Matriks (Matematika Wajib Kelas XI)

$\text{A. Pendahuluan}$

Dalam kehidupan sehari-hari kita sering menjumpai atau mendapatkan informasi yang tersaji dalam bentuk daftar atau tabel. Sebagai misal ketika seorang wali kelas merekap data absensi kelas waliannya selama satu semester sebagaimana diperlihatkan dalam tabel berikut ini

$\begin{array}{l}\\ \begin{array}{|lccc|}\hline \text{Nama}& \text{Sakit}& \text{Izin}& \text{Tanpa}\\ \text{Siswa}& & & \text{Keterangan}\\ \text{Andi}& 2& 1& 3\\ \text{Budi}& 1& 4& 2\\ \text{Carli}& 1& 1& 5\\ \text{Dodi}& 2& 2& 1\\ \hline\end{array}& \begin{array}{rl}& \Leftarrow \text{Judul baris}\\ & \\ & \\ & \\ & \end{array}\\ \begin{array}{rl}& \Uparrow \\ & \text{Judul kolom}\end{array}& \end{array}$

Dari tabel di atas, jika kita tuliskan bilangannya saja maka akan kita dapatkan bilangan yang seolah-olah tersusun berbentuk persegi atau persegi panjang dan oleh kareanya sebagaimana ilsutrasi tabel di atas bilangan terbut juga tersusun dalam baris dan kolom sebagaimana berikut ini

$\begin{array}{l}\\ & 2& 1& 3\\ \\ & 1& 4& 2\\ \\ & 1& 1& 5\\ \\ & 2& 2& 1\end{array}$

Selanjutnya kaitanya dengan matriks apa bila susunan bilangan-bilangan di atas, diberikan tanda kurung tertentu jadil bentuk matriks. Dan dari paparan tersebut kita dapat mengakatakan matriks adalah susunan bilangan dalam bentuk persegi atau persegi panjang yang di atur menurut baris dan kolom dalam tanda kurung tertentu.

Selanjutnya bilangan yang diatus menurut baris dan kolom disebut unsur atau elemen atau entri dari suatu matri

Perhatikanlah ilustrasi berikut

Sebagai tambahan nama sebuah matrik adalah sebuah huruf besar dan memiliki ukuran sebuah matrik(selanjutnya dapat isebut sebagai ordo) = baris x kolom.

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Diketahui}M=\left[\begin{array}{ccccc}-2& 3& 0& 0& -3\\ -1& 2& 1& 1& 2\\ 0& 1& -3& 2& -2\\ 1& 0& -5& 3& 1\end{array}\right],\text{tentukanlah}\\ \\ & \text{a}.\text{ordo dari matrik tersebut}\\ & \text{b}.\text{elemen penyusun kolom pertama}\\ & \text{c}.\text{elemen penyusun baris pertama}\\ & \text{d}.\text{elemen penyusun kolom kedua}\\ & \text{e}.\text{elemen penyusun baris kedua}\\ & \text{f}.\text{elemen penyusun kolom ketiga}\\ & \text{g}.\text{elemen penyusun baris ketiga}\\ & \text{h}.\text{elemen penyusun kolom keempat}\\ & \text{i}.\text{elemen penyusun baris keempat}\\ & \text{j}.\text{elemen penyusun kolom kelima}\\ & \text{k}.\text{elemen penyusun baris kelima}\\ & \text{l}.\text{elemen yang terletak pada baris kelima dan kolom kelima}\\ & \text{m}.\text{elemen yang terletak pada baris pertama dan kolom kelima}\\ & \text{n}.\text{elemen yang terletak pada baris kedua dan kolom keempat}\\ & \text{o}.\text{elemen yang terletak pada baris ketiga dan kolom ketiga}\\ & \text{p}.\text{elemen yang terletak pada baris kedua dan kolom kedua}\\ & \text{q}.\text{elemen yang terletak pada baris pertama dan kolom pertama}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\\ & \text{a}.\text{ordo matriknya}4\times 5\\ & \begin{array}{|cccc|}\hline \text{b}.& -2,-1,0,1& \text{c}.& -2,3,0,0,-3\\ \text{d}.& 3,2,1,0& \text{e}.& -1,2,1,1,2\\ \text{f}.& 0,1,-3,-5& \text{g}.& 0,1,-3,2,-2\\ \text{h}.& 0,1,2,3& \text{i}.& 1,0,-5,3,1\\ \text{j}& -3,2,-2,1& \text{k}& \text{tidak ada}\\ \text{l}.& \text{tidak ada}& \text{m}& m_{15}=-3\\ \text{n}.& m_{24}=1& \text{o}.& m_{33}=-3\\ \text{p}.& m_{22}=2& \text{q}.& m_{11}=-2\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Diketahui}A=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],B=\left[\begin{array}{cc}3& 7\\ 4& 8\\ 5& 9\\ 6& 10\end{array}\right],\\ & C=\left[\begin{array}{ccc}11& 12& 13\\ 14& 15& 16\\ 17& 18& 19\\ 20& 21& 22\end{array}\right]\text{tentukanlah nilai}\\ \\ & \text{a}.a_{11},a_{31},b_{42},\text{dan}{c}_{43}\\ & \text{b}.b_{42}+c_{43}\\ & \text{c}.a_{41}-b_{31}+c_{21}-a_{11}\\ & \text{d}.a_{11}+b_{22}+c_{33}\\ & \text{e}.a_{11}^2+b_{22}^2+c_{33}^2\\ & \text{f}.{(a_{11}+b_{22})}^2-c_{33}^2\end{array}$

$.\begin{array}{rl}& \text{Jawab}\\ & \begin{array}{|clcl|}\hline \text{a}.& \begin{array}{rl}& \text{perhatikanlah}\\ & a_{11}=1,a_{31}=3\\ & b_{42}=10,c_{43}=22\end{array}& \text{b}.& \begin{array}{rl}& b_{42}+c_{43}\\ & =10+22\\ & =32\end{array}\\ \text{c}.& \begin{array}{rl}& a_{41}-b_{31}+c_{21}-a_{11}\\ & =4-5+14-1\\ & =12\end{array}& \text{d}.& \begin{array}{rl}& a_{11}+b_{22}+c_{33}\\ & =1+8+19\\ & =28\end{array}\\ \text{e}.& \begin{array}{rl}& a_{11}^2+b_{22}^2+c_{33}^2\\ & =1^2+8^2+{19}^2\\ & =1+64+361=426\end{array}& \text{f}.& \begin{array}{rl}& {(a_{11}+b_{22})}^2-c_{33}^2\\ & =(1+8{)}^2-{19}^2\\ & =81-361=-280\end{array}\\ \hline\end{array}\end{array}$

Contoh Soal 8 Statistika

$\begin{array}{l}\\ 36.& \text{Diketahui nilai statistik lima serangkai}\\ & \text{dari empat kelompok data seperti terlihat}\\ & \text{dalam tabel berikut}\\ & \begin{array}{|lccccc|}\hline \text{Data}& \text{min}& Q_1& Q_2& Q_3& \text{mak}\\ \text{I}& 74& 80& 88& 92,5& 99\\ \text{II}& 66& 81,5& 86& 90,5& 96\\ \text{III}& 70& 77,5& 85& 92,5& 100\\ \text{IV}& 55& 80& 88& 90& 97,5\\ \hline\end{array}\\ & \text{Data yang memuat pencilan terdapat pada}\\ & \text{tabel}\\ & \begin{array}{l}\\ \text{a}.& \text{I dan II}\\ \text{b}.& \text{II dan III}\\ \text{c}.& \text{I dan III}\\ \text{d}.& \text{III dan IV}\\ \text{e}.& \text{II dan IV}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{Diketa}& \text{hui bahwa}\mathbf{\text{pencilan}}\text{adalah datum yang}\\ \text{bernil}& \text{ai kurang dari pagar dalam dan lebih besar}\\ \text{dari p}& \text{agar luar}\\ \text{Rumu}& \text{s pagar dalam}=Q_1-L\\ & =Q_1-{\displaystyle \frac{3}{2}(Q_3-Q_1)={\displaystyle \frac{5}{2}{Q}_1-\frac{3}{2}{Q}_3}}\\ \text{Rumu}& \text{s pagar luar}=Q_3+L\\ & =Q_3+{\displaystyle \frac{3}{2}(Q_3-Q_1)={\displaystyle \frac{5}{2}{Q}_3-\frac{3}{2}{Q}_1}}\\ \text{Data I}& \text{pagar dalamnya}={\displaystyle \frac{5}{2}(80)-\frac{3}{2}(92,5)}\\ & =200-138,75=61,25\\ \text{Data I}& \text{pagar luarnya}={\displaystyle \frac{5}{2}(92,5)-\frac{3}{2}(80)}\\ & =231,25-120=111,25\\ \text{Jadi, d}& \text{ata I tidak ada pencilan}\\ \text{Data I}& \text{I pagar dalamnya}={\displaystyle \frac{5}{2}(81,5)-\frac{3}{2}(90,5)}\\ & =203,75-135,75=68\\ \text{Data I}& \text{I pagar luarnya}={\displaystyle \frac{5}{2}(90,5)-\frac{3}{2}(81,5)}\\ & =226,25-122,25=104\\ \text{Jadi, d}& \text{ata II ada pencilan, yaitu}66<68\\ & \text{karena}66\text{adalah datum terkecil data II}\\ \text{Data I}& \text{II pagar dalamnya}={\displaystyle \frac{5}{2}(77,5)-\frac{3}{2}(92,5)}\\ & =193,75-138,75=55\\ \text{Data I}& \text{II pagar luarnya}={\displaystyle \frac{5}{2}(92,5)-\frac{3}{2}(77,5)}\\ & =231,25-116,25=115\\ \text{Jadi, d}& \text{ata III tidak ada pencilan}\\ \text{Data I}& \text{V pagar dalamnya}={\displaystyle \frac{5}{2}(80)-\frac{3}{2}(90)}\\ & =200-135=65\\ \text{Data I}& \text{V pagar luarnya}={\displaystyle \frac{5}{2}(90)-\frac{3}{2}(70)}\\ & =225-105=120\\ \text{Jadi, d}& \text{ata IV ada pencilan, yaitu}55<65\\ & \text{karena}55\text{adalah datum terkecil data IV}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastolan, & Sulasim. 2005. Kompetensi Matematika SMA Kelas 2 Semester 1 Program IPA Kurikulum Berbasis Kompetensi. Jakarta: YUDHISTIRA.
2. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI. Bandung: YRAMA WIDYA.
3. Sharma, dkk. 2017. Jelajah Matematika SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA.
4. Tim Supermat. 2007. Cara Mudah MenghadapiS SMBB TELKOM. Jakarta: LITERATUR MEDIA SUKSES.

Contoh Soal 7 Statistika

$\begin{array}{l}\\ 31.& (\mathbf{\text{SMBB TELKOM 06}})\\ & \text{Berikut tidak termasuk ukuran penyebaran}\\ & \text{data adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{rentang}\\ \text{b}.& \text{varians}\\ \text{c}.& \text{jarak antar kuartil}\\ \text{d}.& \text{kuartil}\\ \text{e}.& \text{simpangan baku}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Ingat bahwa mean, median, modus}\\ & \text{adalah bagian ukuran pemusatan data}\\ & \text{sedangkan kuartil adalah ukuran letak data}\end{array}\end{array}$.

