Contoh Soal Fungsi

$\begin{array}{l}\\ 1.& \text{Relasi berikut yang akan berupa fungsi adalah}....\\ & \begin{array}{l}\\ \text{a}.& f(x)=\sqrt{x}& & & \\ \text{b}.& f(x)=1-\sqrt{x}& \\ \text{c}.& f(x)=\sqrt{x}+1& \\ \text{d}.& f(x)=\sqrt{x}-1\\ \text{e}.& f(x)=\left|x\right|\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{|clc|}\hline \text{No}& \text{Fungsi}& \text{Grafik}\\ 1.\text{a}& f(x)=y=\sqrt{x}& y^2=x\\ 1.\text{b}& f(x)=y=1-\sqrt{x}& (1-y{)}^2=x\\ 1.\text{c}& f(x)=y=1+\sqrt{x}& (y-1{)}^2=x\\ 1.\text{d}& f(x)=y=\sqrt{x}-1& (y+1{)}^2=x\\ 1.\text{e}& f(x)=y=\left|x\right|& y=\{\begin{array}{ll}x& \text{ jika}x\ge 0\\ -x& \text{ jika }x<0\end{array}\\ \hline\end{array}\\ & \begin{array}{rl}& \text{Dengan prepeta yang berbeda}\\ & \text{akan menghasilkan peta yang}\\ & \text{berbeda pula (fungsi bijektif)}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Fungsi dari himpunan A ke himpunan B }\\ & \text{berikut termasuk jenis fungsi}\end{array}$.

$\begin{array}{l}\\ ..& \text{Relasi berikut yang akan berupa fungsi adalah}....\\ & \text{a}.\text{fungsi umum}\\ & \text{b}.\text{fungsi satu-satu, tetapi bukan fungsi pada}\\ & \text{c}.\text{fungsi pada, tetapi bukan fungsi satu-satu}\\ & \text{d}.\text{fungsi pada dan satu-satu}\\ & \text{e}.\text{tidak ada jawaban yang benar}\\ \\ & \text{Jawab}:\\ & \begin{array}{|ccl|}\hline \text{No}& \text{Keterangan}& \text{Alasan}\\ 2.\text{a}& \text{Sesuai}& \text{Sesuai definisi fungsi}\\ 2.\text{b}& \text{Salah}& \begin{array}{rl}& \text{Karena bukan fungsi }\\ & \text{satu-satu(fungsi injektif)}\\ & \text{walau benar dikatakan bukan}\\ & \text{fungsi pada (fungsi surjektif)}\end{array}\\ 2.\text{c}& \text{Salah}& \begin{array}{rl}& \text{Karena bukan fungsi }\\ & \text{pada(fungsi surjektif)}\\ & \text{walau benar dikatakan bukan }\\ & \text{fungsi satu-satu (fungsi injektif)}\end{array}\\ 2.\text{d}& \text{Salah}& \begin{array}{rl}& \text{Jelas bukan fungsi pada }\\ & \text{dan satu-satu(fungsi bijektif)}\end{array}\\ 2.\text{e}& \text{Salah}& \text{Tidak sesuai}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Himpunan pasangan terurut }\\ & \text{yang ditunjukkan oleh fungsi}\\ & f:x\mapsto 2-{(x+1)}^2\\ & \text{dari domain}\{-1,0,1,2\}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{(-1,2),(0,3),(1,5),(2,7)\}\\ \text{b}.& \{(-1,2),(0,1),(1,-2),(2,-7)\}\\ \text{c}.& \{(-1,1),(0,-1),(1,-4),(2,7)\}\\ \text{d}.& \{(-1,0),(0,3),(1,-2),(2,7)\}\\ \text{e}.& \{(-1,0),(0,-4),(1,5),(2,-7)\}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f:x& \mapsto 2-{(x+1)}^2\\ -1& \mapsto 2-{(-1+1)}^2=2-0=2& ....(-1,2)\\ 0& \mapsto 2-{(0+1)}^2=2-1=1& ....(0,1)\\ 1& \mapsto 2-{(1+1)}^2=2-4=-2& ....(1,-2)\\ 2& \mapsto 2-{(2+1)}^2=2-9=-7& ....(2,-7)\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Dari beberapa fungsi berikut yang }\\ & \text{merupakan fungsi genap adalah}....\\ & \begin{array}{l}\\ \text{a}.& f(x)=x^2+\left|x\right|-1\\ \text{b}.& f(x)=x^3-\left|x\right|+x\\ \text{c}.& f(x)=x\left|x\right|+x\\ \text{d}.& f(x)=\sqrt{x-1}\\ \text{e}.& f(x)=4-2x\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \\ \text{Suatu fungsi}& \text{dinamakan fungsi genap}\\ & \text{jika}f(x)=f(-x)\\ & \end{array}\\ & \begin{array}{|cllc|}\hline \text{No}& f(x)& f(-x)& \text{Keterangan}\\ 4.\text{a}& x^2+\left|x\right|-1& x^2+\left|x\right|-1& f(x)=f(-x)\\ 4.\text{b}& x^3-\left|x\right|+x& -x^3-\left|x\right|-x& f(x)\ne f(-x)\\ 4.\text{c}& x\left|x\right|+x& -x\left|x\right|-x& f(x)\ne f(-x)\\ 4.\text{d}& \sqrt{x-1}& \sqrt{-x-1}& f(x)\ne f(-x)\\ 4.\text{e}.& 4-2x& 4+2x& f(x)\ne f(-x)\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Diketahui himpunan}\\ & A=\left\{x|x\text{adalah faktor prima dari}16\right\}\\ & B=\left\{x|x\text{adalah faktor dari}16\right\}\\ & \text{Banyaknya pemetaan dari}AkeB\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1& & & \text{d}.& 25\\ \\ \text{b}.& 2& \text{c}.& 5& \text{e}.& 32\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}A& =\left\{x|x\text{adalah faktor prima dari}16\right\}\\ & =\left\{2\right\}\Rightarrow n(A)=1\\ B& =\left\{x|x\text{adalah faktor dari}16\right\}\\ & =\{1,2,4,8,16\}\Rightarrow n(B)=5\\ \text{B}& \text{anyaknya pemetaan dari}A\text{ke}B\text{adalah}:\\ & =n(B{)}^{n(A)}\\ & =5^1\\ & =5\end{array}\end{array}$

Lanjutan Materi Fungsi

$\text{F. Domain, Kodomain dan Range Fungsi}$

Suatu fungsi  $f$ dari himpunan $\text{A}$ ke himpunan $\text{B}$  dituliskan dengan bentuk  $f:\text{A}\to \text{B}$. Jika fungsi  $f$  memetakan  $x\in \text{A}$  ke  $y\in \text{B}$, maka dituliskan dengan  $f:x\to y$  atau  $f:x\to f(x)$.

Perhatikan gambar berikut sebagai ilustrasinya

• Himpunan  $\text{A}$  sebagai Domain/daerah asal/prapeta dari fungsi  $f$
• Himpunan  $\text{B}$  sebagai Kodomain/daerah kawan dari fungsi  $f$
• Himpunan semua bayangan (bagian dari peta) disebut sebagai Range/daerah hasil  dari fungsi  $f$.

$\text{CONTOH SOAL}$

1. Perhatikanlah gambar berikut

dan tentukanlah domain, kodomain, serta range fungsinya

$\begin{array}{rl}\text{Dari}& \text{ilustrasi di atas diperoleh bahwa}:\\ & \text{Domain}:D_{{}_f}=A=\{a,b,c,d\}\\ & \text{Kodomain}:K_{{}_f}=B=\{1,2,3,4,5\}\\ & \text{Range}:R_{{}_f}=\{1,2,3,5\}\subseteq B\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah domain dan range dari fungsi}\\ & f(x)=\sqrt{x^2-x}\\ \\ & \text{Jawab}:\\ & \begin{array}{|ll|}\hline \text{Domain}& \text{Range}\\ \begin{array}{rl}& \text{Kumpulan nilai}x\\ & \text{yang mungkin, yaitu:}\\ & x^2-x\ge 0\\ & x(x-1)\ge 0\\ & \text{dengan garis bilangan}\\ & \begin{array}{l}\\ +& +& +& -& -& -& -& +& +& +\\ & & 0& & & & & 1& & \end{array}\\ & \text{Jadi},D_{{}_f}=\{x|x\le 0\text{atau}x\ge 1,x\in \mathbb{R}\}\end{array}& \begin{array}{rl}& \text{Hasil akar pangkat 2}\\ & \text{tidak pernah negatif}\\ & \text{Jadi},R_{{}_f}=\{y|y\ge 0\}\\ & \\ & \\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah domain dari}\\ & \begin{array}{l}\\ \text{a}.f(x)=2x+3& \\ \text{b}.f(x)={\displaystyle \frac{2}{3x-15}}& \\ \text{c}.g(x)={\displaystyle \frac{x-1}{x^2-x-6}}& \\ \text{d}.g(x)=\sqrt{x^2-1}& \\ \text{e}.h(x)=\sqrt{3x+2}\\ \text{f}.h(x)=\sqrt{{\displaystyle \frac{x-1}{x^2-x-6}}}\\ \text{g}.k(x)={}^{{}^2}\log x^2-2x-15\\ \text{h}.k(x)={}^{{}^{{}^{\mathbf{\text{(x+2)}}}}}\log (x^2-2x-3)\end{array}\\ \\ & \text{Jawab}:\end{array}.$

$.\begin{array}{|l|}\hline \begin{array}{rl}\text{a}.f(x)& =2x+3\\ D_{{}_f}& =\{x|x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{b}.f(x)& ={\displaystyle \frac{2}{3x-15}}\\ & \text{supaya terdefinisi}\\ & \text{maka},\\ & 3x-15\ne 0\\ & x\ne 5,\text{sehingga}\\ D_{{}_f}& =\{x|x\ne 5,x\in \mathbb{R}\}\\ & \end{array}\\ \begin{array}{rl}\text{c}.g(x)& ={\displaystyle \frac{x-1}{x^2-x-6}}\\ & \text{supaya terdefinisi}\\ & \text{maka},\\ & x^2-x-6\ne 0\\ & x\ne 3\text{dan}x\ne -2,\\ & \text{sehingga}\\ D_{{}_g}& =\{x|x\ne 3\text{dan}x\ne -2,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{d}.g(x)& =\sqrt{x^2-1}\\ \text{ma}& \text{ka}{x}^2-1\ge 0\\ & (x+1)(x-1)\ge 0\\ D_{{}_g}& =\{x|x\le -1\text{atau}x\ge 1,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{e}.h(x)& =\sqrt{3x+2}\\ \text{ma}& \text{ka}3x+2\ge 0\\ & x\ge -{\displaystyle \frac{2}{3}}\\ D_{{}_h}& =\{x|x\ge -{\displaystyle \frac{2}{3},x\in \mathbb{R}}\}\end{array}\\ \begin{array}{rl}\text{f}.h(x)& =\sqrt{{\displaystyle \frac{x-1}{x^2-x-6}}}\\ & =\sqrt{{\displaystyle \frac{(x-1)}{(x-3)(x+2)}}}\\ & =\sqrt{{\displaystyle \frac{x-1}{(x-3)(x+2)}}}\\ & \text{maka},{\displaystyle \frac{x-1}{(x-3)(x+2)}\ge 0}\\ D_{{}_h}& =\{x|-2<x\le 1\text{atau}x>3,x\in \mathbb{R}\}\end{array}\\ \hline\end{array}$.

$.\begin{array}{|l|}\hline \begin{array}{rl}\text{g}.k(x)& ={}^{{}^{{}^2}}\log (x^2-2x-15)\\ \text{sya}& \text{rat}(x^2-2x-15)>0\\ & (x-5)(x+3)>0\\ D_{{}_k}& =\{x|x<-3\text{atau}x>5,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{h}.k(x)& ={}^{{}^{{}^{\mathbf{\text{(x+2)}}}}}\log (x^2-2x-3)\\ \text{sya}& \text{rat}1.\{\begin{array}{ll}(x+2)& >0\Rightarrow x>-2\\ (x+2)& \ne 0\Rightarrow x\ne -2\end{array}\\ & 2.(x^2-2x-3)>0\Rightarrow (x-3)(x+1)>0\\ D_{{}_k}& =\{x|-2<x<-1\text{atau}x>3,x\in \mathbb{R}\}\end{array}\\ \hline\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui bahwa 2 buah fungsi}\\ & f(x)=2x+1\text{dan}g(x)=\sqrt{1-x}\\ & \begin{array}{l}\\ \text{a}.(f+g)(x)& \text{c}.(f.g)(x)\\ \text{b}.(f-g)(x)& \text{d}.\left({\displaystyle \frac{f}{g}}\right)(x)\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{|l|}\hline \begin{array}{rl}\text{a}.(f+g)(x)& =f(x)+g(x)\\ & =(2x+1)+\sqrt{1-x}\\ D_{{}_{(f+g)}}& =\{x|x\le 1,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{b}.(f-g)(x)& =f(x)-g(x)\\ & =(2x+1)-\sqrt{1-x}\\ D_{{}_{(f-g)}}& =\{x|x\le 1,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{c}.(f.g)(x)& =f(x).g(x)\\ & =(2x+1)\sqrt{1-x}\\ & =\sqrt{(2x+1{)}^2(1-x)}\\ & (2x+1{)}^2(1-x)\ge 0\\ D_{{}_{(f.g)}}& =\{x|x\le 1,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{d}.\left({\displaystyle \frac{f}{g}}\right)(x)& =....................\\ & ....................\\ & ....................\end{array}\\ \hline\end{array}\end{array}$

DAFTAR PUSTAKA

1. Sodyarto. Nugroho, Maryanto. 2008. Matematika 2 untuk SMA dan MA Kelas XI Program IPA. Jakarta: Pusat Perbukuan Departemen Pendidikan Nasional.
2. Sunardi, Waluyo, S., Sutrisno, & Subagya. 2005. Matematika 2 untuk SMA Kelas 2 IPA. Jakarta: BUMI AKSARA.

