Transformasi Geometri (XI Matematika Wajib)

 A. Pengertian 

Transformasi Geometri adalah suatu perubahan objek geometri atau suatu pemetaan dari suatu titik-titik ke himpunan titik-titik yang lain pada bidang kartesius.

Dari pengertian di atas jelas bahwa aturan transformasi sebagaimana fungsi atau pemetaan dan transformasi ini selanjutnya dapat disimbolkan dengan sebuah huruf kapital, misal M, T, R, dan lain sebagainya. Sebagai misal titik P(x,y) oleh transformasi T menghasilkan titik baru yaitu P'(x',y') dan operasi ini dapat dituliskan dengan:

$\begin{aligned}&P(x,y)\overset{T}{\rightarrow}P'(x',y') \end{aligned}$.

B. Matriks Transformasi

Misalkan suatu transfomasi T memetakan sebuah titik A(x,y) ke A'(x',y') 

selanjutnya perhatikan ilustrasi berikut:

$\boxed{\begin{aligned}A(x,y)&\xrightarrow[.]{\color{red}Transformasi\, =\: T}A'(x',y')=A'\left ( ax+by,cx+dy \right )\\\\ \Rightarrow &\begin{pmatrix} x'\\ y' \end{pmatrix}=\underset{\underset{transformasi}{Matriks}}{\underbrace{\color{blue}\begin{pmatrix} a & b\\ c & d \end{pmatrix}}}\begin{pmatrix} x\\ y \end{pmatrix} \end{aligned}}$.

C. Jenis-Jenis Transformasi dengan matriks yang sesuaian

1. Translasi (Geseran)

$\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Translasi}&(x,y)\xrightarrow[.]{\begin{pmatrix} a\\ b \end{pmatrix}}(x+a,y+b)&\begin{pmatrix} a\\ b \end{pmatrix}\\\hline \end{array}$.

2. Rotasi (Perputaran)

$\begin{aligned}&\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Rotasi}&&\\\hline \begin{aligned}&\textrm{Pusat rotasi}\\ & \left [ O,\alpha \right ] \end{aligned}&\begin{aligned}&\begin{cases} x' =... \\ y' = ... \end{cases}\\ &\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{biru} \end{aligned} \end{aligned}&\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\\\hline \begin{aligned}&\textrm{Pusat}\\ & (a,b)\: \textrm{sudut}\: \alpha \end{aligned}&\begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=&\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{merah} \end{aligned}\\\hline \end{array}\\ &\color{blue}\begin{cases} x' =x\cos \alpha -y\sin \alpha \\ y' = x\sin \alpha +y\cos \alpha \end{cases}\\ &\color{red}\triangleright \triangleright \triangleright \triangleright \begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}.\begin{pmatrix} x-a\\ y-b \end{pmatrix} \end{aligned}$.

3. Refleksi (Pencerminan)

$\begin{array}{|l|c|c|}\hline \textrm{Refleksi}&&\\\hline \textrm{terhadap sumbu}-\textrm{X}&(x,y)\rightarrow (x,-y)&\begin{pmatrix} 1 &0 \\ 0 & -1 \end{pmatrix}\\\hline \textrm{terhadap sumbu}-\textrm{Y}&(x,y)\rightarrow (-x,y)&\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\\\hline \textrm{terhadap garis y = x}&(x,y)\rightarrow (y,x)&\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = -x}&(x,y)\rightarrow (-y,-x)&\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis x = h}&(x,y)\rightarrow (2h-x,y)&\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} 2h\\ 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = x}&(x,y)\rightarrow (y,x)&\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = -x}&(x,y)\rightarrow (-y,-x)&\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis x = h}&(x,y)\rightarrow (2h-x,y)&\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} 2h\\ 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = k}&(x,y)\rightarrow (x,2k-y)&\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} 0\\ 2k \end{pmatrix}\\\hline \textrm{pusat}\: (0,0)\begin{cases} y=mx \\ m=\tan \alpha \end{cases}&&\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}\\\hline \end{array}$.

4. Dilatasi (Perkalian)

$\begin{aligned}&\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Dilatasi}&&\\\hline \textrm{Pusat}\: \left [ O,k \right ]&(x,y)\rightarrow (kx,ky)&\begin{pmatrix} k & 0\\ 0 & k \end{pmatrix}\\\hline \begin{aligned}&\textrm{Pusat}\: (a,b)\\ & \textrm{faktor skala}\: k \end{aligned}&\begin{pmatrix} x'-a\\ y'-b \end{pmatrix}&\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{merah} \end{aligned}\\\hline \begin{aligned}&\textrm{Luas bangun}\\ &\textrm{ datar} \end{aligned}&\textrm{Misal bangun A}&\textrm{T}=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\\hline &\textbf{Bangun A}'&= \textrm{det T}\times \textrm{A}\\\hline \end{array}\\ &\color{red}\triangleright \triangleright \triangleright \triangleright \triangleright \triangleright \begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}.\begin{pmatrix} x-a\\ y-b \end{pmatrix} \end{aligned}$.

Catatan:

Translasi, refleksi, dan rotasi suatu objek adalah bagian dari transformasi yang hanya mengubah posisi objek saja, sehingga jenis transformasi-transformasi ini juga disebut dengan transformasi isometri