$\begin{array}{l}\\ 32.& \text{Perhatikanlah barisan}1,-2,3,-4,5,-6,...\\ & \text{dengan suku ke}-n\text{adalah}(-1{)}^{n-1}\times n.\\ & \text{Rata-rata 200 suku pertama barisan tersebut}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& -0,5\\ \text{c}.& 0\\ \text{d}.& 0,5\\ \text{e}.& 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}:1,-2,3,-4,5,-6,...,199,-200\\ & {\bar{x}}_{200}={\displaystyle \frac{1-2+3-4+5-6+...+199-200}{200}}\\ & ={\displaystyle \frac{-100}{200}=-0,5}\end{array}\end{array}$

$\begin{array}{l}\\ 33.& (\mathbf{\text{STT TELKOM 2005}})\\ & \text{Seratus mahasiswa telah mengikuti ujian psikotes}\\ & \text{dan rata-rata skornya yang diperoleh adalah 100}.\\ & \text{Banyaknya mahasiswa junior yang mengikuti}\\ & \text{ujian psikotes}50\mathrm{\%}\text{lebih besar dari banyknya ma}-\\ & \text{hasiswa senior. Jika rata-rata skor dari mahasiswa}\\ & \text{senior}50\mathrm{\%}\text{lebih tinggi dari mahasiswa junior,}\\ & \text{maka rata-rata skor mahasiswa senior adalah}....\\ & \begin{array}{l}\\ \text{a}.& 110\\ \text{b}.& 115\\ \text{c}.& 120\\ \text{d}.& 125\\ \text{e}.& 150\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Perhatikanlah tabel berikut}\\ & \begin{array}{|ccc|}\hline \text{Mahasiswa}& \text{Mahasiswa}& \text{Total}\\ \text{senior}& \text{junior}& \\ n_{senior}& n_{junior}& 100\\ n_s& {\displaystyle \frac{3}{2}{n}_s}& 100\\ {\bar{x}}_{senior}& {\bar{x}}_{junior}& 100\\ {\bar{x}}_s& {\displaystyle \frac{2}{3}{\bar{x}}_s}& 100\\ \hline\end{array}\\ & n_s+{\displaystyle \frac{3}{2}{n}_s=100}\\ & n_s=40,\\ & \text{maka}{n}_j=60.\\ & \text{Selanjutnya kita tentukan nilai rata-rata}\\ & \text{mahasiswa senior, dengan}\\ & {\displaystyle \frac{40{\bar{x}}_s+(60){\displaystyle \frac{2}{3}{\bar{x}}_s}}{100}=100}\\ & 80{\bar{x}}_s=10000\\ & {\bar{x}}_s={\displaystyle \frac{10000}{80}}\\ & {\bar{x}}_s=125\end{array}\end{array}$

$\begin{array}{l}\\ 34.& (\mathbf{\text{SPMB 04}})\\ & \text{Median, rata-rata, dan modus dari data yang}\\ & \text{terdiri atas empat bilangan asli adalah 7. Jika}\\ & \text{selisih antara data terbesar dan terkecil adalah}\\ & \text{6, maka hasil kali keempat datum tersebut}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1.864\\ \text{b}.& 1.932\\ \text{c}.& 1.960\\ \text{d}.& 1.976\\ \text{e}.& 1.983\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}:x_1,x_2,x_3,\text{dan}{x}_4\text{adalah}\\ & \text{asli dengan}{x}_4-x_1=6........(1)\\ & \mathrm{\&}\text{modusnya adalah 7, maka data dapat dituliskan}\\ & :x_1,7,7,x_4.\text{Jawaban ini sesuai karena median}=7.\\ & \text{Karena mean}=7,\text{maka dapat dituliskan}\\ & {\displaystyle \frac{x_1+x_2+x_3+x_4}{4}=7\iff x_1+x_2+x_3+x_4=28}\\ & \iff x_1+14+x_4=28\iff x_4+x_1=14........(2)\\ & \text{Selanjutnya kita eliminasi}(1)\mathrm{\&}(2)\\ & \begin{array}{l}\\ x_4-x_1& =6& \\ x_4+x_1& =14+\\ 2x_4& =20& \\ x_4& =10.......(3)\\ \text{maka}{x}_1& =4.........(4)\end{array}\\ & \text{Jadi},x_1\times x_2\times x_3\times x_4=(4).(7).(7).(10)=1960\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Desil ke-8}\left(D_8\right)\text{dari data berikut adalah}....\\ & \begin{array}{|cc|}\hline \text{Nilai}& f\\ 41-45& 7\\ 46-50& 12\\ 51-55& 9\\ 56-60& 8\\ 61-65& 4\\ \hline\end{array}\\ & \begin{array}{l}\\ \text{a}.& 58\\ \text{b}.& 57,5\\ \text{c}.& 57\\ \text{d}.& 56,75\\ \text{e}.& 56,25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui}& \text{desil ke}-8=D_8,\text{dengan}n=\sum f=40\\ D_i& =t_b+p\left({\displaystyle \frac{{\displaystyle \frac{i\times n}{10}-F}}{f}}\right)\\ D_8& =\text{datum ke}-\left({\displaystyle \frac{8n}{10}}\right)=x_{\frac{8\times 40}{10}}=x_{32}\\ \text{Dan}{x}_{32}& \text{terletak di kelas interval}:56-60\\ D_8& =55,5+5\left({\displaystyle \frac{32-28}{8}}\right)\\ & =55,5+2,5\\ & =58\end{array}\end{array}$

Contoh Soal 6 Statistika

$\begin{array}{l}\\ 26.& \text{Jika rata-rata dari}{x}_1,x_2,x_3,x_4,...,x_{10}\\ & \text{adalah}{x}_0,\text{maka rata-rata dari data}\\ & (x_1-1),(x_2+2),(x_3-3),(x_4+4),..\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x_0+5,5\\ \text{b}.& x_0+25\\ \text{c}.& x_0+0,5\\ \text{d}.& x_0-0,5\\ \text{e}.& x_0-2,5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Rata}-& \text{ratanya adalah}:\\ \bar{x}=x_0& ={\displaystyle \frac{x_1+x_2+x_3+...+x_{10}}{10}}\\ \iff 10x_0& =x_1+x_2+x_3+...+x_{10}\\ \text{Selanju}& \text{tnya penghitungan rata-rata yang data baru},\\ {\bar{x}}_{baru}& ={\displaystyle \frac{(x_1-1)+(x_2+2)+(x_3-3)+...+(x_{10}+10)}{10}}\\ & ={\displaystyle \frac{x_1+x_2+...+x_{10}+(2-1+4-3+..+10-9)}{10}}\\ & ={\displaystyle \frac{10x_0+5}{10}}\\ & =x_0+0,5\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Nilai rata-rata dari 20 bilangan adalah}14,2\\ & \text{Jika rata-rata dari 12 bilangan pertama adalah}\\ & \text{12,6 dan rata-rata dari 6 bilangan berikutnya}\\ & \text{adalah 18,2, rata-rata 2 bilangan tersisa}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 10,4\\ \text{b}.& 11,8\\ \text{c}.& 12,2\\ \text{d}.& 12,8\\ \text{e}.& 13,8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\bar{x}}_{total}& ={\displaystyle \frac{{\bar{x}}_{12}\times (12)+{\bar{x}}_6\times (6)+{\bar{x}}_2\times (2)}{20}}\\ 14,2& ={\displaystyle \frac{(12,6\times 12)+(18,2\times 6)+{\bar{x}}_2\times (2)}{20}}\\ 284& =151,2+109,2+2{\bar{x}}_2\\ 2{\bar{x}}_2& =284-(151,2+109,2)=23,6\\ {\bar{x}}_2& ={\displaystyle \frac{23,6}{2}}\\ & =11,8\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Dari 3 bilangan yang terkecil adalah 39}\\ & \text{dan terbesarnya adalah 75, maka rata-rata}\\ & \text{hitung ketiga bilangan tersebut tidak}\\ & \text{mungkin sama dengan}....\\ & \begin{array}{l}\\ \text{a}.& 49\\ \text{b}.& 52\\ \text{c}.& 53\\ \text{d}.& 59\\ \text{e}.& 60\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\bar{x}}_3& ={\displaystyle \frac{39+A+75}{3}}\\ \text{Se}& \text{lanjutnya rentang nilai}A\text{akan berada di}\\ :& 39\le A\le 75\\ \text{Se}& \text{hingga},\\ \text{un}& \text{tuk}A=39,\text{maka}\\ {\bar{x}}_3& ={\displaystyle \frac{39+39+75}{3}=\frac{153}{3}=51,\text{dan}}\\ \text{un}& \text{tuk}A=75,\text{maka}\\ {\bar{x}}_3& ={\displaystyle \frac{39+75+75}{3}=\frac{189}{3}=63}\end{array}\end{array}$

$\begin{array}{l}\\ 29.& (\mathbf{\text{SPMB 04}})\\ & \text{Nilai rata-rata tes Matematika dari kelompok}\\ & \text{siswa dan kelompok siswi di suatu kelas berturut-}\\ & \text{turut adalah 5 dan 7. Jika nilai rata-rata di kelas}\\ & \text{tersebut adalah 6,2, maka perbandingan banyaknya}\\ & \text{siswa dan siswi adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2:3\\ \text{b}.& 3:4\\ \text{c}.& 2:5\\ \text{d}.& 3:5\\ \text{e}.& 4:5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\bar{x}}_{gabungan}& ={\displaystyle \frac{n_1{\bar{x}}_1+n_2{\bar{x}}_2}{n_1+n_2}}\\ 6,2& ={\displaystyle \frac{n_1(5)+n_2(7)}{n_1+n_2}}\\ 6,2(n_1+n_2)& =5n_1+7n_2\\ 6,2n_1-5n_1& =7n_2-6,2n_2\\ 1,2n_1& =0,8n_2\\ {\displaystyle \frac{n_1}{n_2}}& ={\displaystyle \frac{0,8}{1,2}=\frac{2}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 30.& (\mathbf{\text{SPMB 05}})\\ & \text{Nilai rata-rata ulangan kelas A adalah}{\bar{x}}_A\text{dan}\\ & \text{kelas B adalah}{\bar{x}}_B.\text{Setelah kedua kelas digabungkan}\\ & \text{nilai rata-ratanya adalah}\bar{x}.\text{Perbandingan nilai}\\ & \text{kelas A dan B adalah}10:9.\text{Jika perbandingan nilai}\\ & \text{rata-rata kedua kelas dan kelas B adalah}85:81,\\ & \text{maka perbandinganbanyaknya siswa kelas A dan B}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 8:9\\ \text{b}.& 4:5\\ \text{c}.& 3:4\\ \text{d}.& 3:5\\ \text{e}.& 9:10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\bar{x}}_A:{\bar{x}}_B& =10:9=90:81\\ \bar{x}:{\bar{x}}_B& =85:81,\text{maka}\\ \bar{x}:{\bar{x}}_A:{\bar{x}}_B& =85:90:81\\ (n_A+n_B)\bar{x}& =n_A\times {\bar{x}}_A+n_B\times {\bar{x}}_B\\ {\displaystyle \frac{n_A}{n_B}}& ={\displaystyle \frac{\bar{x}-{\bar{x}}_B}{{\bar{x}}_A-\bar{x}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{85}{81}{\bar{x}}_B-{\bar{x}}_B}}{{\displaystyle \frac{10}{9}{\bar{x}}_B-{\displaystyle \frac{85}{81}{\bar{x}}_B}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{4}{81}{\bar{x}}_B}}{{\displaystyle \frac{5}{81}{\bar{x}}_B}}}\\ & ={\displaystyle \frac{4}{5}}\end{array}\end{array}$