Fungsi Pembangkit untuk Kombinasi (Pengayaan)

 $\text{F. Fungsi Pembangkit}$

$\begin{array}{rl}& \text{Pandang}\left(a_n\right)=\left(a_0,a_1,a_2,a_3,\cdots \right)\mathbf{\text{suatu barisan}}\\ & \text{Fungsi Pembangkit Biasa (FPB) dari barisan}\left(a_n\right)\\ & {\displaystyle \sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots }\\ & \underset{―}{\text{CONTOH 1}}\\ & \text{Diketahui}f(x)={\displaystyle \frac{1}{1-x}.\text{Tentukanlah barisan}\left(a_n\right)}\\ & \text{yang terkait pada FPB di atas}\\ & \underset{―}{\mathbf{\text{Jawab}}}\\ & \text{Diketahui bahwa}f(x)={\displaystyle \frac{1}{1-x}}\\ & \text{karena}{\displaystyle \frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+\cdots }\\ & \text{maka}{\displaystyle \frac{1}{1-x}\text{adalah fungsi pembangkit dari barisan}}\\ & \left(a_n\right)=(1,1,1,1,1,\cdots )\end{array}$

$\begin{array}{rl}& \underset{―}{\text{CONTOH 2}}\\ & f(x)=2+3x^2+{\displaystyle \frac{2}{1-4x}.\text{Tentukanlah barisan}\left(a_n\right)}\\ & \text{yang terkait pada FPB di atas}\\ & \underset{―}{\mathbf{\text{Jawab}}}\\ & \text{Diketahui bahwa}f(x)=2+3x^2+{\displaystyle \frac{2}{1-4x}}\\ & \begin{array}{rl}f(x)& =2+3x^2+2\left({\displaystyle \frac{1}{1-4x}}\right)\\ & =2+3x^2+2{\displaystyle \sum _{n=0}^{\mathrm{\infty }}(4x{)}^n}\\ & =2+3x^2+2{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{4}^n{x}^n}\\ & =2+3x^2+2(4^0+4^{1}x+4^2{x}^2+4^3{x}^3+{\displaystyle \sum _{n=4}^{\mathrm{\infty }}(4n{)}^n})\\ & =2+3x^2+28x+32x^2+128x^3+2{\displaystyle \sum _{n=4}^{\mathrm{\infty }}(4x{)}^n}\\ & =4+8x+35x^2+128x^3+2{\displaystyle \sum _{n=4}^{\mathrm{\infty }}(4x{)}^n}\end{array}\\ & \text{maka}f(x)\text{adalah fungsi pembangkit dari barisan}\\ & \left(a_n\right)=(4,8,35,128,\cdots ,2(4{)}^n)\end{array}$

$\text{G. Fungsi Pembangkit untuk Kombinasi}$

Perhatikanlah permasalahan berikut:

$\begin{array}{rl}& \text{Ada berapa cara jika ada 3 objek}a,b,\text{dan}c\\ & \text{diambil}k\text{objek dengan ketentuan bahwa}\\ & a\text{terambil maksimum 1},b\text{terambil maksimum}\\ & 2,c\text{terambil maksimum 1}\\ & \mathbf{\text{Pembahasan}}\\ & \text{Permasalahan di atas dapat dimodelkan}\\ & ((ax{)}^0+(ax{)}^1)((bx{)}^0+(bx{)}^1+(bx{)}^2)((cx{)}^0+(cx{)}^1)\\ & =(a^0{x}^0+a^1{x}^1)(b^0{x}^0+b^1{x}^1+b^2{x}^2)(c^0{x}^0+c^1{x}^1)\\ & =(1+ax)(1+bx+b^2{x}^2)(1+cx)\\ & =1+(a+b+c)x+(ab+ac+bc+b^2)x^2\\ & +(abc+a{b}^2+b^{2}c)x^3+a{b}^{2}c{x}^4\\ & \text{Sebagai misal Untuk 2 huruf, maka koefisien dari}{x}^2\\ & \text{yaitu}:ab,ac,bc,\text{dan}bb\text{ada 4 cara}.\\ & \text{Sehingga}\\ & \begin{array}{rl}(a+b+c)x& \text{ada 3 cara untuk 1 huruf}\\ (ab+ac+bc+b^2)x^2& \text{ada 4 cara untuk 2 huruf}\\ (abc+a{b}^2+b^{2}c)x^3& \text{ada 3 cara untuk 3 huruf}\\ a{b}^{2}c{x}^4& \text{ada 1 cara untuk 4 huruf}\end{array}\\ & \text{Selanjutnya coba kita bandingkan dengan FPB}.\\ & \text{FPB dari permasalahan di atas adalah}:\\ & f(x)=(1+x)(1+x+x^2)(1+x)\\ & =1-3x-4x^2-3x^3-x^4\end{array}$

$\begin{array}{rl}& \text{Sehingga secara UMUM}\\ & \text{Misalkan terdapat}p\text{sekian objek}.\\ & \text{Jika ada}{n}_1\text{objek pada tipe ke}-1,\\ & \text{dan ada}{n}_2\text{objek pada tipe ke}-2,\\ & \text{lalu ada}{n}_3\text{objek pada tipe ke}-3,\\ & ⋮\\ & dst\\ & ⋮\\ & \text{dan ada}{n}_p\text{objek pada tipe ke}-p,\\ & \text{Dan misalkan}{t}_r\text{menyatakan banyaknya}\\ & \text{cara mengambil}r\text{objek dengan membolehkan}\\ & \text{mengambil sembarang banyak objek pada tipa tipe},\\ & \text{fungsi pembangkit untuk}{t}_r\text{adalah}:\\ & f(x)={\displaystyle \sum t_r{x}^r,\text{di mana}}\\ & f(x)=(1+x+x^2+\cdots +x^{n_1})(1+x+x^2+\cdots +x^{n_2})\cdots (1+x+x^2+\cdots +x^{n_p})\\ & \text{Nantinya}{t}_r\text{adalah koefisien}{x}^r\text{dalam}f(x)\\ \\ & \begin{array}{rl}\text{Sela}& \text{njutnya dalam peruntukannya dibagi menjadi}\\ \text{a}.& \text{Fungsi Pembangkit Tanpa Pengulangan}\\ & f(x)=\underset{n\mathbf{\text{faktor}}}{\underbrace{(1+x)(1+x)(1+x)\cdots (1+x)}}\\ & f(x)=(1+x{)}^n={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^r(\text{teorema binom})}\\ \text{b}.& \text{Fungsi Pembangkit Dengan Pengulangan}\\ & f(x)=\underset{n\mathbf{\text{faktor}}}{\underbrace{(1+x+x^2+\cdots )(1+x+x^2+\cdots )\cdots (1+x+x^2+\cdots )}}\\ & f(x)=(1+x+x^2+\cdots {)}^n\\ & ={\left({\displaystyle \frac{1}{1-x}}\right)}^n\\ & =(1-x{)}^{-n}\\ & ={\displaystyle \sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}-n\\ r\end{array}\right)(-1{)}^r{x}^r(\text{teorema binom})}\\ & \text{Selanjutnya, untuk}r>0\text{koefisien}{x}^r\\ & \text{pada}f(x)\text{di atas adalah}:\\ & \end{array}\end{array}$

$\begin{array}{rl}\left(\begin{array}{c}-n\\ r\end{array}\right)(-1{)}^r& ={\displaystyle \frac{(-n)(-n-1)\cdots (-n-r+1)(-n-r)!}{r!(-n-r)!}(-1{)}^r}\\ & ={\displaystyle \frac{(-n)(-n-1)\cdots (-n-r+1)}{r!}(-1{)}^r}\\ & ={\displaystyle \frac{n(n+1)\cdots (n+r-1)}{r!}}\\ & ={\displaystyle \frac{(n+r-1)(n+r-2)\cdots (n+1)n}{r!}}\\ & ={\displaystyle \frac{(n+r-1)(n+r-2)\cdots (n+1)n(n-1)!}{r!(n-1)!}}\\ & ={\displaystyle \frac{(n+r-1)!}{r!(n-1)!}}\\ & =\left(\begin{array}{c}n+r-1\\ r\end{array}\right)\\ \text{Sehingga}& \\ f(x)& ={\displaystyle \sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}-n\\ r\end{array}\right)(-1{)}^r{x}^r={\displaystyle \sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}n+r-1\\ r\end{array}\right)x^r={\displaystyle \sum t_r{x}^r}}}\\ \text{maka}& \\ t_r& =\left(\begin{array}{c}n+r-1\\ r\end{array}\right)\end{array}$

$\begin{array}{rl}& \text{Dan yang perlu diingat untuk selanjutnya}\\ & \text{adalah bahwa untuk}x\ne 0\mathbf{\text{dan}}n\text{bilangan}\\ & \text{cacah berlaku identitas}\\ & {\displaystyle \frac{1-x^{n+1}}{1-x}=1+x+x^2+x^3+\cdots +x^n}\end{array}$

$\text{CONTOH SOAL}$

1. Rasiman, Rahmawati, N., D. 2012. Matematika Diskrit. Semarang: IKIP PGRI Semarang Press.

Binomial Newton

 Pengayaan:

$\text{E. Binomial Newton}$

$\text{E. 1 Binomial Newton}$

$\begin{array}{rl}& \text{Perhatikanlah susunan bilangan berikut}\\ \\ & \begin{array}{|cl|}\hline & \\ 1=C_0^{1}1=C_1^1& (a+b{)}^1\\ & \\ 1=C_0^{2}2=C_1^{2}1=C_2^2& (a+b{)}^2\\ & \\ 1=C_0^{3}3=C_1^{3}3=C_2^{3}1=C_3^3& (a+b{)}^3\\ & \\ 1=C_0^{4}4=C_1^{4}6=C_2^{4}4=C_3^{4}1=C_4^4& (a+b{)}^4\\ ⋮& ⋮\\ dst& (a+b{)}^{\cdots }\\ & \\ ⋮& ⋮\\ & (a+b{)}^n\\ \hline\end{array}\\ \\ & \text{Susunan bilangan-bilangan di atas selanjutnya}\\ & \text{dinamakan}\mathbf{\text{Segitiga Pascal}}\\ & \end{array}$

$\begin{array}{rl}& \text{Bilangan}{C}_r^n=\left(\begin{array}{c}n\\ r\end{array}\right)\text{merupakan koefisien}\\ & \text{dari binomial}(a+b{)}^n\\ & \text{Selanjutnya perhatikanlah bahwa untuk}\\ & n=1,2,3,4,\cdots \text{berlaku}\\ & \begin{array}{rl}(a+b{)}^n=& C_0^n{a}^n{b}^0+C_1^n{a}^{n-1}{b}^1+C_2^n{a}^{n-2}{b}^2\\ & +C_3^n{a}^{n-3}{b}^3+\cdots +C_{n-3}^n{a}^3{b}^{n-3}\\ & +C_{n-2}^n{a}^2{b}^{n-2}+C_{n-1}^n{a}^1{b}^{n-1}+C_n^n{a}^0{b}^n\\ & ={\displaystyle \sum _{r=0}^n{C}_r^n{a}^{n-r}{b}^r}\end{array}\\ & \end{array}$

$\text{E. 2 Perluasan Binomial Newton}$

$\begin{array}{rl}& \text{Untuk bilangan real}n\text{dan bilangan}\\ & \text{non negatif}r,\text{serta}\left|A\right|<1,\text{berlaku}:\\ & (1+A{)}^n={\displaystyle \sum _{r=0}^n{C}_r^n{A}^r}\end{array}$

$\text{E. 3 Teorema Multinomial}$

Pada bentuk multinomial dengan ekspresi  $(x_1+x_2+x_3+\cdots +x_r{)}^n$  dengan n dan r bilangan bulat positif, maka koefisien dari  $x_1^{n_1}{x}_2^{n_2}{x}_3^{n_3}\cdots x_r^{n_r}$   adalah  ${\displaystyle \frac{n!}{n_1!n_2!n_3!\cdots n_r!}}$  dinotasikan dengan  $\left(\begin{array}{c}n\\ \\ n_1,n_2,n_3,\cdots ,n_r\end{array}\right)$

