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

$\begin{array}{ll}\\ 6.&\textrm{Bentuk sederhana dari}\\ & 2y-5>2x+4y+3\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y-x>4\\ \textrm{b}.&y-x<4\\ \textrm{c}.&y+x+4>0\\ \color{red}\textrm{d}.&y+x+4<0\\ \textrm{e}.&y+x<1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&2y-5>2x+4y+3\\ &2y-4y-2x-5-3>0\\ &-2y-2x-8>0\: \: \color{black}\textrm{dibagi}\: \left ( -\displaystyle \frac{1}{2} \right )\\ &y+x+4<0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 7.&\textrm{Jika}\: \: 3x-4>5x-17\\ &\textrm{maka sebuah bilangan prima}\\ &\textrm{yang mungkin adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&3\\ \textrm{b}.&7\\ \textrm{c}.&11\\ \textrm{d}.&13\\ \textrm{e}.&17 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&3x-4>5x-17\\ &\Leftrightarrow 3x-5x>-17+4\\ &\Leftrightarrow -2x>-13\quad \color{black}\textrm{tiap ruas}\: (\times -1)\\ &\Leftrightarrow 2x<13\\ &\Leftrightarrow x<\displaystyle \frac{13}{2}=6\frac{1}{2}\\ &\color{black}\textrm{Jadi, yang memenuhi adalah 3 dan 5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: \displaystyle \frac{1}{5}<\frac{1}{x}\: \: \textrm{dan}\: \: x<0\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{1}{5}\\ \color{red}\textrm{b}.&-5<x<0\\ \textrm{c}.&0<x<5\\ \textrm{d}.&x<-5\\ \textrm{e}.&-\displaystyle \frac{1}{5}<x<0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\\ \displaystyle \frac{1}{5}&<\frac{1}{x}\: \: \: \textrm{dan}\: \: x<0\\ \displaystyle \frac{1}{5}&<\displaystyle \frac{1}{x}\\ x&<5 \\ x&>-5\qquad \color{black}\textrm{karena}\: \: x<0\\ \textrm{Sehi}&\textrm{ngga}\\ \color{red}-5<&\color{red}x<0 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: -2<y<3\: \: \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&9<(y-2)^{2}<16\\ \textrm{b}.&4<(y-2)^{2}<16\\ \textrm{c}.&1<(y-2)^{2}<16\\ \color{red}\textrm{d}.&0\leq (y-2)^{2}<16\\ \textrm{e}.&-1<(y-2)^{2}<16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}:\: -2<y<3\\ &\color{red}\bullet \quad \textrm{saat dikurangi}\: \: 2\\ &\qquad \Leftrightarrow \: -2-2<y-2<3-2\\ &\qquad -4<y-2<1\\ &\color{red}\bullet \quad \textrm{Saat}\: \: -4<y-2<0\\ &\qquad (-4)^{2}<(y-2)^{2}<0^{2}\quad \textrm{dikuadratkan}\\ &\qquad 16>(y-2)^{2}>0\\ &\qquad 0<(y-2)^{2}<16\\ &\color{red}\bullet \quad \textrm{Saat}\: \: 0\leq y-2<1\\ &\qquad 0^{2}\leq (y-2)^{2}<1^{2}\\ &\qquad 0<(y-2)^{2}<1\\ &\textrm{Jadi}\: ,\: \: \color{red}0\leq (y-2)<16 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 1.&(\textrm{Soal SNMPTN})\\ &\textrm{Jika}\: \: x>5\: \: \textrm{dan}\: \: y<3,\: \: \textrm{maka}\\ &\textrm{nilai }\: \: x-y\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{lebih besar dari pada 1}\\ \textrm{b}.&\textrm{lebih besar dari pada 3}\\ \textrm{c}.&\textrm{lebih besar dari pada 8}\\ \textrm{d}.&\textrm{lebih besar dari pada 5}\\ \color{red}\textrm{e}.&\textrm{lebih besar dari pada 2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}\\ x&>5\: \: \color{red}\&\: \: \color{blue}y<3\\ \textrm{m}&\textrm{aka}\\ &\begin{array}{llllll}\\ x>5&\Rightarrow &x&>5\\ y<3&\Rightarrow &\color{black}-y&\color{black}>-3&\color{red}+\\\hline &&\color{red}x-y&>\color{red}2 \end{array} \end{aligned} \end{array}$

$\begin{array}{l}\\ 2.&\textrm{Batas pertidaksamaan}\: \: 5x-7>13\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-4\\ \color{red}\textrm{b}.&x>4\\ \textrm{c}.&x>-4\\ \textrm{d}.&x<4\\ \textrm{e}.&-4<x<4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}5x&-7>13\\ 5x&>13+7\\ 5x&>20\\ x&\color{red}>4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Penyelesaian dari pertidaksamaan}\\ & 2x+3>5x-7\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<3\\ \color{red}\textrm{b}.&x<3\displaystyle \frac{1}{3}\\ \textrm{c}.&x>3\displaystyle \frac{1}{3}\\ \textrm{d}.&x>3\\ \textrm{e}.&\textrm{Semua pilihan jawaban salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}2x+3&>5x-7\\ 2x-5x&>-7-3\\ -3x&>-10\quad \color{black}\textrm{dikali (-1)}\\ 3x&<10\\ x&<\color{red}\displaystyle \frac{10}{3}=3\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&(\textbf{UMPTN 01})\textrm{Jika pertidaksamaan}\\ & 2x-3a>\displaystyle \frac{3x-1}{2}+ax\: \: \textrm{mempunyai}\\ &\textrm{penyelesaian}\: \: x>5,\: \: \textrm{maka nilai}\: \: a\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{3}{4}\\ \textrm{b}.&-\displaystyle \frac{3}{8}\\ \color{red}\textrm{c}.&\displaystyle \frac{3}{8}\\ \textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}2x-3a&>\displaystyle \frac{3x-1}{2}+ax\quad \color{black}\textrm{tiap ruas}\: (\times 2)\\ 4x-6a&>3x-1+2ax\\ 4x-3x&-2ax>-1+6a\\ x-2a&x>-1+6a\\ (1-2a)&x>-1+6a\\ x&>\displaystyle \frac{-1+6a}{1-2a}\\ \textrm{Diketa}&\textrm{hui}:\: \: x>5\: \: \color{red}\textrm{adalah penyelesaian}\\ \color{red}\textrm{maka}\: \: &\\ 5&=\displaystyle \frac{-1+6a}{1-2a}\\ 5-10a&=-1+6a\\ -6a-10&a=-1-5\\ -16a&=-6\\ a&=\displaystyle \frac{-6}{-16}\\ &=\color{red}\displaystyle \frac{3}{8} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&(\textbf{UMPTN 94})\\ &\textrm{Apabila}\: \: a<x<b\: \: \textrm{dan}\: \: a<y<b\\ & \textrm{maka berlaku}\: \: \: ....\\ &\begin{array}{llll}\\ \textrm{a}.&a<x-y<b\\ \textrm{b}.&b-a<x-y<a-b\\ \color{red}\textrm{c}.&a-b<x-y<b-a\\ \textrm{d}.&\displaystyle \frac{1}{2}(b-a)<x-y<\frac{1}{2}(a-b)\\ \textrm{e}.&\displaystyle \frac{1}{2}(a-b)<x-y<\frac{1}{2}(b-a) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{array}{llllll}\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &-a>-y>-b&&\\\hline &\color{purple}\textrm{saat}&\color{black}\textrm{di susun ulang}&\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &\color{black}-b<-y<-a&\color{red}+&\\\hline &&\color{red}a-b\color{blue}<\color{red}x-y\color{blue}<&\color{red}b-a \end{array} \end{array}$

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

$\begin{array}{ll}\\ &\textbf{BENTUK UMUM}\\ &\color{blue}\begin{cases} ax+by<c \\ ax+by\leq c \\ ax+by>c \\ ax+by\geq c \end{cases}\\\\ &\textbf{LANGKAH-LANGKAH}\\ &\color{purple}\textrm{dalam membuat gambar grafik persamaan linear}\\ &\: \textrm{adalah sebagai berikut}:\\ &\bullet\quad \textrm{membuat gambar grafik}\: \: \color{red}ax+by=c\\ &\quad \: \: \textrm{untuk batas wilayahnya}\\ &\bullet \quad \textrm{menyelidiki wilayah yang dimaksud di sekitar}\\ &\quad \: \: \textrm{garis} \: \: \color{red}ax+by=c\\ &\bullet \quad \textrm{ambillah sebuah titik}\: \color{red}\left ( x_{0},y_{0} \right )\: \color{black}\textrm{sembarang}\\ &\: \: \quad \textrm{kemudian substitusikan ke pertidaksamaan}\\ &\quad \: \: \color{red}ax+by\: ....\: c\\ &\bullet \quad \textrm{jika diperoleh nilai ketaksamaan yang benar},\\ &\: \: \quad \textrm{maka daerah di mana titik uji}\: \color{red}\left ( x_{0},y_{0} \right )\\ &\: \: \quad \textrm{berada merupakan wilayah penyelesaiannya}\\ &\: \: \quad \textrm{demikian juga sebaliknya} \end{array}$