Contoh Soal 5 Statistika

$\begin{array}{l}\\ 21.& \text{Simpangan kuartil dari data}\\ & 71,70,68,40,45,48,52,53,53,67,62\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 8\\ \text{b}.& 10\\ \text{c}.& 15\\ \text{d}.& 18\\ \text{e}.& 20\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}(\text{degan total datum ganjil})\\ :& 71,70,68,40,45,48,52,53,53,67,62\\ \text{Sete}& \text{lah data diurutkan menjadi}\\ :& 40,45,48,52,53,53,62,67,68,70,71\\ \text{Dike}& \text{tahui}n=11\mathbf{\text{ganjil}}\\ Q_1& =x_{\frac{1}{4}(n+1)}=x_{\frac{1}{4}.12}=x_3=48\\ Q_2& =x_{\frac{2}{4}(n+1)}=x_{\frac{2}{4}.12}=x_6=53\\ Q_3& =x_{\frac{3}{4}(n+1)}=x_{\frac{3}{4}.12}=x_9=68\\ \text{Sela}& \text{njutnya data dapat dituliskan}\\ & 40,45,\underset{\begin{array}{c}\Downarrow \\ Q_1\end{array}}{\underbrace{48}},52,53,53,62,67,\underset{\begin{array}{c}\Downarrow \\ Q_3\end{array}}{\underbrace{68}},70,71\\ \text{Simp}& \text{angan kuartil data tunggal adalah}:\\ & ={\displaystyle \frac{1}{2}(Q_3-Q_1)}\\ & ={\displaystyle \frac{1}{2}(68-48)}\\ & ={\displaystyle \frac{1}{2}.20=10}\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Data penjualan suatu barang setiap bulan}\\ & \text{di sebuah toko pada tahun 2019 adalah}:\\ & 20,3,9,11,4,12,1,9,9,12,8,10.\\ & \text{Median, kuartil bawah, dan kuartil atasnya}\\ & \text{berturut-turut adalah}....\\ & \begin{array}{l}\\ \text{a}.& 6{\displaystyle \frac{1}{2},3\frac{1}{2},\text{dan}9\frac{1}{2}}\\ \text{b}.& 9,6,\text{dan}11{\displaystyle \frac{1}{2}}\\ \text{c}.& 6{\displaystyle \frac{1}{2},9,\text{dan}12}\\ \text{d}.& 9,4,\text{dan}12\\ \text{e}.& 9,3{\displaystyle \frac{1}{2},\text{dan}12}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}\\ :& 20,3,9,11,4,12,1,9,9,12,8,10\\ \text{Sete}& \text{lah data diurutkan}\\ :& 1,3,4,8,9,9,9,10,11,12,12,20\\ \text{Dike}& \text{tahui}n=12\mathbf{\text{genap}}\\ Q_1& =x_{\frac{1}{4}n+\frac{1}{2}}=x_{\frac{1}{4}.12+\frac{1}{2}}=x_{3,5}=6\\ Q_2& =x_{\frac{2}{4}n+\frac{1}{2}}=x_{\frac{2}{4}.12+\frac{1}{2}}=x_{6,5}=9=M_e\\ Q_3& =x_{\frac{3}{4}n+\frac{1}{2}}=x_{\frac{3}{4}.12+\frac{1}{2}}=x_{9,5}=11{\displaystyle \frac{1}{2}}\\ \text{Sela}& \text{njutnya data dapat dituliskan}\\ & 1,3,\underset{\begin{array}{c}\Downarrow \\ Q_1\end{array}}{\underbrace{4,8}},9,\underset{\begin{array}{c}\Downarrow \\ Q_2=M_e\end{array}}{\underbrace{9,9}},10,\underset{\begin{array}{c}\Downarrow \\ Q_3\end{array}}{\underbrace{11,12}},12,20\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Ragam(varians) dari data}\\ & 6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 1{\displaystyle \frac{3}{8}}\\ \text{c}.& 1{\displaystyle \frac{1}{8}}\\ \text{d}.& {\displaystyle \frac{7}{8}}\\ \text{e}.& {\displaystyle \frac{5}{8}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}\\ :& 6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \text{Sete}& \text{lah data diurutkan(untuk memudahkan)}\\ :& 5,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9\\ \text{Dike}& \text{tahui}n=16.\text{Selanjutnya kita cari}\\ \bar{x}& ={\displaystyle \frac{5.1+6.4+7.6+8.4+9.1}{16}}\\ & =\frac{112}{16}=7\\ \text{Dan}& \text{rumus untuk menghitung ragam adalah}:\\ S^2& ={\displaystyle \frac{{\displaystyle \sum _{i=1}^{16}{(x_i-\bar{x})}^2}}{n}}\\ & ={\displaystyle \frac{(5-7{)}^2+4(6-7{)}^2+6(7-7{)}^2+4(8-7{)}^2+(9-7{)}^2}{16}}\\ & ={\displaystyle \frac{4+4.1+6.0+4.1+4}{16}}\\ & ={\displaystyle \frac{16}{16}}\\ & =1\end{array}\end{array}$

$\begin{array}{l}\\ 24.& \text{Diketahui}{x}_1=2,x_2=3,5,x_3=5,x_4=7,\\ & \text{dan}{x}_5=7,5.\text{Deviasi rata-rata data di atas}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& 1,8\\ \text{d}.& 2,6\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}\\ :& 2,3{\displaystyle \frac{1}{2},5,7,7{\displaystyle \frac{1}{2}}}\\ \text{Dike}& \text{tahui}n=5.\text{Selanjutnya kita cari}\\ \bar{x}& ={\displaystyle \frac{2+3,5+5+7+7,5}{5}=\frac{25}{5}=5}\\ \text{Dan}& \text{rumus simpangan rata-rata adalah}:\\ SR& ={\displaystyle \frac{{\displaystyle \sum _{i=1}^5|x_i-\bar{x}|}}{n}}\\ & ={\displaystyle \frac{|2-5|+|3,5-5|+|5-5|+|7-5|+|7,5-5|}{5}}\\ & ={\displaystyle \frac{|-3|+|-1,5|+\left|0\right|+\left|2\right|+|2,5|}{5}}\\ & ={\displaystyle \frac{3+1,5+0+2+2,5}{5}=\frac{9}{5}}\\ & =1,8\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Jumlah rataan dan median dari}\\ & (x-6),(x+5),(x+4),(x-7),\\ & (x+9),\text{dan}(x-2)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2x-1,5\\ \text{b}.& 2x-0,5\\ \text{c}.& 2x+1,5\\ \text{d}.& 2x+2,5\\ \text{e}.& 2x+3,5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}\\ :& (x-6),(x+5),(x+4),(x-7),(x+9),(x-2)\\ \text{Dike}& \text{tahui}n=6.\text{Selanjutnya kita urutkan datanya}\\ & (x-7),(x-6),(x-2),(x+4),(x+5),(x+9)\\ \text{Rata}& \text{annya}:\\ \bar{x}& ={\displaystyle \frac{6x+3}{6}=x+0,5}\\ \text{Medi}& \text{annya}:\\ M_e& =x_{3,4}={\displaystyle \frac{x_3+x_4}{2}=\frac{(x-2)+(x+4)}{2}}\\ & ={\displaystyle \frac{2x+2}{2}=x+1}\\ \text{Rata}& \text{an+median}=x+0,5+x+1=2x+1,5\end{array}\end{array}$

Contoh Soal 4 Statistika

$\begin{array}{l}\\ 16.& (\mathbf{\text{UN IPA 2014}})\\ & \text{Kuartil atas dari data pada tabel berikut}\\ & \text{adalah}....\\ & \begin{array}{|cc|}\hline \text{Data}& f\\ 20-25& 4\\ 26-31& 6\\ 32-37& 6\\ 38-43& 10\\ 44-49& 12\\ 50-55& 8\\ 56-61& 4\\ \hline\end{array}\\ & \begin{array}{l}\\ \text{a}.& 49,25\\ \text{b}.& 48,75\\ \text{c}.& 48,25\\ \text{d}.& 47,75\\ \text{e}.& 47,25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Kuartil atas}=Q_3,\text{dengan}n=\sum f=50\\ & \text{Kita sertakan lagi tabel di atas berikut}\\ & \begin{array}{|cc|}\hline \text{Data}& f\\ 20-25& 4\\ 26-31& 6\\ 32-37& 6\\ 38-43& 10\\ \text{44-49}& \text{12}\\ 50-55& 8\\ 56-61& 4\\ \hline\end{array}\\ & Q_3=\text{Datum ke}-\left({\displaystyle \frac{3n}{4}}\right)=x_{\frac{3.50}{4}}=x_{37,5}\\ & \text{dan}{x}_{37,5}\text{terletak di kelas interval}44-49\\ & Q_3=t_b+p\left({\displaystyle \frac{{\displaystyle \frac{3n}{4}-F}}{f}}\right)\\ & =43,5+6\left({\displaystyle \frac{37,5-26}{12}}\right)\\ & =43,5+{\displaystyle \frac{11,5}{2}}\\ & =49,5+5,75\\ & =49,25\end{array}\end{array}$

$\begin{array}{l}\\ 17.& (\mathbf{\text{UN IPA 2014}})\\ & \text{Perhatikanlah histrogram berikut}\\ & \end{array}$

