تمارين (2-9)

(1)- إذا كانت:

$\mathrm{A}=\left[\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\end{array}\right] \mathrm{B}=\left[\begin{array}{r}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\end{array}\right] \mathrm{C}=\left[\begin{array}{r}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1\end{array}\right]$

فجد:

أ- $\mathrm{A} \mathrm{x} \mathrm{B}$

$\mathrm{A}\times \mathrm{B}=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\times \left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=\mathrm{C}$

ب- $\mathrm{A} \mathrm{x} \mathrm{C}$

$\mathrm{A}\times \mathrm{C}=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]=\mathrm{B}$

ج- $\mathrm{B} \mathrm{x} \mathrm{C}$

$\mathrm{B}\times \mathrm{C}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]=\mathrm{A}$

د- $\mathrm{B} \mathrm{x} \mathrm{A}$

$\mathrm{B}\times \mathrm{A}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]$

هـ- $\mathrm{C} \mathrm{x} \mathrm{A}$

$\mathrm{C}\times \mathrm{A}=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$

و- $\mathrm{C} \mathrm{x} \mathrm{B}$

$\mathrm{C}\times \mathrm{B}=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]$

ز- $(\mathrm{A}\times \mathrm{B})\times \mathrm{C}$

$(\mathrm{A}\times \mathrm{B})\times \mathrm{C}=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\mathrm{I}$

ح- $\mathrm{A}\times (\mathrm{B}\times \mathrm{C})$

$\mathrm{A}\times (\mathrm{B}\times \mathrm{C})=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\mathrm{I}$

(2)- إذا كانت A, B, C كما في التمرين السابق وكانت I مصفوفة الوحدة فأثبت أن:

أ- $\mathrm{A}\times \mathrm{B}=-(\mathrm{B}\times \mathrm{A})$

$$\begin{gathered} \left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]=-\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right] \\ \left[\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right] \\ -\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\end{array}\right]=-\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right] \end{gathered}$$

الطرف الأيسر يساوي الطرف الأيمن.

ب- ${\mathrm{A}}^2={\mathrm{C}}^2=\mathrm{I}$

$\begin{array}{rl}& {\mathrm{A}}^2=\mathrm{A}\times \mathrm{A}=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\mathrm{I}\\ & {\mathrm{C}}^2=\mathrm{C}\times \mathrm{C}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\mathrm{I}\end{array}$

ج- ${\mathrm{B}}^2=-\mathrm{I}$

$\begin{array}{rl}& {\mathrm{B}}^2=\mathrm{B}\times \mathrm{B}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\\ & -\mathrm{I}=-\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=\end{array}$

الطرف الأيسر يساوي الطرف الأيمن.

د- $(\mathrm{A}\times \mathrm{B})\times \mathrm{C}=\mathrm{A}\times (\mathrm{B}\times \mathrm{C})$

$\begin{array}{rl}& \left(\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\right)\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\mathrm{I}\\ & \left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\mathrm{I}\end{array}$

الطرف الأيسر يساوي الطرف الأيمن.

(3)- إذا كانت A مصفوفة 2 × 3 وB مصفوفة 3 × 3 وC مصفوفة 3 × 4 وD مصفوفة 2 × 3، فبين نوع كل من المصفوفات الآتية:

أ- $\mathrm{A} \mathrm{x} \mathrm{B}$

تكون المصفوفة بقدر صفوف A وأعمدة B أي 3×2.

ب- $\mathrm{D} \mathrm{x} \mathrm{A}$

تكون المصفوفة بقدر صفوف D وأعمدة A أي 3×3.

ج- $\mathrm{A} \mathrm{x} \mathrm{D}$

تكون المصفوفة بقدر صفوف A وأعمدة B أي 2×2.

د- $\mathrm{C} \mathrm{x} \mathrm{B}$

تكون المصفوفة بقدر صفوف C وأعمدة B أي 3×4.

هـ- $\mathrm{B} \mathrm{x} \mathrm{D}$

تكون المصفوفة بقدر صفوف B وأعمدة D أي 2×3.

و- $\mathrm{D}\times (\mathrm{A}\times \mathrm{B})$

تكون المصفوفة بقدر صفوف D وأعمدة (A×B) أي 3×3.

ز- $(\mathrm{C}\times \mathrm{B})\times \mathrm{D}$

تكون المصفوفة بقدر صفوف (C×B) وأعمدة D أي 2×4.

ح- $(\mathrm{A}\times \mathrm{D})\times \mathrm{A}$

تكون المصفوفة بقدر صفوف (D×A) وأعمدة A أي 3×3.

