جواب تمرین صفحه 30 درس 1 هندسه دوازدهم (ماتریس و کاربردها)

\(\begin{array}{l}AB = \\\\\left[ {\begin{array}{*{20}{c}}1&2&{ - 3}\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 1}\\3\end{array}} \right] = \left[ { - 13} \right] = - 13\\\\ \Rightarrow \left| {AB} \right| = - 13\\\\BA = \\\\\left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 1}\\3\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}1&2&{ - 3}\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 4}&6\\{ - 1}&{ - 2}&3\\3&6&{ - 9}\end{array}} \right]\\\\ \Rightarrow \left| {BA} \right| = \left| {\begin{array}{*{20}{c}}{ - 2}&{ - 4}&6\\{ - 1}&{ - 2}&3\\3&6&{ - 9}\end{array}} \right|\;\begin{array}{*{20}{c}}{ - 2}&4\\{ - 1}&{ - 2}\\3&6\end{array}\\\\ \Rightarrow \left| {BA} \right| = \\\\\left( {\left( { - 2} \right)\left( { - 2} \right)\left( { - 9} \right) + \left( { - 4} \right) \times 3 \times 3 + 6\left( { - 1} \right) \times 6} \right)\\\\ - \left( {6\left( { - 2} \right) \times 3 - 2 \times 3 \times 6 - 4\left( { - 1} \right)\left( { - 9} \right)} \right) = \\\\ - 108 + 108 = 0\end{array}\)

\(\begin{array}{l}{A^2} = A \times A = \\\\\left[ {\begin{array}{*{20}{c}}{ - 2}&0&0\\0&{ - 3}&0\\1&0&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 2}&0&0\\0&{ - 3}&0\\1&0&{ - 5}\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}4&0&0\\0&9&0\\{ - 7}&0&{25}\end{array}} \right]\\\\ \Rightarrow \left| {{A^2}} \right| = \left| {\begin{array}{*{20}{c}}4&0&0\\0&9&0\\{ - 7}&0&{25}\end{array}} \right| = \\\\4 \times 9 \times 25 = 900\end{array}\)

\(\begin{array}{*{20}{l}}{A = \left[ {\begin{array}{*{20}{c}}{5\left| A \right|}&{\left| A \right|}\\5&{4{{\left| A \right|}^2}}\end{array}} \right]}\\{}\\{ \Rightarrow \left| A \right| = (5\left| A \right|)(4{{\left| A \right|}^2}) - 5\left| A \right|}\\{}\\{ \Rightarrow 20{{\left| A \right|}^3} - 6\left| A \right| = 0}\\{}\\{2\left| A \right|(10{{\left| A \right|}^2} - 3) = 0}\\{}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\left| A \right| = 0}\\{}\\{10{{\left| A \right|}^2} - 3 = 0 \Rightarrow \left| A \right| = \pm \frac{{\sqrt {30} }}{{10}}}\end{array}} \right.}\\{}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\left| A \right| = 0}\\{}\\{ \Rightarrow ({{\left| A \right|}^3} - 2) = - 2}\\{}\\{\left| A \right| = \frac{{\sqrt {30} }}{{10}}}\\{}\\{ \Rightarrow ({{\left| A \right|}^3} - 2) = \frac{{3\sqrt {30} }}{{100}} - 2}\\{}\\{\left| A \right| = - \frac{{\sqrt {30} }}{{10}}}\\{}\\{ \Rightarrow ({{\left| A \right|}^3} - 2) = - \frac{{3\sqrt {30} }}{{100}} - 2}\end{array}} \right.}\end{array}\)

\(\begin{array}{l}\left| A \right| = \\\\{( - 1)^{3 + 1}}d\left( {bc - bc} \right) + \\\\{( - 1)^{3 + 2}}e\left( {ac - ac} \right) + \\\\{( - 1)^{3 + 3}}f\left( {ab - ab} \right) = 0\end{array}\)

نتیجه : اگر درایه های دو سطر (یا دو ستون) یک ماتریس مربعی، نظیر به نظیر مساوی باشند، دترمینان آن ماتریس صفر است.