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Misalkan untuk}n\text{bilangan bulat}\\ & \text{Positif. Tunjukklan bahwa}\\ & \text{a}.(1+x{)}^n={\displaystyle \sum _{r=0}^n{C}_r^n{x}^r={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^r}}\\ & \text{b}.\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+\cdots +\left(\begin{array}{c}n\\ n\end{array}\right)=2^n\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}\text{a}.(1+x)& {}^n\\ =& C_0^n{1}^n{x}^0+C_1^n{1}^{n-1}{x}^1+C_2^n{1}^{n-2}{x}^2\\ & +C_3^n{1}^{n-3}{x}^3+\cdots +C_{n-3}^n{1}^3{x}^{n-3}\\ & +C_{n-2}^n{1}^2{x}^{n-2}+C_{n-1}^n{1}^1{x}^{n-1}+C_n^n{1}^0{x}^n\\ =& C_0^n+C_1^{n}x+C_2^n{x}^2+C_3^n{x}^3+\cdots \\ & +C_{n-3}^n{x}^{n-3}+C_{n-2}^n{x}^{n-2}+C_{n-1}^n{x}^{n-1}\\ & +C_n^n{x}^n\\ \text{atau}& \text{dengan bentuk lain}\\ =& \left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)x+\left(\begin{array}{c}n\\ 2\end{array}\right)x^2+\left(\begin{array}{c}n\\ 3\end{array}\right)x^3\\ & +\cdots +\left(\begin{array}{c}n\\ n-3\end{array}\right)x^{n-3}+\left(\begin{array}{c}n\\ n-2\end{array}\right)x^{n-2}\\ & +\left(\begin{array}{c}n\\ n-1\end{array}\right)x^{n-1}+\left(\begin{array}{c}n\\ n\end{array}\right)x^n\\ =& {\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^r}\end{array}\\ & \begin{array}{rl}\text{b}.(1+x)& {}^n\text{lihat jawaban poin}a,\text{saat}x=1\\ (1+1)& {}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)1+\left(\begin{array}{c}n\\ 2\end{array}\right)1^2+\left(\begin{array}{c}n\\ 3\end{array}\right)1^3\\ & +\cdots +\left(\begin{array}{c}n\\ n-3\end{array}\right)1^{n-3}+\left(\begin{array}{c}n\\ n-2\end{array}\right)1^{n-2}\\ & +\left(\begin{array}{c}n\\ n-1\end{array}\right)1^{n-1}+\left(\begin{array}{c}n\\ n\end{array}\right)1^n\\ (2)& {}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+\left(\begin{array}{c}n\\ 3\end{array}\right)\\ & +\cdots +\left(\begin{array}{c}n\\ n-1\end{array}\right)+\left(\begin{array}{c}n\\ n\end{array}\right)\\ =& {\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)}\\ \text{Sehing}& \text{ga}\\ 2^n& ={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Misalkan untuk}n\text{bilangan bulat}\\ & \text{Positif. Tunjukklan bahwa}\\ & \left(\begin{array}{c}n\\ 0\end{array}\right)-\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)-\cdots +(-1{)}^n\left(\begin{array}{c}n\\ n\end{array}\right)=0\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Sebelumnya diketahui bahwa}\\ & \begin{array}{rl}& (a+b{)}^n={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r}\\ & \text{atau}\\ & {\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r=(a+b{)}^n}\\ & ⧫\text{saat}a=b=1,\text{maka}\\ & {\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)1^{n-r}{1}^r=(1+1{)}^n}\\ & \iff {\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)=2^n...(\text{bukti no. 1.b})}\\ & ⧫\text{saat}a=1\mathrm{\&}b=-1\text{maka}\\ & {\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)1^{n-r}(-1{)}^r=(1-1{)}^n=0}\\ & \text{Sehingga}\\ & \left(\begin{array}{c}n\\ 0\end{array}\right)-\left(\begin{array}{c}n\\ 1\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)-\cdots +(-1{)}^n\left(\begin{array}{c}n\\ n\end{array}\right)=0◼\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Untuk}n,r\ge 0,\text{tunjukkan bahwa}\\ & \text{a}.\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)\\ & \text{b}.\left(\begin{array}{c}n\\ r\end{array}\right)={\displaystyle \frac{n}{r}\left(\begin{array}{c}n-1\\ r-1\end{array}\right)}\\ & \text{c}.\left(\begin{array}{c}n\\ r\end{array}\right)={\displaystyle \frac{n-r+1}{r}\left(\begin{array}{c}n\\ r-1\end{array}\right)}\\ & \text{d}.\left(\begin{array}{c}-n\\ r\end{array}\right)=(-1{)}^k\left(\begin{array}{c}n+r-1\\ r\end{array}\right)\\ & \text{e}.\left(\begin{array}{c}n\\ r\end{array}\right)+\left(\begin{array}{c}n\\ r+1\end{array}\right)=\left(\begin{array}{c}n+1\\ r+1\end{array}\right)\\ & \text{f}.\left(\begin{array}{c}n\\ m\end{array}\right)\left(\begin{array}{c}m\\ r\end{array}\right)=\left(\begin{array}{c}n\\ r\end{array}\right)\left(\begin{array}{c}n-r\\ m-r\end{array}\right)\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\text{a}.\left(\begin{array}{c}n\\ r\end{array}\right)& ={\displaystyle \frac{n!}{r!(n-r)!}}\\ & ={\displaystyle \frac{n!}{(n-r)!(n-(n-r))!}}\\ & =\frac{n!}{(n-r)!r!}=\left(\begin{array}{c}n\\ n-r\end{array}\right)\end{array}\\ & \begin{array}{rl}\text{b}.\left(\begin{array}{c}n\\ r\end{array}\right)& ={\displaystyle \frac{n!}{r!(n-r)!}}\\ & ={\displaystyle \frac{n.(n-1)!}{r.(r-1)!((n-1)-(r-1))!}}\\ & ={\displaystyle \frac{n}{r}\frac{(n-1)!}{(r-1)!((n-1)-(r-1))!}}\\ & ={\displaystyle \frac{n}{r}\left(\begin{array}{c}n-1\\ r-1\end{array}\right)}\end{array}\\ & \begin{array}{rl}\text{c}.\left(\begin{array}{c}n\\ r\end{array}\right)& ={\displaystyle \frac{n!}{r!(n-r)!}}\\ & ={\displaystyle \frac{n!}{r.(r-1)!(n-r)!}\times \frac{((n-r)+1)}{((n-r)+1)}}\\ & ={\displaystyle \frac{n-r+1}{r}\times \frac{n!}{(r-1)!((n-r)+1)!}}\\ & ={\displaystyle \frac{n-r+1}{r}\times \frac{n!}{(r-1)!(n-(r-1))!}}\\ & ={\displaystyle \frac{n-r+1}{r}\left(\begin{array}{c}n\\ r-1\end{array}\right)}\end{array}\\ & \text{d}.\text{Silahkan dicoba buat latihan}\\ & \text{e}.\text{Silahkan dicoba buat latihan}\\ & \text{f}.\text{Silahkan dicoba buat latihan}\end{array}$

$\begin{array}{l}\\ 4.& \text{Tentukan nilai dari}\\ & \text{a}.\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{c}3\\ 1\end{array}\right)+\cdots +\left(\begin{array}{c}100\\ 1\end{array}\right)\\ & \text{b}.\left(\begin{array}{c}100\\ 100\end{array}\right)+\left(\begin{array}{c}101\\ 100\end{array}\right)+\left(\begin{array}{c}102\\ 100\end{array}\right)+\cdots +\left(\begin{array}{c}200\\ 100\end{array}\right)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{c}3\\ 1\end{array}\right)+\cdots +\left(\begin{array}{c}100\\ 1\end{array}\right)\\ & \text{Sebelumnya perhatikan}\\ & =\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{c}3\\ 1\end{array}\right)+\cdots +\left(\begin{array}{c}n\\ 1\end{array}\right)\\ & \text{Karena}\\ & \left(\begin{array}{c}n\\ r-1\end{array}\right)+\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n+1\\ r\end{array}\right)\\ & \text{Saat}\\ & \left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}2\\ 1\end{array}\right)=\left(\begin{array}{c}2\\ 2\end{array}\right)+\left(\begin{array}{c}2\\ 1\end{array}\right)=\left(\begin{array}{c}3\\ 2\end{array}\right)\\ & \text{Sehingga}\\ & \underset{\left(\begin{array}{c}5\\ 2\end{array}\right)}{\underbrace{\underset{\left(\begin{array}{c}4\\ 2\end{array}\right)}{\underbrace{\underset{\left(\begin{array}{c}3\\ 2\end{array}\right)}{\underbrace{\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}2\\ 1\end{array}\right)}}+\left(\begin{array}{c}3\\ 1\end{array}\right)}}+\left(\begin{array}{c}4\\ 1\end{array}\right)}}\\ & \text{maka}\\ & =\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{c}3\\ 1\end{array}\right)+\cdots +\left(\begin{array}{c}n\\ 1\end{array}\right)=\left(\begin{array}{c}n+1\\ 2\end{array}\right)\\ & \text{Jadi},\\ & \left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{c}3\\ 1\end{array}\right)+\cdots +\left(\begin{array}{c}100\\ 1\end{array}\right)=\left(\begin{array}{c}101\\ 2\end{array}\right)\end{array}\\ & \text{b}\text{Silahkan coba sendiri}\\ & \text{sebagai latihan}\end{array}$

$\begin{array}{l}\\ 5.& \text{Tentukanlah nilai dari}\\ & \text{a}.{\displaystyle \sum _{r=0}^{1000}\left(\begin{array}{c}1000\\ r\end{array}\right)}\\ & \text{b}.\left(\begin{array}{c}2009\\ 1\end{array}\right)+\left(\begin{array}{c}2009\\ 2\end{array}\right)+\left(\begin{array}{c}2009\\ 3\end{array}\right)+\cdots +\left(\begin{array}{c}2009\\ 2004\end{array}\right)\\ \\ & (\mathbf{\text{OSK 2009}})\\ \\ & \text{Jawab}:\\ & \text{a}.{\displaystyle \sum _{r=0}^{1000}\left(\begin{array}{c}1000\\ r\end{array}\right)={\displaystyle \sum _{r=0}^{1000}\left(\begin{array}{c}1000\\ r\end{array}\right)1^{1000-r}{1}^r}}\\ \\ & =(1+1{)}^{1000}=2^{1000}\\ & \text{b}.{\displaystyle \sum _{r=0}^{2009}\left(\begin{array}{c}2009\\ r\end{array}\right)=2^{2009}}\\ & \text{karena}\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right),\text{maka}\\ & \left(\begin{array}{c}2009\\ 0\end{array}\right)=\left(\begin{array}{c}2009\\ 2009\end{array}\right),\left(\begin{array}{c}2009\\ 1\end{array}\right)=\left(\begin{array}{c}2009\\ 2008\end{array}\right),\\ & \cdots ,\left(\begin{array}{c}2009\\ 1004\end{array}\right)=\left(\begin{array}{c}2009\\ 1005\end{array}\right)\\ & \text{Sehingga}\\ & \left(\begin{array}{c}2009\\ 0\end{array}\right)+\left(\begin{array}{c}2009\\ 1\end{array}\right)+\left(\begin{array}{c}2009\\ 2\end{array}\right)+\cdots +\left(\begin{array}{c}2009\\ 2009\end{array}\right)=2^{2009}\\ & \iff \left(\begin{array}{c}2009\\ 0\end{array}\right)+\left(\begin{array}{c}2009\\ 1\end{array}\right)+\left(\begin{array}{c}2009\\ 2\end{array}\right)+\cdots +\left(\begin{array}{c}2009\\ 1004\end{array}\right)={\displaystyle \frac{2^{2009}}{2}=2^{2008}}\\ & \iff 1+\left(\begin{array}{c}2009\\ 1\end{array}\right)+\left(\begin{array}{c}2009\\ 2\end{array}\right)+\left(\begin{array}{c}2009\\ 3\end{array}\right)+\cdots +\left(\begin{array}{c}2009\\ 1004\end{array}\right)=2^{2008}\\ & \iff \left(\begin{array}{c}2009\\ 1\end{array}\right)+\left(\begin{array}{c}2009\\ 2\end{array}\right)+\left(\begin{array}{c}2009\\ 3\end{array}\right)+\cdots +\left(\begin{array}{c}2009\\ 1004\end{array}\right)=2^{2008}-1\end{array}$

1. Bintari, N. 2009. Master Juara Olimpiade Matematika SMA Nasional dan Internasional. Yogyakarta: PUSTAKA WIDYATAMA.
2. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI. Bandung: SEWU.
3. Rasiman, Rahmawati, N., D. 2012. Matematika Diskrit. Semarang: IKIP PGRI Semarang Press.
4. Sharma, dkk. 2017. Jelajah Matematika SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA.