(4)- أجر عملية الضرب فيما يأتي، إن أمكن واذكر السبب في حالة تعذر إجراء عملية الضرب:

أ- $\left[\begin{array}{l}3\\ 1\\ 0\end{array}\right]\left[\begin{array}{ccc}2& -1& 2\end{array}\right]$

$\left[\begin{array}{l}3\\ 1\\ 0\end{array}\right]\left[\begin{array}{lll}2& -1& 2\end{array}\right]=\left[\begin{array}{ccc}6& -3& 6\\ 2& -1& 2\\ 0& 0& 0\end{array}\right]$

ب- $\left[\begin{array}{ccc}-2& 1& 6\end{array}\right]\left[\begin{array}{l}-5\\ 0\\ 3\end{array}\right]$

$\left[\begin{array}{lll}-2& 1& 6\end{array}\right]\left[\begin{array}{c}-5\\ 0\\ 3\end{array}\right]=\left[\begin{array}{l}10+0+18\end{array}\right]=\left[\begin{array}{ll}2& 8\end{array}\right]$

ج- $\left[\begin{array}{cc}-1& 3\\ 0& 1\\ 2& 4\end{array}\right]\left[\begin{array}{ccc}2& 0& 4\\ 1& 1& 1\end{array}\right]$

$\left[\begin{array}{c}-1 3\\ 0 1\\ 2 4\end{array}\right]\left[\begin{array}{ccc}2& 0& 4\\ 1& 1& 1\end{array}\right]=\left[\begin{array}{ccc}1& 3& -1\\ 1& 1& 1\\ 8& 4& 12\end{array}\right]$

د- $\left[\begin{array}{llll}2& -1& 8& 0\end{array}\right]\left[\begin{array}{l}2\\ 1\\ 4\end{array}\right]$

غير ممكنة لأن عدد أعمدة الأولى لا يساوي عدد صفوف الثانية.

هـ- $\left[\begin{array}{rrr}-1& -3& 2\\ 5& 3& 6\end{array}\right]\left[\begin{array}{ccc}9& 1& 0\\ 2& -4& 5\end{array}\right]$

غير ممكنة لأن عدد أعمدة الأولى لا يساوي عدد صفوف الثانية.

و- $\left[\begin{array}{cc}1& -2\\ -1& 1\end{array}\right]\left[\begin{array}{cccc}1& 2& 0& 3\\ 2& -1& 1& 0\end{array}\right]$

$\left[\begin{array}{cc}1& -2\\ -1& 1\end{array}\right]\left[\begin{array}{cccc}1& 2& 0& 3\\ 2& -1& 1& 0\end{array}\right]=\left[\begin{array}{cccc}-3& 4& -2& 3\\ 1& -3& 1& -3\end{array}\right]$

ز- $\left[\begin{array}{ccc}1& -2& 3\\ 0& 1& 4\\ -3& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& -2& 3\\ 4& 5& 6\end{array}\right]$

غير ممكنة لأن عدد أعمدة الأولى لا يساوي عدد صفوف الثانية.

ح- $\left[\begin{array}{rrr}1& -2& 3\\ 4& 5& 6\end{array}\right]\left[\begin{array}{ccc}1& -2& 3\\ 0& 1& 4\\ -3& 2& 1\end{array}\right]$

$\left[\begin{array}{ccc}1& -2& 3\\ 4& 5& 6\end{array}\right]\left[\begin{array}{ccc}1& -2& 3\\ 0& 1& 4\\ -3& 2& 1\end{array}\right]=\left[\begin{array}{ccc}-8& 2& -2\\ -14& 9& 38\end{array}\right]$

(5)- إذا كانت:

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

بين صحة أو خطأ كل من العبارات الآتية مع ذكر السبب:

أ- $\mathrm{A}\times (\mathrm{B}+\mathrm{C})=\mathrm{A}\times \mathrm{B}+\mathrm{A}\times \mathrm{C}$

$$\begin{gathered} \mathrm{L}.\mathrm{S}=\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right](\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]+\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right])=\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{ll}0& 2\\ 0& 0\end{array}\right]=\left[\begin{array}{ll}0& 2\\ 0& 2\end{array}\right] \\ \mathrm{R}.\mathrm{S}=\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]+\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}-2& 1\\ -3& 1\end{array}\right]+\left[\begin{array}{ll}2& 1\\ 3& 1\end{array}\right]=\left[\begin{array}{ll}0& 2\\ 0& 2\end{array}\right] \end{gathered}$$

صحيحة لأن عملية الضرب توزيعية على الجمع.

ب- $(\mathrm{B}+\mathrm{C})\times \mathrm{A}=\mathrm{B}\times \mathrm{A}+\mathrm{C}\times \mathrm{A}$

$\begin{array}{rl}& \mathrm{L}.\mathrm{S}=(\left[\begin{array}{ll}0& 1\\ -1& 0\end{array}\right]+\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right])\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]=\left[\begin{array}{ll}0& 2\\ 0& 0\end{array}\right]\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]=\left[\begin{array}{ll}2& 6\\ 0& 0\end{array}\right]\\ & \mathrm{R}.\mathrm{S}=\left(\left[\begin{array}{ll}0& 1\\ -1& 0\end{array}\right]\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\right)+\left(\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right])=\left[\begin{array}{cc}1& 3\\ -1& -2\end{array}\right]+\left[\begin{array}{ll}1& 3\\ 1& 2\end{array}\right]=\left[\begin{array}{ll}2& 6\\ 0& 0\end{array}\right]\end{array}$

صحيحة لأن عملية الضرب توزيعية على الجمع.