کافی است یک ماتریس قطری (یا مثلثی) بیابیم که حاصلضرب درایه های قطر اصلی آن 3 باشد؛ مثلاً :

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}3&0&0\\0&{ - 1}&0\\0&0&{ - 1}\end{array}} \right]\\\\ \Rightarrow \left| A \right| = 3 \times ( - 1) \times ( - 1) = 3\end{array}\)

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}4&3\\2&5\end{array}} \right]\\\\ \Rightarrow \left| A \right| = 20 - 6 = 14\\\\ \Rightarrow {A^{ - 1}} = \frac{1}{{14}}\left[ {\begin{array}{*{20}{c}}5&{ - 3}\\{ - 2}&4\end{array}} \right]\\\\ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\frac{5}{{14}}}&{\frac{{ - 3}}{{14}}}\\{\frac{{ - 2}}{{14}}}&{\frac{4}{{14}}}\end{array}} \right]\\\\B = \left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 3}\\5&{ - 1}\end{array}} \right]\\\\ \Rightarrow \left| B \right| = 2 + 15 = 17\\\\ \Rightarrow {B^{ - 1}} = \frac{1}{{17}}\left[ {\begin{array}{*{20}{c}}{ - 1}&3\\{ - 5}&{ - 2}\end{array}} \right]\\\\ \Rightarrow {B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\frac{{ - 1}}{{17}}}&{\frac{3}{{17}}}\\{\frac{{ - 5}}{{17}}}&{\frac{{ - 2}}{{17}}}\end{array}} \right]\\\\2{A^{ - 1}} - 3{B^{ - 1}} = \\\\\left[ {\begin{array}{*{20}{c}}{\frac{5}{7}}&{\frac{{ - 3}}{7}}\\{\frac{{ - 2}}{7}}&{\frac{4}{7}}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{\frac{{ - 3}}{{17}}}&{\frac{9}{{17}}}\\{\frac{{ - 15}}{{17}}}&{\frac{{ - 6}}{{17}}}\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}{\frac{{106}}{{119}}}&{ - \frac{{114}}{{119}}}\\{\frac{{71}}{{119}}}&{\frac{{110}}{{119}}}\end{array}} \right]\end{array}\)

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}5&2\\3&2\end{array}} \right]\\\\ \Rightarrow \left| A \right| = 10 - 6 = 4\\\\ \Rightarrow {A^{ - 1}} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 3}&5\end{array}} \right]\\\\ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&{ - \frac{1}{2}}\\{ - \frac{3}{4}}&{\frac{5}{4}}\end{array}} \right]\\\\ \Rightarrow \left| {{A^{ - 1}}} \right| = \frac{5}{8} - \frac{3}{8} = \frac{2}{8} = \frac{1}{4}\\\\ \Rightarrow \left| {{A^{ - 1}}} \right| = \frac{1}{{\left| A \right|}}\end{array}\)

الف

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right]\\\\ \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right|\;\begin{array}{*{20}{c}}a&b\\d&e\\g&h\end{array}\\\\ = \left( {aei + bfg + cdh} \right) - \left( {bdi + afh + ceg} \right)\\\\B = \left[ {\begin{array}{*{20}{c}}{ka}&{kb}&{kc}\\d&e&f\\g&h&i\end{array}} \right]\\\\ \Rightarrow \left| B \right| = \left| {\begin{array}{*{20}{c}}{ka}&{kb}&{kc}\\d&e&f\\g&h&i\end{array}} \right|\;\begin{array}{*{20}{c}}{ka}&{kb}\\d&e\\g&h\end{array} = \\\\\left( {kaei + kbfg + kcdh} \right) - \left( {kbdi + kafh + kceg} \right)\\\\ \Rightarrow \left| B \right| = k\left| A \right|\end{array}\)

ب

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\\\\ \Rightarrow \left| A \right| = ad - bc\\\\B = \left[ {\begin{array}{*{20}{c}}{ka}&{kb}\\c&d\end{array}} \right]\\\\ \Rightarrow \left| B \right| = kad - kbc\\\\ \Rightarrow \left| B \right| = k\left| A \right|\end{array}\)

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\\\\ \Rightarrow \left| A \right| = ad - bc\\\\kA = \left[ {\begin{array}{*{20}{c}}{ka}&{kb}\\{kc}&{kd}\end{array}} \right]\\\\ \Rightarrow \left| {kA} \right| = (ka)(kd) - (kb)(kc) = \\\\{k^2}ad - {k^2}bc\\\\ \Rightarrow \left| {kA} \right| = {k^2}\left| A \right|\end{array}\)