Contoh 3 Soal dan Pembahasan Materi Permutasi dan Kombinasi

$\begin{array}{l}\\ 11.& \text{Dalam suatu rapat mengelilingi meja bundar}\\ & \text{yang dihadiri sebanyak 7 orang}\\ & \text{a}.\text{ada berapa susunan yang terjadi}?\\ & \text{b}.\text{Jika A dan B bagian dari 7 orang ini}\\ & \text{duduknya selalu berdampingan, maka}\\ & \text{posisi duduk yang terbentuk sejumlah}?\\ & \text{c}.\text{Jika seperti poin b, tetapi yang}\\ & \text{duduk berdampingan atau saling berdekatan}\\ & \text{adalah A, B, dan C}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}n=7\\ & \begin{array}{rl}\text{a}.& \text{Posisi duduk melingkarnya}\\ & =(7-1)!=6!=720\\ & \mathbf{\text{atau}}\\ & n=r=7\text{orang, maka}\\ & ={\displaystyle \frac{P(7,7)}{7}=6!=720}\\ \text{b}.& \text{Ada syarat A dan B berdampingan, maka}\\ & \text{A dan B dihitung 1 objek dulu, sehingga total}\\ & \text{objek ada 1 objek ditambah sisanya = 6 objek}.\\ & \text{Dari 6 objek ini yang dianggap duduk melingkar}\\ & \text{dengan 2 orang (A dan B) bisa gantian posisi}.\\ & \text{sehingga}\\ & (6-1)!\times 2!=5!\times 2!=240\\ & \mathbf{\text{atau}}\\ & ={\displaystyle \frac{P(6,6)}{6}\times P(2,2)}\\ & =5!\times 2!=120\times 2=240\\ \text{b}.& \text{3 orang (A, B, dan C) dianggap 1 objek}\\ & \text{dulu sehigga yang duduk posisi melingkar}\\ & \text{dianggap 5 orang, sehingga perhitungannya}\\ & ={\displaystyle \frac{P(5,5)}{5}\times P(3,3)}\\ & =24\times 6=144\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Suatu kelompok yang terdiri dari 20 remaja}\\ & \text{a}.\text{Jika mereka saling berjabat tangan}\\ & \text{seseorang dengan lainnya hanya satu kali}\\ & \text{maka banyak jabat tangan yang terjadi}?\\ & \text{b}.\text{Jika mereka membentuk regu voly, maka}\\ & \text{berapa banyak regu voly yang terbentuk}?\\ & \text{c}.\text{Jika mereka membentuk regu sepak bola},\\ & \text{maka banyak regu sepak bola yang terbentuk}?\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}n=20\\ & \begin{array}{rl}\text{a}.& \text{Karena jabat tangan dilakukan hanya hanya}\\ & \text{pada dua remaja yang berbeda dan urutan}\\ & \text{tidak diperlukan, maka hal ini persoalan}\\ & \text{kombinasi. Sehingga banyaknya jabat tangan}\\ & \left(\begin{array}{c}n\\ r\end{array}\right)={\displaystyle \frac{n!}{r!(n-r)!}}\\ & \iff \left(\begin{array}{c}20\\ 2\end{array}\right)={\displaystyle \frac{20!}{2!(20-2)!}=\frac{20!}{2!\times 18!}}\\ & \iff \left(\begin{array}{c}20\\ 2\end{array}\right)={\displaystyle \frac{20.19.\text{⧸}18!}{2.\text{⧸}18!}=190}\\ \text{b}.& \text{Karena satu regu voli ada 6 orang, maka}\\ & \left(\begin{array}{c}20\\ 6\end{array}\right)={\displaystyle \frac{20!}{6!(20-6)!}}\\ & \iff \left(\begin{array}{c}20\\ 6\end{array}\right)={\displaystyle \frac{20!}{6!\times 14!}}\\ & \iff \left(\begin{array}{c}20\\ 6\end{array}\right)={\displaystyle \frac{20.19.18.17.16.15.\text{⧸}14!}{720\times \text{⧸}14!}}\\ \text{c}.& \text{Karena satu regu terdiri dari 11 orang},\\ & \text{maka}\\ & \left(\begin{array}{c}20\\ 11\end{array}\right)={\displaystyle \frac{20!}{11!(20-11)!}=\frac{20!}{11!\times 9!}}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Jajargenjang yang dapat dibuat oleh}\\ & \text{himpunan empat garis sejajar yang}\\ & \text{berpotongan dengan garis yang terhimpun}\\ & \text{dalam 7 garis sejajar adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa kombinasi dari dua himpunan}\\ & \text{garis sejajar yang masing-masing berjumlah}\\ & \text{4 dan 7 garis, maka}\text{banyak jajar genjang}\\ & \begin{array}{rl}& =\left(\begin{array}{c}4\\ 2\end{array}\right)\times \left(\begin{array}{c}7\\ 2\end{array}\right)\\ & ={\displaystyle \frac{4!}{2!(4-2)!}\times \frac{7!}{2!\times (7-2)!}}\\ & ={\displaystyle \frac{4\times 3\times \text{⧸}2!}{2\times \text{⧸}2!}\times \frac{7\times 6\times \text{⧸}5!}{2\times \text{⧸}5!}}\\ & =6\times 21\\ & =126\text{jajar genjang}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Diketahui segi enam beraturan. Tentukanlah}\\ & \text{a}.\text{Banyak diagonal dapat dibentuk}?\\ & \text{b}.\text{Banyak segi tiga di dalamnya}?\\ & \text{c}.\text{Banyak perpotongan diagonal-diagonal}\\ & \text{jika tidak ada titik-titik perpotongan}\\ & \text{yang sama}?\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui segi}-n\text{dengan}n=6\\ & \text{Dan perlu diingat bahwa di sini tidak diperlukan}\\ & \text{urutan mana yang perlu didahulukan, maka}\\ & \text{rumus kombinasi yang perlu digunakan, yaitu}\\ & \begin{array}{rl}\text{a}.& \text{Banyak diagonalnya adalah}:\\ & \left(\begin{array}{c}n\\ 2\end{array}\right)-n={\displaystyle \frac{n(n-3)}{2}}\\ & \iff ={\displaystyle \frac{6.(6-3)}{2}=\frac{6.3}{2}=9}\\ \text{b}.& \text{Banyaknya segi tiga, berarti melibatkan}\\ & \text{tiga garis, maka}\\ & \left(\begin{array}{c}6\\ 3\end{array}\right)={\displaystyle \frac{6!}{3!\times (6-3)!}=\frac{6\times 5\times 4\times \text{⧸}3!}{6\times \text{⧸}3!}=20}\\ \text{c}.& \text{Satu buah titik potong dapat dibentuk}\\ & \text{dengan dua garis ekuivalen dengan empat}\\ & \text{buah titik sudut, maka banyaknya titik}\\ & \text{potong adalah}:\\ & \left(\begin{array}{c}6\\ 4\end{array}\right)={\displaystyle \frac{6!}{4!\times (6-4)!}=\frac{6!}{4!\times 2!}=15}\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Perhatikalah dua ilustrasi gambar berikut}\end{array}$

Gambar (1)

Gambar (2)

$\begin{array}{l}\\ .& \text{Tentukanlah}\\ & \text{a}.\text{jalur terpendek dari titik A ke B}\\ & \text{pada gambar (1)}\\ & \text{b}.\text{jalur terpendek dari titik P ke Q}\\ & \text{pada gambar (2)}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \text{Perhatikanlah bahwa langkah dari titik A}\\ & \text{ke titik B harus terdiri dari 8 langkah, yaitu}\\ & \text{3 langkah ke kanan dan 5 langkah ke atas}\\ & \text{Karena yang diinginkan lintasan terpendek}\\ & \text{dan tidak ada kekhususn harus dimulai dari}\\ & \text{mana, maka banyaknya langkah berbdeda}\\ & \text{dan terpendek adalah}:\\ & \left(\begin{array}{c}8\\ 3\end{array}\right)\text{atau}\left(\begin{array}{c}8\\ 5\end{array}\right).\text{Misal kita hitung salah}\\ & \text{satunya saja}:\\ & \left(\begin{array}{c}8\\ 3\end{array}\right)={\displaystyle \frac{8!}{3!(8-5)!}=\frac{8!}{3!\times 5!}=\frac{8.7.6.\text{⧸}5!}{6.\text{⧸}5!}=56}\end{array}\end{array}$

$.\begin{array}{rl}\text{b}.& \text{Untuk poin b, perhatikanlah ilustrasi}\\ & \text{gambar berikut(untuk memudahkan}\\ & \text{perhitungan). Tempatkan titik-titik}\\ & \text{bantu A, B, C, D, E, dan F seperti}\\ & \text{pada gambar berikut}\end{array}$

$.\begin{array}{rl}.& \text{Perhatikanlah untuk setiap lintasan}\\ & \text{terpendek dari titik P ke titik Q}\\ & \text{dapat dipastikan akan melewati}\\ & \text{titik A, B, C, dan D. Sehingga dari}\\ & \text{keempat titik itulah akan diperoleh}\\ & \text{rute PAQ, PBQ, PCQ, dan PDQ}.\\ & \text{Sehingga banyak rute terpendek dari}\\ & \text{titik P ke Q yang selanjutnya kita}\\ & \text{simbolkan dengan}\mathrm{\#}PQ\text{adalah}:\\ & \begin{array}{rl}\mathrm{\#}PQ& =\mathrm{\#}PAQ+\mathrm{\#}PBQ+\mathrm{\#}PCQ+\mathrm{\#}PDQ\\ & =\left(\begin{array}{c}4\\ 0\end{array}\right)\left(\begin{array}{c}5\\ 0\end{array}\right)+\left(\begin{array}{c}4\\ 3\end{array}\right)\left(\begin{array}{c}5\\ 1\end{array}\right)+\mathrm{\#}PECQ+\mathrm{\#}PFCQ+\mathrm{\#}PFDQ\\ & =1.1+4.5+\left(\begin{array}{c}3\\ 2\end{array}\right)\left(\begin{array}{c}3\\ 0\end{array}\right)\left(\begin{array}{c}3\\ 2\end{array}\right)+\left(\begin{array}{c}3\\ 1\end{array}\right)\left(\begin{array}{c}3\\ 1\end{array}\right)\left(\begin{array}{c}3\\ 2\end{array}\right)+\left(\begin{array}{c}3\\ 1\end{array}\right)\left(\begin{array}{c}3\\ 0\end{array}\right)\left(\begin{array}{c}3\\ 0\end{array}\right)\\ & =1+20+3.1.3+3.3.3+3.1.1\\ & =1+20+9+27+3\\ & =60\end{array}\end{array}$

DAFTAR PUSTAKA

1. Bintari, N. 2009. Master Juara Olimpiade Matematika SMA Nasional dan Internasional. Yogyakarta: PUSTAKA WIDYATAMA.
2. Ibrahim, Mussafi, N, S, M. 2013. Pengantar Kombinatorika dan Teori Graf. Yogyakarta: GRAHA ILMU.
3. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: SRIKANDI EMPAT WIDYA UTAMA.
4. Sobirin. 2006. Kompas Matematika: Strategi Praktis Menguasai Tes Matematika (SMA Kelas XI IPA). Jakarta: KAWAN PUSTAKA.
5. Sukino. 2011. Maestro Olimpiade Matematika SMP Seri B. Jakarta: ERLANGGA.
6. Susyanto, N, 2012. Tutor Senior Olimpiade Matematika Lima Benua Tingkat SMP. Yogyakarta: KENDI MAS MEDIA.
7. Tampomas, H. 1999. SeribuPena Matematika SMU Jilid 2 Kelas 2 Berdasarkan Kurikulum 1994 Suplemen CBPP 1999. Jakarta: ERLANGGA.