D. Bayangan Kurva dan Komposisi Transformasi

$\begin{array}{|l|l|}\hline \qquad \color{red}\textrm{Bayangan Kurva}\quad y=f(x)&\qquad\qquad\qquad \color{blue}\textrm{Komposisi Transformasi}\\\hline \begin{aligned}\textrm{Lan}&\textrm{gkah-langkah}:\\ 1.\quad&\textrm{Tentukan bayangan titiknya}\\ &(x,y)\rightarrow \left ( x',y' \right )\\ 2.\quad&\textrm{Salanjutnya tentukan}\: \: x\: \: \textrm{dan}\: \: y\:\\ &\textrm{dalam}\: \: x'\: \: \textrm{dan}\: \: y'\\ 3.\quad&\textrm{Substitusikan}\: \: x\: \: \textrm{dan}\: \: y\\ &\textrm{ke}\: \: \: y=f(x) \end{aligned}&\begin{aligned}\textrm{Lan}&\textrm{gkah-langkah}:\\ 1.\quad&\textrm{Selesaikan sesuai urutan transformasi}\\ &(x,y)\xrightarrow[\qquad.]{T_{1}}(x',y')\xrightarrow[\qquad.]{T_{2}}(x'',y'')\\ 2.\quad&\textrm{Jika dapat disederhanakan kedua transformasi}\\ &\textrm{tersebut di atas, maka cukup dengan}\\ &(x,y)\xrightarrow[\qquad.]{T_{2}\circ T_{1}}(x'',y'') \end{aligned}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah bayangan dari segitiga PQR dengan}\\\ & P(0,4),\: Q(-1,1),\: \textrm{dan}\: \: R(3,6).\\ &\textrm{oleh translasi}\: \: \: T=\begin{pmatrix} 5\\ -2 \end{pmatrix}\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{cases} \begin{pmatrix} x_{P}^{'}\\ y_{P}^{'} \end{pmatrix} &=T+\begin{pmatrix} x_{P}\\ y_{P} \end{pmatrix}=\begin{pmatrix} 5\\ -2 \end{pmatrix}+\begin{pmatrix} 0\\ 4 \end{pmatrix}=\begin{pmatrix} 5+0\\ -2+4 \end{pmatrix}=\begin{pmatrix} 5\\ 2 \end{pmatrix} \\ \begin{pmatrix} x_{Q}^{'}\\ y_{Q}^{'} \end{pmatrix} & =\cdots\qquad \textrm{isilah sendiri} \\ \begin{pmatrix} x_{R}^{'}\\ y_{R}^{'} \end{pmatrix} &= \cdots\qquad \textrm{isilah sendiri} \end{cases} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah bayangan dari garis}\: \: y=2x+4\\ & \textrm{oleh translasi}\: \: T=\begin{pmatrix} -1\\ 2 \end{pmatrix}.\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{array}{|c|c|}\hline \textbf{Bayangan Titik-titik}&\textbf{Bayangan Garis}\\\hline \begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=T+\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -1\\ 2 \end{pmatrix}+\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -1+x\\ 2+y \end{pmatrix}\\ &\begin{cases} x' & =-1+x\Leftrightarrow x=x'+1 \\ y' & =2+y\quad\Leftrightarrow y=y'-2 \end{cases} \end{aligned}&\begin{aligned}y&=2x+4\\ y'-2&=2(x'+1)+4\\ y'&=2x+2+4+2\\ &=2x+8\\ \textrm{Jadi}\, ,&\: \textbf{bayangan garisnya}\\ \textrm{adala}&\textrm{h}:\\ y&=2x+8\\ & \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah bayangan titik A(4,6) oleh rotasi yang berpusat }\\ &\textrm{di titik P(3,-2) dengan sudut putar sebesar}\: \: 90^{\circ} \\\\ &\color{blue}\textbf{Jawab}\\ &\begin{aligned}\textrm{Untuk Ro}&\textrm{tasi yang berpusat di}\: \: (a,b)\: \: \textrm{dengan sudut}\: \: \alpha \: \: \textrm{adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} x-a\\ y-b \end{pmatrix}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ}\\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}\begin{pmatrix} 4-3\\ 6-(-2) \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ 8 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} -8\\ 1 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} -5\\ -1 \end{pmatrix}\\ \textrm{Jadi}\, ,\: &\textrm{bayangan titik A adalah}\: \: \textrm{A}'(-5,-1) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukanlah bayangan titik A(4,6) }\\ &\textrm{oleh dilatasi yang berpusat di titik P(3,-2)}\\ &\textrm{dengan faktor skala}\: \: k=2 \\\\ &\color{blue}\textbf{Jawab}\\ &\begin{aligned}\textrm{Bayangan}&\: \textrm{titik A-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} k & 0\\ 0 & k \end{pmatrix}\begin{pmatrix} x-a\\ y-b \end{pmatrix}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}\begin{pmatrix} 4-3\\ 6-(-2) \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1\\ 8 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2\\ 16 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ 14 \end{pmatrix}\\ \textrm{Jadi}\: ,\: &\textrm{bayangan titik A-nya adalah}\: \: \textrm{A}'(5,14) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tentukanlah bayangan titik A(4,6) }\\ &\textrm{oleh translasi}\: \: t\: \: \textrm{dilanjutkan}\: \: s\: \: \textrm{dengan}\\ &\textrm{matriks transformasi berturut-turut }\\ &\textrm{adalah}\: \: T=\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}\: \: \textrm{dan}\: \: S= \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{aligned}\textrm{Bayangan}&\: \textrm{titik A-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=S\times T\times \begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}\begin{pmatrix} 4\\ 6 \end{pmatrix}\\ &=\begin{pmatrix} 2 & 3\\ 1 & 2 \end{pmatrix}\begin{pmatrix} 4\\ 6 \end{pmatrix}\\ &=\begin{pmatrix} 26\\ 16 \end{pmatrix}\\ \textrm{Jadi}\: ,\: &\textrm{bayangan titik A-nya adalah}\: \: \textrm{A}'(26,16) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Suatu kurva}\: \: y=\, ^{3}\log (2x-2)\: \: \textrm{memiliki bayangan}\\ & y=\, ^{3}\log \left ( \displaystyle \frac{2x+3}{3} \right )\: \: \textrm{oleh translasi}\\ & T=\begin{pmatrix} a\\ b \end{pmatrix}.\: \textrm{Tentukanlah nilai}\: \: a+b\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\\ y&=\, ^{3}\log (2x-2)\quad \Leftrightarrow\quad 3^{y}=2x-2\: (\textrm{benda})\\ y&=\, ^{3}\log \left ( \displaystyle \frac{2x+3}{3} \right )\\ & \Leftrightarrow\quad 3^{y}=\left ( \displaystyle \frac{2x+3}{3} \right )\quad (\color{red}\textbf{bayangan})\\ \textrm{sehingga}&\: \textrm{untuk bayangan}\\ 3^{y'-b}&=2(x'-a)-2\quad \Leftrightarrow \quad 3^{y'}.3^{-b}=2(x'-a)-2\\ & \Leftrightarrow\quad 3^{y'}=\displaystyle \frac{2(x'-a)-2}{3^{-b}}=\displaystyle \frac{2x'+3}{3}\\ \textrm{Jadi}\, ,\: &\begin{cases} a &=\displaystyle \frac{5}{2} \\ b &=-1 \end{cases}\\ \textrm{Sehingga}&\: a+b=\displaystyle \frac{5}{2}+(-1)=\displaystyle \frac{3}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tentukanlah bayangan garis}\: \: ax+by+c=0\: \: \textrm{oleh transformasi}\\ &\textrm{yang bersesuaian dengan matriks}\: \: \: \begin{pmatrix} 1&-2\\ 3&-4 \end{pmatrix}\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{array}{|c|c|}\hline \textbf{Proses Awal}&\textbf{Penentuan Bayangan}\\\hline \begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ 3 & -4 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ 3 & -4 \end{pmatrix}^{-1}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\displaystyle \frac{1}{\begin{vmatrix} 1 & -2\\ 3 & -4 \end{vmatrix}}\begin{pmatrix} -4 & 2\\ -3 & 1 \end{pmatrix}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\displaystyle \frac{1}{-4+6}\begin{pmatrix} -4x'+2y'\\ -3x'+y' \end{pmatrix}\\ &=\displaystyle \frac{1}{2}\begin{pmatrix} -4x'+2y'\\ -3x'+y' \end{pmatrix}\\ &\begin{cases} x &=-2x'+y' \\ y &=-\displaystyle \frac{3}{2}x'+\displaystyle \frac{1}{2}y' \end{cases} \end{aligned}&\begin{aligned}ax+by+c&=0\\ a\left ( -2x'+y' \right )+b\left ( -\displaystyle \frac{3}{2}x'+\frac{1}{2}y' \right )+c&=0\\ -2ax'-\displaystyle \frac{3}{2}bx'+ay'+\displaystyle \frac{1}{2}by'+c&=0\\ (-4a-3b)x'+(2a+b)y'+2c&=0\\ &\\ \textbf{Jadi, bayangan garisnya adalah}:&\\ &\\ \color{red}(-4a-3b)x+(2a+b)y+2c&=0\\ &\\ &\\ &\\ & \end{aligned} \\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Diketahui kurva}\: \: y=4x^{2}-9\: \: \textrm{dicerminkan terhadap sumbu-X kemudian}\\ &\textrm{ditranslasikan dengan}\: \: \begin{pmatrix} -1\\ 2 \end{pmatrix}.\: \textrm{Ordinat titik potong terhadap sumbu-Y adalah}....\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} -1\\ 2 \end{pmatrix}+\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -1\\ 2 \end{pmatrix}+\begin{pmatrix} x\\ -y \end{pmatrix}\\ &=\begin{pmatrix} -1+x\\ 2-y \end{pmatrix}\\ &\begin{cases} x &= x'-1\\ y &= 2-y' \end{cases} \end{aligned}&\begin{aligned}y&=4x^{2}-9\\ (2-y')&=4(x'-1)^{2}-9\\ -y'&=4(x'^{2}-2x'+1)-9-2\\ -y'&=4x'^{2}-8x'+4-11\\ y'&=-4x'^{2}+8x'+7\\ &\\ \textbf{Maka}\, ,&\, \textbf{persamaan kurva bayangannya}:\\ y&=\color{red}-4x^{2}+8x+7 \end{aligned} \\\hline \end{array}\\ &\begin{aligned}\textrm{Sehingga}&\: \textrm{ordinat dari titik potong terhadap sumbu-Y-nya adalah}:\\ y&=-4x^{2}+8x+7,\qquad \textbf{atau}\\ f(x)&=-4x^{2}+8x+7\\ f(0)&=-4(0)^{2}+8(0)+7\qquad\quad \textrm{saat}\: \: x=0\: (\textrm{karena memotong sumbu-Y})\\ &=7\\ \textrm{Jadi}&\: \textrm{ordinatnya adalah}\: \: y=f(0)=\color{red}7 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.

