تدرب وحل المسائل

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

9)

$\underset{\_}{K}=\left[\begin{array}{cc}\begin{array}{ll}1& 2\\ 3& 0\end{array}& \end{array}\right],\underset{\_}{L}=\left[\begin{array}{cc}\begin{array}{cc}0& 1\\ 2& -1\end{array}& \end{array}\right]$

$\underset{\_}{K}\cdot \underset{\_}{L}=\left[\begin{array}{cc}\begin{array}{cc}1& 2\\ 3& 0\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}0& 1\\ 2& -1\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}0+4& 1-2\\ 0+0& 3+0\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}4& -1\\ 0& 3\end{array}& \end{array}\right]\ne \underset{\_}{I}$

لا تمثل نظيراً ضربياً.

10)

$\underset{\_}{M}=\left[\begin{array}{cc}\begin{array}{ll}0& 2\\ 4& 5\end{array}& \end{array}\right],\underset{\_}{N}=\left[\begin{array}{cc}\begin{array}{ll}1& 1\\ 0& 0\end{array}& \end{array}\right]$

$\begin{array}{rl}& \underset{\_}{M}\cdot \underset{\_}{N}=\left[\begin{array}{cc}\begin{array}{ll}0& 2\\ 4& 5\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{ll}1& 1\\ 0& 0\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}0+0& 0+0\\ 4+0& 4+0\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}0& 0\\ 4& 4\end{array}& \end{array}\right]\\ & \underset{\_}{N}\cdot \underset{\_}{M}=\left[\begin{array}{cc}\begin{array}{ll}1& 1\\ 0& 0\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{ll}0& 2\\ 4& 5\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}0+4& 2+5\\ 0+0& 0+0\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}4& 7\\ 0& 0\end{array}& \end{array}\right]\end{array}$

لا تمثل نظيراً ضربياً.

11)

$\underset{\_}{R}=\left[\begin{array}{cc}\begin{array}{rr}\frac{1}{2}& -\frac{1}{4}\\ \frac{1}{4}& -\frac{1}{2}\end{array}& \end{array}\right],\underset{\_}{S}=\left[\begin{array}{cc}\begin{array}{ll}2& 4\\ 4& 2\end{array}& \end{array}\right]$

$\underset{\_}{R}\cdot \underset{\_}{S}=\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{2}& -\frac{1}{4}\\ \frac{1}{4}& -\frac{1}{2}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{ll}2& 4\\ 4& 2\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}1+(-1)& 2+(-\frac{1}{2})\\ \frac{1}{2}+(-2)& 1+(-1)\end{array}& \end{array}\right]\left[\begin{array}{cc}\begin{array}{cc}0& \frac{3}{2}\\ -\frac{3}{2}& 0\end{array}& \end{array}\right]\ne \underset{\_}{I}$

لا تمثل نظيراً ضربياً.

أوجد النظير الضربي لكل مصفوفة فيما يأتي إن وجد:

12)

$\underset{\_}{A}=\left[\begin{array}{cc}\begin{array}{ll}3& 0\\ 0& 2\end{array}& \end{array}\right]$

$\begin{array}{rl}& \left|\begin{array}{c}\begin{array}{ll}3& 0\\ 0& 2\end{array}\end{array}\right|=6-0=6\ne 0\\ & {\underset{\_}{P}}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}\begin{array}{cc}d& -b\\ -c& a\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{\left(3\right)\left(2\right)-\left(0\right)\left(0\right)}\left[\begin{array}{cc}\begin{array}{ll}2& 0\\ 0& 3\end{array}& \end{array}\right]=\frac{1}{6}\left[\begin{array}{cc}\begin{array}{ll}2& 0\\ 0& 3\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}\frac{1}{3}& 0\\ 0& \frac{1}{2}\end{array}& \end{array}\right]\end{array}$

13)

$\underset{\_}{A}=\left[\begin{array}{cc}\begin{array}{ll}2& 3\\ 3& 2\end{array}& \end{array}\right]$

${\underset{\_}{A}}^{-1}=\frac{1}{\left(2\right)\left(2\right)-\left(3\right)\left(3\right)}\left[\begin{array}{cc}\begin{array}{cc}2& -3\\ -3& 2\end{array}& \end{array}\right]=\frac{1}{-5}\left[\begin{array}{cc}\begin{array}{cc}2& -3\\ -3& 2\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}-\frac{2}{5}& \frac{3}{5}\\ \frac{3}{5}& -\frac{2}{5}\end{array}& \end{array}\right]$

14)

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

$\begin{array}{rl}& \underset{\_}{A}=\left|\begin{array}{c}\begin{array}{cc}1& -1\\ -6& -1\end{array}\end{array}\right|=-1-6=-7\\ & {\underset{\_}{A}}^{-1}=\frac{1}{-7}\left[\begin{array}{cc}\begin{array}{cc}-1& 1\\ 6& 1\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{7}& -\frac{1}{7}\\ -\frac{6}{7}& -\frac{1}{7}\end{array}& \end{array}\right]\end{array}$

15)

$\left[\begin{array}{cc}\begin{array}{cc}-5& -4\\ 4& 2\end{array}& \end{array}\right]$

