Jawaban Soal Latihan 1 Rumus Trigonometri Sudut Rangkap

Soal Latihan 1

1. Diketahui Tan A = $\frac{1}{3}$ dan A sudut lancip. Tentukan nilai :
   1. Sin 2A
   2. Cos 2A
   3. Tan 2A
2. Diketahui Cos²A = $\frac{9}{10}$ untuk $0<2A<\frac{\pi }{2}$. Tentukan nilai :
   1. Sin 2A
   2. Cos 2A
   3. Tan 2A
3. Jika Tan x = $\frac{1}{2}$ dan Tan y = $\frac{1}{3}$. Hitunglah :
   1. Tan 2x
   2. Tan 2y
   3. Tan (2x + y)
   4. Tan (x + 2y)

Jawab :

1. Tan A = $\frac{1}{3}$, A sudut lancip
   
   
   
   Sin A = $\frac{1}{\sqrt{10}}$
   Cos A = $\frac{3}{\sqrt{10}}$
   
   1. Sin 2A = 2 Sin A Cos A
      $=2.\frac{1}{\sqrt{10}}.\frac{3}{\sqrt{10}}$
      $=\frac{6}{10}$
      $=\frac{3}{5}$
      
   2. Cos 2A = Cos²A - Sin²A
      $={\left(\left(,\frac{3}{\sqrt{10}},\right)\right)}^2-{\left(\left(,\frac{1}{\sqrt{10}},\right)\right)}^2$
      $=\frac{9}{10}-\frac{1}{10}$
      $=\frac{8}{10}$
      $=\frac{4}{5}$
      
   3. Tan 2A = $\frac{2\mathrm{Tan\; A}}{1-{\mathrm{Tan}}^{2}A}$
      $=\frac{2.\frac{1}{3}}{1-{\left(\left(,\frac{1}{3},\right)\right)}^2}$
      $=\frac{\frac{2}{3}}{1-\frac{1}{9}}$
      $=\frac{\frac{2}{3}}{\frac{8}{9}}$
      $=\frac{2}{3}.\frac{9}{8}$
      $=\frac{3}{4}$
      
2. Cos²A = $\frac{9}{10}$
   Cos A = $\frac{3}{\sqrt{10}}$
   Sin A = $\frac{1}{\sqrt{10}}$
   Tan A = $\frac{1}{3}$
   
   1. Sin 2A = 2 Sin A Cos A
      $=2.\frac{1}{\sqrt{10}}.\frac{3}{\sqrt{10}}$
      $=\frac{6}{10}$
      $=\frac{3}{5}$
      
   2. Cos 2A = Cos²A - Sin²A
      $={\left(\left(,\frac{3}{\sqrt{10}},\right)\right)}^2-{\left(\left(,\frac{1}{\sqrt{10}},\right)\right)}^2$
      $=\frac{9}{10}-\frac{1}{10}$
      $=\frac{8}{10}$
      $=\frac{4}{5}$
      
   3. Tan 2A = $\frac{2\mathrm{Tan\; A}}{1-{\mathrm{Tan}}^{2}A}$
      $=\frac{2.\frac{1}{3}}{1-{\left(\left(,\frac{1}{3},\right)\right)}^2}$
      $=\frac{\frac{2}{3}}{1-\frac{1}{9}}$
      $=\frac{\frac{2}{3}}{\frac{8}{9}}$
      $=\frac{2}{3}.\frac{9}{8}$
      $=\frac{3}{4}$
      
3. Tan x = $\frac{1}{2}$
   Tan y = $\frac{1}{3}$
   
   1. Tan 2x = $\frac{2\mathrm{Tan\; x}}{1-{\mathrm{Tan}}^2\mathrm{x}}$
      $=\frac{2.\frac{1}{2}}{1-{\left(\left(,\frac{1}{2},\right)\right)}^2}$
      $=\frac{1}{1-\frac{1}{4}}$
      $=\frac{1}{\frac{3}{4}}$
      $=\frac{4}{3}$
      
