جد النظير الضربي لكل من المصفوفات الآتية كلما أمكن ذلك: $\left[\begin{array}{ll}4& 0\\ 0& 2\end{array}\right]$

$\begin{array}{rl}& \mathrm{\Delta }=\left|\begin{array}{l}4 0\\ 0 2\end{array}\right|\to \mathrm{\Delta }=4\times 2-0\times 0=8\ne 0\\ & {\mathrm{A}}^{-1}=\frac{1}{\mathrm{\Delta }}\left[\begin{array}{ll}2& 0\\ 0& 4\end{array}\right]=\frac{1}{8}\left[\begin{array}{ll}2& 0\\ 0& 4\end{array}\right]=\left[\begin{array}{ll}\frac{2}{8}& \frac{0}{8}\\ 0& \frac{4}{8}\end{array}\right]=\left[\begin{array}{ll}\frac{1}{4}& 0\\ 0& \frac{1}{2}\end{array}\right]\end{array}$

تمارين (3-9)

(1)- جد النظير الضربي لكل من المصفوفات الآتية كلما أمكن ذلك:

أ- $\left[\begin{array}{ll}4& 0\\ 0& 2\end{array}\right]$

$\begin{array}{rl}& \mathrm{\Delta }=\left|\begin{array}{l}4 0\\ 0 2\end{array}\right|\to \mathrm{\Delta }=4\times 2-0\times 0=8\ne 0\\ & {\mathrm{A}}^{-1}=\frac{1}{\mathrm{\Delta }}\left[\begin{array}{ll}2& 0\\ 0& 4\end{array}\right]=\frac{1}{8}\left[\begin{array}{ll}2& 0\\ 0& 4\end{array}\right]=\left[\begin{array}{ll}\frac{2}{8}& \frac{0}{8}\\ 0& \frac{4}{8}\end{array}\right]=\left[\begin{array}{ll}\frac{1}{4}& 0\\ 0& \frac{1}{2}\end{array}\right]\end{array}$

ب- $\left[\begin{array}{cc}-3& 9\\ 3& -6\end{array}\right]$

$$\begin{gathered} \mathrm{\Delta }\left|\begin{array}{cc}-3& 9\\ 3& -6\end{array}\right|\to \mathrm{\Delta }=(-3)(-6)-3\times 9=18-27=-9\ne 0 \\ {\mathrm{A}}^{-1}=\frac{1}{-9}\left[\begin{array}{cc}-6& -9\\ -3& -3\end{array}\right]=\left[\begin{array}{ll}\frac{{\displaystyle 2}}{{\displaystyle 3}}& 1\\ \frac{{\displaystyle 1}}{{\displaystyle 3}}& \frac{{\displaystyle 1}}{{\displaystyle 3}}\end{array}\right]\begin{array}{}\\ \end{array} \end{gathered}$$

ج- $\left[\begin{array}{ll}3& 3\\ 3& 3\end{array}\right]$

$\mathrm{\Delta }=\left|\begin{array}{l}33\\ 33\end{array}\right|-3\times 3\to 3\times 3=0$

لا يوجد نظير ضربي.

د- $\left[\begin{array}{ll}1& 2\\ 2& 4\end{array}\right]$

$\mathrm{\Delta }=\left|\begin{array}{l}12\\ 24\end{array}\right|\to \mathrm{\Delta }=1\times 4-2\times 2=0$

لا يوجد نظير ضربي.

هـ- $\left[\begin{array}{ll}-1& 3\\ 0& 4\end{array}\right]$

$$\begin{gathered} \mathrm{\Delta }=\left|\begin{array}{cc}-1& 3\\ 0& 4\end{array}\right|=-1\times 4-3\times 0=-4\ne 0 \\ {\mathrm{A}}^{-1}=\frac{1}{-4}\left[\begin{array}{ll}4& -3\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}-1& \frac{3}{4}\\ 0& \frac{1}{4}\end{array}\right] \end{gathered}$$

و- $\left[\begin{array}{cc}3& 6\\ -1& 2\end{array}\right]$

$$\begin{gathered} \mathrm{\Delta }=\left|\begin{array}{cc}3& 6\\ -1& 2\end{array}\right|=3\times 2-(-1\times 6)=6+6=12\ne 0 \\ {\mathrm{A}}^{-1}=\frac{1}{1}\left[\begin{array}{ll}1& 1\\ 1& 2\end{array}\right]=\left[\begin{array}{ll}1& 1\\ 1& 2\end{array}\right] \end{gathered}$$