\(\begin{array}{l}\left| {{A_{3 \times 3}}} \right| = 5 \Rightarrow \\\\\left| {\left| A \right|A} \right| = \left| {5A} \right| = {5^3}\left| A \right| = \\\\{5^3} \times 5 = {5^4} = 625\end{array}\)

\(\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}3x - 5y = 1\\\end{array}\\{4x + 2y = 10}\end{array}} \right.}\\{}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{A = \left[ {\begin{array}{*{20}{c}}3&{ - 5}\\4&2\end{array}} \right]}\\{}\\{B = \left[ {\begin{array}{*{20}{c}}1\\{10}\end{array}} \right]}\end{array}} \right.}\\{}\\{ \Rightarrow \left| A \right| = 26 \Rightarrow {A^{ - 1}} = \frac{1}{{26}}\left[ {\begin{array}{*{20}{c}}2&5\\{ - 4}&3\end{array}} \right]}\\{}\\{ \Rightarrow X = {A^{ - 1}}B = \frac{1}{{26}}\left[ {\begin{array}{*{20}{c}}2&5\\{ - 4}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\{10}\end{array}} \right]}\\{}\\{ = \frac{1}{{26}}\left[ {\begin{array}{*{20}{c}}{52}\\{26}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right] \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x = 2}\\{}\\{y = 1}\end{array}} \right.}\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}kx + 3y = 4\\\\x - 2y = 3\end{array} \right.\\\\A = \left[ {\begin{array}{*{20}{c}}k&3\\1&{ - 2}\end{array}} \right]\\\\ \Rightarrow \left| A \right| \ne 0 \Rightarrow - 2k - 3 \ne 0\\\\ \Rightarrow 2k \ne - 3\\\\ \Rightarrow k \ne - \frac{3}{2}\end{array}\)

الف

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}3x - 5y = - 1\\\end{array}\\{2x + y = 8}\end{array}} \right.\\\\ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}3&{ - 5}\\2&1\end{array}} \right]\,\,\,\,,\,\,B = \,\left[ {\begin{array}{*{20}{c}}{ - 1}\\8\end{array}} \right]\\\\ \Rightarrow \left| A \right| = 3 \times 1 - ( - 5) \times 2 = 13\end{array}\)

دستگاه جواب منحصر به فرد دارد؛ جواب آن برابر است با:

\(\begin{array}{l} \Rightarrow {A^{ - 1}} = \frac{1}{{13}}\left[ {\begin{array}{*{20}{c}}1&5\\{ - 2}&3\end{array}} \right]\\\\ \Rightarrow X = {A^{ - 1}}B = \\\\\frac{1}{{13}}\left[ {\begin{array}{*{20}{c}}1&5\\{ - 2}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 1}\\8\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right] \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x = 2}\\{}\\{y = 1}\end{array}} \right.\end{array}\)

ب

\(\begin{array}{l}\left\{ \begin{array}{l}x + 3y = 5\\\\ - 2x - 6y = 1\end{array} \right.\\\\ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}1&3\\{ - 2}&{ - 6}\end{array}} \right]\,\,\,\,,\,\,B = \,\left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]\\\\ \Rightarrow \left| A \right| = 1 \times ( - 6) - 3 \times ( - 2) = 0\\\\ \Rightarrow - \frac{1}{2} = - \frac{3}{6} \ne \frac{5}{1}\end{array}\)

دستگاه هیچ جوابی ندارد

پ

\(\begin{array}{l}\left\{ \begin{array}{l} - 2x + 3y = 2\\\\4x - 6y = - 4\end{array} \right.\\\\ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}{ - 2}&3\\4&{ - 6}\end{array}} \right]\,\,\,\,,\,\,B = \,\left[ {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right]\\\\ \Rightarrow \left| A \right| = ( - 2) \times ( - 6) - 3 \times 4 = 0\\\\ \Rightarrow - \frac{2}{4} = - \frac{3}{6} = - \frac{2}{4}\end{array}\)

دستگاه بیشمار جواب دارد