   2. Tan 2y = $\frac{2\mathrm{Tan\; y}}{1-{\mathrm{Tan}}^2\mathrm{y}}$
      $=\frac{2.\frac{1}{3}}{1-{\left(\left(,\frac{1}{3},\right)\right)}^2}$
      $=\frac{\frac{2}{3}}{1-\frac{1}{9}}$
      $=\frac{\frac{2}{3}}{\frac{8}{9}}$
      $=\frac{2}{3}.\frac{9}{8}$
      $=\frac{3}{4}$
      
   3. Tan (2x + y) = $\frac{\mathrm{Tan\; 2x}+\mathrm{Tan\; y}}{1-\mathrm{Tan\; 2x}.\mathrm{Tan\; y}}$
      $=\frac{\frac{4}{3}+\frac{1}{3}}{1-\frac{4}{3}.\frac{1}{3}}$
      $=\frac{\frac{5}{3}}{1-\frac{4}{9}}$
      $=\frac{\frac{5}{3}}{\frac{5}{9}}$
      $=\frac{5}{3}.\frac{9}{5}$
      $=3$
      
   4. Tan (x + 2y) = $\frac{\mathrm{Tan\; x}+\mathrm{Tan\; 2y}}{1-\mathrm{Tan\; x}.\mathrm{Tan\; 2y}}$
      $=\frac{\frac{1}{2}+\frac{3}{4}}{1-\frac{1}{2}.\frac{3}{4}}$
      $=\frac{\frac{2}{4}+\frac{3}{4}}{1-\frac{3}{8}}$
      $=\frac{\frac{5}{4}}{\frac{5}{8}}$
      $=\frac{5}{4}.\frac{8}{5}$
      $=2$
      

Jawaban Soal Latihan 1 Rumus Trigonometri Jumlah dan Selisih Dua Sudut

Soal Latihan 1

1. Diketahui Sin a = $\frac{1}{2}$, Cos b = $\frac{\sqrt{3}}{2}$, a dan b sudut lancip. Hitunglah nilai dari :
   1. Cos ( a + b )
   2. Cos ( a - b )
   3. Sin ( a + b )
   4. Sin ( a - b )
   5. Tan ( a + b )
   6. Tan ( a - b )
2. Diketahui Tan a = $-\frac{3}{4}$ dan Tan b = $\frac{7}{24}$, a sudut tumpul dan b sudut lancip. Hitunglah nilai dari :
   1. Cos ( a + b )
   2. Cos ( a - b )
   3. Sin ( a + b )
   4. Sin ( a - b )
   5. Tan ( a + b )
   6. Tan ( a - b )
3. Tentukan nilai dari :
   1. Sin 75º
   2. Cos 75º
   3. Tan 75º
   4. Tan 105º
   5. Cos 105º
   6. Cos 195º
4. Tentukan nilai dari :
   1. Sin 15º
   2. Sec 15º
   3. Cos 15º
   4. Tan 15º
5. Sederhanakanlah :
   1. Cos 200º . Cos 10º - Sin 200º . Sin 10º
   2. Cos ( $\frac{\pi }{2}$ + a ) . Cos a + Sin ( $\frac{\pi }{2}$ + a ) . Sin a
   3. $\frac{Sin\; 2a}{Sin\; a}+\frac{Cos\; 2a}{Cos\; a}$
   4. $\frac{Sin\; 160}{Sin\; 20}-\frac{Cos\; 160}{Cos\; 20}$

Jawab :

1. Sin a = $\frac{1}{2}$, a sudut lancip
   
   
   
   Cos a = $\frac{\sqrt{3}}{2}$
   Tan a = $\frac{1}{\sqrt{3}}$
   
   Cos b = $\frac{\sqrt{3}}{2}$, b sudut lancip
   
   
   