$\LARGE\color{blue}\fbox{CONTOH SOAL}$

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

Dan berikut untuk wilayah dan juga batas-batas untuk pertidalsamaan

$\color{red}3x+2y<6$

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

$\begin{array}{|c|c|c|}\hline \color{black}\color{purple}\textrm{Titik}&\color{black}\textrm{Pengujian}&\color{black}\textrm{Keterangan}\\ &\color{blue}\textrm{Uji}&\color{red}3x+2y<6\\\hline (0,0)&3(0)+2(0)=0<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,1)&3(0)+2(1)=2<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,0)&3(1)+2(0)=3<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,1)&3(1)+2(1)=5<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,2)&3(0)+2(2)=4<\textbf{6}&\textrm{Dalam wilayah}\\\hline (2,0)&3(2)+2(0)=6=\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (2,2)&3(2)+2(2)=10>\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (0,3)&3(0)+2(3)=6=\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (3,0)&3(3)+2(0)=9>\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (3,3)&3(3)+2(3)=15>\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline \vdots &\vdots&\vdots \\\hline \end{array}$

Dan berikut untuk wilayah yang memenuhi  $"\color{red}3x+2y\leq 6$

$\begin{array}{ll}\\ 2.&\textrm{Selesaikanlah pertidaksamaan berikut}\\ &\textrm{a}.\quad 12x+2>4x+6\\ &\textrm{b}.\quad 2-3x<6-x\\ &\textrm{c}.\quad 6x+1\geq 2\\ &\textrm{d}.\quad \displaystyle \frac{2-3x}{2}<\frac{3-x}{3}\\\\ &\textrm{Jawab}\\ &\color{purple}\begin{aligned}\color{black}\textrm{a}.\: \: 12x&+2>4x+6\\ 12x&-4x>6-2\\ 8x&>4\\ x&>\displaystyle \frac{1}{2} \end{aligned}\\ &\color{blue}\begin{aligned}\color{black}\textrm{b}.\: \: &2-3x<6-x\\ &-3x+x<6-2\\ &-2x<4\: \: \color{red}\textrm{dikali}\: \: (-1)\\ &2x>-4\: \: (\color{black}\textrm{tanda berubah})\\ &x>-2 \end{aligned}\\ &\color{purple}\begin{aligned}\color{black}\textrm{c}.\: \: 6x&+1\geq 2\\ 6x&\geq 2-1\\ x&\geq \displaystyle \frac{1}{6} \end{aligned}\\ &\color{blue}\begin{aligned}\color{black}\textrm{d}.\: \: \: \: \color{blue}\displaystyle \frac{2-3x}{2}&<\frac{3-x}{3}\\ 3(2-3x)&<2(3-x)\\ 6-9x&<6-2x\\ -9x+2x&<6-6\\ -7x&<0\: \: \color{red}\textrm{di kali}\: \: (-1)\\ 7x&>0\: \: (\color{black}\textrm{tanda berubah})\\ x&>\displaystyle \frac{0}{7}\\ x&>0 \end{aligned} \end{array}$

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}{ll}\\ 37.&(\textbf{SMBTN 2013})\\ &\textrm{Median dan rata-rata dari data yang}\\ &\textrm{terdiri dari empat bilangan asli yang}\\ &\textrm{telah diurutkan mulai dari yang terkecil}\\ &\textrm{adalah 7. Jika data tersebut tidak}\\ &\textrm{memiliki modus dan selisih antara data}\\ &\textrm{dan data terkecil adalah 8, maka hasil}\\ &\textrm{kali terbesar dari datum kedua dan}\\ &\textrm{keempat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&39\\ \textrm{b}.&44\\ \textrm{c}.&48\\ \textrm{d}.&55\\ \color{red}\textrm{e}.&66 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui data}\: :\: \color{black}x_{1},x_{2},x_{3},\: \: \textrm{dan}\: \: x_{4}\\ &\bullet \quad\textrm{Modus}\: :\: \textrm{tidak ada}\\ &\textrm{berarti}\: \color{red}\textrm{semua datumnya berbeda}\\ &\bullet \quad \textrm{Median} \: :\: 7\\ &\textrm{maka}\: \: \displaystyle \frac{x_{2}+x_{3}}{2}=7\color{black}\Rightarrow x_{2}+x_{3}=14\: ...\color{red}(1)\\ &\bullet \quad\textrm{Rata-rata (Mean)}\: :\: 7\\ &\textrm{berarti}\: \: \displaystyle \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=7\\ &\color{black}\Rightarrow x_{1}+x_{2}+x_{3}+x_{4}=28\: ..........\color{red}(2)\\ &\bullet \quad\textrm{Jangkauan}\: :\: x_{4}-x_{1}=8\: .....\color{red}(3)\\ &\color{black}\textrm{SUBSTITUSI}\\ &\textrm{Dari persamaan (1) ke (2) diperoleh}:\\ &x_{1}+x_{4}=14\: ...........\color{red}(5)\\ &\color{black}\textrm{ELIMINASI}\\ &\textrm{Dari persamaan (3) dan (4) diperoleh}:\\ &x_{1}=3\: \: \textrm{dan}\: \: x_{4}=11\\ &\color{black}\textrm{KEMUNGKINAN}\\ &\textrm{nilai}\: x_{2}\: \textrm{dan}\: x_{3}\: \textrm{adalah}:\: 3< x_{2}\: ;\: x_{3}<11\\ &\blacklozenge \: x_{2}=4\rightarrow x_{3}=10<11\: \: \color{red}\checkmark\\ &\blacklozenge \: x_{2}=5\rightarrow x_{3}=9<11\: \: \color{red}\checkmark\\ &\blacklozenge \: x_{2}=6\rightarrow x_{3}=8<11\: \: \color{red}\checkmark\\ &\blacklozenge \: x_{2}=7\rightarrow x_{3}=7<11\: \: \color{red}\times \\ &\color{black}\textrm{KESIMPULAN}\\ &\textrm{Hasil kali terbesar}\: \: \color{red}x_{2}\times x_{3}=6\times 11=66 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 39.&(\textbf{SIMAK UI})\\ &\textrm{Jika rata-rata 20 bilangan bulat nonnegatif}\\ &\textrm{berbeda adalah 20, maka bilangan terbesar}\\ &\textrm{yang mungkin adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&210\\ \textrm{b}.&229\\ \textrm{c}.&230\\ \textrm{d}.&239\\ \textrm{e}.&240 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\color{black}\displaystyle \frac{x_{1}+x_{2}+x_{3}+...+x_{18}+x_{19}+x_{20}}{20}=\color{red}20\\ &x_{1}+x_{2}+x_{3}+...+x_{18}+x_{19}+x_{20}=\color{red}400\\ &\color{black}\textrm{Yang mungkin bilangan bulat nonnegatif}\\ &\textrm{dan berbeda adalah}\: :\: \color{red}0, 1,2,3,....dst\\ &0+1+2+..+18+x_{20}=\color{red}400\\ &\displaystyle \frac{18\times 19}{2}+x_{20}=\color{red}400\\ &171+x_{20}=\color{red}400\\ &x_{20}=\color{red}400-171=229 \end{aligned} \end{array}$.

$\begin{array}{ll} 40.&\textbf{SPMB 2006}\\ &\textrm{Jika jangkauan dari data terurut}:x-1,\: 2x-1,\: 3x,\\ &5x-3,\: 4x+3,\: 6x+2\: \: \textrm{adalah}\: \: 18,\: \textrm{maka mediannya}\\ &\textrm{ adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{9}&&&\textrm{d}.&\textrm{21}\\ \textrm{b}.&\textrm{10},5\quad &\textrm{c}.&\textrm{12}\quad&\textrm{e}.&\textrm{24},8 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &x-1,\: 2x-1,\: 3x,\: 5x-3,\: 4x+3,\: 6x+2\\ &\textrm{Dan diketahui pula nilai jangkauannya}\: =18\\ &J=x_{_{max}}-x_{_{min}}=18\\ &\Leftrightarrow (6x+2)-(x-1)=18\\ &\Leftrightarrow 5x+3=18\\ &\Leftrightarrow 5x=15\\ &\Leftrightarrow x=3\\ &\textrm{Sehingga datanya}:\: \color{red}2,3,9,12,15,20\\ &\textrm{Selanjutnya ditentukan mediannya}=M_{e}=Q_{_{2}}\\ &\textrm{karena}\: \: n=6,\: \textrm{datum ke}-\displaystyle \frac{2}{4}(6+1)=3,5\\ & M_{e}=Q_{2}=x_{._{3}}+0,5(x_{._{4}}-x_{._{3}})=9+0,5(12-9)\\ &\: \: \: \quad =9+0,5(3)=9+1,5=10,5\\ &\textrm{Jadi},\: J=\color{red}10,5 \end{aligned} \end{array}$.

