استعمل معادلة مصفوفية لحل كل نظام فيما يأتي:

$\begin{array}{rl}& -2x+y=9\\ & x+y=3\end{array}$

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{c}-2x+y\\ x+y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}9\\ 3\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}-2& 1\\ 1& 1\end{array}& \end{array}\right]\left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}9\\ 3\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{(-2)\left(1\right)-\left(1\right)\left(1\right)}\left[\begin{array}{cc}\begin{array}{cc}1& -1\\ -1& -2\end{array}& \end{array}\right]\\ & =\frac{1}{-3}\left[\begin{array}{cc}\begin{array}{cc}1& -1\\ -1& -2\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{3}\cdot 1& -\frac{1}{3}\cdot (-1)\\ -\frac{1}{3}\cdot (-1)& -\frac{1}{3}\cdot (-2)\end{array}& \end{array}\right]\end{array}$

$\begin{array}{rl}& =\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}-2& 1\\ 1& 1\end{array}& \end{array}\right]\left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}9\\ 3\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}\frac{2}{3}+\frac{1}{3}& -\frac{1}{3}+\frac{1}{3}\\ -\frac{2}{3}+\frac{2}{3}& \frac{1}{3}+\frac{2}{3}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-3+1\\ 3+2\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-2\\ 5\end{array}& \end{array}\right]\\ & x=-2\\ & y=5\end{array}$

حل اسئلة تأكد

حدد ما إذا كانت كل من المصفوفتين تمثل نظيراً ضربياً للأخرى أم لا فيما يأتي:

1)

$\underset{\_}{A}=\left[\begin{array}{cc}\begin{array}{cc}2& 1\\ -1& 0\end{array}& \end{array}\right] , \underset{\_}{B}=\left[\begin{array}{cc}\begin{array}{ll}1& 2\\ 2& 1\end{array}& \end{array}\right]$

$\begin{array}{rl}\underset{\_}{A}\cdot \underset{\_}{B}& =\left[\begin{array}{cc}\begin{array}{cc}2& 1\\ -1& 0\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}1& 2\\ 2& 1\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}2+2& 4+1\\ -1+0& -2+0\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}4& 5\\ -1& -2\end{array}& \end{array}\right]\ne \underset{\_}{I}\end{array}$

الإجابة لا.

2)

$F=\left[\begin{array}{cc}\begin{array}{cc}-1& 1\\ 0& -1\end{array}& \end{array}\right],G=\left[\begin{array}{cc}\begin{array}{cc}-1& -1\\ 0& -1\end{array}& \end{array}\right]$

$\begin{array}{rl}\underset{\_}{F}\cdot \underset{\_}{G}& =\left[\begin{array}{cc}\begin{array}{cc}-1& 1\\ 0& -1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}-1& -1\\ 0& -1\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}1+0& 1+(-1)\\ 0+0& 0+1\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]=\underset{\_}{I}\end{array}$

الإجابة نعم.

أوجد النظير الضربي لكل مصفوفة مما يأتي إن وجد:

3)

$\left[\begin{array}{cc}\begin{array}{cc}6& -3\\ -1& 0\end{array}& \end{array}\right]$

$\begin{array}{rl}\left|\begin{array}{c}\begin{array}{cc}6& -1\\ -3& 0\end{array}\end{array}\right|& =0-3=-3\ne 0\\ & \left[\begin{array}{cc}\begin{array}{cc}6& -3\\ -1& 0\end{array}& \end{array}\right]=\frac{1}{\left(6\right)\left(0\right)-(-1)(-3)}\left[\begin{array}{cc}\begin{array}{cc}0& -(-3)\\ -(-1)& 6\end{array}& \end{array}\right]\\ & =\frac{1}{-3}\left[\begin{array}{cc}\begin{array}{ll}0& 3\\ 1& 6\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{ll}0\cdot -\frac{1}{3}& 3\cdot -\frac{1}{3}\\ 1\cdot -\frac{1}{3}& 6\cdot -\frac{1}{3}\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}0& -1\\ -\frac{1}{3}& -2\end{array}& \end{array}\right]\end{array}$

4)

$\left[\begin{array}{cc}\begin{array}{cc}-3& 0\\ 5& 2\end{array}& \end{array}\right]$

$\begin{array}{rl}\left[\begin{array}{cc}\begin{array}{cc}-3& 0\\ 5& 2\end{array}& \end{array}\right]& =-6-0=-6\ne 0\\ A^{-1}& =\frac{1}{(-3)(-2)-\left(5\right)\left(0\right)}\left[\begin{array}{cc}\begin{array}{cc}2& 0\\ -5& -3\end{array}& \end{array}\right]\\ & =\frac{1}{-6}\left[\begin{array}{cc}\begin{array}{cc}2& 0\\ -5& -3\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}2\cdot -\frac{1}{6}& 0\cdot -\frac{1}{6}\\ -5\cdot -\frac{1}{6}& -6\cdot -\frac{1}{6}\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{3}& 0\\ \frac{5}{6}& 1\end{array}& \end{array}\right]\end{array}$

5) نقود: مع أحمد 25 قطعة نقدية من فئة أرباع وأنصاف الريالات بحيث تبلغ قيمة ما معه 5.8 ريالات فما عدد الأرباع والانصاف التي معه؟

نفرض أن x عدد النصف ريال y عدد ربع الريال.

