Contoh Soal 2 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 6.& \text{Bentuk sederhana dari}\\ & 2y-5>2x+4y+3\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y-x>4\\ \text{b}.& y-x<4\\ \text{c}.& y+x+4>0\\ \text{d}.& y+x+4<0\\ \text{e}.& y+x<1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 2y-5>2x+4y+3\\ & 2y-4y-2x-5-3>0\\ & -2y-2x-8>0\text{dibagi}(-{\displaystyle \frac{1}{2}})\\ & y+x+4<0\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Jika}3x-4>5x-17\\ & \text{maka sebuah bilangan prima}\\ & \text{yang mungkin adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 7\\ \text{c}.& 11\\ \text{d}.& 13\\ \text{e}.& 17\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& 3x-4>5x-17\\ & \iff 3x-5x>-17+4\\ & \iff -2x>-13\text{tiap ruas}(\times -1)\\ & \iff 2x<13\\ & \iff x<{\displaystyle \frac{13}{2}=6\frac{1}{2}}\\ & \text{Jadi, yang memenuhi adalah 3 dan 5}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}{\displaystyle \frac{1}{5}<\frac{1}{x}\text{dan}x<0}\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{1}{5}}\\ \text{b}.& -5<x<0\\ \text{c}.& 0<x<5\\ \text{d}.& x<-5\\ \text{e}.& -{\displaystyle \frac{1}{5}<x<0}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}\\ {\displaystyle \frac{1}{5}}& <\frac{1}{x}\text{dan}x<0\\ {\displaystyle \frac{1}{5}}& <{\displaystyle \frac{1}{x}}\\ x& <5\\ x& >-5\text{karena}x<0\\ \text{Sehi}& \text{ngga}\\ -5<& x<0\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika}a,b,c\text{dan}d\text{bilangan real}\\ & \text{dengan}a>b\text{dan}c>d\\ & \text{maka berlaku}\\ & (1)ac>bd\\ & (2)a+c>b+d\\ & (3)ad>bc\\ & (4)ac+bd>ad+bc\\ & \text{Pernyataan-pernyataan di atas}\\ & \text{yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& (1),(2),\text{dan}(3)\\ \text{b}.& (1)\text{dan}(3)\\ \text{c}.& (2)\text{dan}(4)\\ \text{d}.& (4)\\ \text{e}.& \text{Semua benar}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}:a,b,c\text{dan}d\text{bilangan real}\\ & \text{Jelas bahwa baik bilangan positif maupun}\\ & \text{negatif termasuk semunya dibolehkan}\\ & \text{dengan}a>b\text{dan}c>d\\ & \bullet \text{Sehingga pernyataan (1)}ac>bd\\ & \text{salah saat kita coba bilangan negatif}\\ & \bullet \text{Pernyataan (2) benar karena}\\ & \begin{array}{llll}a& >& b& \\ c& >& d& +\\ a+c& >& b+d\end{array}\\ & \bullet \text{Kasusnya sama dengan poin (1)}\\ & \text{saat dicoba dengan bilangan positif}\\ & \text{tidak semuanya memenuhi}\\ & \bullet \text{Pernyataan (4) tepat juga karena}\\ & \begin{array}{l}\\ a-b>0\\ c-d>0\text{Saat dikalikan}\\ (a-b)\times (c-d)>0\\ \iff ac-ad-bc+bd>0\\ \iff ac+bd>ad+bc\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika}-2<y<3\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 9<(y-2{)}^2<16\\ \text{b}.& 4<(y-2{)}^2<16\\ \text{c}.& 1<(y-2{)}^2<16\\ \text{d}.& 0\le (y-2{)}^2<16\\ \text{e}.& -1<(y-2{)}^2<16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}:-2<y<3\\ & \bullet \text{saat dikurangi}2\\ & \iff -2-2<y-2<3-2\\ & -4<y-2<1\\ & \bullet \text{Saat}-4<y-2<0\\ & (-4{)}^2<(y-2{)}^2<0^2\text{dikuadratkan}\\ & 16>(y-2{)}^2>0\\ & 0<(y-2{)}^2<16\\ & \bullet \text{Saat}0\le y-2<1\\ & 0^2\le (y-2{)}^2<1^2\\ & 0<(y-2{)}^2<1\\ & \text{Jadi},0\le (y-2)<16\end{array}\end{array}$

Contoh Soal 1 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 1.& (\text{Soal SNMPTN})\\ & \text{Jika}x>5\text{dan}y<3,\text{maka}\\ & \text{nilai }x-y\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{lebih besar dari pada 1}\\ \text{b}.& \text{lebih besar dari pada 3}\\ \text{c}.& \text{lebih besar dari pada 8}\\ \text{d}.& \text{lebih besar dari pada 5}\\ \text{e}.& \text{lebih besar dari pada 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{D}& \text{iketahui bahwa}\\ x& >5\mathrm{\&}y<3\\ \text{m}& \text{aka}\\ & \begin{array}{l}\\ x>5& \Rightarrow & x& >5\\ y<3& \Rightarrow & -y& >-3& +\\ & & x-y& >2\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Batas pertidaksamaan}5x-7>13\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-4\\ \text{b}.& x>4\\ \text{c}.& x>-4\\ \text{d}.& x<4\\ \text{e}.& -4<x<4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}5x& -7>13\\ 5x& >13+7\\ 5x& >20\\ x& >4\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Penyelesaian dari pertidaksamaan}\\ & 2x+3>5x-7\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<3\\ \text{b}.& x<3{\displaystyle \frac{1}{3}}\\ \text{c}.& x>3{\displaystyle \frac{1}{3}}\\ \text{d}.& x>3\\ \text{e}.& \text{Semua pilihan jawaban salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}2x+3& >5x-7\\ 2x-5x& >-7-3\\ -3x& >-10\text{dikali (-1)}\\ 3x& <10\\ x& <{\displaystyle \frac{10}{3}=3{\displaystyle \frac{1}{3}}}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& (\mathbf{\text{UMPTN 01}})\text{Jika pertidaksamaan}\\ & 2x-3a>{\displaystyle \frac{3x-1}{2}+ax\text{mempunyai}}\\ & \text{penyelesaian}x>5,\text{maka nilai}a\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{3}{4}}\\ \text{b}.& -{\displaystyle \frac{3}{8}}\\ \text{c}.& {\displaystyle \frac{3}{8}}\\ \text{d}.& {\displaystyle \frac{1}{4}}\\ \text{e}.& {\displaystyle \frac{3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}2x-3a& >{\displaystyle \frac{3x-1}{2}+ax\text{tiap ruas}(\times 2)}\\ 4x-6a& >3x-1+2ax\\ 4x-3x& -2ax>-1+6a\\ x-2a& x>-1+6a\\ (1-2a)& x>-1+6a\\ x& >{\displaystyle \frac{-1+6a}{1-2a}}\\ \text{Diketa}& \text{hui}:x>5\text{adalah penyelesaian}\\ \text{maka}& \\ 5& ={\displaystyle \frac{-1+6a}{1-2a}}\\ 5-10a& =-1+6a\\ -6a-10& a=-1-5\\ -16a& =-6\\ a& ={\displaystyle \frac{-6}{-16}}\\ & ={\displaystyle \frac{3}{8}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& (\mathbf{\text{UMPTN 94}})\\ & \text{Apabila}a<x<b\text{dan}a<y<b\\ & \text{maka berlaku}....\\ & \begin{array}{l}\\ \text{a}.& a<x-y<b\\ \text{b}.& b-a<x-y<a-b\\ \text{c}.& a-b<x-y<b-a\\ \text{d}.& {\displaystyle \frac{1}{2}(b-a)<x-y<\frac{1}{2}(a-b)}\\ \text{e}.& {\displaystyle \frac{1}{2}(a-b)<x-y<\frac{1}{2}(b-a)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{l}\\ a<x<b& \Rightarrow & a<x<b& \\ a<y<b& \Rightarrow & -a>-y>-b& & \\ & \text{saat}& \text{di susun ulang}& \\ a<x<b& \Rightarrow & a<x<b& \\ a<y<b& \Rightarrow & -b<-y<-a& +& \\ & & a-b<x-y<& b-a\end{array}\end{array}$

Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ & \mathbf{\text{BENTUK UMUM}}\\ & \{\begin{array}{l}ax+by<c\\ ax+by\le c\\ ax+by>c\\ ax+by\ge c\end{array}\\ \\ & \mathbf{\text{LANGKAH-LANGKAH}}\\ & \text{dalam membuat gambar grafik persamaan linear}\\ & \text{adalah sebagai berikut}:\\ & \bullet \text{membuat gambar grafik}ax+by=c\\ & \text{untuk batas wilayahnya}\\ & \bullet \text{menyelidiki wilayah yang dimaksud di sekitar}\\ & \text{garis}ax+by=c\\ & \bullet \text{ambillah sebuah titik}(x_0,y_0)\text{sembarang}\\ & \text{kemudian substitusikan ke pertidaksamaan}\\ & ax+by....c\\ & \bullet \text{jika diperoleh nilai ketaksamaan yang benar},\\ & \text{maka daerah di mana titik uji}(x_0,y_0)\\ & \text{berada merupakan wilayah penyelesaiannya}\\ & \text{demikian juga sebaliknya}\end{array}$