Materi Lanjutan Distribusi Normal (Matematika Peminatan Kelas XII)

C. Transformasi Suatu Variabel Random Berdistribusi Normal

Dalam menentukan luas suatu variabel berdistribusi normal ke dalam variabel random berdistribusi normal baku dengan jalan mentransformasikannya

Adapun luasnya sama yaitu: 

$\begin{aligned}&\begin{aligned}\textrm{P}(\textrm{x}_{1}< X< \textrm{x}_{2})&=\displaystyle \int_{\textrm{x}_{1}}^{\textrm{x}_{2}}\displaystyle \frac{1}{\sigma \sqrt{2\pi }}e^{-\frac{1}{2}\left (\displaystyle \frac{\textrm{x}-\mu }{\sigma }  \right )^{2}}d\textrm{x}\\ &=\displaystyle \int_{\textrm{z}_{1}}^{\textrm{z}_{2}}\displaystyle \frac{1}{ \sqrt{2\pi }}e^{-\frac{1}{2}\textrm{Z}^{2}}d\textrm{x}\\ &=\textrm{P}(\textrm{z}_{1}<\textrm{Z}<\textrm{z}_{2}) \end{aligned}\\\\ &\textrm{Luas di atas adalah hasil tranformasi}\\ &\textrm{variabel acak X}\sim \textrm{N}(\mu ,\sigma )\: \: \textrm{ke}\: \: \textrm{Z}\sim \textrm{N}(0,1)\\ &\textrm{dengan}\: \: \textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma } \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui variabel acak Z berdistribusi}\\ &\textrm{normal}\: \: \textrm{N}(0,1)\: \textrm{dan X berdistribusi}\\ &\textrm{normal N}(18,5).\: \textrm{Tentukanlah besar}\\ &\textrm{peluang berikut}\\ &\textrm{a}.\quad \textrm{P}(\textrm{Z}>0,68)\\ &\textrm{b}.\quad \textrm{P}(\textrm{X}<20)\\ &\textrm{c}.\quad \textrm{P}(0,36<\textrm{Z}<1,42)\\ &\textrm{d}.\quad \textrm{P}(17<\textrm{X}<18,5)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad \textrm{P}(\textrm{Z}>0,68)&=0,5-\textrm{P}(0<\textrm{Z}<0,68)\\ &=0,5-0,2517=\color{blue}0,2483 \end{aligned}\\&\begin{aligned}\textrm{b}.\quad&\textrm{Transformasi}\: \: \textrm{x}=\color{red}20, \: \: \color{black}\textrm{dengan}\begin{cases} \mu  & =18 \\  \sigma  & =5  \end{cases}\\ &\textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}20\color{black}-18}{5}=0,4,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(\textrm{X}< 20)&=\textrm{P}(\textrm{Z}< 0,4)\\ &=0,5+\textrm{P}(0< \textrm{Z}< 0,4)\\ &=0,5+0,1554=\color{blue}0,6554 \end{aligned} \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad \textrm{P}(0,36<\textrm{Z}<1,42)&=\textrm{P}(0<\textrm{Z}<1,42)-\textrm{P}(0<\textrm{Z}<0,36)\\ &=0,4222-0,1406=\color{blue}0,2816 \end{aligned}\\&\begin{aligned}\textrm{d}.\quad&\textrm{Transformasi}\: \: \textrm{x}_{1}=\color{red}17, \: \: \color{black}\textrm{dan}\: \: \textrm{x}_{2}=\color{red}18,5\\ &\begin{cases} \mu  & =18 \\  \sigma  & =5  \end{cases}\\ &\textrm{z}_{1}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}17\color{black}-18}{5}=-0,2,\: \: \textrm{dan}\\ &\textrm{z}_{2}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}18,5\color{black}-18}{5}=0,1,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(17<\textrm{X}< 18,5)&=\textrm{P}(-0,2<\textrm{Z}< 0,1)\\ &=\textrm{P}(0< \textrm{Z}< 0,2)+\textrm{P}(0< \textrm{Z}< 0,1)\\ &=0,0793+0,0398=\color{blue}0,1191 \end{aligned} \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui variabel acak X berdistribusi}\\ &\textrm{normal memiliki rata-rata 16 dan simpangan}\\ &\textrm{baku}\: \: 1,4.\: \: \textrm{Hitunglah besar peluang dari}\\ &\textrm{a}.\quad \textrm{P}(\textrm{X}\leq 18,8)\\ &\textrm{d}.\quad \textrm{P}(12,1\leq \textrm{X}\leq 16,3)\\\\ &\textbf{Jawab}:\\&\begin{aligned}\textrm{a}.\quad&\textrm{Transformasi}\: \: \textrm{x}=\color{red}18,8, \: \: \color{black}\textrm{dengan}\begin{cases} \mu  & =16 \\  \sigma  & =1,4  \end{cases}\\ &\textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}18,8\color{black}-16}{1,4}=2,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(\textrm{X}< 18,8)&=\textrm{P}(\textrm{Z}< 2)\\ &=0,5+\textrm{P}(0< \textrm{Z}< 2)\\ &=0,5+0,4772=\color{blue}0,9772 \end{aligned} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\textrm{Transformasi}\: \: \textrm{x}_{1}=\color{red}12,1 \: ,\: \color{black}\textrm{dan}\: \: \textrm{x}_{2}=\color{red}16,3\\ &\begin{cases} \mu  & =16 \\  \sigma  & =1,4  \end{cases}\\ &\textrm{z}_{1}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}12,1\color{black}-16}{1,4}=-2,79,\: \: \textrm{dan}\\&\textrm{z}_{2}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}16,3\color{black}-16}{1,4}=0,21,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(12,1<\textrm{X}< 16,3)&=\textrm{P}(-2,79<\textrm{Z}< 0,21)\\ &=\textrm{P}(0< \textrm{Z}< 2,79)+\textrm{P}(0< \textrm{Z}< 0,21)\\ &=0,4974+0,0832=\color{blue}0,5806 \end{aligned} \end{aligned}   \end{array}$ 

