جواب کار در کلاس صفحه 18 درس 1 هندسه دوازدهم (ماتریس و کاربردها)

الف

\(\begin{array}{l}\left\{ \begin{array}{l}A = {\left[ {\begin{array}{*{20}{c}}1&{ - 1}&4\\0&0&0\\1&2&3\end{array}} \right]_{\;3 \times 3}}\\\\B = {\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}1&2&3&{ - 1}\end{array}}\\{\begin{array}{*{20}{c}}{ - 1}&0&1&{ - 2}\end{array}}\\{\begin{array}{*{20}{c}}3&1&{ - 1}&0\end{array}}\end{array}} \right]_{\;3 \times 4}}\;\end{array} \right.\;\\\\ \Rightarrow \;A \times B = \\\\\left[ {\begin{array}{*{20}{c}}1&{ - 1}&4\\0&0&0\\1&2&3\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}1&2&3&{ - 1}\\{ - 1}&0&1&{ - 2}\\3&1&{ - 1}&0\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}{14}&6&{ - 2}&1\\0&0&0&0\\8&5&2&{ - 5}\end{array}} \right]\end{array}\)

ب

\(\begin{array}{l}\left\{ \begin{array}{l}A = {\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{\begin{array}{*{20}{c}}2\\3\end{array}}&{\begin{array}{*{20}{c}}1\\{ - 2}\end{array}}\end{array}} \right]_{\;3 \times 2}}\\\\B = {\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}2\\3\end{array}}&{\begin{array}{*{20}{c}}{ - 1}\\0\end{array}}&{\begin{array}{*{20}{c}}1\\5\end{array}}\end{array}} \right]_{\;2 \times 3}}\end{array} \right.\;\;\\\\ \Rightarrow \;A \times B = \\\\\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&1\\3&{ - 2}\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\3&0&5\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&{ - 4}\\7&{ - 2}&7\\0&{ - 3}&{ - 7}\end{array}} \right]\\\\B \times A = \\\\\left[ {\begin{array}{*{20}{c}}2&{ - 1}&1\\3&0&5\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&1\\3&{ - 2}\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}3&{ - 5}\\{18}&{ - 13}\end{array}} \right]\end{array}\)

پ

\(\begin{array}{l}\left\{ \begin{array}{l}A = {\left[ \begin{array}{l}1\\ - 1\\2\end{array} \right]_{\;3 \times 1}}\\\\B = {\left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]_{\;1 \times 3}}\;\end{array} \right.\\\\ \Rightarrow \;A \times B = \\\\\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\2\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}3&3&4\\{ - 2}&{ - 3}&{ - 4}\\4&6&8\end{array}} \right]\\\\B \times A = \\\\\left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\2\end{array}} \right] = \left[ 7 \right] = 7\end{array}\)

ت

\(\begin{array}{l}\left\{ \begin{array}{l}A = {\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}1&{ - 1}\end{array}}&0\\{\begin{array}{*{20}{c}}2&{ - 1}\end{array}}&{ - 1}\end{array}} \right]_{\;2 \times 3}}\\\\B = {\left[ {\begin{array}{*{20}{c}}{ - 1}&2&{ - 2}\\{ - 1}&2&{ - 2}\\{ - 1}&2&{ - 2}\end{array}} \right]_{\;3 \times 3}}\end{array} \right.\\\\A \times B = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}0\\0\end{array}}&{\begin{array}{*{20}{c}}0\\0\end{array}}\end{array}}&{\begin{array}{*{20}{c}}0\\0\end{array}}\end{array}} \right]\end{array}\)

چنین حکمی در مورد قسمت (ت) صدق نمی کند؛ زیرا همان طور که می بینیم هیچ کدام از ماتریس ها A و B برابر با ماتریس ¯¯¯¯O$\overline{O}$  نمی باشد.