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

$\begin{array}{l}\\ 1.& \text{Gambarlah himpunan penyelesaian (HP)}\\ & \text{dari pertidaksamaan linear berikut}\\ & \text{a}.3x+2y<6\\ & \text{b}.3x+2y\le 6\\ & \text{c}.3x+2y>6\\ & \text{d}.3x+2y\ge 6\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Mula}-\text{mula kita gambar garis}3x+2y=6\\ \\ & \begin{array}{|ccc|}\hline \text{Komponen}& \text{pada}& \text{pada}\\ \text{titik}& \text{sumbu}-y& \text{sumbu}-x\\ x& 0& 2\\ y& 3& 0\\ (x,y)& (0,3)& (2,0)\\ \hline\end{array}\\ \\ & \text{Selanjutnya gambar grafiknya sebagai berikut}.\end{array}\end{array}$

Dan berikut untuk wilayah dan juga batas-batas untuk pertidalsamaan

$3x+2y<6$

Kita dapat menggunakan titik uji untuk memastikan kondisi gambar di atas, yaitu di antaranya

$\begin{array}{|ccc|}\hline \text{Titik}& \text{Pengujian}& \text{Keterangan}\\ & \text{Uji}& 3x+2y<6\\ (0,0)& 3(0)+2(0)=0<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (0,1)& 3(0)+2(1)=2<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (1,0)& 3(1)+2(0)=3<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (1,1)& 3(1)+2(1)=5<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (0,2)& 3(0)+2(2)=4<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (2,0)& 3(2)+2(0)=6=\mathbf{\text{6}}& \text{Di luar wilayah}\\ (2,2)& 3(2)+2(2)=10>\mathbf{\text{6}}& \text{Di luar wilayah}\\ (0,3)& 3(0)+2(3)=6=\mathbf{\text{6}}& \text{Di luar wilayah}\\ (3,0)& 3(3)+2(0)=9>\mathbf{\text{6}}& \text{Di luar wilayah}\\ (3,3)& 3(3)+2(3)=15>\mathbf{\text{6}}& \text{Di luar wilayah}\\ ⋮& ⋮& ⋮\\ \hline\end{array}$

Dan berikut untuk wilayah yang memenuhi  $"3x+2y\le 6$

$\begin{array}{l}\\ 2.& \text{Selesaikanlah pertidaksamaan berikut}\\ & \text{a}.12x+2>4x+6\\ & \text{b}.2-3x<6-x\\ & \text{c}.6x+1\ge 2\\ & \text{d}.{\displaystyle \frac{2-3x}{2}<\frac{3-x}{3}}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{a}.12x& +2>4x+6\\ 12x& -4x>6-2\\ 8x& >4\\ x& >{\displaystyle \frac{1}{2}}\end{array}\\ & \begin{array}{rl}\text{b}.& 2-3x<6-x\\ & -3x+x<6-2\\ & -2x<4\text{dikali}(-1)\\ & 2x>-4(\text{tanda berubah})\\ & x>-2\end{array}\\ & \begin{array}{rl}\text{c}.6x& +1\ge 2\\ 6x& \ge 2-1\\ x& \ge {\displaystyle \frac{1}{6}}\end{array}\\ & \begin{array}{rl}\text{d}.{\displaystyle \frac{2-3x}{2}}& <\frac{3-x}{3}\\ 3(2-3x)& <2(3-x)\\ 6-9x& <6-2x\\ -9x+2x& <6-6\\ -7x& <0\text{di kali}(-1)\\ 7x& >0(\text{tanda berubah})\\ x& >{\displaystyle \frac{0}{7}}\\ x& >0\end{array}\end{array}$

DAFTAR PUSTAKA

1. Heryadi, D. 2007. Modul Matematikauntuk SMK Kelas X. Bogor: YUDHISTIRA.
2. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo. PT. TIGA SERANGKAI PUSTAKA MANDIRI.

Contoh Soal 9 Statistika

$\begin{array}{l}\\ 37.& (\mathbf{\text{SMBTN 2013}})\\ & \text{Median dan rata-rata dari data yang}\\ & \text{terdiri dari empat bilangan asli yang}\\ & \text{telah diurutkan mulai dari yang terkecil}\\ & \text{adalah 7. Jika data tersebut tidak}\\ & \text{memiliki modus dan selisih antara data}\\ & \text{dan data terkecil adalah 8, maka hasil}\\ & \text{kali terbesar dari datum kedua dan}\\ & \text{keempat adalah}....\\ & \begin{array}{l}\\ \text{a}.& 39\\ \text{b}.& 44\\ \text{c}.& 48\\ \text{d}.& 55\\ \text{e}.& 66\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui data}:x_1,x_2,x_3,\text{dan}{x}_4\\ & \bullet \text{Modus}:\text{tidak ada}\\ & \text{berarti}\text{semua datumnya berbeda}\\ & \bullet \text{Median}:7\\ & \text{maka}{\displaystyle \frac{x_2+x_3}{2}=7\Rightarrow x_2+x_3=14...(1)}\\ & \bullet \text{Rata-rata (Mean)}:7\\ & \text{berarti}{\displaystyle \frac{x_1+x_2+x_3+x_4}{4}=7}\\ & \Rightarrow x_1+x_2+x_3+x_4=28..........(2)\\ & \bullet \text{Jangkauan}:x_4-x_1=8.....(3)\\ & \text{SUBSTITUSI}\\ & \text{Dari persamaan (1) ke (2) diperoleh}:\\ & x_1+x_4=14...........(5)\\ & \text{ELIMINASI}\\ & \text{Dari persamaan (3) dan (4) diperoleh}:\\ & x_1=3\text{dan}{x}_4=11\\ & \text{KEMUNGKINAN}\\ & \text{nilai}{x}_2\text{dan}{x}_3\text{adalah}:3<x_2;x_3<11\\ & ⧫x_2=4\to x_3=10<11✓\\ & ⧫x_2=5\to x_3=9<11✓\\ & ⧫x_2=6\to x_3=8<11✓\\ & ⧫x_2=7\to x_3=7<11\times \\ & \text{KESIMPULAN}\\ & \text{Hasil kali terbesar}{x}_2\times x_3=6\times 11=66\end{array}\end{array}$

$\begin{array}{l}\\ 38.& (\mathbf{\text{SIMAK UI 2012}})\\ & \text{Diketahui bahwa jika Deni mendapatkan}\\ & \text{nilai 75 pada ulangan yang akan datang}\\ & \text{maka rata-rata nilai ulangannya adalah 82.}\\ & \text{Jika Deni mendapatkan nilai 93, maka}\\ & \text{rata-rata nilai ulangannya adalah 85.}\\ & \text{Banyak ulangan yang telah diikuti Deni}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 4\\ \text{c}.& 5\\ \text{d}.& 6\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{MISAL}\\ & n=\text{banyak ulangan yang dijalani oleh Deni}\\ & x=\text{Total nilai ulangannya Deni}\\ & \text{MODEL MATEMATIKA}\\ & \bullet {\displaystyle \frac{x+75}{n+1}=82}\\ & \iff x+75=82(n+1)\iff x=82n+82-75\\ & \iff x=82n+7..................(1)\\ & \bullet {\displaystyle \frac{x+93}{n+1}=85}\\ & \iff x+93=85(n+1)\iff x=85n+85-93\\ & \iff x=85n-8..................(2)\\ & \text{KESAMAAN}\\ & \text{Dari persamaan (1) dan (2), maka}\\ & x=x\\ & 85n-8=82n+7\\ & 85n-82n=7+8\iff 3n=15\iff n=5\end{array}\end{array}$

$\begin{array}{l}\\ 39.& (\mathbf{\text{SIMAK UI}})\\ & \text{Jika rata-rata 20 bilangan bulat nonnegatif}\\ & \text{berbeda adalah 20, maka bilangan terbesar}\\ & \text{yang mungkin adalah}....\\ & \begin{array}{l}\\ \text{a}.& 210\\ \text{b}.& 229\\ \text{c}.& 230\\ \text{d}.& 239\\ \text{e}.& 240\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {\displaystyle \frac{x_1+x_2+x_3+...+x_{18}+x_{19}+x_{20}}{20}=20}\\ & x_1+x_2+x_3+...+x_{18}+x_{19}+x_{20}=400\\ & \text{Yang mungkin bilangan bulat nonnegatif}\\ & \text{dan berbeda adalah}:0,1,2,3,....dst\\ & 0+1+2+..+18+x_{20}=400\\ & {\displaystyle \frac{18\times 19}{2}+x_{20}=400}\\ & 171+x_{20}=400\\ & x_{20}=400-171=229\end{array}\end{array}$.

$\begin{array}{ll}40.& \mathbf{\text{SPMB 2006}}\\ & \text{Jika jangkauan dari data terurut}:x-1,2x-1,3x,\\ & 5x-3,4x+3,6x+2\text{adalah}18,\text{maka mediannya}\\ & \text{ adalah}....\\ & \begin{array}{llllll}\text{a}.& \text{9}& & & \text{d}.& \text{21}\\ \text{b}.& \text{10},5& \text{c}.& \text{12}& \text{e}.& \text{24},8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{b}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & x-1,2x-1,3x,5x-3,4x+3,6x+2\\ & \text{Dan diketahui pula nilai jangkauannya}=18\\ & J=x_{{}_{max}}-x_{{}_{min}}=18\\ & \iff (6x+2)-(x-1)=18\\ & \iff 5x+3=18\\ & \iff 5x=15\\ & \iff x=3\\ & \text{Sehingga datanya}:2,3,9,12,15,20\\ & \text{Selanjutnya ditentukan mediannya}=M_e=Q_{{}_2}\\ & \text{karena}n=6,\text{datum ke}-{\displaystyle \frac{2}{4}(6+1)=3,5}\\ & M_e=Q_2=x_{{.}_3}+0,5(x_{{.}_4}-x_{{.}_3})=9+0,5(12-9)\\ & =9+0,5(3)=9+1,5=10,5\\ & \text{Jadi},J=10,5\end{array}\end{array}$.