   Sin b = $\frac{1}{2}$
   Tan b = $\frac{1}{\sqrt{3}}$
   
   
   1. Cos ( a + b ) = Cos a . Cos b - Sin a . Sin b = $\frac{\sqrt{3}}{2}$ . $\frac{\sqrt{3}}{2}$ - $\frac{1}{2}$ . $\frac{1}{2}$ = $\frac{3}{4}$ - $\frac{1}{4}$ = $\frac{2}{4}$ = $\frac{1}{2}$
   2. Cos ( a - b ) = Cos a . Cos b + Sin a . Sin b = $\frac{\sqrt{3}}{2}$ . $\frac{\sqrt{3}}{2}$ + $\frac{1}{2}$ . $\frac{1}{2}$ = $\frac{3}{4}$ + $\frac{1}{4}$ = 1
   3. Sin ( a + b ) = Sin a . Cos b + Cos a . Sin b = $\frac{1}{2}$ . $\frac{\sqrt{3}}{2}$ + $\frac{\sqrt{3}}{2}$ . $\frac{1}{2}$ = $\frac{\sqrt{3}}{4}$ + $\frac{\sqrt{3}}{4}$ = $\frac{2\sqrt{3}}{4}$ = $\frac{1}{2}\sqrt{3}$
   4. Sin ( a - b ) = Sin a . Cos b - Cos a . Sin b = $\frac{1}{2}$ . $\frac{\sqrt{3}}{2}$ - $\frac{\sqrt{3}}{2}$ . $\frac{1}{2}$ = $\frac{\sqrt{3}}{4}$ - $\frac{\sqrt{3}}{4}$ = 0
   5. Tan ( a + b ) = $\frac{\mathrm{Tan\; a}+\mathrm{Tan\; b}}{1-\mathrm{Tan\; a}.\mathrm{Tan\; b}}$ = $\frac{\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}}}$ = $\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$ = $\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$ = $\frac{2}{\sqrt{3}}.\frac{3}{2}$ = $\frac{3}{\sqrt{3}}$ = $\frac{3}{\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}$ = $\frac{3\sqrt{3}}{3}$ = $\sqrt{3}$
   6. Tan ( a - b ) = $\frac{\mathrm{Tan\; a}-\mathrm{Tan\; b}}{1+\mathrm{Tan\; a}.\mathrm{Tan\; b}}$ = $\frac{\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}}}$ = $\frac{0}{1+\frac{1}{3}}$ = 0


2. Tan a = $-\frac{3}{4}$, a sudut tumpul
   
   
   
   Sin a = Sin (180 - a) = Sin a = $\frac{3}{5}$ (Dengan Rumus Trigonometri Sudut Berelasi)
   Cos a = Cos (180 - a) = - Cos a = $-\frac{4}{5}$ (Dengan Rumus Trigonometri Sudut Berelasi)
   
   Tan b = $\frac{7}{24}$, b sudut lancip
   
   
   
   Sin b = $\frac{7}{25}$
   Cos b = $\frac{24}{25}$
   
   
   1. Cos ( a + b ) = Cos a . Cos b - Sin a . Sin b = $-\frac{4}{5}$ . $\frac{24}{25}$ - $\frac{3}{5}$ . $\frac{7}{25}$ = $-\frac{96}{125}$ - $\frac{21}{125}$ = $-\frac{117}{125}$
   2. Cos ( a - b ) = Cos a . Cos b + Sin a . Sin b = $-\frac{4}{5}$ . $\frac{24}{25}$ + $\frac{3}{5}$ . $\frac{7}{25}$ = $-\frac{96}{125}$ + $\frac{21}{125}$ = $-\frac{75}{125}$ = $-\frac{3}{5}$
   3. Sin ( a + b ) = Sin a . Cos b + Cos a . Sin b = $\frac{3}{5}$ . $\frac{24}{25}$ + $-\frac{4}{5}$ . $\frac{7}{25}$ = $\frac{72}{125}$ - $\frac{28}{125}$ = $\frac{44}{125}$
   4. Sin ( a - b ) = Sin a . Cos b - Cos a . Sin b = $\frac{3}{5}$ . $\frac{24}{25}$ - $-\frac{4}{5}$ . $\frac{7}{25}$ = $\frac{72}{125}$ + $\frac{28}{125}$ = $\frac{100}{125}$ = $\frac{4}{5}$
   5. Tan ( a + b ) = $\frac{\mathrm{Tan\; a}+\mathrm{Tan\; b}}{1-\mathrm{Tan\; a}.\mathrm{Tan\; b}}$ = $\frac{-\frac{3}{4}+\frac{7}{24}}{1--\frac{3}{4}.\frac{7}{24}}$ = $\frac{-\frac{18}{24}+\frac{7}{24}}{1+\frac{21}{96}}$ = $\frac{-\frac{11}{24}}{\frac{117}{96}}$ = $-\frac{11}{24}.\frac{96}{117}$ = $-\frac{44}{117}$
   6. Tan ( a - b ) = $\frac{\mathrm{Tan\; a}-\mathrm{Tan\; b}}{1+\mathrm{Tan\; a}.\mathrm{Tan\; b}}$ = $\frac{-\frac{3}{4}-\frac{7}{24}}{1+-\frac{3}{4}.\frac{7}{24}}$ = $\frac{-\frac{18}{24}-\frac{7}{24}}{1-\frac{21}{96}}$ = $\frac{-\frac{25}{24}}{\frac{75}{96}}$ = $-\frac{25}{24}.\frac{96}{75}$ = $-\frac{4}{3}$


