جواب تمرین صفحه 20 درس 1 هندسه دوازدهم (ماتریس و کاربردها)

\(\begin{array}{*{20}{l}}{A = {{\left[ {{a_{ij}}} \right]}_{3 \times 4}} = \left\{ {\begin{array}{*{20}{l}}{7\quad \quad \quad i = j}\\{i + j\,\,\,\,\,\,\,\,\,\,\,i > j}\\{{i^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,i < j}\end{array}} \right.\quad }\\{}\\{ \Rightarrow \quad A = \left[ {\begin{array}{*{20}{c}}7&1&1&1\\3&7&4&4\\4&5&7&9\end{array}} \right]}\end{array}\)

\(\begin{array}{*{20}{l}}{\underline {\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}2x - y = 3\\\end{array}\\{2x + y = 5}\end{array}} \right.} }\\\begin{array}{l}\quad 4x = 8 \Rightarrow x = 2 \Rightarrow y = 1\\\end{array}\\\begin{array}{l}{\mkern 1mu} \;\;\; \Rightarrow z = - 2\\\end{array}\\{\;\;\;\; \Rightarrow x + y + z = 1}\end{array}\)

\(\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{A = \left[ {\begin{array}{*{20}{c}}2&3&0\\4&6&0\\0&0&0\end{array}} \right] \ne \bar O}\\{}\\{B = \left[ {\begin{array}{*{20}{c}}0&3&{ - 6}\\0&{ - 2}&4\\2&4&5\end{array}} \right] \ne \bar O}\end{array}} \right.\quad }\\{}\\{ \Rightarrow \quad A \times B = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = \bar O}\end{array}\)

مثال نقض اول :

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

مثال نقض دوم :

\(\begin{array}{*{20}{l}}{A = \left[ {\begin{array}{*{20}{c}}1&2&1\\1&0&{ - 1}\\1&{ - 2}&{ - 3}\end{array}} \right]}\\{}\\{B = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\0&1&0\\1&1&0\end{array}} \right]}\\{}\\{C = \left[ {\begin{array}{*{20}{c}}2&1&2\\{ - 1}&{ - 1}&{ - 1}\\2&3&1\end{array}} \right]}\\{}\\{ \Rightarrow \;\;\left\{ {\begin{array}{*{20}{l}}{AB = AC = \left[ {\begin{array}{*{20}{c}}2&2&1\\0&{ - 2}&1\\{ - 2}&{ - 6}&1\end{array}} \right]}\\{}\\{B \ne C}\end{array}} \right.}\end{array}\)

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\\\\ \Rightarrow {A^2} = \\\\\left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I\\\\{A^3} = A.{A^2} = \\\\\left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right] = A\\\\\left\{ \begin{array}{l}{A^7} = A^6 \cdot A = (A^2)^3 \cdot A = I^3 \cdot A = I \cdot A = A \\ \text{یا } \\ A = {A^3} = {A^5} = {A^7} = \cdots = \left[ {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right]\\\\{A^2} = {A^4} = {A^6} = \cdots = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I\end{array} \right.\end{array}\)

\(\begin{array}{*{20}{l}}{A \times B = \left[ {\begin{array}{*{20}{c}}4&a\\b&{ - 1}\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\3&2\end{array}} \right]}\\{}\\{ = \left[ {\begin{array}{*{20}{c}}{4 + 3a}&{ - 8 + 2a}\\{b - 3}&{ - 2b + 2}\end{array}} \right]}\\{}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{ - 8 + 2a = 0}\\{b - 3 = 0}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 4}\\{b = 3}\end{array}} \right.}\end{array}\)

\(\begin{array}{*{20}{l}}\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}0&1\\1&3\\2&1\end{array}} \right]\\\\B = \left[ {\begin{array}{*{20}{c}}2&1&0\\2&5&1\end{array}} \right]\quad \end{array}\\{}\\{ \Rightarrow \;\;\left\{ \begin{array}{l}A \times B = \left[ {\begin{array}{*{20}{c}}2&5&1\\8&{16}&3\\6&7&1\end{array}} \right]\\\\B \times A = \left[ {\begin{array}{*{20}{c}}1&5\\7&{18}\end{array}} \right]\end{array} \right.}\end{array}\)

\(\begin{array}{*{20}{l}}\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}{{r_1}}&0&0\\0&{{r_2}}&0\\0&0&{{r_3}}\end{array}} \right]\;\\\\B = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\end{array}\\{}\\{ \Rightarrow \;\;A \times B = \left[ {\begin{array}{*{20}{c}}{{r_1}{a_{11}}}&{{r_1}{a_{12}}}&{{r_1}{a_{13}}}\\{{r_2}{a_{21}}}&{{r_2}{a_{22}}}&{{r_2}{a_{23}}}\\{{r_3}{a_{31}}}&{{r_3}{a_{32}}}&{{r_3}{a_{33}}}\end{array}} \right]}\end{array}\)