$\begin{array}{ll}\\ 3.&\textrm{Di sebuah MA dengan 1000 siswa diperoleh data}\\ &\textrm{rata-rata berat bada siswanya}\: \: 54\: \textrm{Kg}\: \textrm{dan simpangan}\\ &\textrm{baku}\: 8\: \textrm{Kg}.\: \textrm{Jika data tersebut berdistribusi normal}\\ &\textrm{tentukan bsnysk siswa yang memiliki berat badan}\\ &\textrm{a}.\quad \textrm{lebih dari 70 Kg}\\ &\textrm{b}.\quad \textrm{antara 40 Kg sampai 50 Kg}\\\\ &\textbf{Jawab}:\\  &\begin{aligned}\textrm{a}.\quad&\textrm{Transformasi}\: \: \textrm{x}=\color{red}70, \: \: \color{black}\textrm{dengan}\begin{cases} \mu  & =54 \\  \sigma  & =8  \end{cases}\\ &\textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}70\color{black}-54}{8}=\frac{16}{8}=2\\ &\textrm{Dari tabel diperoleh}\\ &P(0< \textrm{Z}< 2)=0,4772,\: \: \textrm{selanjutnya}\\ &\begin{aligned}\textrm{P}(\textrm{X}>70)&=\textrm{P}(\textrm{Z}>2)\\ &=0,5-\textrm{P}(0< \textrm{Z}< 2)\\ &=0,5-0,4772=\color{blue}0,0228 \end{aligned}\\ &\textrm{Selanjutnya frekuensi harapan dalam hal ini}\\ &\begin{aligned}f_{h}(\textrm{X}>70)&=1000\times P(\textrm{X}>70)\\ &=1000\times \color{blue}0,0228\\ &=22,8\\ &\approx 23  \end{aligned}\\ &\textrm{Jadi, ada sebanyak 23 anak dengan berat badan}\\ &>\: \:  \textrm{70 Kg} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\textrm{Transformasi}\: \: \textrm{x}_{1}=\color{red}40 \: ,\: \color{black}\textrm{dan}\: \: \textrm{x}_{2}=\color{red}50\\ &\begin{cases} \mu  & =54 \\  \sigma  & =8  \end{cases}\\ &\textrm{z}_{1}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}40\color{black}-54}{8}=-1,75,\: \: \textrm{dan}\\&\textrm{z}_{2}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}50\color{black}-54}{8}=-0,5,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(40<\textrm{X}<50)&=\textrm{P}(-1,75<\textrm{Z}<-0,5)\\ &=\textrm{P}(0,5<\textrm{Z}<1,75)\\ &=\textrm{P}(0< \textrm{Z}< 1,75)-\textrm{P}(0< \textrm{Z}< 0,5)\\ &=0,4599-0,1915=\color{blue}0,2684 \end{aligned}\\ &\textrm{Selanjutnya frekuensi harapan dalam hal ini}\\ &\begin{aligned}f_{h}(40<\textrm{X}<50)&=1000\times P(40<\textrm{X}<50)\\ &=1000\times \color{blue}0,2684\\&=268,4\\ &\approx 268  \end{aligned}\\ &\textrm{Jadi, ada sebanyak 268 anak dengan berat badan}\\ &\textrm{antara 40 samapi 50 Kg} \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui Nilai-nilai ujian penerimaan pegawai baru}\\&\textrm{diperoleh mean 78 dan deviasi standarnya 6. Jika}\\ &\textrm{hanya}\: \: 12,5\%\: \: \textrm{calon yang akan diterima, maka nilai}\\ &\textrm{terendah yang lolos jika distribusinya normal}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Transformasi}\: \: \textrm{X}=\color{red}\textrm{x}\\ &\color{black}\textrm{dengan}\begin{cases} \mu  & =78 \\  \sigma  & =6,\: \textrm{dan}\\ \textrm{P}(\textrm{X}>x)&=12,5\%=0,125  \end{cases}\\&\begin{aligned}\textrm{P}(\textrm{X}>\textrm{x})\qquad&=\textrm{P}(\textrm{Z}>0)-\textrm{P}(0< \textrm{Z}<\textrm{z})\\ 0,125&=0,5-\textrm{P}(0< \textrm{Z}< \textrm{z})\\ \textrm{P}(0< \textrm{Z}< \textrm{z})\: \, &=0,5-0,125=\color{blue}0,375 \end{aligned}\\ &\textrm{Dari tabel diperoleh}\: \: \textrm{P}(0< \textrm{Z}< \textrm{z})=0,375,\\ &\textrm{selanjutnya didapatkan nilai}\: \: z=\color{red}1,15\\ &\textrm{Sehingga nilai terendah x yang diterima}\\&\textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }\\ &\Leftrightarrow \textrm{x}=\textrm{z}\sigma +\mu =(1,15).6+78=84,9\approx 85 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Tasari, Aksin, N., Miyanto, Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten: INTAN PARIWARA.