3. Dengan menggunakan Tabel Trigonometri Sudut - Sudut Istimewa
   1. Sin 75º = Sin (30º + 45º)
           = Sin 30º . Cos 45º + Cos 30º . Sin 45º
      $=\frac{1}{2}.\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}.\frac{1}{2}\sqrt{2}$
      $=\frac{1}{4}\sqrt{2}+\frac{1}{4}\sqrt{6}$
      $=\frac{1}{4}(\sqrt{2}+\sqrt{6})$
   2. Cos 75º = Cos (30º + 45º)
           = Cos 30º . Cos 45º - Sin 30º . Sin 45º
      $=\frac{1}{2}\sqrt{3}.\frac{1}{2}\sqrt{2}-\frac{1}{2}.\frac{1}{2}\sqrt{2}$
      $=\frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2}$
      $=\frac{1}{4}\left((\sqrt{6}-\sqrt{2})\right)$
   3. Tan 75º = Tan (30º + 45º)
           = $\frac{\mathrm{Tan\; 30°}+\mathrm{Tan\; 45°}}{1-\mathrm{Tan\; 30°}.\mathrm{Tan\; 45°}}$
           = $\frac{\frac{1}{3}\sqrt{3}+1}{1-\frac{1}{3}\sqrt{3}.1}$
           = $\frac{\frac{1}{3}\sqrt{3}+1}{1-\frac{1}{3}\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}$
           = $\frac{1+\sqrt{3}}{\sqrt{3}-1}.\frac{\sqrt{3}+1}{\sqrt{3}+1}$
           = $\frac{\sqrt{3}+1+3+\sqrt{3}}{3+\sqrt{3}-\sqrt{3}-1}$
           = $\frac{4+2\sqrt{3}}{2}$
           = $2+\sqrt{3}$
      
   4. Tan 105º = Tan (60º + 45º)
      $=\frac{\mathrm{Tan\; 60°}+\mathrm{Tan\; 45°}}{1-\mathrm{Tan\; 60°}.\mathrm{Tan\; 45°}}$
      $=\frac{\sqrt{3}+1}{1-\sqrt{3}.1}$
      $=\frac{1+\sqrt{3}}{1-\sqrt{3}}.\frac{1+\sqrt{3}}{1+\sqrt{3}}$
      $=\frac{1+\sqrt{3}+\sqrt{3}+3}{1+\sqrt{3}-\sqrt{3}+3}$
      $=\frac{4+2\sqrt{3}}{4}$
      $=1+\frac{1}{2}\sqrt{3}$
      
   5. Cos 105º = Cos (60º + 45º)
             = Cos 60º . Cos 45º - Sin 60º . Sin 45º
      $=\frac{1}{2}.\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{3}.\frac{1}{2}\sqrt{2}$
      $=\frac{1}{4}\sqrt{2}-\frac{1}{4}\sqrt{6}$
      $=\frac{1}{4}(\sqrt{2}-\sqrt{6})$
      