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}{ll}\\ 41.&\textrm{Jika}\: \: x=\: ^{15}\log 75\: \: \textrm{dan}\: \: y=\: ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125},\\ &\textrm{maka nilai}\: \: 5x+3y-2xy\: \: \textrm{adalah}\: ....\\\\ &\color{blue}\textrm{KOMPETISI HARDIKNAS ONLINE}\\ &\color{blue}\textrm{POSI}(\textrm{Pelatihan Olimpiade Sain Indonesia})\\ &\color{blue}\textrm{Bidang Matematika 2020}\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \textrm{b}.&1\\ \textrm{c}.&3\\ \textrm{d}.&5\\ \color{red}\textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\color{black}5x+3y-2xy\\ &=5\left ( ^{15}\log 75 \right )+3\left ( ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125} \right )\\ &\qquad -2\left ( ^{15}\log 75 \right )\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 )\\ &\qquad -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 )\\ &\qquad -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 )\\ &\qquad -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 )\\ &\qquad -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 )\\\\ &\color{red}\textrm{Misalkan}\: \: \color{black}\log 3=A,\: \: \log 5=B \end{aligned} \end{array}$

$.\qquad\color{purple}\begin{aligned} &\color{red}\textrm{Selanjutnya}\\ &=5\left ( \displaystyle \frac{A+2B}{A+B} \right )+3\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &\qquad -2\left ( \displaystyle \frac{A+2B}{A+B} \right )\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &=\left ( \displaystyle \frac{5A+10B}{A+B} \right )+\left ( \displaystyle \frac{6A-9B}{A-B} \right )\\ &\qquad -\left ( \displaystyle \frac{2A+4B}{A+B} \right )\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &=\displaystyle \frac{(5A+10B)(A-B)+(6A-9B)(A+B)}{A^{2}-B^{2}}\\ &\qquad -\left (\displaystyle \frac{4A^{2}-6AB+8AB-12B^{2}}{A^{2}-B^{2}} \right )\\ &=\displaystyle \frac{5A^{2}-5AB+10AB-10B^{2}}{A^{2}-B^{2}}\\ &\quad +\displaystyle \frac{6A^{2}+6AB-9AB-9B^{2}}{A^{2}-B^{2}}\\ &\qquad -\left (\displaystyle \frac{4A^{2}-6AB+8AB-12B^{2}}{A^{2}-B^{2}} \right )\\ &=\displaystyle \frac{7A^{2}-7B^{2}}{A^{2}-B^{2}}\\ &=\displaystyle \frac{7\left ( A^{2}-B^{2} \right )}{A^{2}-B^{2}}\\ &=7 \end{aligned}$

$\begin{array}{ll}\\ 42.&\textrm{Diberikan}\: \: A=\: ^{6}\log 16\: \: \textrm{dan}\: \: B=\: ^{12}\log 27\\ &\textrm{Terdapat bilangan-bilangan bulat positif}\\ &a,\: b,\: \: \textrm{dan}\: \: c\: \: \textrm{sehingga}\: \: (A+a)(B+b)=c\\ &\textrm{Nilai dari}\: \: a+b+c\: \: \textrm{adalah}\: ....\\\\ &\color{blue}\textrm{KOMPETISI HARDIKNAS ONLINE}\\ &\color{blue}\textrm{POSI}(\textrm{Pelatihan Olimpiade Sain Indonesia})\\ &\color{blue}\textrm{Bidang Matematika 2020}\\ &\begin{array}{llll}\\ \textrm{a}.&23\\ \textrm{b}.&24\\ \textrm{c}.&27\\ \textrm{d}.&30\\ \textrm{e}.&34 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{....}\\ &\color{blue}\begin{aligned}&\color{black}\textrm{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}\\ &\Leftrightarrow \color{black}\log 2+\log 3=\displaystyle \frac{4\log 2}{A}\: ...........\color{red}(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}\\ &\Leftrightarrow \color{black}2\log 2+\log 3=\displaystyle \frac{3\log 3}{B}\: .........\color{red}(2)\\ &\color{black}\textrm{ELIMINASI}\\ &\textrm{Dari persamaan (1) dan (2) diperoleh}:\\ &\bullet \quad \log 2=\displaystyle \frac{3\log 3}{B}-\displaystyle \frac{4\log 2}{A}\\ &\qquad \Leftrightarrow \log 2=\displaystyle \frac{3A\log 3-4B\log 2}{AB}\\ &\qquad \Leftrightarrow AB\log 2=3A\log 3-4B\log 2\\ &\qquad \Leftrightarrow AB\log 2+4B\log 2=3A\log 3\\ &\qquad \Leftrightarrow (AB+4B)\log 2=3A\log 3\\ &\qquad \Leftrightarrow \displaystyle \frac{\log 2}{\log 3}=\color{purple}\displaystyle \frac{3A}{AB+4B}\: ..........\color{red}(3)\\ &\bullet \quad \log 3=\displaystyle \frac{8\log 2}{A}-\displaystyle \frac{3\log 3}{B}\\ &\qquad \Leftrightarrow \log 3=\displaystyle \frac{8B\log 2-3A\log 3}{AB}\\ &\qquad \Leftrightarrow AB\log 3=8B\log 2-3A\log 3\\ &\qquad \Leftrightarrow AB\log 3+3A\log 3=8B\log 2\\ &\qquad \Leftrightarrow (AB+3A)\log 3=8B\log 2\\ &\qquad \Leftrightarrow \frac{\log 2}{\log 3}=\color{purple}\displaystyle \frac{AB+3A}{8B}...........\color{red}(4)\\ &\color{black}\textrm{KESAMAAN}\\ &\qquad\quad \frac{\log 2}{\log 3}=\frac{\log 2}{\log 3}\\ &\color{purple}\displaystyle \frac{AB+3A}{8B}=\color{purple}\displaystyle \frac{3A}{AB+4B}\\ &\qquad \Leftrightarrow (AB+3A)(AB+4B)=(8B).(3A)\\ &\qquad \Leftrightarrow (B+3)(A+4)=24\\ &\qquad \Leftrightarrow (A+4)(B+3)=24\\ &\color{black}\textrm{KESIMPULAN}\\ &a=4,\: b=3,\: \: \textrm{dan}\: \: c=24,\\ &\color{purple}\textrm{maka}\: \: \color{black}a+b+c=4+3+24=\color{red}31 \end{aligned} \end{array}$

Contoh Soal 8 Fungsi Logaritma (Persamaan Logaritma)

$\begin{array}{ll}\\ 36.&\textrm{Persamaan}\: \: ^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\textrm{mempunyai akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2},\: \: \textrm{maka}\\ &\textrm{nilai}\: \: x_{1}+x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&6\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\color{purple}\textrm{Alternatif 1}\\ &^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 2(3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 6x-8=2\\ &\Leftrightarrow \: \color{black}6x-8=x^{2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0,\: \: \color{purple}\textrm{dengan}\: \begin{cases} a &=1 \\ b &=-6 \\ c &=8 \end{cases}\\ &\Leftrightarrow \: x_{1}+x_{2}=-\displaystyle \frac{b}{a}\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}-\displaystyle \frac{-6}{1}=6\\ &\color{purple}\textrm{Alternatif 2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0\\ &\Leftrightarrow \: (x-2)(x-4)\\ &\Leftrightarrow \: x_{1}=2\: \: \textrm{atau}\: \: x_{2}=4\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}2+4=6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{memenuhi}\\ &(\log x)(2\log x-3)=\log 100\\ &\textrm{maka}\: \: x_{1}\times x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&100\\ \color{red}\textrm{b}.&10\sqrt{10}\\ \textrm{c}.&\sqrt{10}\\ \textrm{d}.&-\sqrt{10}\\ \textrm{e}.&-10\sqrt{10} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&(\log x)(2\log x-3)=\color{red}\log 100\\ &\Leftrightarrow (\log x)\left ( 2\log x-3 \right )=\color{red}2\\ &\color{black}\Leftrightarrow 2\log ^{2}x-3\log x-2=0\: \color{purple}\begin{cases} a &=2 \\ b &=-3 \\ c &=-2 \end{cases}\\ &\Leftrightarrow \log x_{1}+\log x_{2}=-\displaystyle \frac{-3}{2}=\frac{3}{2}\\ &\Leftrightarrow \log \left ( x_{1}\times x_{2} \right )=1\displaystyle \frac{1}{2}\\ &\Leftrightarrow \color{red}\left ( x_{1}\times x_{2} \right )=10^{1\frac{1}{2}}=10\sqrt{10} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Persamaan}\\ & 10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ & \textrm{mempunyai dua akar yaitu}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\\ &\textrm{Nilai}\: \: x_{1}\times x_{2}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-5\\ \color{red}\textrm{c}.&2\\ \textrm{d}.&5\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &10^{\, 2^{\, ^{2}}\log x }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &\color{black}\textrm{adalah persamaan kuadrat dalam}\: \: \color{red}10^{\,^{\, ^{2}}\log x }\\ &\color{black}\textrm{Misalkan}\: \: \color{red}p=10^{\,^{\, ^{2}}\log x },\: \: \textrm{maka persamaan}\\ &\color{black}\textrm{menjadi}\: \: \color{purple}p^{2}-7p+10=0\: \begin{cases} a & =1 \\ b & =-7 \\ c & =10 \end{cases}\\ & \color{red}\textrm{Karena nilai}\: \: \color{black}p_{1}\times p_{2}=\displaystyle \frac{c}{a}\: \: \textrm{maka}\\ &10^{\,^{\, ^{2}}\log x_{1} }\times 10^{\,^{\, ^{2}}\log x_{2} }=\displaystyle \frac{10}{1}=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10^{1}\\ &\Leftrightarrow \: ^{2}\log x_{1}\: +\: ^{2}\log x_{2}=1\\ &\Leftrightarrow \: ^{2}\log x_{1}\times x_{2}=1\\ &\Leftrightarrow \: \color{red}x_{1}\times x_{2}=2^{1}=2\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &^{x}\log (x+12)-\: ^{x}\log 4^{3}=-1\\ &^{x}\log \displaystyle \frac{x+12}{64}=-1\\ &\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}\\ &x+12=\displaystyle \frac{64}{x}\\ &\color{purple}x^{2}+12x-64=0\\ &\color{purple}(x+16)(x-4)=0\\ &x=-16\: \: \color{black}\textrm{atau}\: \: \color{red}x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{dan}\: \: 6\\ \textrm{b}.&-2\: \: \textrm{dan}\: \: 6\\ \textrm{c}.&-1\\ \textrm{d}.&-2\\ \color{red}\textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &^{x}\log (2x-3)-\: ^{x}\log (x+6)+\: ^{x}\log (x+2)=1\\ &^{x}\log (2x-3)(x+2)=1+\log (x+6)\\ &^{x}\log \displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=1\\ &\displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=x^{1}\\ &\left ( 2x^{2}+x-6 \right )=x^{2}+6x\\ &\color{purple}x^{2}-5x-6=0\\ &\color{purple}(x+1)(x-6)=0\\ &x=-1\: \: \color{black}\textrm{atau}\: \: \color{red}x=6 \end{aligned} \end{array}$

