تحقق من فهمك

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

1)

$\underset{\_}{X}=\left[\begin{array}{cc}\begin{array}{rr}4& -1\\ 2& -2\end{array}& \end{array}\right], \underset{\_}{Y}=\left[\begin{array}{cc}\begin{array}{rr}\frac{1}{3}& -\frac{1}{6}\\ \frac{1}{3}& -\frac{2}{3}\end{array}& \end{array}\right]$

$\begin{array}{rl}\underset{\_}{X}\cdot \underset{\_}{Y}& =\left[\begin{array}{cc}\begin{array}{cc}4& -1\\ 2& -2\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}\frac{1}{3}& -\frac{1}{6}\\ \frac{1}{3}& -\frac{2}{3}\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{ll}\frac{4}{3}+\frac{-1}{3}& \frac{-4}{6}+\frac{2}{3}\\ \frac{2}{3}+\frac{-2}{3}& \frac{-2}{6}+\frac{4}{3}\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}\frac{4}{3}+\frac{-1}{3}& \frac{-4}{6}+\frac{4}{6}\\ \frac{2}{3}+\frac{-2}{3}& \frac{-2}{6}+\frac{8}{6}\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\end{array}$

$\begin{array}{rl}\underset{\_}{\mathit{Y}}\underset{\_}{\mathit{X}}& =\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{3}& -\frac{1}{6}\\ \frac{1}{3}& -\frac{2}{3}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}\mathbf{4}& -\mathbf{1}\\ \mathbf{2}& -2\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{ll}\frac{4}{3}+\frac{-2}{6}& \frac{-1}{3}+\frac{2}{6}\\ \frac{4}{3}+\frac{-4}{3}& \frac{-1}{3}+\frac{4}{3}\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}\frac{8}{6}+\frac{-2}{6}& \frac{-1}{3}+\frac{1}{3}\\ \frac{4}{3}+\frac{-4}{3}& \frac{-1}{3}+\frac{4}{3}\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\end{array}$

إن كل من المصفوفتين $X,Y$ نظير ضربي للأخرى.

أوجد النظير الضربي لكل مصفوفة فيما يأتي إن وجد:

2A)

$\underset{\_}{D}=\left[\begin{array}{cc}\begin{array}{cc}3& 7\\ 1& -4\end{array}& \end{array}\right]$

$\begin{array}{rl}& \left|\begin{array}{c}\begin{array}{cc}3& 7\\ 1& -4\end{array}\end{array}\right|=3(-4)-1\left(7\right)=-12-7=-19\ne 0\\ & {\underset{\_}{D}}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}\begin{array}{cc}d& -b\\ -c& a\end{array}& \end{array}\right]=\frac{1}{-19}\left[\begin{array}{cc}\begin{array}{cc}-4& -7\\ -1& 3\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}\frac{4}{19}& \frac{7}{19}\\ \frac{1}{19}& \frac{-3}{19}\end{array}& \end{array}\right]\end{array}$

2B)

$\underset{\_}{Q}=\left[\begin{array}{cc}\begin{array}{cc}2& 1\\ -4& 3\end{array}& \end{array}\right]$

$\begin{array}{rl}& \left|\begin{array}{c}\begin{array}{cc}2& 1\\ -4& 3\end{array}\end{array}\right|=6-(-4)=10\ne 0\\ & {\underset{\_}{Q}}^{-1}=\frac{1}{10}\left[\begin{array}{cc}\begin{array}{cc}3& -1\\ 4& 2\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}\frac{3}{10}& \frac{-1}{10}\\ \frac{4}{10}& \frac{2}{10}\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{3}{10}& \frac{-1}{10}\\ \frac{2}{5}& \frac{1}{5}\end{array}& \end{array}\right]\end{array}$