   6. Cos 195º = Cos (105º + 90º) = Cos 105º . Cos 90º - Sin 105º . Sin 90º
      Cos 105º = $\frac{1}{4}(\sqrt{2}-\sqrt{6})$
      Sin 105º = Sin (60º + 45º)
            = Sin 60º . Cos 45º + Cos 60º . Sin 45º
      $=\frac{1}{2}\sqrt{3}.\frac{1}{2}\sqrt{2}+\frac{1}{2}.\frac{1}{2}\sqrt{2}$
      $=\frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2}$
      $=\frac{1}{4}(\sqrt{6}+\sqrt{2})$
      Cos 195º = Cos (105º + 90º)
             = Cos 105º . Cos 90º - Sin 105º . Sin 90º
      $=\frac{1}{4}(\sqrt{2}-\sqrt{6}).0-\frac{1}{4}(\sqrt{6}+\sqrt{2}).1$
      $=0-\frac{1}{4}(\sqrt{6}+\sqrt{2})$
      $=-\frac{1}{4}(\sqrt{6}+\sqrt{2})$
      


4. Dengan menggunakan Tabel Trigonometri Sudut - Sudut Istimewa
   1. Sin 15º = Sin (45º - 30º)
           = Sin 45º . Cos 30º - Cos 45º . Sin 30º
      $=\frac{1}{2}\sqrt{2}.\frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}.\frac{1}{2}$
      $=\frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2}$
      $=\frac{1}{4}(\sqrt{6}-\sqrt{2})$
      
   2. Sec 15º = $\frac{1}{Cos\; 15°}$ (Dengan menggunakan Rumus Identitas Trigonometri)
      Cos 15º = Cos (45º - 30º)
           = Cos 45º . Cos 30º + Sin 45º . Sin 30º
      $=\frac{1}{2}\sqrt{2}.\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{2}.\frac{1}{2}$
      $=\frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2}$
      $=\frac{1}{4}(\sqrt{6}+\sqrt{2})$
      Sec 15º = $\frac{1}{Cos\; 15°}$
      $=\frac{1}{\frac{1}{4}(\sqrt{6}+\sqrt{2})}$
      $=\frac{4}{\sqrt{6}+\sqrt{2}}.\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$
      $=\frac{4\sqrt{6}-4\sqrt{2}}{6-\sqrt{12}+\sqrt{12}-2}$
      $=\frac{4\sqrt{6}-4\sqrt{2}}{4}$
      $=\sqrt{6}-\sqrt{2}$
      
   3. Cos 15º = $\frac{1}{4}(\sqrt{6}+\sqrt{2})$
      
   4. Tan 15º = Tan (45º - 30º)
      $=\frac{\mathrm{Tan\; 45°}-\mathrm{Tan\; 30°}}{1+\mathrm{Tan\; 45°}.\mathrm{Tan\; 30°}}$
      $=\frac{1-\frac{1}{3}\sqrt{3}}{1+1.\frac{1}{3}\sqrt{3}}$
      $=\frac{1-\frac{1}{3}\sqrt{3}}{1+\frac{1}{3}\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}$
      $=\frac{\sqrt{3}-1}{\sqrt{3}+1}.\frac{\sqrt{3}-1}{\sqrt{3}-1}$
      $=\frac{3-\sqrt{3}-\sqrt{3}+1}{3-\sqrt{3}+\sqrt{3}-1}$
      $=\frac{4-2\sqrt{3}}{2}$
      $=2-\sqrt{3}$
      