Contoh Soal 5 Transformasi Geometri

$\begin{array}{ll}\\ 21.&\textrm{Jika setiap titik pada grafik dengan}\\ &\textrm{dengan persamaan}\: \: y=\sqrt{x}\: \: \textrm{dicerminkan} \\ &\textrm{terhadap garis}\: \: y=x\: ,\: \textrm{maka persamaan}\\ &\textrm{grafik yang dihasilkan adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}y=x^{2}\: ,\: x\geq 0&&\\ \textrm{b}.\quad y=-\sqrt{x}\: ,\: x\geq 0&\\ \textrm{c}.\quad y=-x^{2}\: ,\: x\leq 0&\\ \textrm{d}.\quad y=\sqrt{-x}\: ,\: x\leq 0\\ \textrm{e}.\quad y=-\sqrt{-x}\: ,\: x\leq 0 \end{array}\\\\ &\quad\quad\qquad \textbf{UMB Tahun 2011 Kode 152}\\\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ y&=\sqrt{x},\: \: \textrm{atau}\: \: y^{2}=x\\ \textbf{Alt}&\textbf{ernatif 1}\\ \textrm{mak}&\textrm{a}\: \: \textrm{saat dicerminkan terhadap}\\ \textrm{gari}&\textrm{s}\: \: y=x,\: \textrm{adalah}\: \: \color{red}x^{2}=y\\ \textrm{atau}&\: \: \color{red}y=x^{2}.\\ \textbf{Alt}&\textbf{ernatif 2}\\ \textrm{Jika}\: &\textrm{ingin dikerjakan dengan rumus}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=M_{x=y}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} y\\ x \end{pmatrix}\\ \textrm{Sela}&\textrm{njutnya hasilnya disubstitusikan}\\ \textrm{ke p}&\textrm{ersamaan}\: \: y=\sqrt{x}\Rightarrow \color{red}x'=\sqrt{y'}\\ \sqrt{y'} &=x'\: \: \: \textrm{maka}\\ y'&=\left ( x' \right )^{2}\: \: \: \textrm{selanjutnya}\\ y&=x^{2} \end{aligned} \end{array}$.

Sebelum dicerminkan terhadap garis y=x

Gambar kurva/grafik setelah cerminkan terhadap garis y=x

$\begin{array}{ll}\\ 22.&\textrm{Transformasi}\: \: T\: \: \textrm{adalah pencerminan}\\ &\textrm{terhadap garis}\: \: y=\displaystyle \frac{x}{3}\: \: \textrm{dilanjutkan oleh} \\ &\textrm{pencerminan terhadap garis}\: \: y=-3x.\\ &\textrm{Matriks yang bersesuian dengan}\\ &\textrm{transformasi}\: \: T\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}&&\\ \textrm{b}.\quad \color{red}\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}&\\ \textrm{c}.\quad \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}&\\ \textrm{d}.\quad \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}\\ \textrm{e}.\quad \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \end{array}\\\\ &\quad\quad \textbf{SBMPTN Tahun 2013 Kode 433}\\\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \textrm{sebu}&\textrm{ah persamaan garis lurus}\\ \textrm{dapa}&\textrm{t dituliskan dengan}:\: y=\color{red}m\color{black}x\\ \textrm{Dike}&\textrm{tahui pula bahwa ada 2 garis}:\\ y_{1}&=\displaystyle \frac{1}{3}x\quad \textrm{dan}\: \: \: y_{2}=-3x\\ \textrm{seba}&\textrm{gai representasi transformasi}\: \: T.\\ \textrm{Kare}&\textrm{na}\: \: m_{1}\times m_{2}=\left ( \displaystyle \frac{1}{3} \right )(-3)=-1\\ \textrm{bera}&\textrm{rti 2 garis di atas saling tegak}\\ \textrm{luru}&\textrm{s dan hal ini seperti rotasi 2}\\ \textrm{kali}\: \: &90^{\circ}\: \: \textrm{atau}\: \: 180^{\circ}\\ \textrm{Jadi},&\: T=\color{purple}\begin{pmatrix} \cos 180^{\circ} & -\sin 180^{\circ}\\ \sin 180^{\circ} & \cos 180^{\circ} \end{pmatrix}\\ \Leftrightarrow &\: T=\color{red}\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix} \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.
3. Sharma,S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.

Contoh Soal 1 Transformasi Geometri

$\begin{array}{ll}\\ 1.&\textrm{Suatu translasi yang memetakan titik P(9,8) }\\ &\textrm{ke titik}\: \: \textrm{P}'(14,-2)\: \: \textrm{adalah}\: ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\begin{pmatrix} 5\\ -10 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 6\\ 6 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 5\\ 6 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} 23\\ -10 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} 5\\ 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\textrm{T}+\begin{pmatrix} x\\ y \end{pmatrix}\\ \textrm{T}&=\begin{pmatrix} x'-x\\ y'-y \end{pmatrix}\\ &=\begin{pmatrix} 14-9\\ -2-8 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ -10 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ &\textrm{Maka bayangan titik M}(2,-1)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (7,10)&&\textrm{d}.\quad (1,10)\\ \textrm{b}.\quad (10,7)&\textrm{c}.\quad \color{red}(1,4)&\textrm{e}.\quad (4,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ \left.\begin{matrix} x=2\\ y=-1 \end{matrix}\right\}&\Rightarrow \begin{cases} x' & =2(2)+3(-1)=4-3=1 \\ y' & =3(2)+2(-1)=6-2=4 \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Bayangan untuk titik A(1,3) oleh rotasi }\\ &\textrm{dengan pusat}\: \: \textit{O}(0,0)\textrm{sejauh}\: \: 90^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-1,3)&&\textrm{d}.\quad (1,-3)\\ \textrm{b}.\quad (-1,-3)&\textrm{c}.\quad \color{red}(-3,1)&\textrm{e}.\quad (3,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\: \: 90^{\circ},\\ \textrm{maka}\qquad\qquad\: \: &\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ}\\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga}\quad\qquad&\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -3\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Suatu lingkaran dengan jari-jari 4 }\\ &\textrm{dengan pusat di O(0,0) dtranslasikan}\\ &\textrm{oleh}\: \: \textrm{T}=\begin{pmatrix} 2\\ -3 \end{pmatrix},\: \textrm{maka luas }\\ &\textrm{bayangan lingkaran tersebut adalah}\\ & ....\: \textrm{satuan luas}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \pi &&\textrm{d}.\quad 8\pi \\ \textrm{b}.\quad 2\pi &\textrm{c}.\quad 4\pi &\textrm{e}.\quad \color{red}16\pi \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran berpusat}\\ &\textrm{di O dengan}\: \: r=4.\: \textrm{Karena translasi adalah}\\ &\textrm{termasuk transformasi isometri(kongruen)}\\ &\textrm{maka jari-jari lingkaran bayangannya }\\ &\textrm{akan sama dengan bendanya. Sehingga}\\ &\textrm{ luas bayangan lingkarannya}\\ &=\pi r^{2}=\pi \times 4^{2}=\color{red}16\pi \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\textrm{Yang merupakan titik invarian (tidak berubah) }\\ &\textrm{adalah}\: ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (0,0)&&\textrm{d}.\quad (0,-1)\\ \textrm{b}.\quad \color{red}(1,-1)&\textrm{c}.\quad (1,0)&\textrm{e}.\quad (1,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}:\\ &\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\begin{array}{|c|c|c|c|}\hline \textrm{NO}&\textrm{Titik}&\begin{aligned}&\textrm{Disubstitusikan ke}\\ & \color{blue}\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases} \end{aligned}&\begin{aligned}&\textrm{Keterangan}\\ &\quad\textrm{Titik} \end{aligned}\\\hline \textrm{a}.&(0,0)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(0)-4=-4 \end{cases}&\textrm{Varian}\\\hline \textrm{b}&(1,-1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(-1)-4=-1 \end{cases}&\color{red}\textbf{Invarian}\\\hline \textrm{c}&(1,0)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(0)-4=-2 \end{cases}&\textrm{Varian}\\\hline \textrm{d}&(0,-1)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(-1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \textrm{e}&(1,1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \end{array} \end{aligned} \end{array}$