Soal lanjutannya (yaitu Contoh Soal 10 Statistika) silahkan klik di sini

DAFTAR PUSTAKA

1. Kanginan, M., Terzalgi, Y. 2013. Matematika Wajib untuk SMA/MA Kelas X. Bandung: SRIKANDI EMPAT WIDYA UTAMA.

Contoh Soal 9 Fungsi Logaritma (Pemecahan Masalah Olimpiade)

$\begin{array}{l}\\ 41.& \text{Jika}x={}^{15}\log 75\text{dan}y={}^{{\displaystyle {}^{\frac{3}{5}}}}\log {\displaystyle \frac{9}{125},}\\ & \text{maka nilai}5x+3y-2xy\text{adalah}....\\ \\ & \text{KOMPETISI HARDIKNAS ONLINE}\\ & \text{POSI}(\text{Pelatihan Olimpiade Sain Indonesia})\\ & \text{Bidang Matematika 2020}\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& 1\\ \text{c}.& 3\\ \text{d}.& 5\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 5x+3y-2xy\\ & =5({}^{15}\log 75)+3\left({}^{{\displaystyle {}^{\frac{3}{5}}}}\log {\displaystyle \frac{9}{125}}\right)\\ & -2({}^{15}\log 75)\left({}^{{\displaystyle {}^{\frac{3}{5}}}}\log {\displaystyle \frac{9}{125}}\right)\\ & =5\left({\displaystyle \frac{\log 75}{\log 15}}\right)+3\left({\displaystyle \frac{\log {\displaystyle \frac{9}{125}}}{\log {\displaystyle \frac{3}{5}}}}\right)\\ & -2\left({\displaystyle \frac{\log 75}{\log 15}}\right)\left({\displaystyle \frac{\log {\displaystyle \frac{9}{125}}}{\log {\displaystyle \frac{3}{5}}}}\right)\\ & =5\left({\displaystyle \frac{\log {3.5}^2}{\log 3.5}}\right)+3\left({\displaystyle \frac{\log -\log 125}{\log 3-\log 5}}\right)\\ & -2\left({\displaystyle \frac{\log {3.5}^2}{\log 3.5}}\right)\left({\displaystyle \frac{\log 9-\log 125}{\log 3-\log 5}}\right)\\ & =5\left({\displaystyle \frac{\log 3+\log 5^2}{\log 3-\log 5}}\right)+3\left({\displaystyle \frac{\log 3^2-\log 5^3}{\log 3-\log 5}}\right)\\ & -2\left({\displaystyle \frac{\log 3+\log 5^2}{\log 3+\log 5}}\right)\left({\displaystyle \frac{\log 3^2-\log 5^3}{\log 3-\log 5}}\right)\\ & =5\left({\displaystyle \frac{\log 3+2\log 5}{\log 3-\log 5}}\right)+3\left({\displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5}}\right)\\ & -2\left({\displaystyle \frac{\log 3+2\log 5}{\log 3+\log 5}}\right)\left({\displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5}}\right)\\ \\ & \text{Misalkan}\log 3=A,\log 5=B\end{array}\end{array}$

$.\begin{array}{rl}& \text{Selanjutnya}\\ & =5\left({\displaystyle \frac{A+2B}{A+B}}\right)+3\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & -2\left({\displaystyle \frac{A+2B}{A+B}}\right)\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & =\left({\displaystyle \frac{5A+10B}{A+B}}\right)+\left({\displaystyle \frac{6A-9B}{A-B}}\right)\\ & -\left({\displaystyle \frac{2A+4B}{A+B}}\right)\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & ={\displaystyle \frac{(5A+10B)(A-B)+(6A-9B)(A+B)}{A^2-B^2}}\\ & -\left({\displaystyle \frac{4A^2-6AB+8AB-12B^2}{A^2-B^2}}\right)\\ & ={\displaystyle \frac{5A^2-5AB+10AB-10B^2}{A^2-B^2}}\\ & +{\displaystyle \frac{6A^2+6AB-9AB-9B^2}{A^2-B^2}}\\ & -\left({\displaystyle \frac{4A^2-6AB+8AB-12B^2}{A^2-B^2}}\right)\\ & ={\displaystyle \frac{7A^2-7B^2}{A^2-B^2}}\\ & ={\displaystyle \frac{7(A^2-B^2)}{A^2-B^2}}\\ & =7\end{array}$

$\begin{array}{l}\\ 42.& \text{Diberikan}A={}^6\log 16\text{dan}B={}^{12}\log 27\\ & \text{Terdapat bilangan-bilangan bulat positif}\\ & a,b,\text{dan}c\text{sehingga}(A+a)(B+b)=c\\ & \text{Nilai dari}a+b+c\text{adalah}....\\ \\ & \text{KOMPETISI HARDIKNAS ONLINE}\\ & \text{POSI}(\text{Pelatihan Olimpiade Sain Indonesia})\\ & \text{Bidang Matematika 2020}\\ & \begin{array}{l}\\ \text{a}.& 23\\ \text{b}.& 24\\ \text{c}.& 27\\ \text{d}.& 30\\ \text{e}.& 34\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{....}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & A={}^6\log 16={\displaystyle \frac{\log 16}{\log 6}={\displaystyle \frac{\log 2^4}{\log 2.3}=\frac{4\log 2}{\log 2+\log 3}}}\\ & \iff \log 2+\log 3={\displaystyle \frac{4\log 2}{A}...........(1)}\\ & B={}^{12}\log 27={\displaystyle \frac{\log 27}{\log 12}=\frac{\log 3^3}{\log 2^2.3}=\frac{3\log 3}{2\log 2+\log 3}}\\ & \iff 2\log 2+\log 3={\displaystyle \frac{3\log 3}{B}.........(2)}\\ & \text{ELIMINASI}\\ & \text{Dari persamaan (1) dan (2) diperoleh}:\\ & \bullet \log 2={\displaystyle \frac{3\log 3}{B}-{\displaystyle \frac{4\log 2}{A}}}\\ & \iff \log 2={\displaystyle \frac{3A\log 3-4B\log 2}{AB}}\\ & \iff AB\log 2=3A\log 3-4B\log 2\\ & \iff AB\log 2+4B\log 2=3A\log 3\\ & \iff (AB+4B)\log 2=3A\log 3\\ & \iff {\displaystyle \frac{\log 2}{\log 3}={\displaystyle \frac{3A}{AB+4B}..........(3)}}\\ & \bullet \log 3={\displaystyle \frac{8\log 2}{A}-{\displaystyle \frac{3\log 3}{B}}}\\ & \iff \log 3={\displaystyle \frac{8B\log 2-3A\log 3}{AB}}\\ & \iff AB\log 3=8B\log 2-3A\log 3\\ & \iff AB\log 3+3A\log 3=8B\log 2\\ & \iff (AB+3A)\log 3=8B\log 2\\ & \iff \frac{\log 2}{\log 3}={\displaystyle \frac{AB+3A}{8B}...........(4)}\\ & \text{KESAMAAN}\\ & \frac{\log 2}{\log 3}=\frac{\log 2}{\log 3}\\ & {\displaystyle \frac{AB+3A}{8B}={\displaystyle \frac{3A}{AB+4B}}}\\ & \iff (AB+3A)(AB+4B)=(8B).(3A)\\ & \iff (B+3)(A+4)=24\\ & \iff (A+4)(B+3)=24\\ & \text{KESIMPULAN}\\ & a=4,b=3,\text{dan}c=24,\\ & \text{maka}a+b+c=4+3+24=31\end{array}\end{array}$

Contoh Soal 8 Fungsi Logaritma (Persamaan Logaritma)

$\begin{array}{l}\\ 36.& \text{Persamaan}{}^x\log 2+{}^x\log (3x-4)=2\\ & \text{mempunyai akar}{x}_1\text{dan}{x}_2,\text{maka}\\ & \text{nilai}{x}_1+x_2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 6\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & {}^x\log 2+{}^x\log (3x-4)=2\\ & \iff {}^x\log 2(3x-4)=2\\ & \iff {}^x\log 6x-8=2\\ & \iff 6x-8=x^2\\ & \iff x^2-6x+8=0,\text{dengan}\{\begin{array}{ll}a& =1\\ b& =-6\\ c& =8\end{array}\\ & \iff x_1+x_2=-{\displaystyle \frac{b}{a}}\\ & \iff x_1+x_2=-{\displaystyle \frac{-6}{1}=6}\\ & \text{Alternatif 2}\\ & \iff x^2-6x+8=0\\ & \iff (x-2)(x-4)\\ & \iff x_1=2\text{atau}{x}_2=4\\ & \iff x_1+x_2=2+4=6\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \text{Jika}{x}_1\text{dan}{x}_2\text{memenuhi}\\ & (\log x)(2\log x-3)=\log 100\\ & \text{maka}{x}_1\times x_2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 100\\ \text{b}.& 10\sqrt{10}\\ \text{c}.& \sqrt{10}\\ \text{d}.& -\sqrt{10}\\ \text{e}.& -10\sqrt{10}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& (\log x)(2\log x-3)=\log 100\\ & \iff (\log x)(2\log x-3)=2\\ & \iff 2{\log }^{2}x-3\log x-2=0\{\begin{array}{ll}a& =2\\ b& =-3\\ c& =-2\end{array}\\ & \iff \log x_1+\log x_2=-{\displaystyle \frac{-3}{2}=\frac{3}{2}}\\ & \iff \log (x_1\times x_2)=1{\displaystyle \frac{1}{2}}\\ & \iff (x_1\times x_2)={10}^{1\frac{1}{2}}=10\sqrt{10}\end{array}\end{array}$