5. 1. Cos 200º . Cos 10º - Sin 200º . Sin 10º = Cos (200º + 10º) = Cos 210º
   2. Cos ( $\frac{\pi }{2}$ + a ) . Cos a + Sin ( $\frac{\pi }{2}$ + a ) . Sin a = Cos { ($\frac{\pi }{2}$ + a) - a } = Cos $\frac{\pi }{2}$
   3. $\frac{Sin\; 2a}{Sin\; a}+\frac{Cos\; 2a}{Cos\; a}=\frac{\mathrm{Sin\; 2a}.\mathrm{Cos\; a}+\mathrm{Cos\; 2a}.\mathrm{Sin\; a}}{\mathrm{Sin\; a}.\mathrm{Cos\; a}}=\frac{\mathrm{Sin}(\mathrm{2a}+\mathrm{a})}{\mathrm{Sin\; a}.\mathrm{Cos\; a}}=\frac{\mathrm{Sin\; 3a}}{\mathrm{Sin\; a}.\mathrm{Cos\; a}}$
   4. $\frac{Sin\; 160}{Sin\; 20}-\frac{Cos\; 160}{Cos\; 20}=\frac{\mathrm{Sin\; 160}.\mathrm{Cos\; 20}-\mathrm{Cos\; 160}.\mathrm{Sin\; 20}}{\mathrm{Sin\; 20}.\mathrm{Cos\; 20}}=\frac{\mathrm{Sin}(160-20)}{\mathrm{Sin\; 20}.\mathrm{Cos\; 20}}=\frac{\mathrm{Sin}140}{\mathrm{Sin\; 20}.\mathrm{Cos\; 20}}$

Perkalian Matriks

1. Perkalian Matriks Dengan Bilangan Real
   Jika k adalah bilangan real dan A suatu matriks berordo m x n, maka k A adalah matriks berordo m x n yang elemen-elemennya adalah setiap elemen matriks A dikalikan dengan k.
   Dengan demikian,
   
   jika A = $\left(\left(,,\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array},,\right)\right)$, maka k A = k $\left(\left(,,\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}{\mathrm{ka}}_{11}& {\mathrm{ka}}_{12}\\ {\mathrm{ka}}_{21}& {\mathrm{ka}}_{22}\end{array},,\right)\right)$.
   
   Contoh :
   Diberikan matriks A = $\left(\left(,,\begin{array}{cc}3& 5\\ 6& 1\end{array},,\right)\right)$. Tentukan 2A dan -3A.
   Jawab :
   2A = 2 $\left(\left(,,\begin{array}{cc}3& 5\\ 6& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}6& 10\\ 12& 2\end{array},,\right)\right)$
   
   -3A = -3 $\left(\left(,,\begin{array}{cc}3& 5\\ 6& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}-9& -15\\ -18& -3\end{array},,\right)\right)$


2. Perkalian Matriks Dengan Matriks
   Dua buah matriks A dan B dapat dikalikan jika banyaknya kolom matriks A sama dengan banyaknya baris matriks B. Hasilnya adalah matriks baru yang elemen-elemennya diperoleh dengan menjumlahkan hasil kali elemen pada baris matriks A dengan elemen pada kolom matriks B.
   Jika matriks A_{m x n} dan B _{p x q} maka matriks A dan B dapat dikalikan jika n = p dan hasil kalinya adalah matriks C_{m x q}.
   
   Jika A = $\left(\left(,,\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array},,\right)\right)$ dan B = $\left(\left(,,\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array},,\right)\right)$, maka A x B = $\left(\left(,,\begin{array}{cc}{a}_{11}{b}_{11}+a_{12}{b}_{21}& a_{11}{b}_{12}+a_{12}{b}_{22}\\ a_{21}{b}_{11}+a_{22}{b}_{21}& a_{21}{b}_{12}+a_{22}{b}_{22}\end{array},,\right)\right)$.
   
   Contoh :
   
   A = $\left(\left(,,\begin{array}{cc}4& 5\\ 2& 1\end{array},,\right)\right)$, B = $\left(\left(,,\begin{array}{cc}1& 2\\ 3& 1\end{array},,\right)\right)$
   
   AB = $\left(\left(,,\begin{array}{cc}4& 5\\ 2& 1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}1& 2\\ 3& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}4.1+5.3& 4.2+5.1\\ 2.1+1.3& 2.2+1.1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}19& 13\\ 5& 5\end{array},,\right)\right)$
   
   BA = $\left(\left(,,\begin{array}{cc}1& 2\\ 3& 1\end{array},,\right)\right)$ $\left(\left(,,\begin{array}{cc}4& 5\\ 2& 1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}1.4+2.2& 1.5+2.1\\ 3.4+1.2& 3.5+1.1\end{array},,\right)\right)$ = $\left(\left(,,\begin{array}{cc}8& 7\\ 14& 16\end{array},,\right)\right)$