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

1)

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

2)

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

3)

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

4)

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

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

5)

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

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

6)

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

7)

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

8)

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)

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

9)

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

10)

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

11)

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

12)

\(\left\{ \begin{array}{l}kx + 3y = 4\\x - 2y = 3\end{array} \right. \Rightarrow \frac{k}{1} \ne  - \frac{3}{2} \Rightarrow k \ne  - \frac{3}{2}\)

13)

الف)

\(\left\{ \begin{array}{l}3x - 5y =  - 1\\2x + y = 8\end{array} \right. \Rightarrow  - \frac{3}{2} \ne  - \frac{5}{1} \Rightarrow\)

دستگاه جواب منحصر به فرد دارد

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}3&{ - 5}\\2&1\end{array}} \right]\\B = \left[ {\begin{array}{*{20}{c}}{ - 1}\\8\end{array}} \right]\end{array} \right.\\\\ \Rightarrow \left| A \right| = 3 + 10 = 13\\\\ \Rightarrow {A^{ - 1}} = \frac{1}{{13}}\left[ {\begin{array}{*{20}{c}}1&5\\{ - 2}&3\end{array}} \right]\\\\ \Rightarrow X = {A^{ - 1}}B = \frac{1}{{13}}\left[ {\begin{array}{*{20}{c}}1&5\\{ - 2}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 1}\\8\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\end{array}\)

ب)

\(\left\{ \begin{array}{l}x + 3y = 5\\ - 2x - 6y = 1\end{array} \right. \Rightarrow  - \frac{1}{2} =  - \frac{3}{6} \ne \frac{5}{1} \Rightarrow\)

دستگاه هیچ جوابی ندارد

پ)

\(\left\{ \begin{array}{l} - 2x + 3y = 2\\4x - 6y =  - 4\end{array} \right. \Rightarrow  - \frac{2}{4} =  - \frac{3}{6} =  - \frac{2}{4} \Rightarrow\)

دستگاه بیشمار جواب دارد