MCQOPTIONS
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| 1. |
The matrix form of the linear system \(\frac{{dx}}{{dt}} = 3x - 5y\) and \(\frac{{dy}}{{dt}} = 4x + 8y\) is |
| A. | \(\frac{d}{{dt}}\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 3&{ - 5}\\ 4&8 \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\}\) |
| B. | \(\frac{d}{{dt}}\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 3&8\\ 4&{ - 5} \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\}\) |
| C. | \(\frac{d}{{dt}}\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 4&{ - 5}\\ 3&8 \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\}\) |
| D. | \(\frac{d}{{dt}}\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 4&8\\ 3&{ - 5} \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\}\) |
| Answer» B. \(\frac{d}{{dt}}\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 3&8\\ 4&{ - 5} \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right\}\) | |