Contoh Soal 7 Fungsi Logaritma (Pertidaksamaan Logaritma)

 $\begin{array}{ll}\\ 32.&\textrm{Agar}\: \: \log \left ( x^{2}-1 \right )<0\: \: \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<1\\ \textrm{b}.&-\sqrt{2}<x<\sqrt{2}\\ \textrm{c}.&x<-1\: \: \textrm{atau}\: \: x>1\\ \textrm{d}.&x<-\sqrt{2}\: \: \textrm{atau}\: \: x>\sqrt{2}\\ \color{red}\textrm{e}.&-\sqrt{2}<x<-1\: \: \textrm{atau}\: \: 1<x<\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\log \left ( x^{2}-1 \right )<0\\ &\textrm{Diketahui}\: \: \color{red}\log f(x)<0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-1>0\\ &\Leftrightarrow x<-1\: \: \textrm{atau}\: \: x>1\\ &\color{purple}\textrm{Syarat (2)},\: \: \log \left ( x^{2}-1 \right )<0\\ &\log \left ( x^{2}-1 \right )<\log 1\\ &\Leftrightarrow x^{2}-1<1\\ &\Leftrightarrow x^{2}-2<0\\ &\Leftrightarrow x^{2}-\left ( \sqrt{2} \right )^{2}<0\\ &\Leftrightarrow -\sqrt{2}<x<\sqrt{2}\\ &\textrm{Jadi},\: \: \color{red}-\sqrt{2}<x<-1\: \: \textrm{atau}\: \: 1<x<\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Himpunan penyelesaian dari}\\ &^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|-2<x<-\sqrt{3}\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \right \}\\ \textrm{b}.&\left \{ x|-\sqrt{3}<x<-1\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \right \}\\ \textrm{c}.&\left \{ x|-2<x<-\sqrt{3} \right \}\\ \textrm{d}.&\left \{ x|-2<x<-\sqrt{3} \right \}\\ \textrm{e}.&\left \{ x|\sqrt{3}<x<2 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0\\ &\textrm{Diketahui}\: \: \color{red}^{.^{\frac{1}{2}}}\log f(x)>0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-3>0\\ &\Leftrightarrow x<-\sqrt{3}\: \: \textrm{atau}\: \: x>\sqrt{3}\\ &\color{purple}\textrm{Syarat (2)},\: \: ^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0\\ &^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>\: ^{.^{\frac{1}{2}}}\log 1\\ &\Leftrightarrow x^{2}-3<1\quad \left (\color{black}\textrm{karena basisnya}\: \: \displaystyle \frac{1}{2}<1 \right )\\ &\Leftrightarrow x^{2}-4<0\\ &\Leftrightarrow x^{2}-2^{2}>0\\ &\Leftrightarrow -2<x<2\\ &\textrm{Jadi},\: \: \color{red}-2<x<-\sqrt{3}\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &^{2}\log \left ( x^{2}-x \right )\leq 1 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<0\: \: \textrm{atau}\: \: x>1\\ \textrm{b}.&-1\leq x\leq 2,\: x\neq 1\: \: \textrm{atau}\: \: x\neq 0\\ \color{red}\textrm{c}.&-1\leq x< 0\: \: \textrm{atau}\: \: 1<x\leq 2\\ \textrm{d}.&-1< x\leq 0\: \: \textrm{atau}\: \: 1\leq x< 2\\ \textrm{e}.&-1\leq x\leq 0\: \: \textrm{atau}\: \: 1\leq x\leq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{2}\log \left ( x^{2}-x \right )\leq 1\\ &\textrm{Diketahui}\: \: \color{red}^{2}\log f(x)>0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-x>0\Leftrightarrow x(x-1)>0\\ &\Leftrightarrow x<0\: \: \textrm{atau}\: \: x>1\\ &\color{purple}\textrm{Syarat (2)},\: \: ^{2}\log \left ( x^{2}-x \right )\leq 1\\ &^{2}\log \left ( x^{2}-x \right )\leq \: ^{2}\log 2\\ &\Leftrightarrow x^{2}-x\leq 2\\ &\Leftrightarrow x^{2}-x-2\leq 0\\ &\Leftrightarrow (x+1)(x-2)\leq 0\\ &\Leftrightarrow -1\leq x\leq 2\\ &\textrm{Jadi},\: \: \color{red}-1\leq x< 0\: \: \textrm{atau}\: \: 1<x\leq 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\left | \log (x+1) \right |> 1 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<-0,9\: \: \textrm{atau}\: \: x>9\\ \textrm{b}.&x<-9\: \: \textrm{atau}\: \: x>9\\ \color{red}\textrm{c}.&-1<x<-0,9\: \: \textrm{atau}\: \: x>9\\ \textrm{d}.&-9< x<0,9\\ \textrm{e}.&-0,9<x<9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Ingat bahwa}\\ &\left | x \right |>A\Leftrightarrow \color{black}x<-A\: \: \textrm{atau}\: \: x>A,\: \: \color{red}A>0\\ &\Leftrightarrow \log (x+1)<-1\: \: \textrm{atau}\: \: \log (x+1)>1\\ &\color{red}\textrm{Syarat (1) buat keduanya},\: \: \color{red}f(x)>0\\ &(x+1)>0\Leftrightarrow x>-1\\ &\color{red}\textrm{Syarat (2)},\: \: \log \left ( x+1 \right )<-1\\ &\log (x+1)<\log 10^{-1}\\ &x+1<\displaystyle \frac{1}{10}\Leftrightarrow x<-\frac{9}{10}\\ &\color{red}\textrm{Syarat (3)},\: \: \log \left ( x+1 \right )> 1\\ &\log (x+1)>\log 10^{1}\\ &(x+1)>10\Leftrightarrow x>9\\ &\textrm{Jadi},\: \: \color{red}-1<x<-0,9\: \: \textrm{atau}\: \: x>9 \end{aligned} \end{array}$

Lanjutan Materi Fungsi Logaritma (Kelas X Matematika Peminatan)

 MENGINGAT KEMBALI

$\color{blue}\textrm{E. Sifat-Sifat Eksponen dan Logaritma}$

$\color{blue}\textrm{E.1  Persamaan Eksponen dan Logaritma}$

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

Selanjutnya

$\color{blue}\begin{array}{|l|l|l|}\hline \textrm{No}&\qquad\textrm{Bentuk}&\qquad\qquad\color{red}\textrm{Syarat}\\\hline 1.&a^{f(x)}=1&a\neq 0,\quad \textrm{maka}\: \: \color{red}f(x)=0\\\hline 2.&a^{f(x)}=a^{p}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: \color{red}f(x)=p\\\hline 3.&a^{f(x)}=a^{g(x)}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: \color{red}f(x)=g(x)\\\hline 4.&a^{f(x)}=b^{f(x)}&a\neq 0,\: b\neq 0\: ,\quad \textrm{maka}\: \: \color{red}f(x)=0\\\hline 5.&f(x)^{g(x)}=1&\begin{cases} f(x)=1 & \\ g(x)=0, & \textrm{jika}\: \: f(x)\neq 0 \\ f(x)=-1, & \textrm{jika}\: \: g(x)=\: \textrm{genap} \end{cases}\\\hline 6.&\color{black}f(x)^{g(x)}=f(x)^{h(x)}&\color{red}\begin{cases} (i).\quad g(x)=h(x)& \\ (ii).\quad f(x)=1& \\ (iii).\quad f(x)=0,&g(x)>0,\: \: h(x)>0 \\ (iv).\quad f(x)=-1,&g(x)\: \textrm{dan}\: h(x)\: \: \\ &\color{black}\textrm{keduanya ganjil}\\ &\color{black}\textrm{atau genap} \end{cases}\\\hline 7.&g(x)^{f(x)}=h(x)^{f(x)}&\begin{cases} (i).\quad g(x) =h(x)& \\ (ii).\quad f(x)=0, & g(x)\neq 0,\: h(x)\neq 0 \end{cases}\\\hline 8.&\begin{aligned}&\color{black}A\left ( a^{f(x)} \right )^{2}\\ &\quad \color{black}+B\left ( a^{f(x)} \right )\\ &\qquad \color{black}+C=0 \end{aligned}&\color{red}a>0,\: \: a\neq 1\\\hline \end{array}$