ج- $\mathrm{A}\times (\mathrm{B}+\mathrm{A})=\mathrm{A}\times \mathrm{B}+{\mathrm{A}}^2$

$$\begin{gathered} \mathrm{L}.\mathrm{S}=\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]+\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right])=\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{ll}1& 3\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}1& 9\\ 1& 12\end{array}\right] \\ \mathrm{R}.\mathrm{S}=\left(\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\right)+\left(\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right])=\left[\begin{array}{cc}-2& 1\\ -3& 1\end{array}\right]+\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right]=\left[\begin{array}{cc}1& 9\\ 1& 12\end{array}\right] \end{gathered}$$

صحيحة لأن عملية الضرب توزيعية على الجمع.

د- $\mathrm{A}\times (\mathrm{B}+\mathrm{C})=\mathrm{B}\times \mathrm{A}+\mathrm{C}\times \mathrm{A}$

$\begin{array}{rl}& \mathrm{L}.\mathrm{S}=\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right](\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]+\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right])=\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{ll}0& 2\\ 0& 0\end{array}\right]=\left[\begin{array}{ll}0& 2\\ 0& 2\end{array}\right]\\ & \mathrm{R}.\mathrm{S}=\left(\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\right)+\left(\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\right)=\left[\begin{array}{cc}1& 3\\ -1& -2\end{array}\right]+\left[\begin{array}{ll}1& 3\\ 1& 2\end{array}\right]=\left[\begin{array}{ll}2& 6\\ 0& 0\end{array}\right]\end{array}$

خاطئة لأن عملية الضرب المصفوفات غير إبدالية.

هـ- $(\mathrm{A}\times \mathrm{B})\times \mathrm{C}=\mathrm{A}\times (\mathrm{B}\times \mathrm{C})$

$$\begin{gathered} \mathrm{L}.\mathrm{S}=\left(\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right],\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}-2& 1\\ -3& 1\end{array}\right]\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{ll}1& -2\\ 1& -3\end{array}\right] \\ \mathrm{R}.\mathrm{S}=\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left(\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\right)=\left[\begin{array}{ll}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{ll}1& -2\\ 1& -3\end{array}\right] \end{gathered}$$

صحيحة لأن عملية الضرب تجميعية.

(6)- إذا كانت:

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

فأثبت أن:

أ- ${\mathrm{A}}^2-2\mathrm{A}-3\mathrm{I}=0$

$\begin{array}{rl}\mathrm{L}.\mathrm{S}& =\left[\begin{array}{cc}-1& 2\\ 0& 3\end{array}\right]\left[\begin{array}{cc}-1& 2\\ 0& 3\end{array}\right]-2\left[\begin{array}{cc}-1& 2\\ 0& 3\end{array}\right]-3\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ & =\left[\begin{array}{ll}1& 4\\ 0& 9\end{array}\right]+\left[\begin{array}{cc}2& -4\\ 0& -6\end{array}\right]+\left[\begin{array}{cc}-3& 0\\ 0& -3\end{array}\right]=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]=0 \text{R.S}\end{array}$

ب- ${\mathrm{B}}^2-\mathrm{B}+\mathrm{I}=0$

$\begin{array}{rl}\mathrm{L}.\mathrm{S}& =\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]-2\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]+2\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ & =\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]+\left[\begin{array}{cc}-2& -2\\ 2& -2\end{array}\right]+\left[\begin{array}{ll}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]=0 \text{R.S}\end{array}$

ج- $\mathrm{A}\times \mathrm{B}\ne \mathrm{B}\times \mathrm{A}$

$$\begin{gathered} \mathrm{L}.\mathrm{S}=\left[\begin{array}{cc}-1& 2\\ 0& 3\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]=\left[\begin{array}{cc}-3& 1\\ -3& 3\end{array}\right] \\ \mathrm{R}.\mathrm{S}=\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]\left[\begin{array}{cc}-1& 2\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}-1& 5\\ 1& 1\end{array}\right] \end{gathered}$$

(7)- إذا كانت:

$\mathrm{B}=\left[\begin{array}{cc}\frac{1}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right] \mathrm{A}=\left[\begin{array}{rr}1& 1\\ -1& 1\end{array}\right]$

فأثبت أن $\mathrm{A}\times \mathrm{B}=\mathrm{B}\times \mathrm{A}=\mathrm{I}$ لاحظ أن $\mathrm{A}\times \mathrm{B}$ كل منها النظير الضربي للآخر.

$$\begin{gathered} \mathrm{L}.\mathrm{S}=\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\mathrm{I} \\ \mathrm{R}.\mathrm{S}=\left[\begin{array}{cc}\frac{1}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\mathrm{I} \end{gathered}$$