أوجد النظير الضربي لكل مصفوفة فيما يأتي، إن وجد :

$\left[\begin{array}{cc}\begin{array}{cc}2& 5\\ -5& -13\end{array}& \end{array}\right]$

$\begin{array}{rl}& \left|\begin{array}{c}\begin{array}{cc}2& 5\\ -5& -13\end{array}\end{array}\right|=\left(2\right)(-13)-(-5)\left(5\right)=-1\\ & {\underset{\_}{A}}^{-1}=\frac{1}{-1}\left[\begin{array}{cc}\begin{array}{cc}-13& -5\\ 5& 2\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}13& 5\\ -5& -2\end{array}& \end{array}\right]\end{array}$

حل مراجعة درس النظير الضربي للمصفوفة وأنظمة المعادلات الخطية

أوجد النظير الضربي لكل مصفوفة فيما يأتي، إن وجد :

21)

$\left[\begin{array}{cc}\begin{array}{ll}7& 4\\ 3& 2\end{array}& \end{array}\right]$

$\begin{array}{rl}\left|\begin{array}{c}\begin{array}{ll}7& 4\\ 3& 2\end{array}\end{array}\right|& =14-12=2\\ {\underset{\_}{A}}^{-1}& =\frac{1}{\left(7\right)\left(2\right)-\left(3\right)\left(4\right)}\left[\begin{array}{cc}\begin{array}{cc}2& -4\\ -3& 7\end{array}& \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}\begin{array}{cc}2& -4\\ -3& 7\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}1& -2\\ -\frac{3}{2}& \frac{7}{2}\end{array}& \end{array}\right]\end{array}$

22)

$\left[\begin{array}{cc}\begin{array}{cc}2& 5\\ -5& -13\end{array}& \end{array}\right]$

$\begin{array}{rl}& \left|\begin{array}{c}\begin{array}{cc}2& 5\\ -5& -13\end{array}\end{array}\right|=\left(2\right)(-13)-(-5)\left(5\right)=-1\\ & {\underset{\_}{A}}^{-1}=\frac{1}{-1}\left[\begin{array}{cc}\begin{array}{cc}-13& -5\\ 5& 2\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}13& 5\\ -5& -2\end{array}& \end{array}\right]\end{array}$

23)

$\left[\begin{array}{cc}\begin{array}{cc}6& -3\\ -8& 4\end{array}& \end{array}\right]$

$\left[\begin{array}{cc}\begin{array}{cc}6& -3\\ -8& 4\end{array}& \end{array}\right]=24-\left(24\right)=0$

لا يوجد نظير ضربي.

حل المعادلة المصفوفية فيما يأتي:

24)

$\left[\begin{array}{cc}\begin{array}{ll}5& 3\\ 3& 2\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{l}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}4\\ 0\end{array}& \end{array}\right]$

$\begin{array}{rl}& \left|\underset{\_}{C}\right|=\left|\begin{array}{c}\begin{array}{ll}5& 3\\ 3& 2\end{array}\end{array}\right|=10-9=1\\ & {\underset{\_}{A}}^{-1}=\frac{1}{1}\left[\begin{array}{cc}\begin{array}{cc}2& -3\\ -3& 5\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}2& -3\\ -3& 5\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}2& -3\\ -3& 5\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}5& 3\\ 3& 2\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}2& -3\\ -3& 5\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}4\\ 0\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}8\\ -12\end{array}& \end{array}\right]\\ & x=8\\ & y=-12\end{array}$

25)

$\left[\begin{array}{cc}\begin{array}{cc}3& -1\\ 1& 2\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{l}a\\ b\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}5\\ 4\end{array}& \end{array}\right]$

$\begin{array}{rl}& {\underset{\_}{A}}^{-1}=\frac{1}{\left(3\right)\left(2\right)-\left(1\right)(-1)}\left[\begin{array}{cc}\begin{array}{cc}2& 1\\ -1& 3\end{array}& \end{array}\right]=\frac{1}{7}\left[\begin{array}{cc}\begin{array}{cc}2& 1\\ -1& 3\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}\frac{2}{7}& \frac{1}{7}\\ -\frac{1}{7}& \frac{3}{7}\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}\frac{2}{7}& \frac{1}{7}\\ -\frac{1}{7}& \frac{3}{7}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}3& -1\\ 1& 2\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}a\\ b\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{2}{7}& \frac{1}{7}\\ -\frac{1}{7}& \frac{3}{7}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}5\\ 4\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}a\\ b\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \end{array}\right]\end{array}$

$\begin{array}{rl}& a=2\\ & b=1\end{array}$