$\begin{array}{rl}& \begin{array}{c}x+y=25\\ \frac{1}{2}x+\frac{1}{4}y=8.5\end{array}\\ & \left[\underset{\_}{C}\right]=\left|\begin{array}{c}\begin{array}{ll}1& 1\\ \frac{1}{2}& \frac{1}{4}\end{array}\end{array}\right|=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\\ & \left[\begin{array}{cc}\begin{array}{ll}1& 1\\ \frac{1}{2}& \frac{1}{4}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}25\\ 8.5\end{array}& \end{array}\right]\\ & A^{-1}=-\frac{1}{4}\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{4}& -1\\ -\frac{1}{2}& 1\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{16}& \frac{1}{4}\\ \frac{1}{8}& -\frac{1}{4}\end{array}& \end{array}\right]\end{array}$

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{16}& \frac{1}{4}\\ \frac{1}{8}& -\frac{1}{4}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}1& 1\\ \frac{1}{2}& \frac{1}{4}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{16}& \frac{1}{4}\\ \frac{1}{8}& -\frac{1}{4}\end{array}& \end{array}\right]\cdot \left[\begin{array}{c}25\\ 8.5\end{array}\begin{array}{cc}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}\frac{1}{16}& 0\\ 0& \frac{1}{16}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}\frac{9}{16}\\ 1\end{array}& \end{array}\right]\\ & x=\frac{9}{16}\cdot \frac{16}{1}=9\\ & y=1\cdot \frac{16}{1}=16\end{array}$

مع أحمد 16 ربعاً و9 أنصاف.

استعمل معادلة مصفوفية لحل كل نظام فيما يأتي:

6)

$\begin{array}{rl}& -2x+y=9\\ & x+y=3\end{array}$

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{c}-2x+y\\ x+y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}9\\ 3\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}-2& 1\\ 1& 1\end{array}& \end{array}\right]\left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}9\\ 3\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{(-2)\left(1\right)-\left(1\right)\left(1\right)}\left[\begin{array}{cc}\begin{array}{cc}1& -1\\ -1& -2\end{array}& \end{array}\right]\\ & =\frac{1}{-3}\left[\begin{array}{cc}\begin{array}{cc}1& -1\\ -1& -2\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{3}\cdot 1& -\frac{1}{3}\cdot (-1)\\ -\frac{1}{3}\cdot (-1)& -\frac{1}{3}\cdot (-2)\end{array}& \end{array}\right]\end{array}$

$\begin{array}{rl}& =\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}-2& 1\\ 1& 1\end{array}& \end{array}\right]\left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}9\\ 3\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}\frac{2}{3}+\frac{1}{3}& -\frac{1}{3}+\frac{1}{3}\\ -\frac{2}{3}+\frac{2}{3}& \frac{1}{3}+\frac{2}{3}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-3+1\\ 3+2\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-2\\ 5\end{array}& \end{array}\right]\\ & x=-2\\ & y=5\end{array}$

7)

$\begin{array}{rl}& 4x-2y=22\\ & 6x+9y=-3\end{array}$

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{c}4x-2y\\ 6x+9y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}22\\ -3\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}4& -2\\ 6& 9\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}22\\ -3\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{\left(4\right)\left(9\right)-\left(6\right)(-2)}\left[\begin{array}{cc}\begin{array}{cc}9& 2\\ -6& 4\end{array}& \end{array}\right]\\ & =\frac{1}{48}\left[\begin{array}{cc}\begin{array}{cc}9& 2\\ -6& 4\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{48}\cdot 9& \frac{1}{48}\cdot \left(2\right)\\ \frac{1}{48}\cdot (-6)& \frac{1}{48}\cdot \left(4\right)\end{array}& \end{array}\right]\end{array}$

$\begin{array}{rl}& =\left[\begin{array}{cc}\begin{array}{cc}\frac{3}{16}& \frac{1}{24}\\ -\frac{1}{8}& \frac{1}{12}\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}\frac{3}{16}& \frac{1}{24}\\ -\frac{1}{8}& \frac{1}{12}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}4& -2\\ 6& 9\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{3}{16}& \frac{1}{24}\\ -\frac{1}{8}& \frac{1}{12}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}22\\ -3\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}\frac{12}{16}+\frac{6}{24}& \frac{-6}{16}+\frac{9}{24}\\ -\frac{4}{8}+\frac{6}{12}& \frac{2}{8}+\frac{9}{12}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}4\\ -3\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}4\\ -3\end{array}& \end{array}\right]\\ & x=4\\ & y=-3\end{array}$

8)

$\begin{array}{rl}& -2x+y=-4\\ & 3x+y=1\end{array}$

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{c}-2x+y\\ 3x+y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-4\\ 1\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}-2& 1\\ 3& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-4\\ 1\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{(-2)\left(1\right)-\left(3\right)\left(1\right)}\left[\begin{array}{cc}\begin{array}{cc}1& -1\\ -3& -2\end{array}& \end{array}\right]=\frac{1}{-5}\left[\begin{array}{cc}\begin{array}{cc}1& -1\\ -3& -2\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{5}& \frac{1}{5}\\ \frac{3}{5}& \frac{2}{5}\end{array}& \end{array}\right]\end{array}$

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{5}& \frac{1}{5}\\ \frac{3}{5}& \frac{2}{5}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}-2& 1\\ 3& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}-\frac{1}{5}& \frac{1}{5}\\ \frac{3}{5}& \frac{2}{5}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}-4\\ 1\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}\frac{2}{5}+\frac{3}{5}& -\frac{1}{5}+\frac{1}{5}\\ -\frac{6}{5}+\frac{6}{5}& \frac{3}{5}+\frac{2}{5}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}\frac{4}{5}+\frac{1}{5}\\ -\frac{12}{5}+\frac{2}{5}\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}1\\ -2\end{array}& \end{array}\right]\\ & x=1\\ & y=-2\end{array}$