$\color{blue}\textrm{E.2  Pertidaksamaan Eksponen dan Logaritma}$

Berikut sifat pertidaksamaan Eksponen

$\color{blue}\begin{array}{|l|l|}\hline \qquad\qquad \color{red}a>1&\qquad\qquad \color{black}0<a<1\\\hline a^{f(x)}\leq a^{g(x)}\Rightarrow f(x)\leq g(x)&a^{f(x)}\leq a^{g(x)}\Rightarrow f(x)\geq g(x)\\\hline a^{f(x)}< a^{g(x)}\Rightarrow f(x)< g(x)&a^{f(x)}< a^{g(x)}\Rightarrow f(x)> g(x)\\\hline a^{f(x)}\geq a^{g(x)}\Rightarrow f(x)\geq g(x)&a^{f(x)}\geq a^{g(x)}\Rightarrow f(x)\leq g(x)\\\hline a^{f(x)}> a^{g(x)}\Rightarrow f(x)> g(x)&a^{f(x)}> a^{g(x)}\Rightarrow f(x)< g(x)\\\hline \end{array}$

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

$\color{blue}\begin{array}{|l|l|}\hline \qquad\qquad \color{red}a>1&\qquad \color{black}0<a<1\\\hline \begin{aligned}&^a\log f(x)\leq \: ^a\log g(x)\\ &\Rightarrow f(x)\leq g(x) \end{aligned}&\begin{aligned}&^a\log f(x)\leq \: ^a\log g(x)\\ &\Rightarrow f(x)\geq g(x) \end{aligned}\\\hline \begin{aligned}&^a\log f(x)< \: ^a\log g(x)\\ &\Rightarrow f(x)< g(x) \end{aligned}&\begin{aligned}&^a\log f(x)< \: ^a\log g(x)\\ &\Rightarrow f(x)> g(x) \end{aligned}\\\hline \begin{aligned}&^a\log f(x)\geq \: ^a\log g(x)\\ &\Rightarrow f(x)\geq g(x) \end{aligned}&\begin{aligned}&^a\log f(x)\geq \: ^a\log g(x)\\ &\Rightarrow f(x)\leq g(x) \end{aligned}\\\hline \begin{aligned}&^a\log f(x)> \: ^a\log g(x)\\ &\Rightarrow f(x)> g(x) \end{aligned}&\begin{aligned}&^a\log f(x)> \: ^a\log g(x)\\ &\Rightarrow f(x)< g(x) \end{aligned}\\\hline \end{array}$

Contoh Soal 6 Matriks

$\begin{array}{ll}\\ 26.&(\textbf{UM UGM 2004})\\ &\textrm{Nilai-nilai}\: \: x\: \: \textrm{agar matriks}\\ &\qquad\quad\quad\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}\\ &\textrm{tidak memiliki invers adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&4\: \: \textrm{atau}\: \: 5\\ \color{red}\textrm{b}.&-2\: \: \textrm{atau}\: \: 2\\ \textrm{c}.&-4\: \: \textrm{atau}\: \: 5\\ \textrm{d}.&-6\: \: \textrm{atau}\: \: 4\\ \textrm{e}.&0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{supaya matriks}\: \: \color{black}\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}\\ &\textrm{tidak memiliki invers},\: \textrm{maka}\\ &\textrm{determinan matriks}\: \: \color{black}\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}=0\\ &\color{red}\textrm{Sehingga}\\ &\begin{vmatrix} 5x & 5\\ 4 & x \end{vmatrix}=0\\ &\Leftrightarrow 5x^{2}-20=0\\ &\Leftrightarrow x^{2}=\color{red}4\\ &\Leftrightarrow x=\color{red}\pm 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&(\textbf{UM UGM 2005})\\ &\textrm{Matriks}\: \: \begin{pmatrix} x & 1\\ -2 & 1-x \end{pmatrix}\\ &\textrm{tidak memiliki invers untuk}\\ &\textrm{nilai}\: \: x=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{atau}\: \: -2\\ \textrm{b}.&-1\: \: \textrm{atau}\: \: 0\\ \textrm{c}.&-1\: \: \textrm{atau}\: \: 1\\ \color{red}\textrm{d}.&-1\: \: \textrm{atau}\: \: 2\\ \textrm{e}.&1\: \: \textrm{atau}\: \: 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\textrm{Mirip dengan pembahasan no. 26}\\ &\color{blue}\begin{aligned}&\textrm{Nilai}\: \: \color{black}\begin{vmatrix} x & 1\\ -2 & 1-x \end{vmatrix}=0\\ &\Leftrightarrow x-x^{2}-(-2)=0\\ &\Leftrightarrow 2+x-x^{2}=0\\ &\Leftrightarrow x^{2}-x-2=0\\ &\Leftrightarrow (x-2)(x+1)=0\\ &\Leftrightarrow \color{red}x=2\: \: \color{blue}\textrm{atau}\: \: \color{red}x=-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&(\textbf{Mat Das SIMAK UI 2014})\\ &\textrm{Jika matriks}\: \: \textrm{A}\: \: \textrm{adalah invers}\\ &\textrm{dari matriks}\: \: \displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix}\: \: \textrm{dan}\\ &\textrm{A}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 1\\ 3 \end{pmatrix}\: \: \textrm{maka nilai}\: \: 2x+y\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-\displaystyle \frac{10}{3}\\ \color{red}\textrm{b}.&-\displaystyle \frac{1}{3}\\ \textrm{c}.&1\\ \textrm{d}.&\displaystyle \frac{9}{7}\\ \textrm{e}.&\displaystyle \frac{20}{3} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Misalkan diketahui matriks}\\ &\textrm{B}=\displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix},\\ &\textrm{maka}\: \: \textrm{A}=\left ( \displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix} \right )^{-1}\\ &\textrm{selanjutnya}\: \: \textrm{A}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=A^{-1}\begin{pmatrix} 1\\ 3 \end{pmatrix},\\ & \textrm{ingat bahwa}\: \: \left (\textbf{A}^{-1} \right )^{-1}=\textbf{A}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=\left ( \left ( \displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix} \right )^{-1} \right )^{-1}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3}\begin{pmatrix} -1-9\\ 4+15 \end{pmatrix}=\begin{pmatrix} \displaystyle -\frac{10}{3}\\ \displaystyle \frac{19}{3} \end{pmatrix}\\ &2x+y=2\left ( -\displaystyle \frac{10}{3} \right )+\frac{19}{3}\\ &\qquad\: \: \: \, =\color{red}\displaystyle \frac{-20+19}{3}=-\displaystyle \frac{1}{3} \end{aligned} \end{array}$

