer: Option A

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