Contoh 2 Soal dan Pembahasan Materi Permutasi dan Kombinasi

$\begin{array}{l}\\ 6.& \text{Dari angka-angka 2,3,5,6,7, dan 9 dibuat}\\ & \text{susunan bilangan}\\ & \text{a. berapa banyak bilangan yang terdiri dari 4}\\ & \text{angka berlainan}\\ & \text{b. berapa banyak bilangan yang terdiri dari 4}\\ & \text{angka boleh berulang}\\ & \text{c. berapa banyak bilangan ganjil yang terdiri}\\ & \text{dari 4 angka berlainan}\\ & \text{d. berapa banyak bilangan genap yang terdiri}\\ & \text{dari 4 angka berlainan}\\ & \text{e. berapa banyak bilangan yang terdiri dari 4}\\ & \text{angka berlainan yang lebih dari 2021}\\ & \text{f. berapa banyak bilangan yang terdiri dari 4}\\ & \text{angka boleh berulang yang lebih dari 2021}\\ & \text{g. berapa banyak bilangan genap yang terdiri}\\ & \text{dari 4 angka berlainan yang lebih dari 2021}\\ & \text{h. berapa banyak bilangan ganjil yang terdiri}\\ & \text{dari 4 angka berlainan yang lebih dari 2021}\\ \\ & \mathbf{\text{Jawab}}:\end{array}$

$.\begin{array}{rl}\text{a}.& P(6,4)={\displaystyle \frac{6!}{(6-2)!}=\frac{6!}{2!}=6.5.4.3=360}\\ \text{b}.& P(6,1{)}^4=6^4=1296\\ \text{c}.& \text{Untuk digit satuan ditentukan dulu, yaitu}\\ & \text{karena digit ganjil ada 4, maka ada 4 pilihan}\\ & \text{sisanya disebar ke slot ribuan sampai puluhan}\\ & \text{maka}\\ & \begin{array}{|cccc|}\hline \text{kotak}& \text{kotak}& \text{kotak}& \text{kotak}\\ 1& 2& 3& 4\\ \text{digit}& \text{digit}& \text{digit}& \text{digit}\\ \text{ribuan}& \text{ratusan}& \text{puluhan}& \text{satuan}\\ P(5,1)& P(4,1)& P(3,1)& P(4,1)\\ \text{pilihan}& \text{pilihan}& \text{pilihan}& \text{pilihan}\\ \hline\end{array}\\ & \text{Sehingga banyak bilangan yg terjadi}\\ & P(5,1).P(4,1).P(3,1).P(4,1)=5.4.3.4=240\\ \text{d}.& \mathbf{\text{Cara pertama}}\\ & \text{Semisal dengan jawaban poin c, Karena}\\ & \text{digit genap ada 2, maka digit satuan ada}\\ & \text{2 pilihan, sisanya disebar, yaitu}\\ & \begin{array}{|cccc|}\hline \text{kotak}& \text{kotak}& \text{kotak}& \text{kotak}\\ 1& 2& 3& 4\\ \text{digit}& \text{digit}& \text{digit}& \text{digit}\\ \text{ribuan}& \text{ratusan}& \text{puluhan}& \text{satuan}\\ P(5,1)& P(4,1)& P(3,1)& P(2,1)\\ \text{pilihan}& \text{pilihan}& \text{pilihan}& \text{pilihan}\\ \hline\end{array}\\ & \text{Sehingga banyak bilangan yg terjadi}\\ & P(5,1).P(4,1).P(3,1).P(2,1)=5.4.3.2=120\\ & \mathbf{\text{Cara kedua}}\\ & \text{Jawaban poin a dikurangi poin c, yaitu}\\ & 360-240=120\end{array}$

$.\begin{array}{rl}\text{e}.& \mathbf{\text{Cara Pertama}}\\ & \text{Karena digit pilihannya, 2,3,5,6,7, dan 9}\\ & \text{disusun bagaimanapun bilangan 4 digit}\\ & \text{yang diambilkan dari bilangan di atas}\\ & \text{pasti semunya akan lebih besar dari 2021}\\ & \text{maka banyaknya bilangan yang terjadi}\\ & \text{adalah}:\\ & \begin{array}{|cccc|}\hline \text{kotak}& \text{kotak}& \text{kotak}& \text{kotak}\\ 1& 2& 3& 4\\ \text{digit}& \text{digit}& \text{digit}& \text{digit}\\ \text{ribuan}& \text{ratusan}& \text{puluhan}& \text{satuan}\\ P(6,1)& P(5,1)& P(4,1)& P(3,1)\\ \text{pilihan}& \text{pilihan}& \text{pilihan}& \text{pilihan}\\ \hline\end{array}>2021\\ & \text{Sehingga totalnya banyaknya}\\ & P(6,1)\times P(5,1)\times P(4,1)\times P(3,1)\\ & =6.5.4.3=360\\ & \mathbf{\text{Cara Kedua}}\\ & \text{Sama seperti jawaban pada poin a}\\ \text{f}.& \text{Sama persis jawaban poin b, yaitu}\\ & P(6,1{)}^4=6^4=1296\\ & \text{Jika diuraikan adalah sebagai berikut}\\ & \begin{array}{|cccc|}\hline \text{kotak}& \text{kotak}& \text{kotak}& \text{kotak}\\ 1& 2& 3& 4\\ \text{digit}& \text{digit}& \text{digit}& \text{digit}\\ \text{ribuan}& \text{ratusan}& \text{puluhan}& \text{satuan}\\ P(6,1)& P(6,1)& P(6,1)& P(6,1)\\ \text{pilihan}& \text{pilihan}& \text{pilihan}& \text{pilihan}\\ \hline\end{array}>2021\end{array}$

$\begin{array}{l}\\ 7.& \text{Andi akan mengambil 4 buah bola dari}\\ & \text{10 warna yang berbeda. Berapakah banyak}\\ & \text{kombinasi warna yang berbeda yang diambil}\\ & \text{oleh Andi}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}n=10& \text{dan}r=4\\ C(n,r)& ={\displaystyle \frac{n!}{r!(n-r)!}}\\ C(10,4)& ={\displaystyle \frac{10!}{4!(10-4)!}}\\ & ={\displaystyle \frac{10!}{4!\times 6!}}\\ & ={\displaystyle \frac{10\times 9\times 8\times 7\times 6!}{(4\times 3\times 2\times 1)\times 6!}}\\ & =420\text{kombinasi warna bola berbeda}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Berapa banyak cara dapat memilih untuk}\\ & \text{3 perwakilan dari 10 anggota suatu}\\ & \text{kelompok, jika}\\ & \text{a. tanpa perlakuan khusus}\\ & \text{b. salah seorang harus terpilih}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Dengan tanpa perlakuan}\\ & \text{memilih 3 orang dari 10 orang adalah}:\\ & C(10,3)={\displaystyle \frac{10!}{3!(10-3)!}=\frac{10!}{3!\times 7!}=120}\\ \text{b}.& \text{Dengan perlakuan 1 orang terpilih}\\ & (\text{1 orang ini artinya tidak perlu diperhitungkan})\\ & \text{memilih 2 orang dari 9 orang adalah}:\\ & C(9,2)={\displaystyle \frac{9!}{2!(9-2)!}=\frac{9!}{2!\times 8!}=36}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Berapa banyak cara dapat memilih 2 buku}\\ & \text{matematika dan 3 buku fisika serta 4 buku}\\ & \text{ekonomi pada suatu lemari buku yang}\\ & \text{di dalamnya terdapat 10 buku matematika,}\\ & \text{11 buku fisika dan 12 buku ekonomi}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Banyak}& \text{cara pemilihan tersebut adalah}:\\ & =C(10,2)\times C(11,3)\times C(12,4)\\ & ={\displaystyle \frac{10!}{2!\times 8!}\times \frac{11!}{3!\times 8!}\times \frac{12!}{4!\times 8!}}\\ & ={\displaystyle \frac{10\times 9}{1\times 2}\times \frac{11\times 10\times 9}{1\times 2\times 3}\times \frac{12\times 11\times 10\times 9}{1\times 2\times 3\times 4}}\\ & =3675375\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Banyak susunan huruf yang berbeda}\\ & \text{pada satu baris yang dapat dibentuk}\\ & \text{dari huruf-huruf pada kata "MATEMATIKA"}\\ & \text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \{\begin{array}{ll}\text{Jumlah huruf}& n=10\\ \text{Penyusunnya},& \text{ yaitu }:\{\begin{array}{ll}M& \text{ jumlah }=2\\ A& \text{ jumlah }=3\\ T& \text{ jumlah }=2\\ E& \text{ banyak }=1\\ I& \text{ banyak }=1\\ K& \text{ banyak }=1\end{array}\end{array}\\ & \text{Sehingga}\\ & \begin{array}{rl}P(10;2,3,2,1,1,1)& ={\displaystyle \frac{10!}{2!.3!.2!.1!.1!.1!}}\\ & ={\displaystyle \frac{10\times 9\times 8\times 7\times 6\times 5\times 4}{4}}\\ & =10\times 9\times 8\times 7\times 6\times 5\\ & =151200\end{array}\end{array}$

Contoh 1 Soal dan Pembahasan Materi Permutasi dan Kombinasi

$\begin{array}{l}\\ 1.& \text{Bentuk sederhana dari}\\ & \text{a}.{\displaystyle 5!+6!+7!}\\ & \text{b}.{\displaystyle \frac{(n+1)!}{(n-1)!}}\\ & \text{c}.{\displaystyle \frac{(n+2)!}{n!}}\\ & \text{d}.{\displaystyle \frac{(n-2)!}{(n+1)!}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& 5!+6!+7!=5!+6.5!+7.6.5!\\ & =(1+6+42).5!\\ & =49.5!=49.120=5880\\ \text{b}.& {\displaystyle \frac{(n+1)!}{(n-1)!}=\frac{(n+1)n(n-1)!}{(n-1)!}}\\ & =(n+1)n=n^2+n\\ \text{c}.& {\displaystyle \frac{(n+2)!}{n!}=\frac{(n+1)(n+1)n!}{n!}}\\ & =(n+2)(n+1)\\ & =n^2+3n+2\\ \text{d}.& {\displaystyle \frac{(n-2)!}{(n+1)!}=\frac{(n-2)!}{(n+1)n(n-1)(n-2)!}}\\ & ={\displaystyle \frac{1}{(n+1)n(n-1)}}\\ & ={\displaystyle \frac{1}{n^3-n}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai}n\text{yang memenuhi}\\ & \text{persamaan berikut}\\ & \text{a}.{\displaystyle \frac{n!3!}{6!(n-3)!}=\frac{33}{4}}\\ & \text{b}.{\displaystyle \frac{3}{8!}-\frac{2}{7!}+\frac{1}{6!}=\frac{5n+3}{8!}}\\ & \text{c}.{\displaystyle \frac{7!}{5!2!}:\frac{10!}{5!5!}=1:4n}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{n!3!}{6!(n-3)!}=\frac{33}{4}}\\ & \iff {\displaystyle \frac{n(n-1)(n-2)\text{⧸}(n-3)!.\text{⧸}3!}{6.5.\text{⧸}4.\text{⧸}3!\text{⧸}(n-3)!}=\frac{33}{\text{⧸}4}}\\ & \iff n(n-1)(n-2)=33.6.5=11.10.9\\ & \iff n(n-1)(n-2)=11.(11-1).(11-2)\\ & \iff n=11\\ \text{b}.& {\displaystyle \frac{3}{8!}-\frac{2}{7!}+\frac{1}{6!}=\frac{5n+3}{8!}}\\ & \iff {\displaystyle \frac{3-2.8+56}{8!}=\frac{5n+3}{8!}}\\ & \iff \frac{43}{8!}=\frac{5n+3}{8!}\\ & \iff 43=5n+3\iff 5n=40\iff n=8\\ \text{c}.& {\displaystyle \frac{7!}{5!2!}:\frac{10!}{5!5!}=1:4n}\\ & \iff 4n={\displaystyle \frac{5!2!10!}{7!5!5!}}\\ & \iff 4n={\displaystyle \frac{\text{⧸}5!2!10.9.8.\text{⧸}7!}{\text{⧸}7!5!\text{⧸}5!}}\\ & \iff n=3\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai}n\text{yang memenuhi}\\ & \text{persamaan berikut}\\ & \text{a}.P(n,2)=42\\ & \text{b}.7.P(n,3)=6.P(n+1,3)\\ & \text{c}.3.P(n,4)=P(n-1,5)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& P(n,2)=42\\ & \iff {\displaystyle \frac{n!}{(n-2)!}=42}\\ & \iff {\displaystyle \frac{n!}{(n-2)!}={\displaystyle \frac{n\times (n-1)\times (n-2)!}{(n-2)!}=42}}\\ & \iff {\displaystyle n\times (n-1)=7.6=7.(7-1)}\\ & \iff n=7\\ \text{b}.& 7.P(n,3)=6.P(n+1,3)\\ & {\displaystyle \frac{7.n!}{(n-3)!}=\frac{6(n+1)!}{(n+1-3)!}}\\ & {\displaystyle \frac{7\text{⧸}n!}{(n-3)!}=\frac{6.(n+1).\text{⧸}n!}{(n-2)!}}\\ & \iff \frac{7}{\text{⧸}(n-3)!}=\frac{6n+6}{(n-1)\text{⧸}(n-3)!}\\ & \iff 7(n-2)=6n+6\\ & \iff 7n-6n=6+14\iff n=20\\ \text{c}.& 3.P(n,4)=P(n-1,5)\\ & \iff {\displaystyle \frac{3.n!}{(n-4)!}=\frac{(n-1)!}{(n-1-5)!}}\\ & \iff \frac{3.n.\text{⧸}(n-1)!}{(n-4)!}=\frac{\text{⧸}(n-1)!}{(n-6)!}\\ & \iff \frac{3n}{(n-4)(n-5).\text{⧸}(n-6)!}=\frac{1}{\text{⧸}(n-6)!}\\ & \iff 3n=(n-4)(n-5)\\ & \iff 3n=n^2-9n+20\\ & \iff n^2-12n+20=0\\ & \iff (n-2)(n-10)=0\\ & \iff n=2\text{tidak memenuhi}\text{atau}n=10\\ & \text{jadi},n=10\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Jika 10 siswa akan dipilih 4 orang untuk}\\ & \text{menjadi ketua kelas, wakil, sekretaris dan}\\ & \text{seorang bendahara, maka banyak susunan}\\ & \text{terjadi adalah}....\\ \\ & \text{Jawab}:\\ & \text{Penyusunan memerlukan urutan}\\ & \text{maka perlu digunakan permutasi, yaitu}:\\ & P(n,r)={\displaystyle \frac{n!}{(n-r)!}}\\ & \iff P(10,4)={\displaystyle \frac{10!}{(10-4)!}=\frac{10!}{6!}}\\ & \iff ={\displaystyle \frac{10\times 9\times 8\times 7\times \text{⧸}6!}{\text{⧸}6!}}\\ & \iff =5040\end{array}$