ز- $\left[\begin{array}{cr}2& -1\\ -1& 1\end{array}\right]$

ح- $\left[\begin{array}{ll}\mathrm{a}& 0\\ 0& \mathrm{b}\end{array}\right]$

$$\begin{gathered} \mathrm{\Delta }=\left|\begin{array}{ll}\mathrm{a}& 0\\ 0& \mathrm{b}\end{array}\right|=\mathrm{ab}-0=\mathrm{ab}\ne 0 \\ \begin{array}{rl}& \mathrm{a}\ne 0\\ & \mathrm{b}\ne 0\end{array} \\ {\mathrm{A}}^{-1}=\frac{1}{\mathrm{ab}}\left[\begin{array}{ll}\mathrm{b}& 0\\ 0& \mathrm{a}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{\mathrm{a}}& 0\\ 0& \frac{1}{\mathrm{b}}\end{array}\right] \end{gathered}$$

(2)- احسب قيم x التي تجعل كلاً من المصفوفات الآتية ليس لها نظير ضربي:

أ- $\left[\begin{array}{ll}\mathrm{x}& 2\\ 6& 3\end{array}\right]$

$\mathrm{\Delta }\left|\begin{array}{ll}\mathrm{x}& 2\\ 6& 3\end{array}\right|=3\mathrm{x}-12=0\to \mathrm{x}=4$

ب- $\left[\begin{array}{ll}9& \mathrm{x}\\ & \mathrm{x}\end{array}\right]$

$\mathrm{\Delta }=\left|\begin{array}{l}9 \mathrm{x}\\ 4 \mathrm{x}\end{array}\right|=9\mathrm{x}-4\mathrm{x}=0-5\mathrm{x}=0\to \mathrm{x}=0$

ج- $\left[\begin{array}{cc}\mathrm{x}& 4\\ 2& {\mathrm{x}}^{-2}\end{array}\right]$

$\begin{array}{rl}\mathrm{\Delta }=\left|\begin{array}{cc}4& \mathrm{x}\\ 2& {\mathrm{x}}^2\end{array}\right|& =\frac{4}{{\mathrm{x}}^2}-2\mathrm{x}=0\\ & 4-2{\mathrm{x}}^3=0\\ & 2-{\mathrm{x}}^3=0\\ & {\mathrm{x}}^3=2\\ \mathrm{x}& =\sqrt[3]{2}\end{array}$

د- $\left[\begin{array}{cc}{\mathrm{x}}^{-2}& 2\\ 1& {\mathrm{x}}^{-2}\end{array}\right]$

$\begin{array}{ll}\mathrm{\Delta }=\left|\begin{array}{cc}-2& 2\\ 1& {\mathrm{x}}^{-2}\end{array}\right|& =\frac{-2}{{\mathrm{x}}^2}-2=0\\ & \begin{array}{rl}& -2-{\mathrm{x}}^2=0\\ & 2{\mathrm{x}}^2=2\\ & {\mathrm{x}}^2=1\\ & \mathrm{x}=\pm 1\end{array}\end{array}$

(3)- إذا كانت:

$\mathrm{B}=\left[\begin{array}{ll}2& 0\\ 8& 3\end{array}\right] \mathrm{A}=\left[\begin{array}{ll}1& 2\\ 3& 8\end{array}\right]$

فأجب عما يلي:

أ- احسب كلاً من ${\mathrm{A}}^{-1} , {\mathrm{B}}^{-1}$

$$\begin{gathered} \begin{array}{rl}& \mathrm{A}=\left[\begin{array}{ll}1& 2\\ 3& 8\end{array}\right]\to \mathrm{\Delta }\left[\begin{array}{c}1 2\\ 3 8\end{array}\right]=1\times 8-2\times 3=8-6=2\ne 0\\ & {\mathrm{A}}^{-1}=\frac{1}{2}\left[\begin{array}{cc}8& -2\\ -3& 1\end{array}\right]=\left[\begin{array}{cc}4& -1\\ \frac{-3}{4}& \frac{1}{2}\end{array}\right]\end{array} \\ \begin{array}{rl}& \mathrm{B}=\left[\begin{array}{ll}2& 0\\ 8& 3\end{array}\right]\to \mathrm{\Delta }\mid \begin{array}{l}2 0\\ 8 3\end{array}]=6-0=6\ne 0\\ & {\mathrm{B}}^{-1}=\frac{1}{6}\left[\begin{array}{cc}3& 0\\ -8& 2\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ -\frac{3}{4}& \frac{1}{3}\end{array}\right]\end{array} \end{gathered}$$

ب- جد ناتج ${\mathrm{A}}^{-1}\times {\mathrm{B}}^{-1} , {\mathrm{B}}^{-1}\times {\mathrm{A}}^{-1}$

$$\begin{gathered} {\mathrm{A}}^{-1}\times {\mathrm{B}}^{-1}=\left[\begin{array}{cc}4& -1\\ -\frac{3}{2}& \frac{1}{2}\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& 0\\ -\frac{4}{3}& \frac{1}{3}\end{array}\right]=\left[\begin{array}{cc}\frac{10}{3}& -\frac{1}{3}\\ -\frac{17}{12}& \frac{1}{6}\end{array}\right] \\ {\mathrm{B}}^{-1}\times {\mathrm{A}}^{-1}=\left[\begin{array}{cc}\frac{1}{2}& 0\\ -\frac{4}{3}& \frac{1}{3}\end{array}\right]\left[\begin{array}{cc}4& -1\\ -\frac{3}{2}& \frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}2& -\frac{1}{2}\\ -\frac{35}{6}& \frac{3}{2}\end{array}\right] \end{gathered}$$