$\begin{array}{l}\\ 38.& \text{Persamaan}\\ & {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{mempunyai dua akar yaitu}{x}_1\text{dan}{x}_2\\ & \text{Nilai}{x}_1\times x_2=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -5\\ \text{c}.& 2\\ \text{d}.& 5\\ \text{e}.& 10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & {10}^{2^{{}^2}\log x}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{adalah persamaan kuadrat dalam}{10}^{{}^{{}^2}\log x}\\ & \text{Misalkan}p={10}^{{}^{{}^2}\log x},\text{maka persamaan}\\ & \text{menjadi}{p}^2-7p+10=0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =10\end{array}\\ & \text{Karena nilai}{p}_1\times p_2={\displaystyle \frac{c}{a}\text{maka}}\\ & {10}^{{}^{{}^2}\log x_1}\times {10}^{{}^{{}^2}\log x_2}={\displaystyle \frac{10}{1}=10}\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}=10\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}={10}^1\\ & \iff {}^2\log x_1+{}^2\log x_2=1\\ & \iff {}^2\log x_1\times x_2=1\\ & \iff x_1\times x_2=2^1=2\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Nilai}x\text{yang memenuhi}\\ & {}^x\log (x+12)-3.{}^x\log 4+1=0\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 8\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^x\log (x+12)-3.{}^x\log 4+1=0\\ & {}^x\log (x+12)-{}^x\log 4^3=-1\\ & {}^x\log {\displaystyle \frac{x+12}{64}=-1}\\ & {\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}}\\ & x+12={\displaystyle \frac{64}{x}}\\ & x^2+12x-64=0\\ & (x+16)(x-4)=0\\ & x=-16\text{atau}x=4\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{{}^2\log (2x-3)}{{}^2\log x}-{}^x\log (x+6)+{\displaystyle \frac{1}{{}^{(x+2)}\log x}=1}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\text{dan}6\\ \text{b}.& -2\text{dan}6\\ \text{c}.& -1\\ \text{d}.& -2\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{{}^2\log (2x-3)}{{}^2\log x}-{}^x\log (x+6)+{\displaystyle \frac{1}{{}^{(x+2)}\log x}=1}}\\ & {}^x\log (2x-3)-{}^x\log (x+6)+{}^x\log (x+2)=1\\ & {}^x\log (2x-3)(x+2)=1+\log (x+6)\\ & {}^x\log {\displaystyle \frac{(2x^2+x-6)}{(x+6)}=1}\\ & {\displaystyle \frac{(2x^2+x-6)}{(x+6)}=x^1}\\ & (2x^2+x-6)=x^2+6x\\ & x^2-5x-6=0\\ & (x+1)(x-6)=0\\ & x=-1\text{atau}x=6\end{array}\end{array}$

Contoh Soal 7 Fungsi Logaritma (Pertidaksamaan Logaritma)

 $\begin{array}{l}\\ 32.& \text{Agar}\log (x^2-1)<0\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& -1<x<1\\ \text{b}.& -\sqrt{2}<x<\sqrt{2}\\ \text{c}.& x<-1\text{atau}x>1\\ \text{d}.& x<-\sqrt{2}\text{atau}x>\sqrt{2}\\ \text{e}.& -\sqrt{2}<x<-1\text{atau}1<x<\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \log (x^2-1)<0\\ & \text{Diketahui}\log f(x)<0,\text{maka}\\ & \text{Syarat (1)},f(x)>0\\ & \iff x^2-1>0\\ & \iff x<-1\text{atau}x>1\\ & \text{Syarat (2)},\log (x^2-1)<0\\ & \log (x^2-1)<\log 1\\ & \iff x^2-1<1\\ & \iff x^2-2<0\\ & \iff x^2-{\left(\sqrt{2}\right)}^2<0\\ & \iff -\sqrt{2}<x<\sqrt{2}\\ & \text{Jadi},-\sqrt{2}<x<-1\text{atau}1<x<\sqrt{2}\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Himpunan penyelesaian dari}\\ & {}^{{.}^{\frac{1}{2}}}\log (x^2-3)>0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|-2<x<-\sqrt{3}\text{atau}\sqrt{3}<x<2\}\\ \text{b}.& \{x|-\sqrt{3}<x<-1\text{atau}\sqrt{3}<x<2\}\\ \text{c}.& \{x|-2<x<-\sqrt{3}\}\\ \text{d}.& \{x|-2<x<-\sqrt{3}\}\\ \text{e}.& \{x|\sqrt{3}<x<2\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^{{.}^{\frac{1}{2}}}\log (x^2-3)>0\\ & \text{Diketahui}{}^{{.}^{\frac{1}{2}}}\log f(x)>0,\text{maka}\\ & \text{Syarat (1)},f(x)>0\\ & \iff x^2-3>0\\ & \iff x<-\sqrt{3}\text{atau}x>\sqrt{3}\\ & \text{Syarat (2)},{}^{{.}^{\frac{1}{2}}}\log (x^2-3)>0\\ & {}^{{.}^{\frac{1}{2}}}\log (x^2-3)>{}^{{.}^{\frac{1}{2}}}\log 1\\ & \iff x^2-3<1\left(\text{karena basisnya}{\displaystyle \frac{1}{2}<1}\right)\\ & \iff x^2-4<0\\ & \iff x^2-2^2>0\\ & \iff -2<x<2\\ & \text{Jadi},-2<x<-\sqrt{3}\text{atau}\sqrt{3}<x<2\end{array}\end{array}$

$\begin{array}{l}\\ 34.& \text{Nilai}x\text{yang memenuhi}\\ & {}^2\log (x^2-x)\le 1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<0\text{atau}x>1\\ \text{b}.& -1\le x\le 2,x\ne 1\text{atau}x\ne 0\\ \text{c}.& -1\le x<0\text{atau}1<x\le 2\\ \text{d}.& -1<x\le 0\text{atau}1\le x<2\\ \text{e}.& -1\le x\le 0\text{atau}1\le x\le 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^2\log (x^2-x)\le 1\\ & \text{Diketahui}{}^2\log f(x)>0,\text{maka}\\ & \text{Syarat (1)},f(x)>0\\ & \iff x^2-x>0\iff x(x-1)>0\\ & \iff x<0\text{atau}x>1\\ & \text{Syarat (2)},{}^2\log (x^2-x)\le 1\\ & {}^2\log (x^2-x)\le {}^2\log 2\\ & \iff x^2-x\le 2\\ & \iff x^2-x-2\le 0\\ & \iff (x+1)(x-2)\le 0\\ & \iff -1\le x\le 2\\ & \text{Jadi},-1\le x<0\text{atau}1<x\le 2\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Nilai}x\text{yang memenuhi}\\ & |\log (x+1)|>1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<-0,9\text{atau}x>9\\ \text{b}.& x<-9\text{atau}x>9\\ \text{c}.& -1<x<-0,9\text{atau}x>9\\ \text{d}.& -9<x<0,9\\ \text{e}.& -0,9<x<9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Ingat bahwa}\\ & \left|x\right|>A\iff x<-A\text{atau}x>A,A>0\\ & \iff \log (x+1)<-1\text{atau}\log (x+1)>1\\ & \text{Syarat (1) buat keduanya},f(x)>0\\ & (x+1)>0\iff x>-1\\ & \text{Syarat (2)},\log (x+1)<-1\\ & \log (x+1)<\log {10}^{-1}\\ & x+1<{\displaystyle \frac{1}{10}\iff x<-\frac{9}{10}}\\ & \text{Syarat (3)},\log (x+1)>1\\ & \log (x+1)>\log {10}^1\\ & (x+1)>10\iff x>9\\ & \text{Jadi},-1<x<-0,9\text{atau}x>9\end{array}\end{array}$

Lanjutan Materi Fungsi Logaritma (Kelas X Matematika Peminatan)

 MENGINGAT KEMBALI

$\text{E. Sifat-Sifat Eksponen dan Logaritma}$

$\text{E.1 Persamaan Eksponen dan Logaritma}$

$\begin{array}{|ll|}\hline \text{Eksponens}& \text{Logaritma}\\ {\displaystyle a^n=\underset{nfaktor}{\underbrace{a\times a\times a\times \cdots \times a}}}& {}^a\log b=c\Rightarrow a^c=b\\ \bullet a^p\times a^q=a^{p+q}& \bullet {}^a\log x+{}^a\log y={}^a\log xy\\ \bullet a^p:a^q=a^{p-q}& \bullet {}^a\log x-{}^a\log y={}^a\log {\displaystyle \frac{x}{y}}\\ \bullet {\left(a^p\right)}^q=a^{p.q}& \bullet {}^a\log x={\displaystyle \frac{{}^m\log x}{{}^m\log a}}\\ \bullet {\displaystyle \sqrt[q]{a^p}={\displaystyle a^{\left(\frac{p}{q}\right)}}}& \bullet {}^a\log b\times {}^b\log c={}^a\log c\\ \bullet {(a\times b)}^p=a^p\times b^p& \bullet {}^{a^m}\log b^n={\displaystyle \frac{n}{m}\times {}^a\log b}\\ \bullet {\left({\displaystyle \frac{a}{b}}\right)}^p={\displaystyle \frac{{\displaystyle a^p}}{{\displaystyle b^p}}}& \bullet {\displaystyle a^{{}^a\log b}=b}\\ \bullet a^{-p}={\displaystyle \frac{1}{{\displaystyle a^p}}}& \bullet {}^a\log b={\displaystyle \frac{1}{{}^b\log a}}\\ \bullet a^0=1,a\ne 0& \bullet {}^a\log 1=0\\ \bullet a^1=1& \bullet {}^a\log a=1\\ \{\begin{array}{l}a,b\in \mathbb{R}\\ p,q\in \mathbb{Q}\end{array}& \{\begin{array}{ll}a\ne 0& \\ a>0& (\text{bilangan pokok})\\ x,y>0& (\text{numerus})\end{array}\\ \hline\end{array}$