$\begin{array}{l}\\ 5.& \text{Jika dari kota A ke kota B terdapat 3 jalur.}\\ & \text{Dan dari kota B ke kota C terdapat 4 jalur,}\\ & \text{serta dari kota C sampai ke kota D ada 5 jalur}\\ & \text{Banyak jalan dari kota A ke kota D adalah}....\\ \\ & \text{Jawab}:\\ & \text{Jalur yang ada semuanya berbeda}\\ & \text{maka perlu digunakan permutasi, yaitu}:\\ & P(n,r)={\displaystyle \frac{n!}{(n-r)!}}\\ & \begin{array}{rl}\text{a}& \text{dari A ke B ada 3 jalur cukup pilih satu, maka}\\ & \bullet P(3,1)={\displaystyle \frac{3!}{(3-1)!}=\frac{3!}{2!}=3}\\ \text{b}& \text{dari B ke C ada 4 jalur cukup pilih satu, maka}\\ & \bullet P(4,1)={\displaystyle \frac{4!}{(4-1)!}=\frac{4!}{3!}=4}\\ \text{c}& \text{dari C ke D ada 5 jalur cukup pilih satu, maka}\\ & \bullet P(5,1)={\displaystyle \frac{5!}{(5-1)!}=\frac{5!}{4!}=5}\end{array}\\ & \text{Jadi, total jalur yang dapat di lalui dari A sampai D adalah}:\\ & P(3,1)\times P(4,1)\times P(5,1)=3\times 4\times 5=60\end{array}$

Notasi Faktorial, Permutasi dan Kombinasi

 $\text{C. Faktorial}$

Perhatikanlah tabel berikut yang berisi perkalian bilangan terurut pada bilangan asli

$\begin{array}{|c|}\hline n!=1\times 2\times 3\times 4\times \cdots \times (n-2)\times (n-1)\times n\\ \mathbf{\text{atau}}\\ n!=n\times (n-1)\times (n-2)\times \cdots \times 4\times 3\times 2\times 1\\ \text{dengan}\\ (n+1)!=(n+1)\times n!\text{untuk}n\ge 1,n\in \mathbb{N}\\ \text{serta didefinisikan bahwa}\\ 0!=1!=1\\ \text{CONTOH}\\ 0!=1\\ 1!=1\\ 2!=2\times 1=2\\ 3!=3\times 2\times 1=6\\ 4!=4\times 3\times 2\times 1=24\\ 5!=5\times 4\times 3\times 2\times 1=120\\ 6!=6\times 5\times 4\times 3\times 2\times 1=720\\ \\ ⋮\\ \\ n!=n\times (n-1)\times (n-2)\times \cdots \times 4\times 3\times 2\times 1\\ \hline\end{array}$

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai}\\ & \begin{array}{l}\\ \text{a}.3!& \text{e}.{\displaystyle \frac{6!}{4!}}& \text{i}.{\displaystyle \frac{2!}{0!}+\frac{3!}{1!}+\frac{4!}{2!}}\\ \text{b}.5!& \text{f}.{\displaystyle \frac{10!}{6!}}& \text{j}.{\displaystyle \frac{2!}{0!}\times \frac{3!}{1!}+\frac{4!}{2!}}\\ \text{c}.0!+1!+2!+3!& \text{g}.{\displaystyle \frac{7!}{3!\times 4!}}& \text{k}.{\displaystyle \frac{3\times 4!}{3!(5!-5!)}}\\ \text{d}.(2!)!+(3!)!& \text{h}.{\displaystyle \frac{13!}{12!+12!}}& \text{l}.{\displaystyle \frac{3!+5!+7!}{4!+6!}}\end{array}\\ \\ & \text{Jawab}:\\ \\ & \begin{array}{c}\\ \text{a}.3!=3.2.1=6\\ \text{b}.5!=5.4.3.2.1=120\\ \begin{array}{rl}\text{c}.0!+1!+2!+3!& =1+1+2+6\\ & =10\end{array}\\ \begin{array}{rl}\text{d}.(2!)!+(3!)!& =2!+6!\\ & =2+720\\ & =722\end{array}\\ \text{e}.{\displaystyle \frac{6!}{4!}=\frac{720}{24}=30\text{atau}{\displaystyle \frac{6!}{4!}={\displaystyle \frac{6.5.\text{⧸}4.\text{⧸}3.\text{⧸}2.\text{⧸}1}{\text{⧸}4.\text{⧸}3.\text{⧸}2.\text{⧸}1}=6.5=30}}}\\ \text{f}.{\displaystyle \frac{10!}{6!}=\frac{10.9.8.7.6.5.4.3.2.1}{6.5.4.3.2.1}=....(\text{silahkan diselesaikan sendiri})}\\ \text{g}.{\displaystyle \frac{7!}{3!\times 4!}=\frac{7.6.5.4.3.2.1}{(3.2.1)\times (4.3.2.1)}=....(\text{silahkan juga diselesaikan sendiri})}\\ ⋮\\ (\text{silahkan selanjutnya diselesaikan sendiri})\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Sederhanakanlah}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{n!}{(n-1)!}}& \text{e}.{\displaystyle \frac{1}{n!}+\frac{n}{(n+1)!}-\frac{1}{(n-1)!}}\\ \text{b}.{\displaystyle \frac{(n+2)!}{(n+1)!}}& \text{f}.{\displaystyle \frac{(4n)!}{(4n+1)!}+\frac{(4n)!}{(4n-1)!}}\\ \text{c}.{\displaystyle \frac{(2n)!}{(2n+1)!}}& \text{g}.{\displaystyle \frac{1}{n}-\frac{n!}{(n-1).(n-2)!}}\\ \text{d}.{\displaystyle \frac{(n+2)!}{(n^2+3n+2)}}& \text{h}.1.1!+2.2!+3.3!+4.4!+5.5!+...+n.n!\end{array}\\ \\ & \text{Jawab}:\\ \\ & \begin{array}{c}\\ \text{a}.{\displaystyle \frac{n!}{(n-1)!}=\frac{n.(n-1)!}{(n-1)!}=n}\\ \text{b}.{\displaystyle \frac{(n+2)!}{(n+1)!}=\frac{(n+2).(n+1)!}{(n+1)!}=n+2}\\ \text{c}.{\displaystyle \frac{(2n)!}{(2n+1)!}=\frac{(2n)!}{(2n+1).(2n)!}=\frac{1}{2n+1}}\\ \text{d}.{\displaystyle \frac{(n+2)!}{n^2+3n+2}=\frac{(n+2)!}{(n+2).(n+1)}=\frac{(n+2).(n+1).n!}{(n+2).(n+1)}=n!}\\ ⋮\\ (\text{silahkan selanjutnya diselesaikan sendiri sebagai latihan})\\ ⋮\\ \begin{array}{rl}\text{h}.& 1.1!+2.2!+3.3!+4.4!+5.5!+...+n.n!\\ & =(2-1).1!+(3-1).2!+(4-1).3!+(5-1).4!+...+(n+1-1).n!\\ & =2.1!+3.2!+4.3!+5.4!+...+(n+1).n!-1!-2!-3!-4!-...-n!\\ & =2!+3!+4!+5!+...+(n+1)!-(1!+2!+3!+4!+...+n!)\\ & =(n+1)!-1\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Sederhanakanlah bentuk penjumlahan berikut}\\ & {\displaystyle \frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\cdots +{\displaystyle \frac{100}{98!+99!+100!}}}\\ \\ & \text{Jawab}:\\ \\ & \begin{array}{rl}\text{Perhatikan}& \text{bahwa}\\ & {\displaystyle \frac{3}{1!+2!+3!}=\frac{3}{1+2+6}=\frac{3}{9}=\frac{1}{3}\times \frac{2}{2}=\frac{2}{1\times 2\times 3}=\frac{2}{3!}=\frac{3-1}{3!}=\frac{3}{3!}-\frac{1}{3!}=\frac{3}{2!\times 3}-\frac{1}{3!}=\frac{1}{2!}-\frac{1}{3!}}\\ & \text{sehingga}\\ & \frac{3}{1!+2!+3!}=\frac{1}{2!}-\frac{1}{3!}\\ & {\displaystyle \frac{4}{2!+3!+4!}=\cdots =\frac{1}{3!}-\frac{1}{4!}}\\ & {\displaystyle \frac{5}{3!+4!+5!}=\cdots =\frac{1}{4!}-\frac{1}{5!}}\\ & ⋮\\ & {\displaystyle \frac{100}{98!+99!+100!}=\cdots =\frac{1}{99!}-\frac{1}{100!}}\\ & ---------------------------\\ & =\frac{1}{2!}-\frac{1}{100!}\end{array}\end{array}$

$\text{D. Permutasi dan Kombinasi}$

$\begin{array}{|lll|}\hline \text{Istilah}& \text{Permutasi}& \text{Kombinasi}\\ \text{Definisi}& \begin{array}{rl}& \text{Permutasi r unsur dari n unsur adalah}\\ & \text{banyaknya kemungkinan urutan r buah}\\ & \text{unsur yang dipilih dari n unsur}\\ & \text{yang tersedia}.\text{Tiap unsur berbeda dan}\\ & r\le n\end{array}& \begin{array}{rl}& \text{Kombinasi r unsur dan n unsur adalah}\\ & \text{banyaknya kemungkinan tidak terurut}\\ & \text{dalam pemilihan r unsur yang diambil}\\ & \text{dari n unsur yang tersedia}.\text{Tiap unsur}\\ & \text{berbeda dan}r\le n\end{array}\\ \text{Tipe}& \text{Bentuk khusus kaidah perkalian}& \text{Bentuk khusus permutasi}\\ \text{Notasi}& {}_n{P}_r,P_n^r,\text{atau}P(n,k)& {}_n{C}_r,C_r^n,(\genfrac{}{}{0ex}{}{n}{r}),\text{atau}C(n,r)\\ \text{Rumus}& P(n,r)={\displaystyle \frac{n!}{(n-r)!}}& (\genfrac{}{}{0ex}{}{n}{r})=C(n,r)={\displaystyle \frac{n!}{r!(n-r)!}}\\ \hline\end{array}$

Selanjutnya perhatikanlah tabel berikut

$\begin{array}{|cc|}\hline \text{Permutasi}& \text{Permutasi}\\ \text{dengan unsur yang sama}& \text{Siklis}\\ \begin{array}{rl}& P(n;n_1,n_2,n_3,...,n_k)\\ & ={\displaystyle \frac{P(n,n)}{n_1!n_2!n_3!...n_k!}}\\ & ={\displaystyle \frac{n!}{n_1!n_2!n_3!...n_k!}}\end{array}& \begin{array}{rl}& \{\begin{array}{ll}\text{Siklis}& =(n-1)!\\ \\ \text{Kalung}& ={\displaystyle \frac{(n-1)!}{2}}\end{array}\end{array}\\ \hline\end{array}$

dan

$\begin{array}{|cc|}\hline \text{Kombinasi}& \text{Kombinasi dalam}\\ \text{dengan pengulangan}& \text{Binom Newton}\\ \begin{array}{rl}& C(n+r-1,r)\\ & =C(n+r-1,n-1)\\ & (\genfrac{}{}{0ex}{}{n+r-1}{r})\\ & =(\genfrac{}{}{0ex}{}{n+r-1}{n-1})\end{array}& \begin{array}{rl}& (x+y{)}^n\\ & =\sum _{k=o}^n(\genfrac{}{}{0ex}{}{n}{r})x^{n-k}{y}^k\\ \\ & \text{Koefisien untuk}\\ & x^{n-k}{y}^k,\text{yaitu}\\ & \text{suku ke}-(k+1)\\ & \text{adalah}(\genfrac{}{}{0ex}{}{n}{r})\end{array}\\ \hline\end{array}$

serta

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Jika di suatu kelas terdapat 4 orang akan dipilih 3 orang }\\ & \text{untuk menjadi ketua, sekretaris, dan bendahara}.\\ & \text{Tentukanlah banyak cara memilih 3 orang tersebut?}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Karena ada 4 orang, misal A, B, C, dan D yang}\\ & \text{akan dipilih 3 orang untuk menduduki posisi}\\ & \text{ketua, sekretaris, dan bendahara, maka kita tinggal}\\ & \text{buat permutasinya, yaitu posisi ketua dapat dipilih }\\ & \text{dengan 4 cara, sekretaris dapat dipilih dengan 3 cara},\\ & \text{dan bendahara dapat dipilih dengan 2 cara. atau}\\ & P(4,3)={\displaystyle \frac{4!}{(4-3)!}=\frac{4!}{1!}=\frac{4\times 3\times 2\times 1}{1}=24\text{cara}}\\ & \text{Berikut ilustrasinya dengan diagram pohon}\end{array}\end{array}$