ج- جد ناتجهما $\mathrm{A}\times \mathrm{B} , (\mathrm{A}\times \mathrm{B}{)}^{-1}$

$$\begin{gathered} \begin{array}{rl}& \mathrm{A}\times \mathrm{B}=\left[\begin{array}{ll}1& 2\\ 3& 8\end{array}\right]\left[\begin{array}{ll}2& 0\\ 8& 3\end{array}\right]=\left[\begin{array}{cc}18& 6\\ 70& 24\end{array}\right]\\ & \mathrm{A}\times \mathrm{B}=\left[\begin{array}{ll}18& 6\\ 70& 24\end{array}\right]\to \mathrm{\Delta }\left|\begin{array}{cc}18& 6\\ 70& 24\end{array}\right|=18\times 24-6\times 70=12\ne 0\end{array} \\ (\mathrm{A}\times \mathrm{B}{)}^{-1}=\frac{1}{12}\left[\begin{array}{cc}24& -6\\ -70& 18\end{array}\right]=\left[\begin{array}{cc}2& -\frac{1}{2}\\ -\frac{35}{6}& \frac{3}{2}\end{array}\right] \end{gathered}$$

د- تحقق من أن ${\left({\mathrm{A}}^{-1}\right)}^{-1}=\mathrm{A}$

$$\begin{gathered} {\mathrm{A}}^{-1}=\left[\begin{array}{cc}4& -1\\ -\frac{3}{2}& \frac{1}{2}\end{array}\right]\to \mathrm{\Delta }\left|\begin{array}{cc}4& -1\\ \frac{-3}{2}& \frac{1}{2}\end{array}\right|=2-\frac{3}{2}=\frac{1}{2} \\ {\left({\mathrm{A}}^{-1}\right)}^{-1}=(1+\frac{1}{2})\left[\begin{array}{ll}\frac{1}{2}& 1\\ \frac{3}{2}& 4\end{array}\right]=1\times 2\left[\begin{array}{cc}\frac{1}{2}& 1\\ \frac{3}{2}& 4\end{array}\right]=\left[\begin{array}{ll}1& 2\\ 3& 8\end{array}\right]=\mathrm{A} \end{gathered}$$

(4)- حل نظام المعادلتين الآتيتين باستخدام المصفوفات ثم حقق النتائج:

$\begin{array}{rl}& 3\mathrm{x}-4\mathrm{y}=-5\\ & 3\mathrm{y}-5\mathrm{x}=1\end{array}$

نرتب المعادلتين حسب المعادلة القياسية $\mathrm{ax}+\mathrm{by}=\mathrm{k}$

$\begin{array}{rl}& 3\mathrm{x}-4\mathrm{y}=-5\\ & -5\mathrm{x}+3\mathrm{y}=1\end{array}$

$\mathrm{A}=\left[\begin{array}{cc}3& -4\\ -5& 3\end{array}\right]\mathrm{B}=\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{C}=\left[\begin{array}{c}-5\\ 1\end{array}\right]\mathrm{\Delta }=3\times 3-(-5\times -4)=9-20=-11\ne 0$

A لها نظير ويكون الحل $\mathrm{B}={\mathrm{A}}^{-1}\times \mathrm{C}$

$$\begin{gathered} {\mathrm{A}}^{-1}=\frac{1}{\mathrm{\Delta }}\left[\begin{array}{ll}3& 4\\ 5& 3\end{array}\right]=\frac{1}{-11}\left[\begin{array}{ll}3& 4\\ 5& 3\end{array}\right]\left[\begin{array}{cc}-\frac{3}{4}& \frac{-4}{11}\\ -\frac{5}{11}& \frac{-3}{11}\end{array}\right] \\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\left[\begin{array}{cc}\frac{-3}{11}& \frac{-4}{11}\\ \frac{-5}{11}& \frac{-3}{11}\end{array}\right]\left[\begin{array}{c}-5\\ 1\end{array}\right]=\left[\begin{array}{ll}\frac{15}{11}& -\frac{4}{11}\\ \frac{25}{11}& -\frac{3}{11}\end{array}\right]=\left[\begin{array}{l}1\\ 2\end{array}\right] \\ \mathrm{x}=1 , \mathrm{y}=2 \end{gathered}$$

التحقيق:

$$\begin{gathered} 3\times 1-4\times 2=-5 \\ -5\times 1+3\times 2=1 \end{gathered}$$