$.\begin{array}{l}\\ & \text{Modus dari data pada histogram adalah}....\\ & \begin{array}{l}\\ \text{a}.& 23,25\\ \text{b}.& 23,75\\ \text{c}.& 24,00\\ \text{d}.& 25,75\\ \text{e}.& 26,25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui dari data histogram di atas adalah}:\\ & \begin{array}{|cc|}\hline \text{Data}& f\\ 3-7& 4\\ 8-12& 6\\ 13-17& 8\\ 18-22& 10\\ \text{23-27}& \text{12}\\ 28-32& 6\\ 33-37& 4\\ 38-42& 2\\ \hline\end{array}\\ & \text{Saat menentukan batas interval kurang lebih sama}\\ & \text{seperti menentukan panjang interval kelas}\\ & \text{Modus dari histogram di atas}\\ & M_0=t_b+p\left({\displaystyle \frac{\mathrm{△}{f}_1}{\mathrm{△}{f}_1+\mathrm{△}{f}_2}}\right)\\ & =22,5+5\left({\displaystyle \frac{(12-10)}{(12-10)+(12-6)}}\right)\\ & =22,5+\left({\displaystyle \frac{2}{2+6}}\right)=22,5+{\displaystyle \frac{10}{8}}\\ & =22,5+1,25=23,75\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Median dari data}\\ & 3,4,7,5,6,9,9,7,6,5,8\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 5\\ \text{b}.& 6\\ \text{c}.& 7\\ \text{d}.& 8\\ \text{e}.& 9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketah}& \text{ui data adalah ganjil}\\ \text{Median}& \text{ (datum tengah) data tunggal}:\\ \text{Data}& :3,4,7,5,6,9,9,7,6,5,8\\ \text{setel}& \text{ah diurutkan}\\ \text{Data}& :3,4,5,5,6,6,7,7,8,9,9\\ \text{Data}& :3,4,5,5,6,6,7,7,8,9,9\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Dari data berikut yang memiliki}\\ & \mathbf{\text{mean}}7{\displaystyle \frac{1}{2}\text{dan}\mathbf{\text{median}}7\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 2,5,6,9,7,8,5,14,8,11\\ \text{b}.& 6,3,7,8,6,4,11,8,9,8\\ \text{c}.& 3,7,10,7,9,5,10,2,14,11\\ \text{d}.& 4,1,6,12,8,11,4,5,8,2\\ \text{e}.& 2,3,4,3,10,8,12,6,15,12\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{|ll|}\hline \text{Mean}& \bar{x}={\displaystyle \frac{75}{10}=7,5}\\ \text{Median}& 2,5,6,9,7,8,5,14,8,11\\ \text{a}& 2,5,5,6,7,8,8,9,11,14\\ \text{Mean}& \bar{x}={\displaystyle \frac{70}{10}=7}\\ \text{Median}& 6,3,7,8,6,4,11,8,9,8\\ \text{b}& 3,4,6,6,7,8,8,8,9,11\\ \text{Mean}& \bar{x}={\displaystyle \frac{78}{10}=7,8}\\ \text{Median}& 3,7,10,7,9,5,10,2,14,11\\ \text{c}& 2,3,5,7,7,9,10,10,11,14\\ \text{Mean}& \bar{x}={\displaystyle \frac{61}{10}=6,1}\\ \text{Median}& 4,1,6,12,8,11,4,5,8,2\\ \text{d}& 1,2,4,4,5,6,8,8,11,12\\ \text{Mean}& \bar{x}={\displaystyle \frac{75}{10}=7,5}\\ \text{Median}& 2,3,4,3,10,8,12,6,15,12\\ \text{e}& 2,3,3,4,6,8,10,12,12,15\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Berikut adalah daftar nilai matematika}\\ & \text{kelas XII IA1}\\ & \begin{array}{|lcccccccc|}\hline \text{Nilai}& 3& 4& 5& 6& 7& 8& 9& 10\\ \text{Frekuensi}& 3& 5& 5& 9& 8& 6& 2& 2\\ \hline\end{array}\\ & \text{Jika siswa yang nilainya di atas rata-rata}\\ & \text{akan diikutsertakan dalam seleksi}\\ & \text{olimpiade matematika, maka banyak}\\ & \text{siswa yang mengikuti seleksi olimpiade}\\ & \text{adalah}...\text{siswa}\\ & \begin{array}{l}\\ \text{a}.& 6\\ \text{b}.& 8\\ \text{c}.& 10\\ \text{d}.& 18\\ \text{e}.& 27\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \begin{array}{|lcccccccc|}\hline x_i& 3& 4& 5& 6& 7& 8& 9& 10\\ f_i& 3& 5& 5& 9& 8& 6& 2& 2\\ x_i{f}_i& 9& 20& 25& 54& 56& 48& 18& 20\\ \hline\end{array}\\ & \text{Rata-rata nilai matematikanya}\\ & \bar{x}={\displaystyle \frac{\sum x_i{f}_i}{\sum f_i}}\\ & ={\displaystyle \frac{250}{40}=6,25}\\ & \text{Jadi, nilai rata-ratanya}6,25\\ & \text{Sehingga yang bisa ikut selesksi adalah}\\ & \text{nilai di atas rata-rata yaitu}:7,8,9,10\\ & \text{dan totalnya yang mendapatkan nilai}\\ & \text{itu sebanyak}:8+6+2+2=18\text{siswa}\end{array}\end{array}$

Contoh Soal 3 Statistika

$\begin{array}{l}\\ 11.& \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut}\\ & \begin{array}{|cc|}\hline \text{Interval}& f\\ 2-6& 2\\ 7-11& 3\\ 12-16& 3\\ 17-21& 6\\ 22-26& 6\\ \hline\end{array}\\ & \text{Rata-ratanya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 17,5\\ \text{b}.& 17\\ \text{c}.& 16,75\\ \text{d}.& 16,5\\ \text{e}.& 15,5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \begin{array}{|cccc|}\hline \text{Interval}& x_i& f_i& x_i.f_i\\ 2-6& 4& 2& 8\\ 7-11& 9& 3& 27\\ 12-16& 14& 3& 42\\ 17-21& 19& 6& 114\\ 22-26& 24& 6& 144\\ & \sum _{i=1}^5& 20& 335\\ \hline\end{array}\\ & \text{maka rata-ratanya}\\ & \bar{x}={\displaystyle \frac{\sum x_i.f_i}{\sum f_i}=\frac{335}{20}=16,75}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut}\\ & \begin{array}{|cc|}\hline \text{Interval}& f\\ 51-60& 5\\ 61-70& 7\\ 71-80& 14\\ 81-90& 8\\ 91-100& 6\\ \hline\end{array}\\ & \text{Rata-ratanya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 75,50\\ \text{b}.& 76,25\\ \text{c}.& 76,50\\ \text{d}.& 78,25\\ \text{e}.& 80,50\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \begin{array}{|cccc|}\hline \text{Interval}& x_i& f_i& x_i.f_i\\ 51-60& 55,5& 5& 277,5\\ 61-70& 65,5& 7& 458,5\\ 71-80& 75,5& 14& 1057\\ 81-90& 85,5& 8& 684\\ 91-100& 95,5& 6& 573\\ & \sum _{i=1}^5& 40& 3050\\ \hline\end{array}\\ & \text{maka rata-ratanya}\\ & \bar{x}={\displaystyle \frac{\sum x_i.f_i}{\sum f_i}=\frac{3050}{40}=76,25}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut(data sama dengan no.12 di atas)}\\ & \begin{array}{|cc|}\hline \text{Interval}& f\\ 51-60& 5\\ 61-70& 7\\ 71-80& 14\\ 81-90& 8\\ 91-100& 6\\ \hline\end{array}\\ & \text{Mediannya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 75,50\\ \text{b}.& 76,20\\ \text{c}.& 76,21\\ \text{d}.& 77,22\\ \text{e}.& 78,23\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui},n=\sum f=40,\text{Perhatikan tabel}\\ & \text{berikut ini}\\ & \begin{array}{|cc|}\hline \text{Interval}& f_i\\ 51-60& 5\\ 61-70& 7\\ 71-80& 14\\ 81-90& 8\\ 91-100& 6\\ & 40\\ \hline\end{array}\\ & Q_k=\text{Datum ke}-\left({\displaystyle \frac{kn}{4}}\right)\\ & Median=Q_2=\text{Datum ke}-\left({\displaystyle \frac{2.40}{4}}\right)=x_{20}\\ & x_{20}\text{terletak pada kelas interval}:71-80\\ & \text{dengan}f=14,F\text{sebelum}{Q}_2=12,\\ & t_b=70,5,\text{serta}p=10\\ & \text{maka mediannya}\\ & Q_2=t_b+p\left({\displaystyle \frac{{\displaystyle \frac{2n}{4}-F}}{f}}\right)\\ & =70,5+10\left({\displaystyle \frac{20-12}{14}}\right)\\ & =70,5+5,714=76,21\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut}\\ & \begin{array}{|cc|}\hline \text{Skor}& f\\ 40-49& 8\\ 50-59& 9\\ 60-69& 22\\ 70-79& 15\\ 80-89& 6\\ \hline\end{array}\\ & \text{Modusnya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 65,50\\ \text{b}.& 66,00\\ \text{c}.& 66,50\\ \text{d}.& 67,00\\ \text{e}.& 85,50\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui}n=\sum f=60,modusnya\\ & \text{terdapat pada kelas dengan frekuensi terbanyak}\\ & \text{yaitu}:60-69,\text{dengan}p=10\\ & \{\begin{array}{ll}\mathrm{△}{f}_i& =f-f_1=22-9=13\\ \mathrm{△}{f}_{ii}& =f-f_2=22-15=7\end{array}\\ & \text{Sehingga}\\ & M_0=t_b+p\left({\displaystyle \frac{\mathrm{△}{f}_1}{\mathrm{△}{f}_1+\mathrm{△}{f}_2}}\right)\\ & M_0=59,5+10\left({\displaystyle \frac{22-9}{(22-9)+(22-15)}}\right)\\ & =59,5+{\displaystyle \frac{10.13}{13+7}}\\ & =59,5+6,5=66,0\end{array}\end{array}$

$\begin{array}{l}\\ 15.& (\mathbf{\text{UN 2013}})\\ & \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut}\\ & \begin{array}{|cc|}\hline \text{Berat Badan (Kg)}& f\\ 45-49& 3\\ 50-54& 6\\ 55-59& 10\\ 60-64& 12\\ 65-69& 15\\ 70-74& 6\\ 75-79& 4\\ \hline\end{array}\\ & \text{Kuartil atasnya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 66{\displaystyle \frac{5}{6}}\\ \text{b}.& 67{\displaystyle \frac{1}{6}}\\ \text{c}.& 67{\displaystyle \frac{5}{6}}\\ \text{d}.& 68{\displaystyle \frac{1}{6}}\\ \text{e}.& 68{\displaystyle \frac{4}{6}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Kuartil atas}=Q_3,\text{dengan}n=\sum f=56\\ & Q_3=\text{Datum ke}-\left({\displaystyle \frac{3n}{4}}\right)=x_{\frac{3.56}{4}}=x_{42}\\ & \text{dan}{x}_{42}\text{terletak di kelas interval}65-69\\ & Q_3=t_b+p\left({\displaystyle \frac{{\displaystyle \frac{3n}{4}-F}}{f}}\right)\\ & =64,5+5\left({\displaystyle \frac{42-(3+6+10+12)}{15}}\right)\\ & =64{\displaystyle \frac{1}{2}+{\displaystyle \frac{11}{3}}}\\ & =64{\displaystyle \frac{3}{6}+3{\displaystyle \frac{4}{6}=68{\displaystyle \frac{1}{6}}}}\end{array}\end{array}$

Contoh Soal 2 Statistika

$\begin{array}{l}\\ 6.& \text{Tabel berikut adalah nilai tes matematika}\\ & \begin{array}{|cc|}\hline \text{Nilai}& f\\ 21-30& 1\\ 31-40& 2\\ 41-50& 5\\ 51-60& 7\\ 61-70& 8\\ 71-80& 8\\ 81-90& 6\\ 91-100& 1\\ \hline\end{array}\\ & \text{Banyak siswa yang mendapatkan nilai 71}\\ & \text{atau lebih adalah}....\\ & \begin{array}{l}\\ \text{a}.& 16\\ \text{b}.& 15\\ \text{c}.& 12\\ \text{d}.& 12\\ \text{e}.& 10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Perhatikan kembali tabelnya}\\ & \begin{array}{|cc|}\hline \text{Nilai}& f\\ 21-30& 1\\ 31-40& 2\\ 41-50& 5\\ 51-60& 7\\ 61-70& 8\\ 71-80& 8\\ 81-90& 6\\ 91-100& 1\\ \hline\end{array}\\ & \text{Nilai yang lebih dari 71 adalah}:\\ & 8+6+1=17\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Jangkauan dari tabel distribusi}\\ & \text{frekuensi pada no.6 di atas adalah}....\\ & \begin{array}{l}\\ \text{a}.& 60\\ \text{b}.& 70\\ \text{c}.& 79\\ \text{d}.& 89\\ \text{e}.& 100\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Karena data berkelompok, maka}\\ & \text{Jangkauan}=\left(\text{Nilai tengah kelas tertinggi}\right)\\ & -\left(\text{Nilai tengah kelas pertama}\right)\\ & ={\displaystyle \frac{1}{2}((100+91)-(30+21))=70}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}n=\text{banyak data},k=\text{banyak interval}\\ & \text{kelas, maka menurut aturan Sturges, rumus}\\ & \text{untuk menentukan nilai}k\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& k=\log \left(10.n^{3,3}\right)\\ \text{b}.& k=1+3,3\log n\\ \text{c}.& k=1-3,3\log (n-1)\\ \text{d}.& k=\log ({10}^{3,3}.n)\\ \text{e}.& k=\log n^{3,3}+2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Cukup jelas}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Rata-rata data soal no.6}\\ & \text{di atas adalah}....\\ & \begin{array}{l}\\ \text{a}& {\displaystyle 64,45}\\ \text{b}& {\displaystyle 64,55}\\ \text{c}& {\displaystyle 65,45}\\ \text{d}& {\displaystyle 65,55}\\ \text{e}& {\displaystyle 66}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \begin{array}{|cccc|}\hline \text{Nilai}& x_i& f_i& x_i.f_i\\ 21-30& 25,5& 1& 25,5\\ 31-40& 35,5& 2& 71\\ 41-50& 45,5& 5& 227,5\\ 51-60& 55,5& 7& 388,5\\ 61-70& 65,5& 8& 524\\ 71-80& 75,5& 8& 604\\ 81-90& 85,5& 6& 513\\ 91-100& 95,5& 1& 95,5\\ & \sum _{i=1}^8& 38& 2449\\ \hline\end{array}\\ & \text{Sehingga, rata-rata data nilai di atas adalah}\\ & \bar{x}={\displaystyle \frac{\sum x_i{f}_i}{\sum f_i}=\frac{2449}{38}=64,447368421\approx 64,45}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Rumus untuk menentukan}\mathrm{\%}{f}_{\text{rel}}=....\\ & \begin{array}{l}\\ \text{a}& {\displaystyle \frac{f_i+1}{\sum f}\times 100\mathrm{\%}}\\ \text{b}& {\displaystyle \frac{f_i-1}{\sum f}\times 100\mathrm{\%}}\\ \text{c}& {\displaystyle \frac{f_i}{\sum f}\times 100\mathrm{\%}}\\ \text{d}& {\displaystyle \frac{f_i}{\sum f+1}\times 100\mathrm{\%}}\\ \text{e}& {\displaystyle \frac{f_i}{\sum f-1}\times 100\mathrm{\%}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Cukup jelas}\end{array}\end{array}$