Contoh Soal 4 Transformasi Geometri

$\begin{array}{ll}\\ 16.&\textrm{Bayangan titik A(2,4) dicerminkan }\\ &\textrm{terhadap garis}\: \: y-x=0\: \: \textrm{dilanjutkan}\\ &\textrm{ke garis}\: \: x\sqrt{3}-3y=0\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}A'\left ( 2+\sqrt{3},-1+2\sqrt{3} \right )&&\\ \textrm{b}.\quad A'\left ( 2+\sqrt{3},1-2\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left ( 1-\sqrt{3},-2+\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( -2+\sqrt{3},1+2\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( 2-\sqrt{3},1-2\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ &\begin{cases} x\sqrt{3}-3y=0 & \Leftrightarrow y=\displaystyle \frac{1}{3}\sqrt{3}x\\ &\Leftrightarrow y=\tan 30^{\circ}.x\\\\ x-y=0 & \Leftrightarrow y=x \end{cases}\\\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.30^{\circ} & \sin 2.30^{\circ} \\ \sin 2.30^{\circ} & -\cos 2.30^{\circ} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2} & \displaystyle \frac{1}{2}\sqrt{3}\\ \displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ 4 \end{pmatrix}\\ &=\begin{pmatrix} \sqrt{3}+2\\ -1+2\sqrt{3} \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Jika}\: \: T_{1}=\begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}\: \: \textrm{dan}\: \: T_{2}=\begin{pmatrix} -2 & 5\\ -1 & 3 \end{pmatrix}\\ &\textrm{maka bayangan garis}\: \: x+y+1=0\\ &\textrm{oleh}\: \: T_{2}\circ T_{1}\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}x-2y-1=0&&\\ \textrm{b}.\quad x+2y-1=0&\\ \textrm{c}.\quad x+2y+1=0&\\ \textrm{d}.\quad x-2y+1=0\\ \textrm{e}.\quad x+y-1=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=T_{2}\circ T_{1}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -2 & 5\\ -1 & 3 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -2+5 & -4+5\\ -1+3 & -2+3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 3 & 1\\ 2 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 3x+y\\ 2x+y \end{pmatrix}\\ \textrm{Dipe}&\textrm{roleh}\\ &\begin{array}{lllllllll}\\ \quad x'&=3x+y\\ \quad y'&=2x+y\qquad\quad-\\\hline x'-y'&=x\\ \Leftrightarrow \quad x&=\color{red}x'-y'\qquad\color{black}....(1)\\ \color{blue}\textrm{maka}\\ \qquad y&=x'-3x\\ &=x'-3(x'-y')\\ &=\color{red}3y'-2x'\quad \color{black}....(2) \end{array}\\ \textrm{Sehin}&\textrm{gga}\\ x+y&+1=0\\ x'-&y'+3y'-2x'+1=0\\ -x'+&2y'+1=0\\ x'-&2y'-1=0\\ \textrm{maka}&\: \textrm{bayangan garisnya}\\ x-2&y-1=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Garis}\: \: 2x+y+4=0\: \: \textrm{ditranslasikan}\\ &\textrm{oleh}\: \: \begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{dilanjutkan transformasi} \\ &\textrm{oleh}\: \: \begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\: \: \textrm{persamaan bayangannya}\\ &\textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2x+y+3=0&&\\ \textrm{b}.\quad \color{red}2x-3y+3=0&\\ \textrm{c}.\quad 2x+3y+3=0&\\ \textrm{d}.\quad 3x+2y+3=0\\ \textrm{e}.\quad 3x-2y+3=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}\: &=\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} -2\\ 5 \end{pmatrix}=\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x-2+2y+10\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x+2y+8\\ y+5 \end{pmatrix}\\ \textrm{Diper}&\textrm{oleh}\\ &\begin{array}{lllllllll}\\ \quad x''&=x+2y+8\\ \: \: \: 2y''&=2y+10\qquad\quad-\\\hline x''-2y''&=x-2\\ \Leftrightarrow \quad x&=\color{red}x''-2y''+2\: \color{black}....(1)\\ \color{blue}\textrm{maka}\\ \qquad y&=\color{red}y''-5\quad \color{black}\qquad....(2) \end{array}\\ \textrm{sehin}&\textrm{gga}\\ 2x+&y+4=0\\ 2(x''&-2y''+2)+(y''-5)+4=0\\ 2x''-&3y''+3=0\\ \textrm{maka}&\: \textrm{bayangan garisnya}\\ 2x-&3y+3=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Diketahui}\: \: M\: \: \textrm{adalah pencerminan terhadap}\\ &\textrm{garis}\: \: y=-x\: \: \textrm{dan}\: \: T\: \: \textrm{adalah transformasi} \\ &\textrm{yang dinyatakan oleh matriks}\: \: \begin{pmatrix} 2 & 3\\ 0 & -1 \end{pmatrix}\\ &\textrm{Koordinat bayangan titik}\: \: A(2,-8)\: \: \textrm{oleh}\\ &\textrm{transformasi}\: \: M\: \: \textrm{dilanjutkan oleh}\: \: T\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-10,2)&&\\ \textrm{b}.\quad (-2,-10)&\\ \textrm{c}.\quad \color{red}(10,2)&\\ \textrm{d}.\quad (-10,-2)\\ \textrm{e}.\quad (2,10) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix} &=\color{purple}T\circ M\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 2 & 3\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} 0-3 & -2+0\\ 0+1 & 0+0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} -3 & -2\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} -6+16\\ 2+0 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 10\\ 2 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: W\: \: \textrm{adalah transformasi oleh}\\ &\textrm{matriks}\: \: \begin{pmatrix} 1 & 0\\ 3 & 1 \end{pmatrix},\: \: \textrm{maka titik mula}\\ &\textrm{dari}\: \: W'(-2,5)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-11,-2)&&\\ \textrm{b}.\quad (11,-2)&\\ \textrm{c}.\quad \color{red}(-2,11)&\\ \textrm{d}.\quad (2,11)\\ \textrm{e}.\quad (12,11) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dimi}&\textrm{salkan}:\\ A&=\begin{pmatrix} -2\\ 5 \end{pmatrix},\: \: \textrm{dan}\\ W&=\begin{pmatrix} 1 & 0\\ 3 & 1 \end{pmatrix},\: \: \textrm{serta}\: \: X=\begin{pmatrix} x\\ y \end{pmatrix}\\ \textrm{mak}&\textrm{a}\\ &\begin{array}{|c|}\hline \color{red}\begin{aligned}A&=BX\\ B^{-1}A&=B^{-1}BX\\ B^{-1}A&=I.X\\ B^{-1}A&=X\\ X&=B^{-1}A \end{aligned}\\\hline \end{array}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\displaystyle \frac{1}{\begin{vmatrix} 1 &0 \\ 3 & 1 \end{vmatrix}}\begin{pmatrix} 1 & 0\\ -3 & 1 \end{pmatrix}\begin{pmatrix} -2\\ 5 \end{pmatrix}\\ &=1.\begin{pmatrix} -2+0\\ 6+5 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} -2\\ 11 \end{pmatrix} \end{aligned} \end{array}$