Selanjutnya

$\begin{array}{|lll|}\hline \text{No}& \text{Bentuk}& \text{Syarat}\\ 1.& a^{f(x)}=1& a\ne 0,\text{maka}f(x)=0\\ 2.& a^{f(x)}=a^p& a>0,a\ne 1,\text{maka}f(x)=p\\ 3.& a^{f(x)}=a^{g(x)}& a>0,a\ne 1,\text{maka}f(x)=g(x)\\ 4.& a^{f(x)}=b^{f(x)}& a\ne 0,b\ne 0,\text{maka}f(x)=0\\ 5.& f(x{)}^{g(x)}=1& \{\begin{array}{ll}f(x)=1& \\ g(x)=0,& \text{jika}f(x)\ne 0\\ f(x)=-1,& \text{jika}g(x)=\text{genap}\end{array}\\ 6.& f(x{)}^{g(x)}=f(x{)}^{h(x)}& \{\begin{array}{ll}(i).g(x)=h(x)& \\ (ii).f(x)=1& \\ (iii).f(x)=0,& g(x)>0,h(x)>0\\ (iv).f(x)=-1,& g(x)\text{dan}h(x)\\ & \text{keduanya ganjil}\\ & \text{atau genap}\end{array}\\ 7.& g(x{)}^{f(x)}=h(x{)}^{f(x)}& \{\begin{array}{ll}(i).g(x)=h(x)& \\ (ii).f(x)=0,& g(x)\ne 0,h(x)\ne 0\end{array}\\ 8.& \begin{array}{rl}& A{\left(a^{f(x)}\right)}^2\\ & +B\left(a^{f(x)}\right)\\ & +C=0\end{array}& a>0,a\ne 1\\ \hline\end{array}$

$\text{E.2 Pertidaksamaan Eksponen dan Logaritma}$

Berikut sifat pertidaksamaan Eksponen

$\begin{array}{|ll|}\hline a>1& 0<a<1\\ a^{f(x)}\le a^{g(x)}\Rightarrow f(x)\le g(x)& a^{f(x)}\le a^{g(x)}\Rightarrow f(x)\ge g(x)\\ a^{f(x)}<a^{g(x)}\Rightarrow f(x)<g(x)& a^{f(x)}<a^{g(x)}\Rightarrow f(x)>g(x)\\ a^{f(x)}\ge a^{g(x)}\Rightarrow f(x)\ge g(x)& a^{f(x)}\ge a^{g(x)}\Rightarrow f(x)\le g(x)\\ a^{f(x)}>a^{g(x)}\Rightarrow f(x)>g(x)& a^{f(x)}>a^{g(x)}\Rightarrow f(x)<g(x)\\ \hline\end{array}$

Untuk pertidaksamaan logaritma (dengan syarat  $(f(x)>0\text{dan}g(x)>0)$ ) adalah sebagai berikut:

$\begin{array}{|ll|}\hline a>1& 0<a<1\\ \begin{array}{rl}& {}^a\log f(x)\le {}^a\log g(x)\\ & \Rightarrow f(x)\le g(x)\end{array}& \begin{array}{rl}& {}^a\log f(x)\le {}^a\log g(x)\\ & \Rightarrow f(x)\ge g(x)\end{array}\\ \begin{array}{rl}& {}^a\log f(x)<{}^a\log g(x)\\ & \Rightarrow f(x)<g(x)\end{array}& \begin{array}{rl}& {}^a\log f(x)<{}^a\log g(x)\\ & \Rightarrow f(x)>g(x)\end{array}\\ \begin{array}{rl}& {}^a\log f(x)\ge {}^a\log g(x)\\ & \Rightarrow f(x)\ge g(x)\end{array}& \begin{array}{rl}& {}^a\log f(x)\ge {}^a\log g(x)\\ & \Rightarrow f(x)\le g(x)\end{array}\\ \begin{array}{rl}& {}^a\log f(x)>{}^a\log g(x)\\ & \Rightarrow f(x)>g(x)\end{array}& \begin{array}{rl}& {}^a\log f(x)>{}^a\log g(x)\\ & \Rightarrow f(x)<g(x)\end{array}\\ \hline\end{array}$

Contoh Soal 6 Matriks

$\begin{array}{l}\\ 26.& (\mathbf{\text{UM UGM 2004}})\\ & \text{Nilai-nilai}x\text{agar matriks}\\ & \left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)\\ & \text{tidak memiliki invers adalah}....\\ & \begin{array}{l}\\ \text{a}.& 4\text{atau}5\\ \text{b}.& -2\text{atau}2\\ \text{c}.& -4\text{atau}5\\ \text{d}.& -6\text{atau}4\\ \text{e}.& 0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{supaya matriks}\left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)\\ & \text{tidak memiliki invers},\text{maka}\\ & \text{determinan matriks}\left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)=0\\ & \text{Sehingga}\\ & \left|\begin{array}{cc}5x& 5\\ 4& x\end{array}\right|=0\\ & \iff 5x^2-20=0\\ & \iff x^2=4\\ & \iff x=\pm 2\end{array}\end{array}$

$\begin{array}{l}\\ 27.& (\mathbf{\text{UM UGM 2005}})\\ & \text{Matriks}\left(\begin{array}{cc}x& 1\\ -2& 1-x\end{array}\right)\\ & \text{tidak memiliki invers untuk}\\ & \text{nilai}x=....\\ & \begin{array}{l}\\ \text{a}.& -1\text{atau}-2\\ \text{b}.& -1\text{atau}0\\ \text{c}.& -1\text{atau}1\\ \text{d}.& -1\text{atau}2\\ \text{e}.& 1\text{atau}2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \text{Mirip dengan pembahasan no. 26}\\ & \begin{array}{rl}& \text{Nilai}\left|\begin{array}{cc}x& 1\\ -2& 1-x\end{array}\right|=0\\ & \iff x-x^2-(-2)=0\\ & \iff 2+x-x^2=0\\ & \iff x^2-x-2=0\\ & \iff (x-2)(x+1)=0\\ & \iff x=2\text{atau}x=-1\end{array}\end{array}$

$\begin{array}{l}\\ 28.& (\mathbf{\text{Mat Das SIMAK UI 2014}})\\ & \text{Jika matriks}\text{A}\text{adalah invers}\\ & \text{dari matriks}{\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)\text{dan}}\\ & \text{A}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right)\text{maka nilai}2x+y\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{10}{3}}\\ \text{b}.& -{\displaystyle \frac{1}{3}}\\ \text{c}.& 1\\ \text{d}.& {\displaystyle \frac{9}{7}}\\ \text{e}.& {\displaystyle \frac{20}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Misalkan diketahui matriks}\\ & \text{B}={\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right),}\\ & \text{maka}\text{A}={\left({\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)}\right)}^{-1}\\ & \text{selanjutnya}\text{A}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & \left(\begin{array}{c}x\\ y\end{array}\right)=A^{-1}\left(\begin{array}{c}1\\ 3\end{array}\right),\\ & \text{ingat bahwa}{\left({\mathbf{\text{A}}}^{-1}\right)}^{-1}=\mathbf{\text{A}}\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\left({\left({\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)}\right)}^{-1}\right)}^{-1}\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)\left(\begin{array}{c}1\\ 3\end{array}\right)}\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3}\left(\begin{array}{c}-1-9\\ 4+15\end{array}\right)=\left(\begin{array}{c}{\displaystyle -\frac{10}{3}}\\ {\displaystyle \frac{19}{3}}\end{array}\right)}\\ & 2x+y=2(-{\displaystyle \frac{10}{3}})+\frac{19}{3}\\ & ={\displaystyle \frac{-20+19}{3}=-{\displaystyle \frac{1}{3}}}\end{array}\end{array}$