$\{\begin{array}{ll}A& \{\begin{array}{ll}B& \{\begin{array}{ll}C& \to ABC\\ D& \to ABD\end{array}\\ C& \{\begin{array}{ll}B& \to ACB\\ D& \to ACD\end{array}\\ D& \{\begin{array}{ll}B& \to ADB\\ C& \to ADC\end{array}\end{array}\\ \\ B& \{\begin{array}{ll}A& \{\begin{array}{ll}C& \to BAC\\ D& \to BAD\end{array}\\ C& \{\begin{array}{ll}A& \to BCA\\ D& \to BCD\end{array}\\ D& \{\begin{array}{ll}A& \to BDA\\ C& \to BDC\end{array}\end{array}\\ \\ C& \{\begin{array}{ll}A& \{\begin{array}{ll}B& \to CAB\\ D& \to CAD\end{array}\\ B& \{\begin{array}{ll}A& \to CBA\\ D& \to CBD\end{array}\\ D& \{\begin{array}{ll}A& \to CDA\\ B& \to CDB\end{array}\end{array}\\ \\ D& \{\begin{array}{ll}A& \{\begin{array}{ll}B& \to DAB\\ C& \to DAC\end{array}\\ B& \{\begin{array}{ll}A& \to DBA\\ C& \to DBC\end{array}\\ C& \{\begin{array}{ll}A& \to DCA\\ B& \to DCB\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Seorang anak akan mengambil 4 buah bola dari}\\ & \text{10 warna yang berbeda. Berapakah banyak}\\ & \text{kombinasi warna yang berbeda yang diambil}\\ & \text{oleh Andi}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}n=10& \text{dan}r=4\\ C(n,r)& ={\displaystyle \frac{n!}{r!(n-r)!}}\\ C(10,4)& ={\displaystyle \frac{10!}{4!(10-4)!}}\\ & ={\displaystyle \frac{10!}{4!\times 6!}}\\ & ={\displaystyle \frac{10\times 9\times 8\times 7\times 6!}{(4\times 3\times 2\times 1)\times 6!}}\\ & =420\text{kombinasi warna bola berbeda}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Berapa banyak cara dapat memilih untuk}\\ & \text{3 perwakilan dari 10 anggota suatu}\\ & \text{kelompok, jika}\\ & \text{a. tanpa perlakuan khusus}\\ & \text{b. salah seorang harus terpilih}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Dengan tanpa perlakuan}\\ & \text{memilih 3 orang dari 10 orang adalah}:\\ & C(10,3)={\displaystyle \frac{10!}{3!(10-3)!}=\frac{10!}{3!\times 7!}=120}\\ \text{b}.& \text{Dengan perlakuan 1 orang terpilih}\\ & (\text{1 orang ini artinya tidak perlu diperhitungkan})\\ & \text{memilih 2 orang dari 9 orang adalah}:\\ & C(9,2)={\displaystyle \frac{9!}{2!(9-2)!}=\frac{9!}{2!\times 8!}=36}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Berapa banyak cara dapat memilih 2 buku}\\ & \text{matematika dan 3 buku fisika serta 4 buku}\\ & \text{ekonomi pada suatu lemari buku yang}\\ & \text{di dalamnya terdapat 10 buku matematika,}\\ & \text{11 buku fisika dan 12 buku ekonomi}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Banyak}& \text{cara pemilihan tersebut adalah}:\\ & =C(10,2)\times C(11,3)\times C(12,4)\\ & ={\displaystyle \frac{10!}{2!\times 8!}\times \frac{11!}{3!\times 8!}\times \frac{12!}{4!\times 8!}}\\ & ={\displaystyle \frac{10\times 9}{1\times 2}\times \frac{11\times 10\times 9}{1\times 2\times 3}\times \frac{12\times 11\times 10\times 9}{1\times 2\times 3\times 4}}\\ & =3675375\end{array}\end{array}$

DAFTAR PUSTAKA

1. Johnaes, Kastolan, & Sulasim. 2004. Kompetensi Matematika SMA Kelas 2 Semester 1 Program Ilmu Sosial KBK 2004. Jakarta: YUDHISTIRA.
2. Kartini, Suprapto, Subandi, & Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
3. Sobirin. 2006. Kompas Matematika Strategi Praktis Menguasai Tes Matematika. Jakarta: KAWAN PUSTAKA.

Vektor di Dimensi Tiga

$\text{A. Vektor Di Ruang}$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{|cc|}\hline \text{Nama}& {\mathbf{\text{R}}}^3\\ \text{Vektor Satuan}& \text{Ruang (Bidang XYZ)}\\ {\hat{e}}_{\overline{a}}={\displaystyle \frac{\overline{a}}{\left|\overline{a}\right|}}& \{\begin{array}{ll}i=& \text{vektor satuan}\\ & \text{yang searah sumbu X}\\ j=& \text{Vektor satuan}\\ & \text{yang searah sumbu Y}\\ k=& \text{Vektor satuan}\\ & \text{searah sumbu Z}\end{array}\\ \text{Vektor nol}& \overrightarrow{O}=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\\ \text{Vektor posisi}& \overrightarrow{OP}=\overrightarrow{p}=\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\end{array}\right)=p_1\overline{i}+p_2\overline{j}+p_3\overline{j}\\ \text{Besar Vektor}& \overrightarrow{OP}=\sqrt{p_1^2+p_2^2+p_3^2}\\ \hline\end{array}$

$\text{B. Operasi Vektor}$

$\text{1. Sifat-Sifat Aljabar Vektor}$

$\begin{array}{|lll|}\hline 1.& \text{Komutatif penjumlahan}& \overrightarrow{a}+\overrightarrow{b}=\overrightarrow{b}+\overrightarrow{a}\\ 2.& \text{Asosiatif penjumlahan}& (\overrightarrow{a}+\overrightarrow{b})+\overrightarrow{c}=\overrightarrow{a}+(\overrightarrow{b}+\overrightarrow{c})\\ 3.& \text{Elemen Identitas}& \overrightarrow{a}+\overrightarrow{0}=\overrightarrow{0}+\overrightarrow{a}=\overrightarrow{a}\\ 4.& \text{Invers Penjumlahan}& \overrightarrow{a}+(-\overrightarrow{a})=(-\overrightarrow{a})+\overrightarrow{a}=\overrightarrow{0}\\ 5.& \text{Perkalian dengan skalar}& k\left(l\overrightarrow{a}\right)=\left(kl\right)\overrightarrow{a}\\ & & k(\overrightarrow{a}+\overrightarrow{b})=k\overrightarrow{a}+k\overrightarrow{b}\\ & & k(\overrightarrow{a}-\overrightarrow{b})=k\overrightarrow{a}-k\overrightarrow{b}\\ 6.& \begin{array}{rl}& \text{Jika A, B, dan C segaris }\\ & \text{(Kolinear)}\end{array}& \{\begin{array}{l}\overrightarrow{AB}=k\overrightarrow{BC}\\ \overrightarrow{AC}=k\overrightarrow{AB}\\ dll\end{array}\\ \hline\end{array}$.

$\begin{array}{|cc|}\hline \text{Vektor}& \text{Contoh}\\ \overrightarrow{z}=a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k}& \begin{array}{rl}& \text{diketahui}\overrightarrow{p}=\overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k}\\ & \text{maka pangjang vektor}\overrightarrow{p}\text{adalah}\\ & \left|\overrightarrow{p}\right|=\sqrt{1^2+(-2{)}^2+2^2}\\ & =\sqrt{1+4+4}=\sqrt{9}=3\end{array}\\ & \begin{array}{rl}& \text{Vektor satuan dari}\overrightarrow{p}\text{adalah}\\ & {\overrightarrow{e}}_{\overrightarrow{p}}=\frac{\overrightarrow{p}}{\left|\overrightarrow{p}\right|}={\displaystyle \frac{\left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)}{3}}\\ & ={\displaystyle \frac{1}{3}\left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)={\displaystyle \left(\begin{array}{c}\frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3}\end{array}\right)}}\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui vektor-vektor}\overrightarrow{a}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\\ & \overrightarrow{b}=\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right),\text{dan}\overrightarrow{c}=\left(\begin{array}{c}7\\ 0\\ 4\end{array}\right),\\ & \text{tentukanlah hasil dari}\\ & \text{a}.\overrightarrow{a}+\overrightarrow{b}\\ & \text{b}.6\overrightarrow{a}+2\overrightarrow{b}\\ & \text{c}.2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ & \text{d}.{\displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+{\displaystyle \frac{3}{4}\overrightarrow{b}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}& \overrightarrow{a}+\overrightarrow{b}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)+\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}2+(-3)\\ 1+(-5)\\ (-4)+2\end{array}\right)\\ & =\left(\begin{array}{c}2-3\\ 1-5\\ -4+2\end{array}\right)=\left(\begin{array}{c}-1\\ -4\\ -2\end{array}\right)\\ \text{b}.& 6\overrightarrow{a}+2\overrightarrow{b}=6\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)+2\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}18-6\\ 6-10\\ -24+4\end{array}\right)=\left(\begin{array}{c}12\\ -4\\ -20\end{array}\right)\\ \text{c}.& 2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ & 2\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)-\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)+\left(\begin{array}{c}7\\ 0\\ 4\end{array}\right)\\ & =\left(\begin{array}{c}4+3+7\\ 2+5+0\\ -8-2+4\end{array}\right)=\left(\begin{array}{c}17\\ 7\\ -6\end{array}\right)\\ \text{d}.& {\displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+{\displaystyle \frac{3}{4}\overrightarrow{b}=\cdots }}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui vektor-vektor}\overrightarrow{a}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\\ & \text{tentukanlah}\left|\overrightarrow{a}\right|\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\left|\overrightarrow{a}\right|& =\sqrt{2^2+1^2+(-4{)}^2}\\ & =\sqrt{4+1+16}\\ & =\sqrt{21}\end{array}\end{array}$

$\text{2. Perkalian Skalar Dua Vektor}$

Konsep perkalian skalar dua buah vektor di ruang sama persis dengan konsep di bidang, yaitu:

$\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$.

Misalkan diketahui

$\begin{array}{rl}& \overrightarrow{a}=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right),\overrightarrow{b}=\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right),\text{maka}\\ & \overrightarrow{a}\cdot \overrightarrow{b}=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\cdot \left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right)\\ & =a_1{b}_1+a_2{b}_2+a_3{b}_3\end{array}$

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Diketahui vektor-vektor}\overrightarrow{a}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\\ & \overrightarrow{b}=\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right),\text{dan}\overrightarrow{c}=\left(\begin{array}{c}7\\ 0\\ 4\end{array}\right),\\ & \text{tentukanlah hasil dari}\\ & \text{a}.\overrightarrow{a}\cdot \overrightarrow{b}\\ & \text{b}.\overrightarrow{a}\cdot \overrightarrow{c}\\ & \text{c}.\overrightarrow{b}\cdot \overrightarrow{c}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}& \overrightarrow{a}\cdot \overrightarrow{b}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\cdot \left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)\\ & =(2)(-3)+(1)(-5)+(-4)(2)\\ & =-6-5-8=-19\\ \text{b}& \overrightarrow{a}\cdot \overrightarrow{c}=\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)\cdot \left(\begin{array}{c}7\\ 0\\ 4\end{array}\right)\\ & =(2)(7)+(1)(0)+(-4)(4)\\ & =14+0-16=-2\\ \text{c}& \overrightarrow{b}\cdot \overrightarrow{c}=\left(\begin{array}{c}-3\\ -5\\ 2\end{array}\right)\cdot \left(\begin{array}{c}7\\ 0\\ 4\end{array}\right)\\ & =(-3)(7)+(-5)(0)+(2)(4)\\ & =-21+0+8=-13\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai}t\text{jika}\\ & \overrightarrow{p}=3\overline{i}+t\overline{j}+\overline{k}\text{dan}\overrightarrow{p}\cdot \overrightarrow{p}=13\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \overrightarrow{p}\cdot \overrightarrow{p}=13\\ & \overrightarrow{p}\cdot \overrightarrow{p}=\left|\overrightarrow{p}\right|\left|\overrightarrow{p}\right|\cos 0^{\circ }=13,\\ & \text{ingat bahwa}\mathrm{\angle }(\overrightarrow{p},\overrightarrow{p})=0^{\circ }\\ & \text{dan nilai}\cos 0^{\circ }=1,\\ & \text{maka}\\ & \overrightarrow{p}\cdot \overrightarrow{p}={\left|\overrightarrow{p}\right|}^2.1=13\iff {\left|\overrightarrow{p}\right|}^2=13\\ & \iff {\left(\sqrt{3^2+t^2+1^2}\right)}^2=13\\ & \iff 3^2+t^2+1^2=13\\ & \iff 9+t^2+1=13\\ & \iff t^2=13-9-1=10\\ & \iff t^2=3\\ & \iff t=\pm \sqrt{3}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Diketahui}\overrightarrow{p}=\left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}4\\ -1\\ t\end{array}\right)\\ & \text{Jika}\overrightarrow{p}\text{tegak lurus}\overrightarrow{q}\text{maka}\\ & \text{tentukanlah nilai}t\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \overrightarrow{p}=\left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}4\\ -1\\ t\end{array}\right)\\ & \text{dengan}\overrightarrow{p}\text{dan}\overrightarrow{q}\text{tegak lurus}\\ & \text{artinya}\mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})={90}^{\circ }.\text{Sehingga}\\ & \text{nilai}\cos {90}^{\circ }=0\\ & \text{maka}\\ & \overrightarrow{p}\cdot \overrightarrow{q}=\left|\overrightarrow{p}\right|\left|\overrightarrow{q}\right|\cos \theta \\ & \overrightarrow{p}\cdot \overrightarrow{q}=\left|\overrightarrow{p}\right|\left|\overrightarrow{q}\right|\cdot 0=0\\ & \iff \left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)\cdot \left(\begin{array}{c}4\\ -1\\ t\end{array}\right)=0\\ & \iff (-2)(4)+(1)(-1)+(3)(t)=0\\ & \iff -8-1+3t=0\\ & \iff 3t=9\\ & \iff t=3\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Yuana, R.A., Indriyastuti. 2017. Persektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI MANDIRI.