Contoh Soal 5 Matriks

$\begin{array}{ll}\\ 21.&(\textbf{SPMB 2003})\\ &\textrm{Diketahu matriks}\: \: \textrm{A}=\begin{pmatrix}a&b\\ c&d \end{pmatrix}.\\ &\textrm{Jika}\: \: \: \textrm{A}^{t}=\textrm{A}^{-1}\: \: \textrm{dengan}\: \: \textrm{A}^{t}\\ &\textrm{adalah transpose matriks A},\\ &\textrm{maka nilai}\: \: ad-bc=....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1\: \: \textrm{atau}\: \: -\sqrt{2}\\ \textrm{b}.&1\: \: \textrm{atau}\: \: \sqrt{2}\\ \textrm{c}.&-\sqrt{2}\: \: \textrm{atau}\: \: -\sqrt{2}\\ \color{red}\textrm{d}.&-1\: \: \textrm{atau}\: \: 1\\ \textrm{e}.&1\: \: \textrm{atau}\: \: -\sqrt{2}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \textrm{matriks}\: \: \textrm{A}=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{A}^{t}=\textrm{A}^{-1},\: \textrm{maka}\\ &\textrm{A}^{t}=\textrm{A}^{-1}\\ &\begin{pmatrix} a & b\\ c & d \end{pmatrix}^{t}=\displaystyle \frac{1}{ad-bc}\times \color{red}\textrm{Adjoin Matriks}\: \: \textrm{A}\\ &\begin{pmatrix} a & c\\ b & d \end{pmatrix}=\displaystyle \frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix},\\ & \color{red}\textrm{didapatkan hubungan}\\ &c=\displaystyle \frac{-b}{ad-bc}\quad ...............(1)\\ &b=\displaystyle \frac{-c}{ad-bc}\quad ...............(2)\\ &\textrm{Persamaan}\: \:  (2)\: \: \textrm{disubstitusikan ke persamaan}\: \: (1)\\ &c=\displaystyle \displaystyle \frac{-\displaystyle \frac{-c}{ad-bc}}{ad-bc}\\ &1=(ad-bc)^{2}\\ &\color{red}(ad-bc)= -1\: \: \textrm{atau}\: \: 1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Diketahu matriks}\: \: \textrm{H}\\ &\textrm{yang memenuhi persamaan}\\ &\textrm{H}\begin{pmatrix} 3 & 2\\ 1 & 4 \end{pmatrix}=\begin{pmatrix} 7 & 8\\ 4 & 6 \end{pmatrix},\\ &\textrm{maka nilai dari}\: \: \: det\: \textrm{H}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-3\\ \textrm{b}.&-2\\ \textrm{c}.&-1\\ \color{red}\textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{array}{|c|}\hline \color{red}\textbf{Alternatif 1}\\\hline \begin{aligned}\textrm{H.A}&=\textrm{B}\\ \textrm{H.A.A}^{-1}&=\textrm{B.A}^{-1}\\ \textrm{H}&=\textrm{B.A}^{-1}\\ &=\begin{pmatrix} 7 & 8\\ 4 & 6 \end{pmatrix}.\displaystyle \frac{1}{\begin{vmatrix} 3 & 2\\ 1 & 4 \end{vmatrix}}\begin{pmatrix} 4 & -2\\ -1 & 3 \end{pmatrix}\\ &=\begin{pmatrix} 7 & 8\\ 4 & 6 \end{pmatrix}.\displaystyle \frac{1}{12-2}\begin{pmatrix} 4 & -2\\ -1 & 3 \end{pmatrix}\\ &=\displaystyle \frac{1}{10}\begin{pmatrix} 28+(-8) & (-14)+24\\ 16+(-6) & (-8)+18 \end{pmatrix}\\ \textrm{H}&=\displaystyle \frac{1}{10}\begin{pmatrix} 20 & 10\\ 10 & 10 \end{pmatrix}=\begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix}\\ det\: \textrm{H}&=\begin{vmatrix} 2 & 1\\ 1 & 1 \end{vmatrix}=2.1-1.1=2-1=\color{purple}1 \end{aligned}\\\hline \color{red}\textbf{Alternatif 2}\\\hline \color{purple}\begin{aligned}\textrm{H.A}&=\textrm{B}\begin{cases} det\: \textrm{H} &=\left | \textrm{H} \right | \\ det\: \textrm{A} &=\left | \textrm{A} \right |=\begin{vmatrix} 3 & 4\\ 2 & 1 \end{vmatrix}\\ &=12-2=10 \\ det\: \textrm{B} &=\left | \textrm{B} \right |=\begin{vmatrix} 7 & 8\\ 4 & 6 \end{vmatrix}\\ &=42-32=10 \end{cases}\\ \left | \textrm{H} \right |.\left | \textrm{A} \right |&=\left | \textrm{B} \right |\\ \left | \textrm{H} \right |&=\displaystyle \frac{\left | \textrm{B} \right |}{\left | \textrm{A} \right |}\\ &=\displaystyle \frac{10}{10}\\ &=\color{red}1 \end{aligned} \\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 23.&(\textbf{UM UGM 2006})\\ &\textrm{Apabila}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{memenuhi}\\ &\textrm{persamaan matriks}\\ &\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -1\\ 2 \end{pmatrix},\\ &\textrm{maka}\: \: x+y=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \color{red}\textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \color{red}A.X&=B\\ \color{red}A^{-1}.A.X&=\color{red}A^{-1}.\color{blue}B\\ \color{red}A^{0}.X&=\color{red}A^{-1}.\color{blue}B\\ \color{red}X&=\color{red}A^{-1}.\color{blue}B\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}^{-1}\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{\begin{vmatrix} 1 & -2\\ -1 & 3 \end{vmatrix}}&\begin{pmatrix} 3 & 2\\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3-2}&\begin{pmatrix} 3.(-1)+2.2 \\ 1.(-1)+1.2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1\\ 1 \end{pmatrix}\\ x+y&=\color{red}1+1=2 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 25.&(\textbf{UM UGM 2005})\textrm{Jika}\\ &\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{pmatrix}=\begin{pmatrix} \sin A & \cos A \end{pmatrix}\\ &\textrm{dan}\: \: A\: \: \textrm{suatu konstanta, maka}\: \: x+y=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \color{red}\textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{pmatrix}=\begin{pmatrix} \sin A & \cos A \end{pmatrix}\\ &\begin{pmatrix} x\sin \alpha -y\cos \alpha & x\cos \alpha +y\sin \alpha \end{pmatrix}=\begin{pmatrix} \sin A & \cos A \end{pmatrix}\\ &\begin{cases} \sin A & =x\sin \alpha -y\cos \alpha =\sqrt{x^{2}+y^{2}}\cos \left ( \alpha -\tan ^{-1}\displaystyle \frac{x}{-y} \right ) \\ \cos A & =x\cos \alpha +y\sin \alpha =\sqrt{x^{2}+y^{2}}\cos \left (\alpha -\tan ^{-1}\displaystyle \frac{y}{x} \right ) \end{cases}\\ &\color{red}\textrm{Supaya}\: \: \color{blue}\cos A=\sqrt{x^{2}+y^{2}}\cos \left (\alpha -\tan ^{-1}\displaystyle \frac{y}{x} \right ),\: \: \color{red}\textrm{maka}\\ &\begin{cases} \sqrt{x^{2}+y^{2}} & =1 \\ \tan ^{-1}\displaystyle \frac{y}{x} & =0\Rightarrow \begin{cases} y & =0 \\ x & =1 \end{cases} \end{cases}\\ &\color{red}\textrm{Sehingga}\: \: \color{black}x+y=1+0=1 \end{aligned} \end{array}$

Contoh Soal 4 Matriks

$\begin{array}{ll}\\ 16.&\textrm{Determinan untuk matriks}\: \: \begin{pmatrix} 2 & -5\\ 3 & -1 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-17\\ \textrm{b}.&-13\\ \textrm{c}.&11\\ \color{red}\textrm{d}.&13\\ \textrm{e}.&17 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Determinan}&\: \textrm{dari matriks}\: \: \begin{pmatrix} 2 & -5\\ 3 & -1 \end{pmatrix}\\ &=\begin{vmatrix} 2 & -5\\ 3 & -1 \end{vmatrix}=2(-1)-3(-5)\\ &=-2+15\\ &=13 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Determinan untuk matriks}\\ &\begin{pmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&10\\ \textrm{b}.&18\\ \textrm{c}.&22\\ \textrm{d}.&30\\ \textrm{e}.&36 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Determinan}\: \: \textrm{dari matriks}\\ &\begin{pmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{pmatrix}\\ &=\begin{vmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{vmatrix}\\ &=+(2.4.3)+(-1.-1.1)+(-1.1.-2)\\ &\quad -(1.4.-1)-(-2.-1.2)-(3.1.-1)\\ &=24+1+2+4-24+3\\ &=10 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika diketahu matriks}\\ &\textrm{A}=\begin{pmatrix} x+3&-2\\ -16&2x-6 \end{pmatrix},\\ &\textrm{maka nilai dari}\: \: \: x\: \: \textrm{supaya matriks}\\ &\textrm{A tidak memiliki invers adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \color{red}\textrm{e}.&5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\textrm{Invers}&\: \textrm{dari matriks A adalah}\: \: \: \textrm{A}^{-1}.\\ \textrm{A}^{-1}&=\displaystyle \frac{1}{det\: \textrm{A}}\times \color{red}Adjoin\: \textrm{A}.\\ \textrm{Karen}&\textrm{a}\: \textrm{tidak memiliki invers},\\ \textrm{maka}\: \, & det\: \textrm{A}=0,\: \textrm{sehingga}\\ det\: \textrm{A}&=\begin{vmatrix} x+3 & -2\\ -16 & 2x-6 \end{vmatrix}=0\\ &\Leftrightarrow (x+3)(2x-6)-(-16.-2)=0\\ &(\color{red}\textrm{masing-masing ruas dibagi 2})\\ &\Leftrightarrow (x+3)(x-3)-16=0\\ &\Leftrightarrow x^{2}-9-16=0\\ &\Leftrightarrow x^{2}-25=0\\ &\Leftrightarrow (x+5)(x-5)=0\\ &\Leftrightarrow x+5=0\quad \textrm{atau}\quad x-5=0\\ &\Leftrightarrow \: \: \: \, \, \color{red}x=-5\quad \textrm{atau}\quad x=5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: \begin{vmatrix} 5^{2x} & -5\\ 1 & 1 \end{vmatrix}=6.5^{x}\\ &\textrm{maka}\: \: 5^{2x}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&625\: \: \textrm{atau}\: \: 1\\ \color{red}\textrm{b}.&25\: \: \textrm{atau}\: \: 1\\ \textrm{c}.&25\: \: \textrm{atau}\: \: 0\\ \textrm{d}.&5\: \: \textrm{atau}\: \: 1\\ \textrm{e}.&5\: \: \textrm{atau}\: \: 0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&5^{2x}+5=6.5^{x}\\ &5^{2x}-6.5^{x}+5=0\\ &\left ( 5^{x}-1 \right )\left ( 5^{x}-5 \right )=0\\ &5^{x}-1=0\: \: \textrm{atau}\: \: 5^{x}-5=0\\ &5^{x}=1\: \: \textrm{atau}\: \: 5^{x}=5\\ &5^{x}=5^{0}\: \: \textrm{atau}\: \: 5^{x}=5^{1}\\ &x=0\: \: \textrm{atau}\: \: x=1\\ &\color{red}\textrm{maka}\\ &5^{2x}=\begin{cases} 5^{2.1} &=5^{2}=25 \\ 5^{2.0} &=5^{0}=1 \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Diketahu determinan suatu}\\ &\textrm{matriks adalah}\: \: \begin{vmatrix} x & 1 & 2\\ x & 1 & x\\ 5 & -3 & 7 \end{vmatrix}=0.\\ &\textrm{Jika}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{adalah akar-akar}\\ &\textrm{yang memenuhi persamaan tersebut}\\ &\textrm{maka nilai dari}\: \: \: p+q\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-3\\ \textrm{b}.&-\displaystyle \frac{1}{3}\\ \textrm{c}.&-1\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{3}\\ \textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Diketahui ba}&\textrm{hwa}:\\ \begin{vmatrix} x & 1 & 2\\ x & 1 & x\\ 5 & -3 & 7 \end{vmatrix}&=0\\ +(x.1.7)+&(1.x.5)+(2.x.-3)\\ -(5.1.2)&-(-3.x.x)-(7.x.1)=0\\ 7x+5x-6x&-10+3x^{2}-7x=0\\ 3x^{2}-x-10&=0\begin{cases} p & \textrm{salah satu akar} \\ q & \textrm{salah satu akar yang lain}, \end{cases}\\ \color{red}\textrm{dengan}\: \: \: &\begin{cases} a &=3 \\ b &=-1 \\ c &=-10 \end{cases}.\\ \textrm{maka}\: \: \: p+q\: \: &=-\displaystyle \frac{b}{a}=-\displaystyle \frac{-1}{3}\\ &=\color{red}\displaystyle \frac{1}{3} \end{aligned} \end{array}$