Contoh Soal 5 Matriks

$\begin{array}{l}\\ 21.& (\mathbf{\text{SPMB 2003}})\\ & \text{Diketahu matriks}\text{A}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right).\\ & \text{Jika}{\text{A}}^t={\text{A}}^{-1}\text{dengan}{\text{A}}^t\\ & \text{adalah transpose matriks A},\\ & \text{maka nilai}ad-bc=....\\ & \begin{array}{l}\\ \text{a}.& -1\text{atau}-\sqrt{2}\\ \text{b}.& 1\text{atau}\sqrt{2}\\ \text{c}.& -\sqrt{2}\text{atau}-\sqrt{2}\\ \text{d}.& -1\text{atau}1\\ \text{e}.& 1\text{atau}-\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\text{matriks}\text{A}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ & \text{dan}{\text{A}}^t={\text{A}}^{-1},\text{maka}\\ & {\text{A}}^t={\text{A}}^{-1}\\ & {\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)}^t={\displaystyle \frac{1}{ad-bc}\times \text{Adjoin Matriks}\text{A}}\\ & \left(\begin{array}{cc}a& c\\ b& d\end{array}\right)={\displaystyle \frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right),}\\ & \text{didapatkan hubungan}\\ & c={\displaystyle \frac{-b}{ad-bc}...............(1)}\\ & b={\displaystyle \frac{-c}{ad-bc}...............(2)}\\ & \text{Persamaan}(2)\text{disubstitusikan ke persamaan}(1)\\ & c={\displaystyle {\displaystyle \frac{-{\displaystyle \frac{-c}{ad-bc}}}{ad-bc}}}\\ & 1=(ad-bc{)}^2\\ & (ad-bc)=-1\text{atau}1\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Diketahu matriks}\text{H}\\ & \text{yang memenuhi persamaan}\\ & \text{H}\left(\begin{array}{cc}3& 2\\ 1& 4\end{array}\right)=\left(\begin{array}{cc}7& 8\\ 4& 6\end{array}\right),\\ & \text{maka nilai dari}det\text{H}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -3\\ \text{b}.& -2\\ \text{c}.& -1\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{|c|}\hline \mathbf{\text{Alternatif 1}}\\ \begin{array}{rl}\text{H.A}& =\text{B}\\ {\text{H.A.A}}^{-1}& ={\text{B.A}}^{-1}\\ \text{H}& ={\text{B.A}}^{-1}\\ & =\left(\begin{array}{cc}7& 8\\ 4& 6\end{array}\right).{\displaystyle \frac{1}{\left|\begin{array}{cc}3& 2\\ 1& 4\end{array}\right|}\left(\begin{array}{cc}4& -2\\ -1& 3\end{array}\right)}\\ & =\left(\begin{array}{cc}7& 8\\ 4& 6\end{array}\right).{\displaystyle \frac{1}{12-2}\left(\begin{array}{cc}4& -2\\ -1& 3\end{array}\right)}\\ & ={\displaystyle \frac{1}{10}\left(\begin{array}{cc}28+(-8)& (-14)+24\\ 16+(-6)& (-8)+18\end{array}\right)}\\ \text{H}& ={\displaystyle \frac{1}{10}\left(\begin{array}{cc}20& 10\\ 10& 10\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)}\\ det\text{H}& =\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|=2.1-1.1=2-1=1\end{array}\\ \mathbf{\text{Alternatif 2}}\\ \begin{array}{rl}\text{H.A}& =\text{B}\{\begin{array}{ll}det\text{H}& =\left|\text{H}\right|\\ det\text{A}& =\left|\text{A}\right|=\left|\begin{array}{cc}3& 4\\ 2& 1\end{array}\right|\\ & =12-2=10\\ det\text{B}& =\left|\text{B}\right|=\left|\begin{array}{cc}7& 8\\ 4& 6\end{array}\right|\\ & =42-32=10\end{array}\\ \left|\text{H}\right|.\left|\text{A}\right|& =\left|\text{B}\right|\\ \left|\text{H}\right|& ={\displaystyle \frac{\left|\text{B}\right|}{\left|\text{A}\right|}}\\ & ={\displaystyle \frac{10}{10}}\\ & =1\end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 23.& (\mathbf{\text{UM UGM 2006}})\\ & \text{Apabila}x\text{dan}y\text{memenuhi}\\ & \text{persamaan matriks}\\ & \left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-1\\ 2\end{array}\right),\\ & \text{maka}x+y=....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ A.X& =B\\ A^{-1}.A.X& =A^{-1}.B\\ A^0.X& =A^{-1}.B\\ X& =A^{-1}.B\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)}^{-1}\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{\left|\begin{array}{cc}1& -2\\ -1& 3\end{array}\right|}}& \left(\begin{array}{cc}3& 2\\ 1& 1\end{array}\right)\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3-2}}& \left(\begin{array}{c}3.(-1)+2.2\\ 1.(-1)+1.2\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}1\\ 1\end{array}\right)\\ x+y& =1+1=2\end{array}\end{array}$

$\begin{array}{l}\\ 24.& (\mathbf{\text{KSM Matematika Kabupten 2019}})\\ & \text{Matriks}A\text{dengan entri bulat dan}\\ & \text{berukuran 2x2},\text{dikalikan dengan matriks}\\ & \left(\begin{array}{cc}1& 2\\ 2& 2\end{array}\right)\text{dari kanan menghasilkan matriks}\\ & \text{yang semua entrinya bilangan prima}.\\ & \text{Jika determinan dari matriks}A\text{juga}\\ & \text{bilangan prima, maka nilai minimum dari}\\ & detA\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 5\\ \text{d}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left(\begin{array}{cc}1& 2\\ 2& 2\end{array}\right)& \times A_{2\times 2}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\\ \left|\begin{array}{cc}1& 2\\ 2& 2\end{array}\right|& \times \left|A_{2\times 2}\right|=\left|\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right|\\ & \left|A_{2\times 2}\right|={\displaystyle \frac{\left|\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right|}{\left|\begin{array}{cc}1& 2\\ 2& 2\end{array}\right|}}\\ & \left|A_{2\times 2}\right|={\displaystyle \frac{(\alpha \delta -\beta \gamma )}{-2}}\\ & \left|A_{2\times 2}\right|={\displaystyle \frac{(\beta \gamma -\alpha \delta )}{2}}\\ \text{Karena}& \left|A_{2\times 2}\right|\text{bilangan prima}\\ \text{akan m}& \text{engakibatkan}(\beta \gamma -\alpha \delta )\\ \text{harus h}& \text{abis dibagi}2,\text{oleh karenanya}\\ \text{menyeb}& \text{abkan}(\beta \gamma -\alpha \delta )\text{berupa bilangan}\\ \text{genap.}& \text{Dan karena}(\beta \gamma -\alpha \delta )\text{genap},\\ \text{maka p}& \text{astilah}\left|A_{2\times 2}\right|\text{juga bernilai genap}\\ \text{sehingg}& \text{a nilai}\left|A_{2\times 2}\right|\text{pastilah 2}\end{array}\end{array}$

$\begin{array}{l}\\ 25.& (\mathbf{\text{UM UGM 2005}})\text{Jika}\\ & \left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right)=\left(\begin{array}{cc}\sin A& \cos A\end{array}\right)\\ & \text{dan}A\text{suatu konstanta, maka}x+y=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right)=\left(\begin{array}{cc}\sin A& \cos A\end{array}\right)\\ & \left(\begin{array}{cc}x\sin \alpha -y\cos \alpha & x\cos \alpha +y\sin \alpha \end{array}\right)=\left(\begin{array}{cc}\sin A& \cos A\end{array}\right)\\ & \{\begin{array}{ll}\sin A& =x\sin \alpha -y\cos \alpha =\sqrt{x^2+y^2}\cos (\alpha -{\tan }^{-1}{\displaystyle \frac{x}{-y}})\\ \cos A& =x\cos \alpha +y\sin \alpha =\sqrt{x^2+y^2}\cos (\alpha -{\tan }^{-1}{\displaystyle \frac{y}{x}})\end{array}\\ & \text{Supaya}\cos A=\sqrt{x^2+y^2}\cos (\alpha -{\tan }^{-1}{\displaystyle \frac{y}{x}}),\text{maka}\\ & \{\begin{array}{ll}\sqrt{x^2+y^2}& =1\\ {\tan }^{-1}{\displaystyle \frac{y}{x}}& =0\Rightarrow \{\begin{array}{ll}y& =0\\ x& =1\end{array}\end{array}\\ & \text{Sehingga}x+y=1+0=1\end{array}\end{array}$