Contoh Soal 1 Statistika

$\begin{array}{rl}& \text{Perhatikanlah tabel berikut}\\ & \text{Nilai ulangan kelas XII}\\ & \text{untuk menjawab soal no. 1 sampai 5}\\ & \begin{array}{|ccc|}\hline \text{Nilai}& \text{Matematika}& \text{Bahasa Inggris}\\ 30-39& 1& 0\\ 40-49& 4& 2\\ 50-59& 6& 7\\ 60-69& 17& 18\\ 70-79& 10& 12\\ 80-89& 7& 6\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 1.& \text{Total datum pada data tabel}\\ & \text{distribusi frekuensi di atas adalah}....\\ & \begin{array}{l}\\ \text{a}.& 30\\ \text{b}.& 35\\ \text{c}.& 40\\ \text{d}.& 45\\ \text{e}.& 50\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Total datum pada tabel di atas}\\ & \text{sama dengan total frekuensi yaitu}:\\ & =1+4+6+17+10+7=45,\\ & \text{atau}\\ & =0+2+7+18+12+6=45\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Banyak kelas interval pada tabel}\\ & \text{distribusi frekuensi tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& 4\\ \text{b}.& 5\\ \text{c}.& 6\\ \text{d}.& 7\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Data di atas terbagai dalam 6}\\ & \text{kelas intervalnya, yaitu}:\\ & \text{kelas pertama : 30-39}\\ & \text{kelas kedua : 40-49}\\ & \text{kelas ketiga : 50-59}\\ & \text{kelas keempat : 60-69}\\ & \text{kelas kelima : 70-79}\\ & \text{kelas keenam : 80-89}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Panjang kelas interval pada tabel}\\ & \text{di atas adalah}....\\ & \begin{array}{l}\\ \text{a}.& 9\\ \text{b}.& 10\\ \text{c}.& 11\\ \text{d}.& 12\\ \text{e}.& 13\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Panjnag kelas interval pada}\\ & \text{tabel distribusi frekuensi di atas}\\ & \text{ambil contoh kelas pertama yaitu}:\\ & \text{pada}30-39\text{ada}\\ & =(39-30)+1=10\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Titik tengah dari kelas interval ke enam}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 84\\ \text{b}.& 84,5\\ \text{c}.& 85\\ \text{d}.& 85,5\\ \text{e}.& 86\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Titik tengah interval kelas keenam}\\ & \text{pada tabel distribusi frekuensi di atas}\\ & \text{yaitu}:\\ & ={\displaystyle \frac{1}{2}(80+89)=\frac{169}{2}=84,5}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Tepi bawah dan tepi atas dari kelas}\\ & \text{interval dari tabel distribusi frekuensi}\\ & \text{di atas yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& 30,5\text{dan}39,5\\ \text{b}.& 39,5\text{dan}49,5\\ \text{c}.& 50,5\text{dan}59,5\\ \text{d}.& 60,5\text{dan}70,5\\ \text{e}.& 79,05\text{dan}89,05\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Tepi bawah dan tepi atas dari kelas}\\ & \text{interval pada tabel distribusi frekuensi}\\ & \text{di atas yaitu}:\\ & \{\begin{array}{ll}\text{tepi bawah}& =x_i-0,5\\ \text{tepi atas}& =x_i+0,5\end{array}\\ & \text{Berikut tabelnya}\\ & \begin{array}{rl}& \begin{array}{|ccc|}\hline \text{Nilai}& \text{tepi bawah}& \text{tepi atas}\\ 30-39& 30-0,5=29,5& 39+0,5=39,5\\ 40-49,5& 40-0,5=39,5& 49+0,5=49,5\\ 50-59& .....=49,5& ....=59,5\\ 60-69& ....=59,5& ....=69,5\\ 70-79& ....=69,5& ....=79,5\\ 80-89& ....=79,5& ....=89,5\\ \hline\end{array}\end{array}\end{array}\end{array}$

Contoh Soal 4 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{l}\\ 16.& \text{Diketahui suatu fungsi kuadrat}\\ & f(x)=a{x}^2+bx+c.\text{Jika fungsi}\\ & (-1,0),(1,4),\text{dan}(2,9),\text{maka}\\ & \text{fungsi yang dimaksud adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle f(x)=x^2-2x+3}\\ \text{b}.& f(x)=x^2+2x+3\\ \text{c}.& f(x)=x^2+2x-3\\ \text{d}.& f(x)=x^2-2x-3\\ \text{e}.& f(x)=x^2+2x+1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}(-1,0)\Rightarrow f(-1)=a-b+c=0....(1)\\ (1,4)\Rightarrow f(1)=a+b+c=4....(2)\\ (2,9)\Rightarrow f(2)=4a+2b+c=9....(3)\end{array}\\ & \text{Saat}(1)\mathrm{\&}(2),\text{didapatkan}\\ & b=2...............(4)\\ & \text{Saat}(1)\mathrm{\&}(3),\text{didapatkan}\\ & \begin{array}{l}\\ 4a+2b+c& =9& \\ a-b+c& =0& -\\ 3a+3b& =9& \\ a+b& =3& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \{\begin{array}{ll}a& =1\\ c& =1\end{array}\\ & \text{Jadi},f(x)=a{x}^2+bx+c=x^2+2x+1\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Diketahui persamaan}\{\begin{array}{ll}x-y& =2\\ kx+y& =3\end{array}\\ & \text{memiliki solusi}(x,y)\text{di kuadran I}\\ & \text{Jika dan hanya jika nilai}k\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle k=-1}\\ \text{b}.& k>-1\\ \text{c}.& k<{\displaystyle \frac{3}{2}}\\ \text{d}.& 0<k<{\displaystyle \frac{3}{2}}\\ \text{e}.& -1<k<{\displaystyle \frac{3}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x-y=2....(1)\\ kx+y=3....(2)\end{array}\\ & \text{Dengan metode matriks didapatkan}\\ & x={\displaystyle \frac{\left|\begin{array}{cc}2& -1\\ 3& 1\end{array}\right|}{\left|\begin{array}{cc}1& -1\\ k& 1\end{array}\right|}={\displaystyle \frac{2-(-3)}{1+k}=\frac{5}{k+1}}}\\ & \text{Dengan cara yang sama pula}\\ & y={\displaystyle \frac{\left|\begin{array}{cc}1& 2\\ k& 3\end{array}\right|}{\left|\begin{array}{cc}1& -1\\ k& 1\end{array}\right|}={\displaystyle \frac{3-2k}{k+1}}}\\ & \text{Supaya memiliki solusi di kwadran I},\\ & \text{maka baik}x\text{maupun}y\\ & \text{haruslah positif, akibatnya}:\\ & k+1>0\Rightarrow k>-1\\ & \text{Sebagai akibat yang lain adalah}:\\ & 3-2k>0\Rightarrow k<{\displaystyle \frac{3}{2}}\\ & \text{Jadi},-1<k<{\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Diketahui sistem persamaan}\\ & y+{\displaystyle \frac{2}{x+z}=4}\\ & 5y+{\displaystyle \frac{18}{2x+y+z}=18}\\ & {\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3}\\ & \text{Nilai}y+\sqrt{x^2-2xz+y^2}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 3}\\ \text{b}.& 5\\ \text{c}.& 7\\ \text{d}.& 9\\ \text{e}.& 11\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}y+{\displaystyle \frac{2}{x+z}=4}\\ 5y+{\displaystyle \frac{18}{2x+y+z}=18}\\ {\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3}\end{array}\\ & \text{Jika disederhanakan beberapa bagian}\\ & \{\begin{array}{ll}y+2A& =4....(1)\\ 5y+18B& =18....(2)\\ 8A-6B& =3....(3)\end{array}\\ & \text{Saat}(1)+(2)\mathrm{\&}(3),\text{maka}\\ & \begin{array}{l}\\ y+2A& =4& |\times 5|& 5y+10A=20\\ 5y+3(8A-3)& =18& |\times 1|& 5y+24A=27& -\\ & & & -14A=-7\\ & & & A={\displaystyle \frac{1}{2}...(4)}\\ \text{maka}B={\displaystyle \frac{1}{6}\mathrm{\&}}& y=3& & \\ \text{akibatnya}\\ \{\begin{array}{ll}x& =1\\ z& =1\end{array}\end{array}\\ & \text{Jadi},y+\sqrt{x^2-2xz+z^2}=3+0=3\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Diberikan}a,b,\text{dan}c\text{adalah angka-angka}\\ & \text{dari bilangan 3 digit yang memenuhi}\\ & 49a+7b+c=286.\text{Nilai dari}a+b+c\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 16\\ \text{b}.& 17\\ \text{c}.& 18\\ \text{d}.& 19\\ \text{e}.& 20\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & 49a+7b+c=286\\ & \text{Nilai maksimum}a\text{adalah}5\\ & 49\times 5=245,\text{akibatnya}:\\ & 245+7b+c=286\Rightarrow 7b+c=286-245=41\\ & \text{Nilai maksimum}b\text{adalah}5\\ & 7\times 5=35,\text{akibatnya}:\\ & 35+c=41\Rightarrow c=41-35=6\\ & \text{Sehingga}a,b,\text{dan}c\text{adalah}5,5,\text{dan}6\\ & \text{Jadi},\text{nilai}a+b+c=5+5+6=16\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Diketahui sistem persamaan}\\ & (2x+3y{)}^{{.}^{\log (x-y+2z)}}=1\\ & 3^{2x+y+z}\times {27}^{3z+2y+x}=81\\ & 5x+3y+8z=2\\ & \text{Himpunan penyelesaian yang}\\ & \text{memenuhi adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6}}\right\}\\ \text{b}.& \{-{\displaystyle \frac{17}{12},\frac{1}{2},\frac{7}{6}}\}\\ \text{c}.& \{-{\displaystyle \frac{17}{12},-\frac{1}{2},-\frac{7}{6}}\}\\ \text{d}.& \left\{{\displaystyle \frac{17}{12},\frac{1}{12},\frac{7}{6}}\right\}\\ \text{e}.& \{-{\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6}}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Untuk persamaan}(1)\\ & (2x+3y{)}^{{.}^{\log (x-y+2z)}}=(2x+3y{)}^0\\ & \iff (x-y+2z)={10}^0=1\\ & \text{Untuk persamaan}(2)\\ & 3^{2x+y+z}\times {27}^{3z+2y+x}=81\\ & \iff 3^{2x+y+z+3(3z+2y+x)}=3^4\\ & \iff 5x+7y+10z=4\\ & \text{Sehingga sistem persamaan akan terlihat}\\ & \{\begin{array}{c}x-y+2z=1....(1)\\ 5x+7y+10z=4....(2)\\ 5x+3y+8z=2....(3)\end{array}\\ & \text{Saat}(2)\mathrm{\&}(3),\text{maka}\\ & \begin{array}{l}\\ 5x+7y+10z& =4& \\ 5x+3y+8z& =2& -\\ 4y+2z& =2\\ 2y+z& =1...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ 5x-5y+10z& =5& \\ 5x+7y+10z& =4& -\\ -12y& =1\\ y& =-{\displaystyle \frac{1}{12}...(5)}\end{array}\\ & \text{Dari persamaan}(5)\text{disubstistusikan ke}(4)\\ & \begin{array}{rl}2y+z& =1\\ 2(-{\displaystyle \frac{1}{12}})+z& =1\\ z& =1+{\displaystyle \frac{1}{6}}\\ z& ={\displaystyle \frac{7}{6}}\end{array}\\ & \text{Cukup jelas juga}x=....\\ & \text{Jadi},\text{pilihannya adalah}e\end{array}\end{array}$