Lanjutan 4 Materi Matriks (Matematika Wajib Kelas XI)

$\color{blue}\textrm{F. Invers Matriks ordo 2x2}$

Perhatikanlah kembali materi sebelumnya berkaitan determinan matriks 2x2, yaitu

$\color{blue}\begin{array}{|c|}\hline \begin{aligned}&\color{black}\textrm{Jika matriks}\: \: \color{red}A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\ &\color{black}\textrm{maka determinan matriks}\: \: \color{red}A\\ &\textrm{ditentukan dengan}\\ &det\: \: \color{red}A=\begin{vmatrix} a & b\\ c & d \end{vmatrix}=ad-bc \end{aligned}\\\hline \end{array}$

Jika  $\color{blue}det\: \: \color{red}A$  bernilai tidak sama dengan nol, maka invers matriks ordo 2x2 yang selanjutnya dilambangkan dengan  $\color{red}A^{\color{black}-1}$  dapat ditentukan dengan formula:

$\LARGE\color{blue}\boxed{\color{red}A^{\color{black}-1}=\color{blue}\frac{1}{\color{red}ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix}}$

$\begin{aligned}&\textbf{Sebagai}\: \: \color{blue}\textrm{CONTOH}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 2x2,\: \: \textrm{yaitu}:\\ &\color{red}A=\color{black}\begin{pmatrix} -3 & 2\\ -1 & 4 \end{pmatrix},\: \: \textrm{maka}\: \: \color{red}A^{^{\color{black}-1}}\: \: \color{black}\textrm{adalah}:\\ &\color{red}A^{^{\color{black}-1}}=\displaystyle \frac{1}{\color{black}\begin{vmatrix} \color{red}-3 & 2\\ -1 & \color{red}4 \end{vmatrix}}\color{blue}\begin{pmatrix} 4 & -2\\ 1 & -3 \end{pmatrix}\\ &\: \qquad =\displaystyle \frac{1}{\color{red}12\color{black}-(-2)}\color{blue}\begin{pmatrix} 4 & -2\\ 1 & -3 \end{pmatrix}\\ &\: \qquad=\displaystyle \frac{1}{14}\color{blue}\begin{pmatrix} 4 & -2\\ 1 & -3 \end{pmatrix}\\ &\: \qquad =\color{blue}\begin{pmatrix} \displaystyle \frac{4}{14} & \displaystyle \frac{-2}{14}\\ \displaystyle \frac{1}{14} & \displaystyle \frac{-3}{14} \end{pmatrix}\\ &\: \qquad=\color{blue}\begin{pmatrix} \displaystyle \frac{2}{7} & -\displaystyle \frac{1}{7}\\ \displaystyle \frac{1}{14} & -\displaystyle \frac{3}{14} \end{pmatrix} \end{aligned}$

$\LARGE\color{purple}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah invers matriks berikut}\\ &\textrm{a}.\quad B=\begin{pmatrix} 5 & -3\\ 4 & -2 \end{pmatrix}\\ &\textrm{b}.\quad C=\begin{pmatrix} -3 & -5\\ 6 & 9 \end{pmatrix}\\ &\textrm{c}.\quad P=\begin{pmatrix} -1 & 2\\ -3 & 6 \end{pmatrix}\\ &\textrm{d}.\quad Q=\begin{pmatrix} 6 & 9\\ 2 & 3 \end{pmatrix}\\\\ &\textrm{Jawab}:\: \color{purple}\textrm{yang dibahas poin a saja}\\ &B^{-1}=\displaystyle \frac{1}{det\: B}\begin{pmatrix} -2 & 3\\ -4 & 5 \end{pmatrix}\\ &\qquad=\displaystyle \frac{1}{-10-(-12)}\begin{pmatrix} -2 & 3\\ -4 & 5 \end{pmatrix}=\displaystyle \frac{1}{2}\begin{pmatrix} -2 & 3\\ -4 & 5 \end{pmatrix}\\ &\qquad =\color{blue}\begin{pmatrix} -1 & \displaystyle \frac{3}{2}\\ -2 & \displaystyle \frac{5}{2} \end{pmatrix}\\ &\color{purple}\textrm{b. Silahkan dicoba sendiri}\\ &\color{purple}\textrm{c. Silahkan dicoba sendiri}\\ &\color{purple}\textrm{d. Silahkan dicoba sendiri} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui matriks}\\ & E=\begin{pmatrix} 4 & 2\\ 5 & 3 \end{pmatrix}\\ &\textrm{a}.\quad \textrm{Tentukanlah}\\ &\qquad (\textrm{i})\: \: E^{-1}\qquad\qquad\qquad (\textrm{iii})\: \: \left ( E^{-1} \right )^{t}\\ &\qquad (\textrm{i})\: \: E^{t}\qquad\qquad\qquad\: \: \: (\textrm{iv})\: \: \left ( E^{t} \right )^{-1}\\ &\textrm{b}.\quad \textrm{Dengan menggunakan hasil-hasil}\\ &\qquad \textrm{pada a. apakah}\: \: \left ( E^{-1} \right )^{t}=\left ( E^{t} \right )^{-1}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\textrm{a}.\quad(\textrm{i})\: \: E^{-1}&=\displaystyle \frac{1}{2}\begin{pmatrix} 3 & -2\\ -5 & 4 \end{pmatrix}=\color{red}\begin{pmatrix} \displaystyle \frac{3}{2} & -1\\ -\displaystyle \frac{5}{2} & 2 \end{pmatrix}\\ (\textrm{ii})\: \: \: \: E^{t}&=\color{purple}\begin{pmatrix} 4 & 5\\ 2 & 3 \end{pmatrix}\\ (\textrm{iii})\: \: \: \quad&\left (E^{-1} \right )^{t}=\begin{pmatrix} \displaystyle \frac{3}{2} & -\displaystyle \frac{5}{2}\\ -1 & 2 \end{pmatrix}\\ (\textrm{iv})\: \: \: \quad&\left (E^{t} \right )^{-1}=\color{purple}\displaystyle \frac{1}{2}\begin{pmatrix} 3 & -5\\ -2 & 4 \end{pmatrix}=\begin{pmatrix} \displaystyle \frac{3}{2} & -\displaystyle \frac{5}{2}\\ -1 & 2 \end{pmatrix}\\ \textrm{b}.\quad \textrm{Dari ha}&\textrm{sil yang didapat dapat disimpulkan}\\ &\color{red}\left ( E^{-1} \right )^{t}=\left ( E^{t} \right )^{-1} \end{aligned} \end{array}$

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)

$\color{blue}\textrm{E. Determinan Matriks}$

$\color{red}\textrm{1. Ordo 2x2}$

Misalkan A adalah matriks persegi berordo 2x2 dan dituliskan dengan  $A=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}$  dengan  $\color{red}a_{11}\: \: \color{purple}\textrm{dan}\: \: \color{red}a_{22}$ sebagai elemen dari diagonal utama dan $\color{blue}a_{12}\: \: \color{purple}\textrm{dan}\: \: \color{blue}a_{21}$ adalah elemen yang menempati diagonal samping, perhatikan lagi matriks A berikut:

$A=\begin{pmatrix} \color{red}a_{11} & \color{blue}a_{12}\\ \color{blue}a_{21} & \color{red}a_{22} \end{pmatrix}$

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:

$\textrm{det}.\: A=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}$ sama dengan

$\color{blue}\textrm{det}.\: A=\begin{vmatrix} \color{red}a_{11} & \color{black}a_{12}\\ \color{black}a_{21} & \color{red}a_{22} \end{vmatrix}=\color{red}a_{11}\times a_{22}-\color{blue}a_{12}\times a_{21}$

$\begin{aligned}&\textbf{Sebagai}\: \: \color{blue}\textrm{CONTOH}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 2x2,\: \: \textrm{yaitu}:\\ &\color{red}A=\color{black}\begin{pmatrix} -3 & 2\\ -1 & 4 \end{pmatrix},\: \: \textrm{maka}\: \: \color{blue}det\: A\: \: \color{black}\textrm{adalah}:\\ &\color{blue}det\: A=\color{black}\begin{vmatrix} \color{red}-3 & 2\\ -1 & \color{red}4 \end{vmatrix}=\color{red}(-3)\times (-4)\color{black}-(-1)\times (2)\\ &\: \qquad =\color{red}12\color{black}-(-2)=\color{red}12\color{black}+2=\color{blue}14 \end{aligned}$

$\color{red}\textrm{2. Ordo 3x3}$

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

$\begin{aligned}&\color{black}\textrm{Misalkan diberikan matriks ordo}\: 3x3\\ &\color{blue}A=\color{black}\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{pmatrix}\\ \end{aligned}$

$\color{red}\textrm{2.1 Menjabarkan mengikuti baris atau kolom}$

$\begin{aligned}\textrm{det A}&=a_{11}\begin{vmatrix} a_{22} & a_{23}\\ a_{32} & a_{33} \end{vmatrix}-a_{12}\begin{vmatrix} a_{21} & a_{23}\\ a_{31} & a_{33} \end{vmatrix}+a_{13}\begin{vmatrix} a_{21} & a_{22}\\ a_{31} & a_{32} \end{vmatrix}\\ &\\ &\textbf{Catatan}:\\ &\textrm{tanda}\: a_{ij}=\color{blue}\textrm{positif jika}\: i+j\: \textrm{genap}\\ &\textrm{tanda}\: a_{ij}=\color{red}\textrm{negatif jika}\: i+j\: \textrm{ganjil} \end{aligned}$

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.