Contoh Soal 4 Matriks

$\begin{array}{l}\\ 16.& \text{Determinan untuk matriks}\left(\begin{array}{cc}2& -5\\ 3& -1\end{array}\right)=....\\ & \begin{array}{l}\\ \text{a}.& -17\\ \text{b}.& -13\\ \text{c}.& 11\\ \text{d}.& 13\\ \text{e}.& 17\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Determinan}& \text{dari matriks}\left(\begin{array}{cc}2& -5\\ 3& -1\end{array}\right)\\ & =\left|\begin{array}{cc}2& -5\\ 3& -1\end{array}\right|=2(-1)-3(-5)\\ & =-2+15\\ & =13\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Determinan untuk matriks}\\ & \left(\begin{array}{ccc}2& -1& -1\\ 1& 4& -1\\ 1& -2& 3\end{array}\right)=....\\ & \begin{array}{l}\\ \text{a}.& 10\\ \text{b}.& 18\\ \text{c}.& 22\\ \text{d}.& 30\\ \text{e}.& 36\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Determinan}\text{dari matriks}\\ & \left(\begin{array}{ccc}2& -1& -1\\ 1& 4& -1\\ 1& -2& 3\end{array}\right)\\ & =\left|\begin{array}{ccc}2& -1& -1\\ 1& 4& -1\\ 1& -2& 3\end{array}\right|\\ & =+(2.4.3)+(-1.-1.1)+(-1.1.-2)\\ & -(1.4.-1)-(-2.-1.2)-(3.1.-1)\\ & =24+1+2+4-24+3\\ & =10\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Jika diketahu matriks}\\ & \text{A}=\left(\begin{array}{cc}x+3& -2\\ -16& 2x-6\end{array}\right),\\ & \text{maka nilai dari}x\text{supaya matriks}\\ & \text{A tidak memiliki invers adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{Invers}& \text{dari matriks A adalah}{\text{A}}^{-1}.\\ {\text{A}}^{-1}& ={\displaystyle \frac{1}{det\text{A}}\times Adjoin\text{A}.}\\ \text{Karen}& \text{a}\text{tidak memiliki invers},\\ \text{maka}& det\text{A}=0,\text{sehingga}\\ det\text{A}& =\left|\begin{array}{cc}x+3& -2\\ -16& 2x-6\end{array}\right|=0\\ & \iff (x+3)(2x-6)-(-16.-2)=0\\ & (\text{masing-masing ruas dibagi 2})\\ & \iff (x+3)(x-3)-16=0\\ & \iff x^2-9-16=0\\ & \iff x^2-25=0\\ & \iff (x+5)(x-5)=0\\ & \iff x+5=0\text{atau}x-5=0\\ & \iff x=-5\text{atau}x=5\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Jika}\left|\begin{array}{cc}{5}^{2x}& -5\\ 1& 1\end{array}\right|={6.5}^x\\ & \text{maka}{5}^{2x}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 625\text{atau}1\\ \text{b}.& 25\text{atau}1\\ \text{c}.& 25\text{atau}0\\ \text{d}.& 5\text{atau}1\\ \text{e}.& 5\text{atau}0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& 5^{2x}+5={6.5}^x\\ & 5^{2x}-{6.5}^x+5=0\\ & (5^x-1)(5^x-5)=0\\ & 5^x-1=0\text{atau}{5}^x-5=0\\ & 5^x=1\text{atau}{5}^x=5\\ & 5^x=5^0\text{atau}{5}^x=5^1\\ & x=0\text{atau}x=1\\ & \text{maka}\\ & 5^{2x}=\{\begin{array}{ll}{5}^{2.1}& =5^2=25\\ 5^{2.0}& =5^0=1\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Diketahu determinan suatu}\\ & \text{matriks adalah}\left|\begin{array}{ccc}x& 1& 2\\ x& 1& x\\ 5& -3& 7\end{array}\right|=0.\\ & \text{Jika}p\text{dan}q\text{adalah akar-akar}\\ & \text{yang memenuhi persamaan tersebut}\\ & \text{maka nilai dari}p+q\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -3\\ \text{b}.& -{\displaystyle \frac{1}{3}}\\ \text{c}.& -1\\ \text{d}.& {\displaystyle \frac{1}{3}}\\ \text{e}.& 3\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Diketahui ba}& \text{hwa}:\\ \left|\begin{array}{ccc}x& 1& 2\\ x& 1& x\\ 5& -3& 7\end{array}\right|& =0\\ +(x.1.7)+& (1.x.5)+(2.x.-3)\\ -(5.1.2)& -(-3.x.x)-(7.x.1)=0\\ 7x+5x-6x& -10+3x^2-7x=0\\ 3x^2-x-10& =0\{\begin{array}{ll}p& \text{salah satu akar}\\ q& \text{salah satu akar yang lain},\end{array}\\ \text{dengan}& \{\begin{array}{ll}a& =3\\ b& =-1\\ c& =-10\end{array}.\\ \text{maka}p+q& =-{\displaystyle \frac{b}{a}=-{\displaystyle \frac{-1}{3}}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$

Lanjutan 4 Materi Matriks (Matematika Wajib Kelas XI)

$\text{F. Invers Matriks ordo 2x2}$

Perhatikanlah kembali materi sebelumnya berkaitan determinan matriks 2x2, yaitu

$\begin{array}{|c|}\hline \begin{array}{rl}& \text{Jika matriks}A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ & \text{maka determinan matriks}A\\ & \text{ditentukan dengan}\\ & detA=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc\end{array}\\ \hline\end{array}$

Jika  $detA$  bernilai tidak sama dengan nol, maka invers matriks ordo 2x2 yang selanjutnya dilambangkan dengan  $A^{-1}$  dapat ditentukan dengan formula:

$\boxed{{\displaystyle A^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)}}$

$\begin{array}{rl}& \mathbf{\text{Sebagai}}\text{CONTOH}\\ & \text{Diketahui sebuah matrik ordo}2x2,\text{yaitu}:\\ & A=\left(\begin{array}{cc}-3& 2\\ -1& 4\end{array}\right),\text{maka}{A}^{{}^{-1}}\text{adalah}:\\ & A^{{}^{-1}}={\displaystyle \frac{1}{\left|\begin{array}{cc}-3& 2\\ -1& 4\end{array}\right|}\left(\begin{array}{cc}4& -2\\ 1& -3\end{array}\right)}\\ & ={\displaystyle \frac{1}{12-(-2)}\left(\begin{array}{cc}4& -2\\ 1& -3\end{array}\right)}\\ & ={\displaystyle \frac{1}{14}\left(\begin{array}{cc}4& -2\\ 1& -3\end{array}\right)}\\ & =\left(\begin{array}{cc}{\displaystyle \frac{4}{14}}& {\displaystyle \frac{-2}{14}}\\ {\displaystyle \frac{1}{14}}& {\displaystyle \frac{-3}{14}}\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{2}{7}}& -{\displaystyle \frac{1}{7}}\\ {\displaystyle \frac{1}{14}}& -{\displaystyle \frac{3}{14}}\end{array}\right)\end{array}$

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

$\begin{array}{l}\\ 1.& \text{Tentukanlah invers matriks berikut}\\ & \text{a}.B=\left(\begin{array}{cc}5& -3\\ 4& -2\end{array}\right)\\ & \text{b}.C=\left(\begin{array}{cc}-3& -5\\ 6& 9\end{array}\right)\\ & \text{c}.P=\left(\begin{array}{cc}-1& 2\\ -3& 6\end{array}\right)\\ & \text{d}.Q=\left(\begin{array}{cc}6& 9\\ 2& 3\end{array}\right)\\ \\ & \text{Jawab}:\text{yang dibahas poin a saja}\\ & B^{-1}={\displaystyle \frac{1}{detB}\left(\begin{array}{cc}-2& 3\\ -4& 5\end{array}\right)}\\ & ={\displaystyle \frac{1}{-10-(-12)}\left(\begin{array}{cc}-2& 3\\ -4& 5\end{array}\right)={\displaystyle \frac{1}{2}\left(\begin{array}{cc}-2& 3\\ -4& 5\end{array}\right)}}\\ & =\left(\begin{array}{cc}-1& {\displaystyle \frac{3}{2}}\\ -2& {\displaystyle \frac{5}{2}}\end{array}\right)\\ & \text{b. Silahkan dicoba sendiri}\\ & \text{c. Silahkan dicoba sendiri}\\ & \text{d. Silahkan dicoba sendiri}\end{array}$

$\begin{array}{l}\\ 2.& \text{Diketahui matriks}\\ & E=\left(\begin{array}{cc}4& 2\\ 5& 3\end{array}\right)\\ & \text{a}.\text{Tentukanlah}\\ & (\text{i})E^{-1}(\text{iii}){\left(E^{-1}\right)}^t\\ & (\text{i})E^t(\text{iv}){\left(E^t\right)}^{-1}\\ & \text{b}.\text{Dengan menggunakan hasil-hasil}\\ & \text{pada a. apakah}{\left(E^{-1}\right)}^t={\left(E^t\right)}^{-1}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.(\text{i})E^{-1}& ={\displaystyle \frac{1}{2}\left(\begin{array}{cc}3& -2\\ -5& 4\end{array}\right)=\left(\begin{array}{cc}{\displaystyle \frac{3}{2}}& -1\\ -{\displaystyle \frac{5}{2}}& 2\end{array}\right)}\\ (\text{ii})E^t& =\left(\begin{array}{cc}4& 5\\ 2& 3\end{array}\right)\\ (\text{iii})& {\left(E^{-1}\right)}^t=\left(\begin{array}{cc}{\displaystyle \frac{3}{2}}& -{\displaystyle \frac{5}{2}}\\ -1& 2\end{array}\right)\\ (\text{iv})& {\left(E^t\right)}^{-1}={\displaystyle \frac{1}{2}\left(\begin{array}{cc}3& -5\\ -2& 4\end{array}\right)=\left(\begin{array}{cc}{\displaystyle \frac{3}{2}}& -{\displaystyle \frac{5}{2}}\\ -1& 2\end{array}\right)}\\ \text{b}.\text{Dari ha}& \text{sil yang didapat dapat disimpulkan}\\ & {\left(E^{-1}\right)}^t={\left(E^t\right)}^{-1}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 2 Kelas 1 Semester 2. Jakarta: ERLANGGA

Lanjutan 3 Materi Matriks (Matematika Wajib Kelas XI)

$\text{E. Determinan Matriks}$

$\text{1. Ordo 2x2}$

Misalkan A adalah matriks persegi berordo 2x2 dan dituliskan dengan  $A=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)$  dengan  $a_{11}\text{dan}{a}_{22}$ sebagai elemen dari diagonal utama dan $a_{12}\text{dan}{a}_{21}$ adalah elemen yang menempati diagonal samping, perhatikan lagi matriks A berikut:

$A=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)$

maka determinan dari matriks A yang berordo 2x2 adalah perkalian elemen pada diagonal utama dikurangi dengan hasil kali perkalian diagonal samping dan di tuliskan dengan det A atau tanda |...|. Sehingga dari pengertian tersebut kita dapat menuliskan  bahwa determinan dari matriks A dalah:

$\text{det}.A=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)$ sama dengan

$\text{det}.A=\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|=a_{11}\times a_{22}-a_{12}\times a_{21}$

$\begin{array}{rl}& \mathbf{\text{Sebagai}}\text{CONTOH}\\ & \text{Diketahui sebuah matrik ordo}2x2,\text{yaitu}:\\ & A=\left(\begin{array}{cc}-3& 2\\ -1& 4\end{array}\right),\text{maka}detA\text{adalah}:\\ & detA=\left|\begin{array}{cc}-3& 2\\ -1& 4\end{array}\right|=(-3)\times (-4)-(-1)\times (2)\\ & =12-(-2)=12+2=14\end{array}$

$\text{2. Ordo 3x3}$

Ada dua buah cara minimal dalam menentukan determinan matriks ordo 3x3, yaitu:

• cara menjabarkan mengikuti baris atau kolom(ekspansi kofaktor)
• aturan Sarrus

Adapun penjelasan lebih lanjut adalah sebagai berikut

$\begin{array}{rl}& \text{Misalkan diberikan matriks ordo}3x3\\ & A=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right)\end{array}$

$\text{2.1 Menjabarkan mengikuti baris atau kolom}$

$\begin{array}{rl}\text{det A}& =a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|\\ & \\ & \mathbf{\text{Catatan}}:\\ & \text{tanda}{a}_{ij}=\text{positif jika}i+j\text{genap}\\ & \text{tanda}{a}_{ij}=\text{negatif jika}i+j\text{ganjil}\end{array}$

Anda juga bisa menjabarkan mengikuti baris yang lain termasuk juga menjabarkan mengikuti kolom. Sehingga total cara menjabarkan ini, karena ada 3 baris dan 3 kolom total akan ada sebanyak 6 cara menentukan determinan dari matriks A tersebut.

$\text{2.2 Aturan Sarrus}$

$\begin{array}{rl}\text{det A}& =a_{11}.a_{22}.a_{33}\\ & +a_{12}.a_{23}.a_{31}\\ & +a_{13}.a_{21}.a_{32}\\ & -a_{31}{a}_{22}.a_{13}\\ & -a_{32}.a_{23}.a_{11}\\ & -a_{33}.a_{21}.a_{12}\end{array}$

$\begin{array}{rl}& \mathbf{\text{Sebagai}}\text{CONTOH MENJABARKAN}\\ & \text{Diketahui sebuah matrik ordo}3x3,\text{yaitu}:\\ & A=\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 4\\ 1& 4& 3\end{array}\right),\text{maka}detA\text{dengan}\\ & \text{menjabarkan baris pertama adalah}:\\ & detA=1\left|\begin{array}{cc}3& 4\\ 4& 3\end{array}\right|-2\left|\begin{array}{cc}1& 4\\ 1& 3\end{array}\right|+3\left|\begin{array}{cc}1& 3\\ 1& 4\end{array}\right|\\ & =(9-16)-2(3-4)+3(4-3)\\ & =-7+2+3\\ & =-2\end{array}$

$\begin{array}{rl}& \mathbf{\text{Dan berikut}}\text{CONTOH aturan SARRUS}\\ & \text{Diketahui sebuah matrik ordo}3x3,\text{yaitu}:\\ & B=\left(\begin{array}{ccc}2& 1& 3\\ 3& 1& 4\\ 4& 1& 3\end{array}\right),\text{maka}detB\text{dengan}\\ & \text{metode SARRUS adalah}:\\ & detB=(2.1.3)+(1.4.4)+(3.1.3)\\ & -(4.1.3)-(1.4.2)-(3.1.3)\\ & =6+16+9-12-8-9=2\end{array}$

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

$\begin{array}{l}\\ 1.& \text{Diketahui matriks-matriks persegi berikut}\\ & \text{a}.\left(\begin{array}{cc}2& 3\\ 6& 7\end{array}\right)\text{c}.\left(\begin{array}{cc}-2& -3\\ 6& 7\end{array}\right)\\ & \text{b}.\left(\begin{array}{cc}0& 4\\ -3& 6\end{array}\right)\text{d}.\left(\begin{array}{cc}\sqrt{3}& 3\sqrt{3}\\ \sqrt{2}& -2\sqrt{2}\end{array}\right)\\ & \\ & \text{Tentukanlah determinan dari}\\ & \text{matriks-matriks persegi di atas}\\ \\ & \text{Jawab}:\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{a}.& \left|\begin{array}{cc}2& 3\\ 6& 7\end{array}\right|\\ & =(2).(7)-(3).(6)\\ & =14-18\\ & =-4\\ & \end{array}& \begin{array}{rl}\text{b}.& \left|\begin{array}{cc}0& 4\\ -3& 6\end{array}\right|\\ & =(0).(6)-(4).(-3)\\ & =0-(-12)\\ & =12\\ & \end{array}\\ \begin{array}{rl}\text{c}.& \left|\begin{array}{cc}-2& -3\\ 6& 7\end{array}\right|\\ & =(-2).(7)-(-3).(6)\\ & =(-14)-(-18)\\ & =-14+18\\ & =4\end{array}& \begin{array}{rl}\text{d}.& \left|\begin{array}{cc}\sqrt{3}& 3\sqrt{3}\\ \sqrt{2}& -2\sqrt{2}\end{array}\right|\\ & =(\sqrt{3}).(-2\sqrt{2})\\ & -(3\sqrt{3}).(\sqrt{2})\\ & =-2\sqrt{6}-3\sqrt{6}\\ & =-5\sqrt{6}\\ & \end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai}x\text{yang memenuhi persamaan}\\ & \left|\begin{array}{cc}1-x& 3\\ 2& 3-x\end{array}\right|=2\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\left|\begin{array}{cc}1-x& 3\\ 2& 3-x\end{array}\right|& =2\\ (1-x)(3-x)-(3)(2)& =2\\ 3-x-3x+x^2-6& =2\\ x^2-4x-3& =2\\ x^2-4x-5& =0\\ (x-5)(x+1)& =0\\ x-5=0\text{atau}x+1& =0\\ x=5\text{atau}x=-1& \end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Diketahui matriks-matriks persegi berikut}\\ & (\text{i}).\left(\begin{array}{ccc}1& 2& 3\\ 2& 4& 5\\ 3& 5& 4\end{array}\right)(\text{iii}).\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)\\ & (\text{ii}).\left(\begin{array}{ccc}-1& -2& -3\\ -2& 6& 0\\ -3& 0& 6\end{array}\right)(\text{iv}).\left(\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right)\\ & \\ & \text{Tentukanlah determinan matriks-matriks}\\ & \text{di atas dengan cara}\\ & \text{a}.Sarrus\\ & \text{b}.\text{Menjabarkan baris pertama}\\ & \text{c}.\text{Menjabarkan baris kedua}\\ & \text{d}.\text{Menjabarkan baris ketiga}\\ & \text{e}.\text{Menjabarkan kolom pertama}\\ & \text{f}.\text{Menjabarkan kolom kedua}\\ & \text{g}.\text{Menjabarkan kolom ketiga}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\\ & \begin{array}{|ll|}\hline (\text{i}).\left(\begin{array}{ccc}1& 1& 3\\ 2& 4& 5\\ 3& 5& 4\end{array}\right)& (\text{ii}).\left(\begin{array}{ccc}-1& -2& -3\\ -2& 6& 0\\ -3& 0& 6\end{array}\right)\\ (\text{iii}).\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)& (\text{iv}).\left(\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right)\\ \begin{array}{rl}(\text{i}).& \left|\begin{array}{ccc}1& 1& 3\\ 2& 4& 5\\ 3& 5& 4\end{array}\right|\\ & =(1)(4)(4)+\\ & (1)(5)(3)+\\ & (3)(2)(5)+\\ & -(3)(4)(3)\\ & -(5)(5)(1)\\ & -(4)(2)(1)\\ & =16+15+30\\ & -36-25-8\\ & =-8\end{array}& \begin{array}{rl}(\text{ii}).& \left|\begin{array}{ccc}-1& -2& -3\\ -2& 6& 0\\ -3& 0& 6\end{array}\right|\\ & =(-1)(6)(6)+\\ & (-2)(0)(-3)+\\ & (-3)(-2)(0)+\\ & -(-3)(6)(-3)\\ & -(0)(0)(-1)\\ & -(6)(-2)(-2)\\ & =-36+0+0\\ & -54-0-24\\ & =-114\end{array}\\ \begin{array}{rl}(\text{iii}).& \left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right|\\ & =(1)(5)(9)+\\ & (2)(6)(7)+\\ & (3)(4)(8)+\\ & -(7)(5)(3)\\ & -(8)(6)(1)\\ & -(9)(4)(2)\\ & =45+84+96\\ & -105-48-72\\ & =0\end{array}& \begin{array}{rl}(\text{iv}).& \left|\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right|\\ & =(2)(2)(2)+\\ & (1)(1)(1)+\\ & (1)(1)(1)+\\ & -(1)(2)(1)\\ & -(1)(1)(2)\\ & -(2)(1)(1)\\ & =8+1+1\\ & -2-2-2\\ & =4\end{array}\\ \hline\end{array}\\ & \text{yang belum dibahas silahkan dibuat latihan}\end{array}$

DAFTAR PUSTAKA

1. Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 2 Kelas 1 Semester 2. Jakarta: ERLANGGA