DAFTAR PUSTAKA

1. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade MAtematika Nasional dan Internasional. Yogyakarta: INDONESIA CERDAS.
2. Kanginan, M. 2016. Matematika untuk SMA-MA/SMK-MAK Kelas X. Bandung: SRIKANDI EMPAT WIDYA UTAMA
3. Kurnianingsih, S. 2008. SPM Matematika SMA dan MA Program IPS Siap Tuntas Menghadapi Ujian. Jakarta: ESIS
4. Susianto, B. 2011. Soal dan Pembahasan Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO
5. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI

Contoh Soal 3 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{l}\\ 11.& \text{Suatu bilangan terdiri atas 3 angka. Jumlah}\\ & \text{ketiga angka tersebut adalah 9. Angka kedua}\\ & \text{dikurangi angka pertama dan angka ketiga }\\ & \text{sama dengan 1. Dua kali angka pertama sama}\\ & \text{dengan jumlah angka kedua dan angka ketiga.}\\ & \text{Angka puluhan pada bilangan tersebut adalah}\\ & ....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 4\\ \text{c}.& 5\\ \text{d}.& 6\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Model matematikanya}\\ & \{\begin{array}{c}A+B+C=9....(1)\\ 2B-A-C=1....(2)\\ 2A=B+C....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle A+B+C}& =9\\ {\displaystyle -A+B-C}& =1& +\\ 2B& =10\\ B& =5& ...(4)\end{array}\\ & \text{Jadi},\text{bilangan kedua adalah}=B=5\end{array}\end{array}$

$\begin{array}{l}\\ 12.& (\mathbf{\text{SIMAK UI 2010}})\\ & \text{Jika}x+y+2z=K,x+2y+z=K,\\ & 2x+y+z=K\text{dengan}K\ne 0,\text{maka}\\ & x^2+y^2+z^2\text{bila dinyatakan dalam}K\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{16}{K}^2}\\ \text{b}.& {\displaystyle \frac{3}{16}{K}^2}\\ \text{c}.& {\displaystyle \frac{4}{17}{K}^2}\\ \text{d}.& {\displaystyle \frac{3}{8}{K}^2}\\ \text{e}.& {\displaystyle \frac{2}{3}{K}^2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+2z=K....(1)\\ x+2y+z=K....(2)\\ 2x+y+z=K....(3)\end{array}\\ & \text{maka}\\ & \{\begin{array}{c}z+(x+y+z)=K....(1)\\ y+(x+y+z)=K....(2)\\ x+(x+y+z)=K....(3)\end{array}\\ & \text{Saat}(1)+(2)+(3),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle x+y+2z}& =K\\ {\displaystyle x+2y+z}& =K& \\ {\displaystyle 2x+y+z}& =K& +\\ 4x+4y+4z& =3K\\ x+y+z& ={\displaystyle \frac{3}{4}K}& ...(4)\end{array}\\ & \text{Saat}(4)\text{disubstitusikan ke}(1),(2),\text{dan}(3)\\ & \text{Jelas bahwa akan didapatkan}\\ & x=y=z={\displaystyle \frac{1}{4}K}\\ & \text{Jadi},x^2+y^2+y^2=3{\left({\displaystyle \frac{1}{4}K}\right)}^2={\displaystyle \frac{3}{16}{K}^2}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Diketahui}0,15252525252...={\displaystyle \frac{p}{2q+r}}\\ & \text{Jika jumlah}p\text{dan}q=\text{3 kali}r,\text{maka}\\ & \text{masing-masing harga}p,q,\text{dan}r=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 152,2819,2584}\\ \text{b}.& {\displaystyle 252,{\displaystyle \frac{5638}{7},\frac{8102}{21}}}\\ \text{c}.& {\displaystyle 151,\frac{2819}{7},\frac{1292}{7}}\\ \text{d}.& {\displaystyle 151,{\displaystyle \frac{2819}{7},\frac{2584}{7}}}\\ \text{e}.& {\displaystyle 152,\frac{2819}{14},\frac{1292}{7}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Dari soal diketahui}\\ & \{\begin{array}{ll}0,1\overline{5252}& ={\displaystyle \frac{p}{2q+r}.....(1)}\\ p+q& =3r............(2)\end{array}\\ & \text{dan}\\ & \begin{array}{l}\\ {\displaystyle x}& =0,15252525252...\\ {\displaystyle 1000x}& =152,5252525252...\\ {\displaystyle 10x}& =1,5252525252...& -\\ 990x& =151\\ x& ={\displaystyle \frac{151}{990},\text{maka}}\\ {\displaystyle \frac{p}{2q+r}}& ={\displaystyle \frac{151}{990}}\\ & \{\begin{array}{ll}p& =151.......(3)\\ 2p+r& =990.......(4)\end{array}\end{array}\\ & \text{Dari}(3)\text{diperoleh}:q=3r-p=3r-151....(5)\\ & \text{Dari}(5)\text{disubstitusikan ke}(4)\\ & \begin{array}{rl}2q+r& =990\\ 2(3r-151)+r& =990\\ 6r-302+r& =990\\ 7r& =990+302=1292\\ r& ={\displaystyle \frac{1292}{7}.....(6)}\end{array}\\ & \text{Dari}(3)\mathrm{\&}(6)\text{disubstitusikan ke}(2)\\ & \begin{array}{rl}p+q& =3r\\ 151+q& =3\left({\displaystyle \frac{1292}{7}}\right)\\ q& ={\displaystyle \frac{3876}{7}-151}\\ & ={\displaystyle \frac{3876-1057}{7}}\\ & ={\displaystyle \frac{2819}{7}.....(7)}\end{array}\\ & \text{Jadi},p,q,r\text{adalah}:{\displaystyle 151,\frac{2819}{7},\frac{1292}{7}}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Perhatikanlah sistem persamaan berikut}\\ & \{\begin{array}{ll}3x+2y-5z& =3\\ 2x-6y+kz& =9\\ 5x-4y-z& =5\end{array}\\ & \text{agar sistem persamaan ini tidak}\\ & \text{memiliki penyelesaian, maka nilai}k=....\\ & \begin{array}{l}\\ \text{a}.& -4\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Agar sistem persamaan}\\ & \{\begin{array}{ll}3x+2y-5z& =3\\ 2x-6y+kz& =9\\ 5x-4y-z& =5\end{array}\\ & \text{tidak berpenyelesaian, maka}\\ & \text{ingat penyelesaian metode matrik}\\ & \text{buatlah penyebutnya}=0,\text{yaitu}:\\ & \left|\begin{array}{ccc}3& 2& -5\\ 2& -6& k\\ 5& -4& -1\end{array}\right|=0\\ & \text{Selanjutnya}\\ & 3\left|\begin{array}{cc}-6& k\\ -4& -1\end{array}\right|-2\left|\begin{array}{cc}2& k\\ 5& -1\end{array}\right|-5\left|\begin{array}{cc}2& -6\\ 5& -4\end{array}\right|=0\\ & 3(6+4k)-2(-2-5k)-5(-8+30)=0\\ & 18+12k+4+10k+40-150=0\\ & 22x=88\\ & x=4\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Diketahui}\\ & \left(\begin{array}{ccc}{\displaystyle \frac{1}{5}}& {\displaystyle \frac{1}{5}}& {\displaystyle \frac{1}{5}}\\ {\displaystyle \frac{1}{5}}& {\displaystyle \frac{1}{5}}& -{\displaystyle \frac{4}{5}}\\ -{\displaystyle \frac{2}{5}}& {\displaystyle \frac{1}{10}}& {\displaystyle \frac{1}{10}}\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}1\\ 2\\ 0\end{array}\right)\\ & \text{Nilai}x,y,\text{dan}z\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{5},\frac{4}{5},-\frac{1}{10}}\\ \text{b}.& -1,5,1\\ \text{c}.& 1,5,-1\\ \text{d}.& -1,1,5\\ \text{e}.& 5,1,-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}{\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1....(1)}\\ {\displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z=2....(2)}\\ -{\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z=0....(3)}\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z}& =1& \\ {\displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z}& =2& -\\ {\displaystyle \frac{5}{5}z}& =-1& \\ {\displaystyle z}& =-1& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z}& =1& |\times 1|& {\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1}\\ -{\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z}& =0& |\times 2|& -{\displaystyle \frac{4}{5}+\frac{1}{5}y+\frac{1}{5}z=0}& -\\ & & & {\displaystyle \frac{5}{5}x=1}& \\ & & & x=-1........(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{akan didapatkan}\\ & y=5\\ & \text{Jadi},(x,y,z)=(1,5,-1)\end{array}\end{array}$