Latihan Soal 10 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{ll}\\ 91.&\textrm{Titik A(4,-4) dicerminkan terhadap}\\ &\textrm{garis}\: \: y=x\tan 15^{\circ}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: A'(a,b)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}&&\textrm{d}.\quad \color{red}4\sqrt{3}\\ \textrm{b}.\quad 2\sqrt{3}&\textrm{c}.\quad 3\sqrt{3}&\textrm{e}.\quad 6\sqrt{3} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} a\\ b \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.15^{\circ}& \sin 2.15^{\circ}\\ \sin 2.15^{\circ} & -\cos 2.15^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \cos 30^{\circ}&\sin 30^{\circ}\\ \sin 30^{\circ}&-\cos 30^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2}\sqrt{3} & \displaystyle \frac{1}{2}\\ \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} 2\sqrt{3}-2\\ 2+2\sqrt{3} \end{pmatrix}\\ &\begin{cases} a &=2\sqrt{3}-2 \\ b &=2+2\sqrt{3} \end{cases}\\ \textrm{mak}&\textrm{a nilai dari}\\ a+b&=\left ( 2\sqrt{3}-2+2+2\sqrt{3} \right )\\ &=4\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 92.&\textrm{Lingkaran}\: \: x^{2}+y^{2}-5x+8y+7=0\\ & \textrm{ditranslasikan oleh}\: \: T=\begin{pmatrix} m\\ n \end{pmatrix}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: x^{2}+y^{2}-9x+2y+6=0.\\ & \textrm{Nilai}\: \: m+n=\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2&&\textrm{d}.\quad \color{red}5\\ \textrm{b}.\quad 3&\qquad\textrm{c}.\quad 4\qquad&\textrm{e}.\quad 6 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui sebuah lingkaran dengan persamaan}:\\ & \color{blue}x^{2}+y^{2}-5x+8y+7=0\\ \textrm{kar}&\textrm{ena akibat translasi, maka}\\ &\begin{cases} x & =x'-m \\ y & =y'-n \end{cases}\\ &x^{2}+y^{2}-5x+8y+7=0\\ \textrm{seh}&\textrm{ingga}\\ &\Leftrightarrow \color{purple}(x'-m)^{2}+(y'-n)^{2}-5(x'-m)+8(y'-n)+7=0\\ &\Leftrightarrow \color{purple}x'^{2}+y'^{2}-2mx'-2ny'+m^{2}+n^{2}-5x'+5m+8y'-8n+7=0\\ &\Leftrightarrow \color{purple}x'^{2}+y'^{2}-(2m+5)x'+(8-2n)y'+m^{2}+n^{2}+5m-8n+7=0\\ &\qquad \equiv \: \color{purple}x'^{2}+y'^{2}-9x'+2y'+6=0\qquad (\color{black}\textbf{akhir bayangan})\\ &\begin{cases} 9 &=2m+5 \Rightarrow m=2\\ 2 & =8-2n \: \Rightarrow \, \: n=3 \end{cases}\\ \textrm{Jad}&\textrm{i , nilai}\: \: m+n=2+3=5\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 93.&\textrm{Jika titik A(-2,1) dicerminkan terhadap garis}\\ & y=-\displaystyle \frac{1}{3}x\sqrt{3}\: ,\: \textrm{maka bayangan dari}\\ &\textrm{titik \textit{A} tersebut adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&&\\ \textrm{b}.\quad \color{red}A'\left ( -1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left (-1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( -1+\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ y&=-\displaystyle \frac{1}{3}x\sqrt{3}=\left ( -\displaystyle \frac{1}{3}\sqrt{3} \right )x\\ &=\left (-\tan 30^{\circ} \right )x=\tan \left ( 180^{\circ}-30^{\circ} \right )x\\ &=\tan 150^{\circ}.x\\ \textrm{maka}\: \: \theta &=150^{\circ}\quad \Rightarrow \quad 2\theta =300^{\circ}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\color{purple}\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\color{purple}\begin{pmatrix} \cos 300^{\circ} & \sin 300^{\circ} \\ \sin 300^{\circ} & -\cos 300^{\circ} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\color{purple}\begin{pmatrix} \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3}\\ -\displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\begin{pmatrix} -1-\displaystyle \frac{1}{2}\sqrt{3}\\ \sqrt{3}-\displaystyle \frac{1}{2} \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 94.&\textrm{Bayangan titik A(2,4) dicerminkan }\\ &\textrm{terhadap garis}\: \: y-x=0\: \: \textrm{dilanjutkan}\\ &\textrm{ke garis}\: \: x\sqrt{3}-3y=0\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}A'\left ( 2+\sqrt{3},-1+2\sqrt{3} \right )&&\\ \textrm{b}.\quad A'\left ( 2+\sqrt{3},1-2\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left ( 1-\sqrt{3},-2+\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( -2+\sqrt{3},1+2\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( 2-\sqrt{3},1-2\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ &\begin{cases} x\sqrt{3}-3y=0 & \Leftrightarrow y=\displaystyle \frac{1}{3}\sqrt{3}x\\ &\Leftrightarrow y=\tan 30^{\circ}.x\\\\ x-y=0 & \Leftrightarrow y=x \end{cases}\\\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.30^{\circ} & \sin 2.30^{\circ} \\ \sin 2.30^{\circ} & -\cos 2.30^{\circ} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2} & \displaystyle \frac{1}{2}\sqrt{3}\\ \displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ 4 \end{pmatrix}\\ &=\begin{pmatrix} \sqrt{3}+2\\ -1+2\sqrt{3} \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 95.