Lanjutan Contoh Soal Vektor Dimensi Dua (Bagian 3)

$\begin{array}{l}\\ 7& \text{Jika}\overline{a}=\left(\begin{array}{c}4\\ -3\end{array}\right)\text{dan}\overline{b}=\left(\begin{array}{c}1\\ 2\end{array}\right)\\ & \text{tentukanlah}\\ & \text{a}.\overline{a}+\overline{b}\text{d}.(3\overline{a}+3\overline{b})\\ & \text{b}.\overline{b}+\overline{a}\text{e}.2(\overline{a}-\overline{b})\\ & \text{c}.3(\overline{a}+\overline{b})\text{f}.2\overline{a}-2\overline{b}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}& \overline{a}+\overline{b}=\left(\begin{array}{c}4\\ -3\end{array}\right)+\left(\begin{array}{c}1\\ 2\end{array}\right)=\left(\begin{array}{c}4+1\\ -3+2\end{array}\right)=\left(\begin{array}{c}5\\ -1\end{array}\right)\\ \text{b}& \overline{b}+\overline{a}=\left(\begin{array}{c}1\\ 2\end{array}\right)+\left(\begin{array}{c}4\\ -3\end{array}\right)=\left(\begin{array}{c}1+4\\ 2+(-3)\end{array}\right)=\left(\begin{array}{c}5\\ -1\end{array}\right)\\ \text{c}& \cdots \\ \text{d}& 3\overline{a}+3\overline{b}=3\left(\begin{array}{c}4\\ -3\end{array}\right)+3\left(\begin{array}{c}1\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}12\\ -9\end{array}\right)+\left(\begin{array}{c}3\\ 6\end{array}\right)=\left(\begin{array}{c}12+3\\ -9+6\end{array}\right)=\left(\begin{array}{c}15\\ -3\end{array}\right)\\ \text{e}& \cdots \\ \text{f}& \cdots \end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Pada Balok ABCD.EFGH diatas diketahui }\\ & \text{DA = 4 cm, DC = 5 cm, dan DH 3 cm}.\\ & \text{Misalkan}\overrightarrow{i}\text{adalah vektor satuan dengan arah }\\ & \text{sama dengan}\overrightarrow{DA},\overrightarrow{j}\text{adalah vektor satuan }\\ & \text{dengan arah sama}\overrightarrow{DC},\text{dan}\overrightarrow{k}\\ & \text{adalah vektor satuan dengan arah sama dengan}\overrightarrow{DH}.\\ & \text{Nyatakanlah vektor-vektor berikut dalam vektor}\overrightarrow{i},\overrightarrow{j}\text{dan}\overrightarrow{k}.\\ & \begin{array}{l}\\ \text{a}.& \overrightarrow{DA},\overrightarrow{DC}\text{dan}\overrightarrow{DH}\\ \text{b}.& \overrightarrow{DB}\text{dan}\overrightarrow{DF}\\ \text{c}.& \overrightarrow{DP}\text{jika}P\text{titik tengan}EF\\ \text{d}.& \overrightarrow{DQ}\text{jika}Q\text{titik pada perpanjangan}\\ & FG\text{dengan}FG=GQ\end{array}\end{array}$.

$.\begin{array}{rl}& \text{Jawab}:\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{a}.& \{\begin{array}{l}\overrightarrow{DA}=4\overrightarrow{i}\\ \overrightarrow{DC}=5\overrightarrow{j}\\ \overrightarrow{DH}=3\overrightarrow{k}\end{array}\\ & \\ & \end{array}& \begin{array}{rl}\text{b}.\overrightarrow{DB}& =\overrightarrow{DC}+\overrightarrow{CB}\\ & =\overrightarrow{DC}+\overrightarrow{DA}\\ & =5\overrightarrow{j}+4\overrightarrow{i}\\ \overrightarrow{DF}& =\overrightarrow{DB}+\overrightarrow{BF}\\ & =4\overrightarrow{i}+5\overrightarrow{j}+3\overrightarrow{k}\end{array}\\ \hline\end{array}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{c}.\overrightarrow{DP}& =\overrightarrow{DE}+\overrightarrow{EP}\\ & =\overrightarrow{DA}+\overrightarrow{AE}+\overrightarrow{EP}\\ & =4\overrightarrow{i}+3\overrightarrow{k}+\frac{1}{2}\overrightarrow{EF}\\ & =4\overrightarrow{i}+3\overrightarrow{k}+\frac{1}{2}\overrightarrow{DC}\\ & =4\overrightarrow{i}+3\overrightarrow{k}+\frac{1}{2}\left(5\overrightarrow{j}\right)\\ & =4\overrightarrow{i}+\frac{5}{2}\overrightarrow{j}+3\overrightarrow{k}\end{array}& \begin{array}{rl}\text{d}.\overrightarrow{DQ}& =\overrightarrow{DG}+\overrightarrow{GQ}\\ & =\overrightarrow{DC}+\overrightarrow{CG}+\overrightarrow{GQ}\\ & =\overrightarrow{DC}+\overrightarrow{DH}+\overrightarrow{AD}\\ & =\overrightarrow{DC}+\overrightarrow{DH}-\overrightarrow{DA}\\ & =5\overrightarrow{j}+3\overrightarrow{k}-4\overrightarrow{i}\\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Diketahui}\overrightarrow{a}=\left(\begin{array}{c}-5\\ 3\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}6\\ -2\end{array}\right).\\ & \text{Tentukanlah}\\ & \text{a}.\overrightarrow{a}-\overrightarrow{b}\text{b}.\overrightarrow{b}-\overrightarrow{a}\\ & \text{c}.4(\overrightarrow{a}-\overrightarrow{b})\text{d}.4\overrightarrow{a}-4\overrightarrow{b}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \begin{array}{rl}a.\overrightarrow{a}-\overrightarrow{b}& =\overrightarrow{a}+(-\overrightarrow{b})\\ & =\left(\begin{array}{c}-5\\ 3\end{array}\right)+\left(\begin{array}{c}-6\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}-11\\ 5\end{array}\right)\end{array}& \\ \begin{array}{rl}b.\overrightarrow{b}-\overrightarrow{a}& =\overrightarrow{b}+(-\overrightarrow{a})\\ & =\left(\begin{array}{c}6\\ -2\end{array}\right)+\left(\begin{array}{c}5\\ -3\end{array}\right)\\ & =\left(\begin{array}{c}11\\ -5\end{array}\right)\end{array}\\ \begin{array}{rl}c.4(\overrightarrow{a}-\overrightarrow{b})& =4\left(\begin{array}{c}-11\\ 5\end{array}\right)\\ & =\left(\begin{array}{c}-44\\ 20\end{array}\right)\\ & \end{array}& \end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Diketahui}\overrightarrow{p}=\left(\begin{array}{c}4\\ -3\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}1\\ 2\end{array}\right).\\ & \text{Tentukanlah}\\ & \text{a}.\overrightarrow{p}+\overrightarrow{q}\text{b}.\overrightarrow{q}+\overrightarrow{p}\\ & \text{c}.4(\overrightarrow{p}+\overrightarrow{q})\text{d}.4\overrightarrow{p}+4\overrightarrow{q}\\ & \text{e}.4(\overrightarrow{p}-\overrightarrow{q})\text{f}.4\overrightarrow{p}-4\overrightarrow{q}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \begin{array}{rl}a.\overrightarrow{p}+\overrightarrow{q}& =\left(\begin{array}{c}4\\ -3\end{array}\right)+\left(\begin{array}{c}1\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}5\\ -1\end{array}\right)\end{array}& \begin{array}{rl}b.\overrightarrow{q}+\overrightarrow{p}& =\left(\begin{array}{c}1\\ 2\end{array}\right)+\left(\begin{array}{c}4\\ -3\end{array}\right)\\ & =\left(\begin{array}{c}5\\ -1\end{array}\right)\end{array}& & \end{array}\end{array}$

$\begin{array}{l}\\ 11.& \text{Pada contoh soal No. 8 tentukanlah }\\ & \text{panjang vektor}\overrightarrow{DA},\overrightarrow{DP}\text{dan}\overrightarrow{DQ}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \left|\overrightarrow{DA}\right|=4cm,\\ & {\left|\overrightarrow{DP}\right|}^2={\left|\overrightarrow{DE}\right|}^2+{\left|\overrightarrow{EP}\right|}^2\\ & ={\left|\overrightarrow{DA}\right|}^2+{\left|\overrightarrow{AE}\right|}^2+{\left|\overrightarrow{EP}\right|}^2\\ & =4^2+3^2+{\left(\frac{5}{2}\right)}^2\\ & =16+9+\frac{25}{4}\\ & \left|\overrightarrow{DP}\right|=\sqrt{\frac{125}{4}}\\ & ={\displaystyle \frac{5}{2}\sqrt{5}cm,\text{dan}}\\ & {\left|\overrightarrow{DQ}\right|}^2={\left|\overrightarrow{DG}\right|}^2+{\left|\overrightarrow{GQ}\right|}^2\\ & ={\left|\overrightarrow{DC}\right|}^2+{\left|\overrightarrow{CG}\right|}^2+{\left|\overrightarrow{GQ}\right|}^2\\ & =5^2+3^2+4^2\\ & =25+9+16\\ & =50\\ & \left|\overrightarrow{DQ}\right|& =\sqrt{50}\\ & =5\sqrt{2}cm\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Perhatikanlah gambar di bawah ini}\\ & \text{dan nyatakan titik-titik pada gambar }\\ & \text{tersebut dalam vektor posisi}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\\ & \text{Perhatikanlah gambar berikut}\end{array}$

$.\begin{array}{|ll|}\hline \text{Titik}& \text{Vektor Posisi}\\ P(5,3)& \overrightarrow{p}=\left(\begin{array}{c}5\\ 3\end{array}\right)\\ Q(2,-3)& \overrightarrow{q}=\left(\begin{array}{c}2\\ -3\end{array}\right)\\ R(-5,-1)& \overrightarrow{r}=\left(\begin{array}{c}-5\\ -1\end{array}\right)\\ S(-3,7)& \overrightarrow{s}=\left(\begin{array}{c}-3\\ 7\end{array}\right)\\ \hline\end{array}$

$\begin{array}{l}\\ 13.& \text{Perhatikanlah gambar pada soal No. 12 di atas. }\\ & \text{Tentukanlah vektor-vektor berikut}\\ & \text{a}.\overrightarrow{PQ}\\ & \text{b}.\overrightarrow{PS}\\ & \text{c}.\overrightarrow{QS}\\ & \text{d}.\overrightarrow{QP}+\overrightarrow{PS}\\ \\ & \text{Jawab}:\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{a}.\overrightarrow{PQ}& =\overrightarrow{PO}+\overrightarrow{OQ}\\ & =-\overrightarrow{OP}+\overrightarrow{OQ}\\ & =\overrightarrow{OQ}-\overrightarrow{OP}\\ & =\overrightarrow{q}-\overrightarrow{p}\\ & =\left(\begin{array}{c}2\\ -3\end{array}\right)-\left(\begin{array}{c}5\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-3\\ -6\end{array}\right)\end{array}& \begin{array}{rl}\text{b}.\overrightarrow{PS}& =\overrightarrow{s}-\overrightarrow{p}\\ \text{dengan }& \text{cara semisal poin a}\\ \overrightarrow{PS}& =\left(\begin{array}{c}-3\\ 7\end{array}\right)-\left(\begin{array}{c}5\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-8\\ 4\end{array}\right)\\ & \\ & \end{array}\\ \hline\end{array}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{c}.\overrightarrow{QS}& =\overrightarrow{s}-\overrightarrow{q}\\ & =\left(\begin{array}{c}-3\\ 7\end{array}\right)-\left(\begin{array}{c}2\\ -3\end{array}\right)\\ & =\left(\begin{array}{c}-5\\ 10\end{array}\right)\end{array}& \begin{array}{rl}\text{d}.\overrightarrow{QP}+\overrightarrow{PS}& =\overrightarrow{QS}\\ & =\left(\begin{array}{c}-5\\ 10\end{array}\right)\\ \text{lihat jawaban}& \text{ poin c}\end{array}\\ \hline\end{array}\end{array}$