Contoh Soal 2 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{l}\\ 6.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y-z=1\\ 2x-y+2z=9\\ x+3y-z=7\end{array}\\ & \text{Nilai}{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}}\\ \text{b}.& {\displaystyle \frac{3}{4}}\\ \text{c}.& {\displaystyle \frac{13}{12}}\\ \text{d}.& {\displaystyle \frac{5}{4}}\\ \text{e}.& {\displaystyle \frac{7}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y-z=1....(1)\\ 2x-y+2z=9....(2)\\ x+3y-z=7....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y-z& =1& \\ 2x-y+2z& =9& +\\ 3x+z& =10& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y-z& =1& |\times 3|& 3x+3y-3z& =3& \\ x+3y-z& =7& |\times 1|& x+3y-z& =7& -\\ & & & 2x-2z& =-4& \\ & & & x-z& =2& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ 3x+z& =10& \\ x-z& =-2& +\\ 4x& =8& \\ x& =2& .....(6)\\ \text{didapat pula}& z& =4......(7)\end{array}\\ & \text{Dari persamaan}(1)\mathrm{\&}(3)\text{didapatkan juga}\\ & \begin{array}{l}\\ x+y-z& =1& \\ x+3y-z& =-7& -\\ -2y& =-6& \\ y& =3& ....(8)\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}={\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}}}\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y+z=5\\ x+y-4z=10\\ -2x+y+z=0\end{array}\\ & \text{Nilai dari}{\displaystyle \frac{xz}{y}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\frac{6}{13}}\\ \text{b}.& {\displaystyle -\frac{5}{13}}\\ \text{c}.& {\displaystyle -\frac{1}{13}}\\ \text{d}.& {\displaystyle \frac{1}{13}}\\ \text{e}.& {\displaystyle \frac{7}{13}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=5.....(1)\\ x+y-4z=10.....(2)\\ -2x+y+z=0.....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =5& \\ x+y-4z& =10& -\\ 5z& =-5& \\ z& =-1& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =5& \\ -2x+y+z& =0& -\\ 3x& =5& \\ x& ={\displaystyle \frac{5}{3}}& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{rl}x+y+z& =5\\ {\displaystyle \frac{5}{3}+y-1}& =5\\ y& =5+1-{\displaystyle \frac{5}{3}=\frac{13}{3}}\end{array}\\ & \text{Jadi},{\displaystyle \frac{xz}{y}={\displaystyle \frac{\left({\displaystyle \frac{5}{3}}\right).(-1)}{{\displaystyle \frac{13}{3}}}=-{\displaystyle \frac{5}{13}}}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Himpunan penyelesaian dari}\\ & \{\begin{array}{ll}{\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4\end{array}\\ & \text{adalah}\{(x,y,z)\},\text{maka}x+3z=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& {\displaystyle \frac{1}{3}}\\ \text{c}.& 1\\ \text{d}.& 3\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{ll}{\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8....(1)\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10.....(2)\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4...........(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}}& =8\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =10& -\\ -{\displaystyle \frac{1}{x}-\frac{1}{z}}& =-2\\ {\displaystyle \frac{1}{x}+\frac{1}{z}}& =2& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ {\displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}}& =8& |\times 2|& {\displaystyle \frac{4}{x}+\frac{4}{y}+\frac{8}{z}}& =16\\ {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4& |\times 1|& {\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}}& =4& -\\ & & & {\displaystyle \frac{2}{x}+\frac{6}{z}}& =12\\ & & \iff & {\displaystyle \frac{1}{x}+\frac{3}{z}}& =6& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ {\displaystyle \frac{1}{x}+\frac{3}{z}}& =6\\ {\displaystyle \frac{1}{x}+\frac{1}{z}}& =2-\\ {\displaystyle \frac{2}{z}}& =4& \\ z& ={\displaystyle \frac{1}{2}......(6)}\\ x& =2-{\displaystyle \frac{1}{2}={\displaystyle \frac{3}{2}}}\end{array}\\ & \text{Jadi},x+3z={\displaystyle \frac{3}{2}+3.\frac{1}{2}=3}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Diketahui tiga buah bilangan berturut-turut}\\ & a,b,\text{dan}c.\text{Rata-rata dari ke tiga bilangan}\\ & \text{itu adalah 12. Bilangan kedua sama dengan}\\ & \text{jumlah bilangan yang lain dikurangi 12}.\\ & \text{Jika bilangan ke tiga sama dengan jumlah}\\ & \text{bilangan yang lain, maka nilai}2a+b-c=....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle 42}\\ \text{b}.& -{\displaystyle 36}\\ \text{c}.& -{\displaystyle 18}\\ \text{d}.& -{\displaystyle 12}\\ \text{e}.& -{\displaystyle 6}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Model matematika dari persamaan di atas}\\ & \{\begin{array}{c}a+b+c=36....(1)\\ -a+b-x=12....(2)\\ a+b-c=0....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ a+b+c& =36& \\ -a+b-c& =12& +\\ 2b& =48& \\ b& =24& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ a+b+c& =36& \\ a+b-c& =0& -\\ 2c& =36& \\ c& =18& ....(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{rl}a+b+c& =36\\ a+24+18& =36\\ a& =36-42\\ & =-6\end{array}\\ & \text{Jadi},2a+b-c=2(-6)+24-18=-6\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jumlah uang terdiri atas koin pecahan}Rp500,00\\ & Rp200,00danRp100,00\text{dengan nilai total}\\ & Rp100.000,00.\text{Jika nilai uang pecahan 500-an}\\ & \text{setengah dari nilai uang pecahan 200-an, tetapi}\\ & \text{tiga kali uang pecahan 100-an, maka banyak koin}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 460\\ \text{b}.& 440\\ \text{c}.& 420\\ \text{d}.& 380\\ \text{e}.& 350\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Model matematika dari kasus di atas}\\ & \{\begin{array}{c}A(500)+B(200)+C(100)=100.000....(1)\\ A(500)={\displaystyle \frac{1}{2}B(200)....(2)}\\ A(500)=3C(100)....(3)\end{array}\\ & \text{Dari persamaan}(2)\text{didapatkan}\\ & 2A(500)=B(200)\\ & \text{Dari persamaan}(3)\text{akan didapatkan}\\ & {\displaystyle \frac{1}{3}A(500)=C(100)}\\ & \text{Dari persamaan}(1)\text{maka},\\ & A(500)+B(200)+C(100)=100.000\\ & A(500)+2A(500)+{\displaystyle \frac{1}{3}A(500)=100.000}\\ & {\displaystyle \frac{10}{3}A(500)=100.000\iff A(500)=30.000}\\ & \text{maka akan didapatkan}\\ & B(200)=2(30.000)=60.000\\ & C(100)={\displaystyle \frac{1}{3}(30.000)=10.000}\\ & \{\begin{array}{ll}A(500)& =30.000\Rightarrow A={\displaystyle \frac{30.000}{500}=60}\\ B(200)& =60.000\Rightarrow B={\displaystyle \frac{60.000}{200}=300}\\ C(100)& =10.000\Rightarrow C={\displaystyle \frac{10.000}{100}=100}\end{array}\\ & \text{Jadi},A+B+C=60+300+100=460\end{array}\end{array}$

Contoh Soal 1 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{l}\\ 1.& \text{Suatu unit pekerjaan dapat diselesaikan oleh A}\\ & \text{B, dan C bersama-sama dalam 2 jam saja.}\\ & \text{Jika pekerjaan itu dapat diselesaikan oleh A dan}\\ & \text{B bersama-sama dalam 2 jam 24 menit, dan oleh}\\ & \text{B dan C bersama-sama dalam waktu 3 jam,}\\ & \text{maka sistem persamaan berikut yang memenuhi}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{\begin{array}{ll}A+B+C& =2\\ A+B& ={\displaystyle \frac{12}{5}}\\ B+C& =3\end{array}\\ \text{b}.& \{\begin{array}{ll}A+B+C& ={\displaystyle \frac{1}{2}}\\ A+B& ={\displaystyle \frac{5}{12}}\\ B+C& ={\displaystyle \frac{1}{3}}\end{array}\\ \text{c}.& \{\begin{array}{ll}{\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}}& =2\\ {\displaystyle \frac{1}{A}+\frac{1}{B}}& ={\displaystyle \frac{12}{5}}\\ {\displaystyle \frac{1}{B}+\frac{1}{C}}& =3\end{array}\\ \text{d}.& \{\begin{array}{ll}{\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}}& ={\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{A}+\frac{1}{B}}& ={\displaystyle \frac{5}{12}}\\ {\displaystyle \frac{1}{B}+\frac{1}{C}}& ={\displaystyle \frac{1}{3}}\end{array}\\ \text{e}.& \{\begin{array}{ll}{\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}}& =2\\ {\displaystyle \frac{1}{A}+\frac{1}{B}-\frac{1}{C}}& ={\displaystyle \frac{12}{5}}\\ {\displaystyle -\frac{1}{A}+\frac{1}{B}+\frac{1}{C}}& =3\end{array}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Per}& \text{hatikan bahwa}:\text{Waktu penyelesaian}\\ \text{sua}& \text{tu pekerjaan adalah termasuk}\\ \text{per}& \text{bandingan berbalik nilai},\text{maka}\\ \bullet & A,B,\text{dan}C\text{dalam 2 jam, artinya}:\\ & {\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{2},\text{demikian juga}}\\ \bullet & A\text{dan}B\text{bersama-sama selesai dalam}\\ & \text{2 jam 24 menit atau}{\displaystyle \frac{12}{5}\text{jam}:}\\ & {\displaystyle \frac{1}{A}+\frac{1}{B}=\frac{5}{12}}\\ \bullet & B\text{dan}C\text{selesai dalam 3 jam}:\\ & {\displaystyle \frac{1}{B}+\frac{1}{C}=\frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Himpunan penyelesaian dari}\\ & \{\begin{array}{c}x+y+4z=15\\ x-y+z=2\\ x+2y-3z=-4\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{(-1,1,3)\}\\ \text{b}.& \{(1,2,3)\}\\ \text{c}.& \{(-2,1,1)\}\\ \text{d}.& \{(3,2,-1)\}\\ \text{e}.& \{(1,-2,3)\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Semunya dikerjakan dengan metode}\\ & \text{matriks}(\mathbf{\text{Cara Cramer}})\\ & \begin{array}{rl}x& ={\displaystyle \frac{\left|\begin{array}{ccc}15& 1& 4\\ 2& -1& 1\\ -4& 2& -3\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 1\\ 1& 2& -3\end{array}\right|}}\\ & ={\displaystyle \frac{15\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}2& 1\\ -4& -3\end{array}\right|+4\left|\begin{array}{cc}2& -1\\ -4& 2\end{array}\right|}{1\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}}\\ & ={\displaystyle \frac{15(3-2)-1(-6+4)+4(4-4)}{1(3-2)-1(-3-1)+4(2+1)}}\\ & ={\displaystyle \frac{15(1)-1(-2)+4(0)}{1(1)-1(-4)+4(3)}=\frac{17}{17}=1}\\ y& ={\displaystyle \frac{\left|\begin{array}{ccc}1& 15& 4\\ 1& 2& 1\\ 1& -4& -3\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 1\\ 1& 2& -3\end{array}\right|}}\\ & ={\displaystyle \frac{1\left|\begin{array}{cc}2& 1\\ -4& -3\end{array}\right|-15\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& 2\\ 1& -4\end{array}\right|}{1\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}}\\ & ={\displaystyle \frac{1(-6+4)-15(-3-1)+4(-4-2)}{1(3-2)-1(-3-1)+4(2+1)}}\\ & ={\displaystyle \frac{1(-2)-15(-4)+4(-6)}{1(1)-1(-4)+4(3)}=\frac{34}{17}=2}\\ z& ={\displaystyle \frac{\left|\begin{array}{ccc}1& 1& 15\\ 1& -1& 2\\ 1& 2& -4\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 1\\ 1& 2& -3\end{array}\right|}}\\ & ={\displaystyle \frac{1\left|\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right|-1\left|\begin{array}{cc}1& 2\\ 1& -4\end{array}\right|+15\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}{1\left|\begin{array}{cc}-1& 1\\ 2& -3\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& -3\end{array}\right|+4\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|}}\\ & ={\displaystyle \frac{1(4-4)-1(-4-2)+15(2+1)}{1(3-2)-1(-3-1)+4(2+1)}}\\ & ={\displaystyle \frac{1(0)-1(-6)+15(3)}{1(1)-1(-4)+4(3)}=\frac{51}{17}=3}\end{array}\end{array}$