$\begin{array}{rl}& \underset{\_}{A}=\left|\begin{array}{c}\begin{array}{cc}-5& -4\\ 4& 2\end{array}\end{array}\right|=-10+16=6\\ & {\underset{\_}{A}}^{-1}=\frac{1}{6}\left[\begin{array}{cc}\begin{array}{cc}2& 4\\ -4& -5\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{2}{6}& \frac{4}{6}\\ \frac{-4}{6}& \frac{-5}{6}\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{3}& \frac{2}{3}\\ -\frac{2}{3}& -\frac{5}{6}\end{array}& \end{array}\right]\end{array}$

16) مشتريات: دفع سعد 225 ريالاً لشراء 5 أقلام حبر جافة و6 أقلام حبر سائلة بينما دفع ماجد 120 ريالاً لشراء 3 أقلام جافة من النوع نفسه و3 أقلام سائلة من النوع نفسه فإذا كان سعر القلم الجاف x وسعر القلم السائل y فاستعمل معادلة مصفوفية لإيجاد سعر القلم الجاف.

$\begin{array}{rl}& \begin{array}{r}5x+6y=250\\ 3x+3y=120\end{array}\\ & \left|\underset{\_}{C}\right|=\left|\begin{array}{c}\begin{array}{cc}5& 6\\ 3& 3\end{array}\end{array}\right|=15-18=-3\\ & {\underset{\_}{A}}^{-1}=\frac{1}{-3}\left[\begin{array}{cc}\begin{array}{cc}3& -6\\ -3& 5\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}-1& 2\\ 1& -\frac{5}{3}\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}-1& 2\\ 1& -\frac{5}{3}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}5& 6\\ 3& 3\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}-1& 2\\ 1& -\frac{5}{3}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}250\\ 120\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}-5+6& -6+6\\ 5-5& 6-5\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-250+240\\ 250-200\end{array}& \end{array}\right]\end{array}$

$\left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{l}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-10\\ 50\end{array}& \end{array}\right]x=-10,y=50$

17) سكان: يبين الشكل الآتي نسبة الهجرة السنوية بين إحدى المدن وضواحيها.

اكتب مصفوفة تبين نسبة المهاجرين والباقين في كل من المدينة والضواحي.

$\begin{array}{r}\\ \text{المدينة}\\ \text{الضواحي}\end{array}\left[\begin{array}{cc}\begin{array}{ll}0.95& 0.03\\ 0.05& 0.97\end{array}& \end{array}\right]$

استعمل معادلة مصفوفية لحل كل نظام إن أمكن فيما يأتي:

18)

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

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{l}-x+y\\ -x+y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}4\\ -4\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}-1& 1\\ -1& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}4\\ -4\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{(-1)\left(1\right)-(-1)\left(1\right)}\left[\begin{array}{cc}\begin{array}{ll}1& -1\\ 1& -1\end{array}& \end{array}\right]=\frac{1}{0}\left[\begin{array}{cc}\begin{array}{ll}1& -1\\ 1& -1\end{array}& \end{array}\right]\end{array}$

ليس لها حل.

19)

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

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{c}-x+y\\ -2x+y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}3\\ 6\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}-1& 1\\ -2& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}3\\ 6\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{(-1)\left(1\right)-(-2)\left(1\right)}\left[\begin{array}{cc}\begin{array}{ll}1& -1\\ 2& -1\end{array}& \end{array}\right]=\frac{1}{1}\left[\begin{array}{cc}\begin{array}{ll}1& -1\\ 2& -1\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}1& -1\\ 2& -1\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}1& -1\\ 2& -1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{ll}-1& 1\\ -2& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{ll}1& -1\\ 2& -1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}3\\ 6\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-3\\ 0\end{array}& \end{array}\right]\\ & x=-3\\ & y=0\end{array}$

20)

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

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{c}x+y\\ -4x+y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}4\\ 9\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}1& 1\\ -4& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\begin{array}{c}4\\ 9\end{array}\\ & {\underset{\_}{A}}^{-1}=\frac{1}{\left(1\right)\left(1\right)-(-4)\left(1\right)}\left[\begin{array}{cc}\begin{array}{cc}1& -1\\ 4& 1\end{array}& \end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}\begin{array}{cc}1& -1\\ 4& 1\end{array}& \end{array}\right]\\ & =\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{5}& -\frac{1}{5}\\ \frac{4}{5}& \frac{1}{5}\end{array}& \end{array}\right]\\ & [\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{5}& -\frac{1}{5}\\ \frac{4}{5}& \frac{1}{5}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}1& 1\\ -4& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{5}& -\frac{1}{5}\\ \frac{4}{5}& \frac{1}{5}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}4\\ 9\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}\frac{1}{5}+\frac{4}{5}& \frac{1}{5}-\frac{1}{5}\\ \frac{4}{5}-\frac{4}{5}& \frac{4}{5}+\frac{1}{5}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}\frac{4}{5}-\frac{9}{5}\\ \frac{16}{5}+\frac{9}{5}\end{array}& \end{array}\right]\end{array}$