$\color{red}\textrm{2.2 Aturan Sarrus}$

$\begin{aligned}\textrm{det A}&=\color{blue}a_{11}.a_{22}.a_{33}\\ &\quad\: \color{blue}+a_{12}.a_{23}.a_{31}\\ &\quad\: \color{blue}+a_{13}.a_{21}.a_{32}\\ &\quad \color{red}-a_{31}a_{22}.a_{13}\\ &\quad \color{red}-a_{32}.a_{23}.a_{11}\\ &\quad \color{red}-a_{33}.a_{21}.a_{12} \end{aligned}$

$\begin{aligned}&\textbf{Sebagai}\: \: \color{blue}\textrm{CONTOH MENJABARKAN}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 3x3,\: \: \textrm{yaitu}:\\ &\color{red}A=\color{black}\begin{pmatrix} 1 & 2&3\\ 1 &3& 4\\ 1&4&3 \end{pmatrix},\: \: \textrm{maka}\: \: \color{blue}det\: A\: \: \color{black}\textrm{dengan}\\ &\textrm{menjabarkan baris pertama adalah}:\\ &\color{blue}det\: A=1\begin{vmatrix} 3 & 4\\ 4 & 3 \end{vmatrix}-2\begin{vmatrix} 1 & 4\\ 1 & 3 \end{vmatrix}+3\begin{vmatrix} 1 & 3\\ 1 & 4 \end{vmatrix}\\ &\: \qquad=(9-16)-2(3-4)+3(4-3)\\ &\: \qquad=-7+2+3\\ &\: \qquad=\color{red}-2 \end{aligned}$

$\begin{aligned}&\textbf{Dan berikut}\: \: \color{blue}\textrm{CONTOH aturan SARRUS}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 3x3,\: \: \textrm{yaitu}:\\ &\color{red}B=\color{black}\begin{pmatrix} 2 & 1&3\\ 3 &1& 4\\ 4&1&3 \end{pmatrix},\: \: \textrm{maka}\: \: \color{blue}det\: B\: \: \color{black}\textrm{dengan}\\ &\textrm{metode SARRUS adalah}:\\ &\color{blue}det\: B=(2.1.3)+(1.4.4)+(3.1.3)\\ &\qquad\: \: \: \: \: \: \: -(4.1.3)-(1.4.2)-(3.1.3)\\ &\qquad\: \, =6+16+9-12-8-9=\color{red}2 \end{aligned}$

$\LARGE\color{purple}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Diketahui matriks-matriks persegi berikut}\\ &\textrm{a}.\: \: \begin{pmatrix} 2 & 3\\ 6 & 7 \end{pmatrix}\qquad\qquad \textrm{c}.\: \: \begin{pmatrix} -2 & -3\\ 6 & 7 \end{pmatrix}\\ &\textrm{b}.\: \: \begin{pmatrix} 0 & 4\\ -3 & 6 \end{pmatrix}\: \quad\quad\quad \textrm{d}.\: \: \begin{pmatrix} \sqrt{3} & 3\sqrt{3}\\ \sqrt{2} & -2\sqrt{2} \end{pmatrix}\\ &\\ &\textrm{Tentukanlah determinan dari}\\ &\textrm{matriks-matriks persegi di atas}\\\\ &\color{black}\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad &\color{red}\begin{vmatrix} 2 & 3\\ 6 & 7 \end{vmatrix}\\ &=(2).(7)-(3).(6)\\ &=14-18\\ &=-4\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad &\color{blue}\begin{vmatrix} 0 & 4\\ -3 & 6 \end{vmatrix}\\ &=(0).(6)-(4).(-3)\\ &=0-(-12)\\ &=12\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad &\color{red}\begin{vmatrix} -2 & -3\\ 6 & 7 \end{vmatrix}\\ &=(-2).(7)-(-3).(6)\\ &=(-14)-(-18)\\ &=-14+18\\ &=4 \end{aligned}&\begin{aligned}\textrm{d}.\quad &\color{blue}\begin{vmatrix} \sqrt{3} & 3\sqrt{3}\\ \sqrt{2} & -2\sqrt{2} \end{vmatrix}\\ &=(\sqrt{3}).(-2\sqrt{2})\\ &\quad-(3\sqrt{3}).(\sqrt{2})\\ &=-2\sqrt{6}-3\sqrt{6}\\ &=-5\sqrt{6}\\ & \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\begin{vmatrix} 1-x & 3\\ 2 & 3-x \end{vmatrix}=2\\\\ &\color{black}\textrm{Jawab}:\\ &\begin{aligned}\color{blue}\begin{vmatrix} 1-x & 3\\ 2 & 3-x \end{vmatrix}&=\color{red}2\\ \left ( 1-x \right )\left ( 3-x \right )-(3)(2)&=\color{red}2\\ 3-x-3x+x^{2}-6&=\color{red}2\\ x^{2}-4x-3&=\color{red}2\\ x^{2}-4x-5&=\color{red}0\\ \left ( x-5 \right )\left ( x+1 \right )&=0\\ x-5=0\: \: \textrm{atau}\: \:x+1&=0\\ \color{purple}x=5\: \: \textrm{atau}\: \: x=-1& \end{aligned} \end{array}$

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

$.\qquad\:  \begin{aligned}&\color{purple}\textrm{Jawab}:\\ &\color{blue}\begin{array}{|l|l|}\hline (\textrm{i}).\quad \begin{pmatrix} 1 & 1&3\\ 2 & 4&5\\ 3&5&4 \end{pmatrix}&(\textrm{ii}).\quad \begin{pmatrix} -1 & -2&-3\\ -2 & 6&0\\ -3&0&6 \end{pmatrix}\\\hline (\textrm{iii}).\quad \begin{pmatrix} 1 & 2&3\\ 4 & 5&6\\ 7&8&9 \end{pmatrix}&(\textrm{iv}).\quad \begin{pmatrix} 2 & 1&1\\ 1 & 2&1\\ 1&1&2 \end{pmatrix}\\\hline \begin{aligned}(\textrm{i}).\quad &\begin{vmatrix} 1 & 1&3\\ 2 & 4&5\\ 3&5&4 \end{vmatrix}\\ &=(1)(4)(4)+\\ &\: \: \quad (1)(5)(3)+\\ &\: \: \quad (3)(2)(5)+\\ &\: \: \quad -(3)(4)(3)\\ &\: \: \quad -(5)(5)(1)\\ &\: \: \quad -(4)(2)(1)\\ &=16+15+30\\ &\: \: \: -36-25-8\\ &=-8 \\\end{aligned} &\begin{aligned}(\textrm{ii}).\quad &\begin{vmatrix} -1 & -2&-3\\ -2 & 6&0\\ -3&0&6 \end{vmatrix}\\ &=(-1)(6)(6)+\\ &\: \: \quad (-2)(0)(-3)+\\ &\: \: \quad (-3)(-2)(0)+\\ &\: \: \quad -(-3)(6)(-3)\\ &\: \: \quad -(0)(0)(-1)\\ &\: \: \quad -(6)(-2)(-2)\\ &=-36+0+0\\ &\: \: \: -54-0-24\\ &=-114 \\\end{aligned} \\\hline \begin{aligned}(\textrm{iii}).\quad &\begin{vmatrix} 1 & 2&3\\ 4 & 5&6\\ 7&8&9 \end{vmatrix}\\ &=(1)(5)(9)+\\ &\: \: \quad (2)(6)(7)+\\ &\: \: \quad (3)(4)(8)+\\ &\: \: \quad -(7)(5)(3)\\ &\: \: \quad -(8)(6)(1)\\ &\: \: \quad -(9)(4)(2)\\ &=45+84+96\\ &\: \: \: -105-48-72\\ &=0 \\\end{aligned} &\begin{aligned}(\textrm{iv}).\quad &\begin{vmatrix} 2 & 1&1\\ 1 & 2&1\\ 1&1&2 \end{vmatrix}\\ &=(2)(2)(2)+\\ &\: \: \quad (1)(1)(1)+\\ &\: \: \quad (1)(1)(1)+\\ &\: \: \quad -(1)(2)(1)\\ &\: \: \quad -(1)(1)(2)\\ &\: \: \quad -(2)(1)(1)\\ &=8+1+1\\ &\: \: \: -2-2-2\\ &=4 \\\end{aligned} \\\hline \end{array}\\ &\color{purple}\textrm{yang belum dibahas silahkan dibuat latihan}  \end{aligned}$

DAFTAR PUSTAKA

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