$.\text{Cara di atas}$  full matriks-Cramer

$\begin{array}{l}\\ 3.& \text{Hasil dari}xyz\text{yang memenuhi}\\ & \{\begin{array}{c}x+y+z=2\\ x-y+z=-2\\ x-y-z=2\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -8\\ \text{b}.& -4\\ \text{c}.& 2\\ \text{d}.& 4\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=2.....(1)\\ x-y+z=-2.....(2)\\ x-y-z=2.....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =2& \\ x-y+z& =-2& -\\ 2y& =4& \\ y& =2& ....(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =2& \\ x-y-z& =2& +\\ 2x& =4& \\ x& =2& ....(5)\end{array}\\ & \text{Persamaan}(4)\mathrm{\&}(5)\text{ke}(1)\\ & \begin{array}{rl}x+y+z& =2\\ (2)+(2)+z& =2\\ z& =-2\end{array}\\ & \text{Jadi},xyz=(2).(2).(-2)=-8\end{array}\end{array}$

$.\text{Cara di atas}$  full eliminasi-substitusi

$\begin{array}{l}\\ 4.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+y+z=-6\\ x-2y+z=3\\ -2x+y+z=9\end{array}\\ & \text{Nilai}xyz=....\\ & \begin{array}{l}\\ \text{a}.& -30\\ \text{b}.& -15\\ \text{c}.& 5\\ \text{d}.& 30\\ \text{e}.& 35\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+y+z=-6....(1)\\ x-2y+z=3....(2)\\ -2x+y+z=9....(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =-6& \\ x-2y+z& =3& -\\ 3y& =-9& \\ y& =-3& ....(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+y+z& =-6& \\ -2x+y+z& =9& -\\ 3x& =-15& \\ x& =-5& ....(5)\end{array}\\ & \text{Persamaan}(4)\mathrm{\&}(5)\text{ke}(1)\\ & \begin{array}{rl}x+y+z& =2\\ (-5)+(-3)+z& =-6\\ z& =2\end{array}\\ & \text{Jadi},xyz=(-5).(-3).(2)=30\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Diketahui sistem persamaan berikut}\\ & \{\begin{array}{c}x+2y+z=4\\ 3x+y+2z=-5\\ x-2y+2z=-6\end{array}\\ & \text{Nilai}xyz=....\\ & \begin{array}{l}\\ \text{a}.& -96\\ \text{b}.& -24\\ \text{c}.& 24\\ \text{d}.& 32\\ \text{e}.& 96\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui sistem persamaan}\\ & \{\begin{array}{c}x+2y+z=4.......(1)\\ 3x+y+2z=-5......(2)\\ x-2y+2z=-6.......(3)\end{array}\\ & \text{Saat}(1)+(2),\text{maka}\\ & \begin{array}{l}\\ x+2y+z& =4& |\times 1|& x+2y+z& =4\\ 3x+y+2z& =-5& |\times 2|& 6x+2y+4z& =-10& -\\ & & & -5x-3z& =14& ...(4)\end{array}\\ & \text{Saat}(1)+(3),\text{maka}\\ & \begin{array}{l}\\ x+2y+z& =4& \\ x-2y+2z& =-6& +\\ 2x+3z& =-2& ...(5)\end{array}\\ & \text{Dari persamaan}(4)\mathrm{\&}(5)\text{maka},\\ & \begin{array}{l}\\ -5x-3z& =14& \\ 2x+3z& =-2& +\\ -3x& =12& \\ x& =-4& .....(6)\\ \text{didapat pula}& z& =2......(7)\end{array}\\ & \text{Dari persamaan}(6)\mathrm{\&}(7)\text{didapatkan}\\ & \begin{array}{rl}x+2y+z& =4\\ (-4)+2y+2& =4\\ y& =3\end{array}\\ & \text{Jadi},xyz=(-4).(3).(2)=-24\end{array}\end{array}$

Lanjutan Materi (6) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI

$\text{E. Persamaan Garis Singgung}$

1. Fungsi Aljabar

Perhatikanlah gambar berikut!

Perhatikanlah kurva di atas, yaitu sebuah gambar grafik fungsi kuadrat  $f(x)=x^2-2x+1$. Misalkan kita menginginkan garis mana yang merupakan persamaan garis singgung di titik $(2,1)$?

Ada 2 unsur penting dalam menentukan persamaan garis singgung, yaitu:

• titik singgung
• gradien (kemiringan) dari garis singgung itu sendiri, yaitu : $m={\displaystyle \frac{dy}{dx}}$

Karena salah satu unsur penentuan persamaan garis singgung telah diketahui, yaitu sebuah titik singgung, langkah berikutnya kita tinggal mencari gradien. Dalam hal ini gradien dari garis singgung diperoleh dengan memasukkan absis seteleh kurva singgung itu diturunkan pertama dan kadang dituliskan dengan notasi  Leibniz  $m={\left({\displaystyle \frac{dy}{dx}}\right)}_{x=a}$ atau kadang juga dituliskan dengan bentuk notasi $m={\begin{array}{c}{\displaystyle \frac{dy}{dx}}\end{array}|}_{x=a}$. Untuk mempermudah, oerhatikanlah kurva di atas, dari keempat garis lurus yang ada, tidak semunya menyinggung. Karena sebagian bahkan berpotongan dengan kurva. Walaupun antara titik potong dan titik singgung sama, tetapi cara mendapatkannya berbeda. Sementara kita fokus pada aplikasi turunan pertama pada suatu kurva. Coba kita perjelas lagi dengan menyertakan persamaan keempat garis lurusnya berikut

Mari kita tentukan persamaan garis singgung kurva di atas dari keempat garis lurus itu, garis yang mana?

Persamaan Garis Singgung kurva dituliskan sebagai: $y=m(x-a)+b$, dengan  $(a,b)$  adalah titik singgung. Kada titik singgung juga dituliskan dengan  $(a,f(a))$.

Sehingga persamaan garis singgung kurva di atas adalah:

$\begin{array}{rl}f(x)=y& =x^2-2x+1=(x-1{)}^2\\ m& =2x-2\\ {\left({\displaystyle \frac{dy}{dx}}\right)}_{x=2}& =m=2(2)-2=4-2=2\\ \text{maka}& \text{persamaan garis singgung kurvanya}\\ y& =m(x-a)+b\\ & =2(x-2)+1\\ & =2x-4+1\\ y& =2x-3\end{array}$.

Jadi, garis pada gambar di atas yang merupakan garis singgung kurva yang dimaksud adalah garis $g_3:y=2x-3$.

2. Fungsi Trigonometri

Tidak jauh berbeda dengan fungsi aljabra, maka pada fungsi trigonometri berlaku sifat yang sama yang membedakan hanyanya kurvanya serta sumbu X (letak absis).

Sebagai misal kita diberikan sebuah fungsi trigonometri  $f(x)=y=\sin 2x$. Jika dituntut untuk menunjukkan persamaan garis singgung di titik yang berabsis  ${\displaystyle \frac{\pi }{2}}$, maka kita juga dapat dengan mudah menentukannya.

Perhatikan uraian berikut sebagai pembahasan dari permasalahan di atas.

$\begin{array}{rl}\text{Diketahui}& x=a={\displaystyle \frac{\pi }{2}}\\ \text{Kita men}& \text{cari titik singgungnya dulu, yaitu}\\ f(a)& =\sin 2\left({\displaystyle \frac{\pi }{2}}\right)=\sin \pi =0,\\ (a,f(a))& =\left({\displaystyle \frac{\pi }{2},0}\right)\\ f(x)=y& =\sin 2x\\ m& =2\cos 2x......(\mathbf{\text{turunan pertama}})\\ {\left({\displaystyle \frac{dy}{dx}}\right)}_{x=\frac{\pi }{2}}& =m=2\cos 2\left({\displaystyle \frac{\pi }{2}}\right)\\ & =2\cos \pi =2.(-1)=-2\\ \text{maka}& \text{persamaan garis singgung kurvanya}\\ y& =m(x-a)+b\\ & =-2(x-{\displaystyle \frac{\pi }{2}})+0\\ & =-2x+\pi \end{array}$

DAFTAR PUSTAKA

1. Kurnia, N. 2018. Jelajah Matematika 3 SMA Kelas XII Peminatan MIPA. Bogor: Yudhistira.
2. Tampomas, H. 1999. SeribuPena Matematika SMU Kelas 2. Jakarta: ERLANGGA
3. Wirodikromo, S. 2007. Matematika Jilid 2 IPA untuk Kelas XI. Jakarta: ERLANGGA.

 

Contoh Soal 5 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{l}\\ 21.& \text{Turunan pertama dari fungsi}\\ & g(x)={\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}}\\ \text{b}.& {\displaystyle \frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}}\\ \text{c}.& {\displaystyle \frac{1}{{\sin }^{2}x{\cos }^{2}x}}\\ \text{d}.& {\displaystyle \frac{-1}{{\sin }^{2}x{\cos }^{2}x}}\\ \text{e}.& {\displaystyle {\sin }^{2}x{\cos }^{2}x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ g(x)& ={\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}\\ & =\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}={\displaystyle \frac{1}{\sin x\cos x}}\\ \text{maka}& \\ g^{\prime }(x)& ={\displaystyle \frac{0.(\sin x\cos x)-1.({\cos }^{2}x-{\sin }^{2}x)}{(\sin x\cos x{)}^2}}\\ & ={\displaystyle \frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}}\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Diketahui}h(x)=\cos \left({\displaystyle \frac{3}{x}}\right),\\ & \text{maka}{\displaystyle \frac{dh}{dx}}\\ & \begin{array}{l}\\ \text{a}.& -3\sin {\displaystyle \frac{3}{x}}\\ \text{b}.& -{\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\\ \text{c}.& -{\displaystyle \frac{3}{x}\sin \frac{3}{x}}\\ \text{d}.& {\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\\ \text{e}.& {\displaystyle \frac{3}{x}\sin \frac{3}{x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\cos {\displaystyle \frac{3}{x}}& =-\sin {\displaystyle \frac{3}{x}\left({\displaystyle \frac{0.(x)-3.1}{x^2}}\right)}\\ & ={\displaystyle \frac{-(-3)}{x^2}\sin \frac{3}{x}}\\ & ={\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Turunan pertama dari}\tan (\cos x),\\ & \text{terhadap}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\sec }^2(\cos x)\sin x\\ \text{b}.& {\sec }^2(\cos x)\sin x\\ \text{c}.& {\sec }^2(\sin x)\cos x\\ \text{d}.& {\displaystyle \sin x}\\ \text{e}.& {\displaystyle -\sin x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Misal}& \text{kan}\\ y& =\tan x(\cos x)\\ y^{\prime }& ={\sec }^2(\cos x)\times (-\sin x)\\ & =-{\sec }^2(\cos x).\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 24.& (\mathbf{\text{UN 2005}})\text{Turunan pertama dari}\\ & f(x)=\sqrt[3]{{\cos }^2(3x^2+5x)}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ \text{b}.& {\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)}\\ \text{c}.& -{\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ \text{d}.& -{\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\\ \text{e}.& {\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Misal}& \text{kan}\\ f(x)& =\sqrt[3]{{\cos }^2(3x^2+5x)}\\ f^{\prime }(x)& ={\cos }^{{.}^{\frac{2}{3}}}(3x^2+5x)\\ & ={\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{2}}}(3x^2+5x)\times (-\sin (3x^2+5x))}\\ & \times (6x+5)\\ & =-{\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ & =-{\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{\frac{2}{3}}}(3x^2+5x)}\\ & \times {\cos }^{-1}(3x^2+5x)\times \sin (3x^2+5x)\\ & =-{\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\end{array}\end{array}$