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

21)

$\begin{array}{rl}& 3x+y=3\\ & 5x+3y=6\end{array}$

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{c}3x+y\\ 5x+3y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}3\\ 6\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}3& 1\\ 5& 3\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}3\\ 6\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{9-5}\left[\begin{array}{cc}\begin{array}{cc}3& -1\\ -5& 3\end{array}& \end{array}\right]=\frac{1}{4}\left[\begin{array}{cc}\begin{array}{cc}3& -1\\ -5& 3\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{3}{4}& -\frac{1}{4}\\ -\frac{5}{4}& \frac{3}{4}\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}\frac{3}{4}& -\frac{1}{4}\\ -\frac{5}{4}& \frac{3}{4}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{ll}3& 1\\ 5& 3\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{3}{4}& -\frac{1}{4}\\ -\frac{5}{4}& \frac{3}{4}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}3\\ 6\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}\frac{3}{4}\\ \frac{3}{4}\end{array}& \end{array}\right]\end{array}$

$\begin{array}{rl}& x=\frac{3}{4}\\ & y=\frac{3}{4}\end{array}$

22)

$\begin{array}{rl}& \begin{array}{c}y-x=5\\ 2y-2x=8\end{array}\\ & -x+y=5\\ & -2x+2y=8\\ & \left[\begin{array}{cc}\begin{array}{c}-x+y\\ -2x+2y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}5\\ 8\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}-1& 1\\ -2& 2\end{array}& \end{array}\right]\left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}5\\ 8\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{-2+2}\left[\begin{array}{cc}\begin{array}{ll}2& -1\\ 2& -1\end{array}& \end{array}\right]=\frac{1}{0}\left[\begin{array}{cc}\begin{array}{ll}2& -1\\ 2& -1\end{array}& \end{array}\right]\end{array}$

ليس لها حل.

23)

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

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{cc}4& 2\\ 6& -3\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}6\\ 9\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{-12-12}\left[\begin{array}{cc}\begin{array}{cc}-3& -2\\ -6& 4\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{8}& \frac{1}{12}\\ \frac{1}{4}& -\frac{1}{6}\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}\frac{1}{8}& \frac{1}{12}\\ \frac{1}{4}& -\frac{1}{6}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}4& 2\\ 6& -3\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}\frac{1}{8}& \frac{1}{12}\\ \frac{1}{4}& -\frac{1}{6}\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}6\\ 9\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}\frac{3}{2}\\ 0\end{array}& \end{array}\right]x=\frac{3}{2},y=0\end{array}$

24)

$\begin{array}{rl}& 1.6y-0.2x=1\\ & 0.4y-0.1x=0.5\end{array}$

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{ll}-0.2& 1.6\\ -0.1& 0.4\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}1\\ 0.5\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{(-0.2)\left(0.4\right)-(-0.1)\left(1.6\right)}\left[\begin{array}{cc}\begin{array}{ll}0.4& -1.6\\ 0.1& -0.2\end{array}& \end{array}\right]=\frac{1}{0.08}\left[\begin{array}{cc}\begin{array}{ll}0.4& -1.6\\ 0.1& -0.2\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}\frac{0.4}{0.08}& \frac{-1.6}{0.08}\\ \frac{0.1}{0.08}& \frac{-0.2}{0.08}\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}5& -20\\ 1.25& -2.5\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}5& -20\\ 1.25& -2.5\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{ll}-0.2& 1.6\\ -0.1& 0.4\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{cc}5& -20\\ 1.25& -2.5\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}1\\ 0.5\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}-1+2& 8-8\\ -0.25+0.25& 2-1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}5-10\\ 1.25-1.25\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-5\\ 0\end{array}& \end{array}\right]\end{array}$

25)

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

$\left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{l}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}-30\\ -8\end{array}& \end{array}\right]x=-30,y=-8$

26)

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

$\begin{array}{rl}& \left[\begin{array}{cc}\begin{array}{cc}-4& 2\\ 4& -3\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}3\\ -6\end{array}& \end{array}\right]\\ & {\underset{\_}{A}}^{-1}=\frac{1}{(-4)(-3)-\left(4\right)\left(2\right)}\left[\begin{array}{cc}\begin{array}{ll}-3& -2\\ -4& -4\end{array}& \end{array}\right]=\frac{1}{4}\left[\begin{array}{cc}\begin{array}{ll}-3& -2\\ -4& -4\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{rr}-\frac{3}{4}& -\frac{1}{2}\\ -1& -1\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{rr}-\frac{3}{4}& -\frac{1}{2}\\ -1& -1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{cc}-4& 2\\ 4& -3\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{rr}-\frac{3}{4}& -\frac{1}{2}\\ -1& -1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}3\\ -6\end{array}& \end{array}\right]\\ & \left[\begin{array}{cc}\begin{array}{ll}1& 0\\ 0& 1\end{array}& \end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{c}x\\ y\end{array}& \end{array}\right]=\left[\begin{array}{cc}\begin{array}{c}\frac{3}{4}\\ 3\end{array}& \end{array}\right]\\ & x=\frac{3}{4},y=3\end{array}$