اعمال ریاضی در بردارها

\(\begin{array}{l}\left. \begin{array}{l}A = \left[ \begin{array}{l}5\\4\end{array} \right] = \left[ \begin{array}{l}{x_A}\\{y_A}\end{array} \right]\\\\B = \left[ \begin{array}{l} - 3\\6\end{array} \right] = \left[ \begin{array}{l}{x_B}\\{y_B}\end{array} \right]\end{array} \right\} \Rightarrow M = \left[ \begin{array}{l}{x_M}\\{y_M}\end{array} \right]\\\\\left. \begin{array}{l}{x_M} = \frac{{{x_A} + {x_B}}}{2} = \frac{{5 + ( - 3)}}{2} = \frac{2}{2} = 1\\\\{y_M} = \frac{{{y_A} + {y_B}}}{2} = \frac{{6 + 4}}{2} = \frac{{10}}{2} = 5\end{array} \right\} \Rightarrow M = \left[ \begin{array}{l}1\\5\end{array} \right]\end{array}\)

فقط بردار \(\overrightarrow O = \left[ {\begin{array}{*{20}{c}} \circ \\ \circ \end{array}} \right]\) ابتدا و انتهای آن روی هم قرار دارند؛ بنابراین:

\(\begin{array}{l}\left[ \begin{array}{l}9 - 3x\\1 - \frac{y}{3}\end{array} \right] = \left[ {\begin{array}{*{20}{c}} \circ \\ \circ \end{array}} \right] \Rightarrow \left\{ \begin{array}{l}9 - 3x = 0 \Rightarrow 9 = 3x \Rightarrow x = 3\\\\1 - \frac{y}{3} = 0 \Rightarrow 1 = \frac{y}{3} \Rightarrow y = 3\end{array} \right.\\\\ \Rightarrow x + y = 6\end{array}\)

\(\begin{array}{l}1)\,\,\left[ \begin{array}{l} - 2\\4\end{array} \right]\,\,\,\,\,\,y \to \,\,\,\,\,\,\left[ \begin{array}{l}2\\4\end{array} \right]\\\\2)\,\,\left[ \begin{array}{l}5\\ - 3\end{array} \right]\,\,\,\,\,\,\,x \to \,\,\,\,\left[ \begin{array}{l}5\\3\end{array} \right]\\\\3)\,\,\left[ \begin{array}{l} - 15\\ - 20\end{array} \right]\,\,\,\,\,\,\,\,\,y \to \,\,\,\,\,\left[ \begin{array}{l}15\\ - 20\end{array} \right]\\\\4)\,\,\left[ \begin{array}{l} - 15\\ - 20\end{array} \right]\,\,\,\,\,\,\,x \to \,\,\,\,\,\left[ \begin{array}{l} - 15\\20\end{array} \right]\\\\5)\,\,\left[ \begin{array}{l}15\\ - 20\end{array} \right]\,\,\,\,\,\,\,\left[ \begin{array}{l} \circ \\ \circ \end{array} \right] \to \,\,\,\,\left[ \begin{array}{l} - 15\\20\end{array} \right]\\\\6)\,\,\left[ \begin{array}{l} - 10\\ + 3\end{array} \right]\,\,\,\,\,\,\,\left[ \begin{array}{l} \circ \\ \circ \end{array} \right] \to \,\,\,\,\left[ \begin{array}{l} + 10\\ - 3\end{array} \right]\end{array}\)