نتیجه: اگر \(A\) ماتریسی قطری و \(B\) یک ماتریس مربعی هم مرتبه با \(A\) باشد. برای محاسبه \(A \times B\) کافی است درایه های قطر اصلی \(A\) را در درایه های سطرهای نظیر آن ها در ماتریس \(B\) ضرب کنیم. (یعنی هر سطر \(B\) در عنصر قطری متناظر از \(A\) ضرب می شود).

الف

اگر \(B = \left[ {\begin{array}{*{20}{c}}{{b_{11}}}&{{b_{12}}}&{{b_{13}}}\\{{b_{21}}}&{{b_{22}}}&{{b_{23}}}\\{{b_{31}}}&{{b_{32}}}&{{b_{33}}}\end{array}} \right]\) و \(A = \left[ {\begin{array}{*{20}{c}}r&0&0\\0&r&0\\0&0&r\end{array}} \right]\)، آنگاه :

\(\begin{array}{*{20}{l}}{A \times B = \left[ {\begin{array}{*{20}{c}}{r{a_{11}}}&{r{a_{12}}}&{r{a_{13}}}\\{r{a_{21}}}&{r{a_{22}}}&{r{a_{23}}}\\{r{a_{31}}}&{r{a_{32}}}&{r{a_{33}}}\end{array}} \right] = rB}\\{}\\{B \times A = \left[ {\begin{array}{*{20}{c}}{r{a_{11}}}&{r{a_{12}}}&{r{a_{13}}}\\{r{a_{21}}}&{r{a_{22}}}&{r{a_{23}}}\\{r{a_{31}}}&{r{a_{32}}}&{r{a_{33}}}\end{array}} \right] = rB}\\{}\\{ \Rightarrow A \times B = B \times A = rB}\end{array}\)

ب

بله؛ این تساوی همواره برقرار است.

الف

\(\begin{array}{*{20}{l}}{{{(A + B)}^2} = (A + B)(A + B) = }\\{}\\{(A + B)A + (A + B)B = }\\{}\\{{A^2} + BA + AB + {B^2}\; = }\\{}\\{{A^2} + AB + AB + {B^2} = }\\{}\\{{A^2} + 2AB + {B^2}}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{(A + B)(A - B) = }\\{}\\{(A + B)A - (A + B)B = }\\{}\\{{A^2} + BA - AB - {B^2} = }\\{}\\{{A^2} + AB - AB - {B^2} = }\\{}\\{{A^2} - {B^2}}\end{array}\)

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}{ - 2}&0&0\\0&3&0\\0&0&4\end{array}} \right]\\\\{A^2} = A \times A = \\\\\left[ {\begin{array}{*{20}{c}}{ - 2}&0&0\\0&3&0\\0&0&4\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}{ - 2}&0&0\\0&3&0\\0&0&4\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}4&0&0\\0&9&0\\0&0&{16}\end{array}} \right]\\\\ \Rightarrow {A^3} = A \times A \times A = \\\\\left[ {\begin{array}{*{20}{c}}4&0&0\\0&9&0\\0&0&{16}\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}{ - 2}&0&0\\0&3&0\\0&0&4\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}{ - 8}&0&0\\0&{27}&0\\0&0&{64}\end{array}} \right]\end{array}\)

نتیجه می گیریم که اگر \(A\) یک ماتریس قطری باشد، توان \(k\) ام آن (\(A^k\)) نیز یک ماتریس قطری است که هر یک از درایه های قطر اصلی آن برابر توان \(k\) ام درایه متناظر در ماتریس \(A\) است.

\(\begin{array}{l}A = {\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&0& \cdots &0\\0&{{a_{22}}}& \cdots &0\\ \vdots & \vdots & \ddots & \vdots \\0&0& \cdots &{{a_{nn}}}\end{array}} \right]_{n \times n}}\quad \\\\\mathop \Rightarrow \limits^{k \in \mathbb{N}} \;\;\;{A^k} = {\left[ {\begin{array}{*{20}{c}}{{a_{11}}^k}&0& \cdots &0\\0&{{a_{22}}^k}& \cdots &0\\ \vdots & \vdots & \ddots & \vdots \\0&0& \cdots &{{a_{nn}}^k}\end{array}} \right]_{n \times n}}\end{array}\)