&\textrm{Jika}\: \: T_{1}=\begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}\: \: \textrm{dan}\: \: T_{2}=\begin{pmatrix} -2 & 5\\ -1 & 3 \end{pmatrix}\\ &\textrm{maka bayangan garis}\: \: x+y+1=0\\ &\textrm{oleh}\: \: T_{2}\circ T_{1}\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}x-2y-1=0&&\\ \textrm{b}.\quad x+2y-1=0&\\ \textrm{c}.\quad x+2y+1=0&\\ \textrm{d}.\quad x-2y+1=0\\ \textrm{e}.\quad x+y-1=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=T_{2}\circ T_{1}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -2 & 5\\ -1 & 3 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -2+5 & -4+5\\ -1+3 & -2+3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 3 & 1\\ 2 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 3x+y\\ 2x+y \end{pmatrix}\\ \textrm{Dipe}&\textrm{roleh}\\ &\begin{array}{lllllllll}\\ \quad x'&=3x+y\\ \quad y'&=2x+y\qquad\quad-\\\hline x'-y'&=x\\ \Leftrightarrow \quad x&=\color{red}x'-y'\qquad\color{black}....(1)\\ \color{blue}\textrm{maka}\\ \qquad y&=x'-3x\\ &=x'-3(x'-y')\\ &=\color{red}3y'-2x'\quad \color{black}....(2) \end{array}\\ \textrm{Sehin}&\textrm{gga}\\ x+y&+1=0\\ x'-&y'+3y'-2x'+1=0\\ -x'+&2y'+1=0\\ x'-&2y'-1=0\\ \textrm{maka}&\: \textrm{bayangan garisnya}\\ x-2&y-1=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 96.&\textrm{Garis}\: \: 2x+y+4=0\: \: \textrm{ditranslasikan}\\ &\textrm{oleh}\: \: \begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{dilanjutkan transformasi} \\ &\textrm{oleh}\: \: \begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\: \: \textrm{persamaan bayangannya}\\ &\textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2x+y+3=0&&\\ \textrm{b}.\quad \color{red}2x-3y+3=0&\\ \textrm{c}.\quad 2x+3y+3=0&\\ \textrm{d}.\quad 3x+2y+3=0\\ \textrm{e}.\quad 3x-2y+3=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}\: &=\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} -2\\ 5 \end{pmatrix}=\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x-2+2y+10\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x+2y+8\\ y+5 \end{pmatrix}\\ \textrm{Diper}&\textrm{oleh}\\ &\begin{array}{lllllllll}\\ \quad x''&=x+2y+8\\ \: \: \: 2y''&=2y+10\qquad\quad-\\\hline x''-2y''&=x-2\\ \Leftrightarrow \quad x&=\color{red}x''-2y''+2\: \color{black}....(1)\\ \color{blue}\textrm{maka}\\ \qquad y&=\color{red}y''-5\quad \color{black}\qquad....(2) \end{array}\\ \textrm{sehin}&\textrm{gga}\\ 2x+&y+4=0\\ 2(x''&-2y''+2)+(y''-5)+4=0\\ 2x''-&3y''+3=0\\ \textrm{maka}&\: \textrm{bayangan garisnya}\\ 2x-&3y+3=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 97.&\textrm{Diketahui}\: \: M\: \: \textrm{adalah pencerminan terhadap}\\ &\textrm{garis}\: \: y=-x\: \: \textrm{dan}\: \: T\: \: \textrm{adalah transformasi} \\ &\textrm{yang dinyatakan oleh matriks}\: \: \begin{pmatrix} 2 & 3\\ 0 & -1 \end{pmatrix}\\ &\textrm{Koordinat bayangan titik}\: \: A(2,-8)\: \: \textrm{oleh}\\ &\textrm{transformasi}\: \: M\: \: \textrm{dilanjutkan oleh}\: \: T\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-10,2)&&\\ \textrm{b}.\quad (-2,-10)&\\ \textrm{c}.\quad \color{red}(10,2)&\\ \textrm{d}.\quad (-10,-2)\\ \textrm{e}.\quad (2,10) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix} &=\color{purple}T\circ M\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 2 & 3\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} 0-3 & -2+0\\ 0+1 & 0+0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} -3 & -2\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} -6+16\\ 2+0 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 10\\ 2 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 98.&\textrm{Jika}\: \: W\: \: \textrm{adalah transformasi oleh}\\ &\textrm{matriks}\: \: \begin{pmatrix} 1 & 0\\ 3 & 1 \end{pmatrix},\: \: \textrm{maka titik mula}\\ &\textrm{dari}\: \: W'(-2,5)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-11,-2)&&\\ \textrm{b}.\quad (11,-2)&\\ \textrm{c}.\quad \color{red}(-2,11)&\\ \textrm{d}.\quad (2,11)\\ \textrm{e}.\quad (12,11) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dimi}&\textrm{salkan}:\\ A&=\begin{pmatrix} -2\\ 5 \end{pmatrix},\: \: \textrm{dan}\\ W&=\begin{pmatrix} 1 & 0\\ 3 & 1 \end{pmatrix},\: \: \textrm{serta}\: \: X=\begin{pmatrix} x\\ y \end{pmatrix}\\ \textrm{mak}&\textrm{a}\\ &\begin{array}{|c|}\hline \color{red}\begin{aligned}A&=BX\\ B^{-1}A&=B^{-1}BX\\ B^{-1}A&=I.X\\ B^{-1}A&=X\\ X&=B^{-1}A \end{aligned}\\\hline \end{array}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\displaystyle \frac{1}{\begin{vmatrix} 1 &0 \\ 3 & 1 \end{vmatrix}}\begin{pmatrix} 1 & 0\\ -3 & 1 \end{pmatrix}\begin{pmatrix} -2\\ 5 \end{pmatrix}\\ &=1.\begin{pmatrix} -2+0\\ 6+5 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} -2\\ 11 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 99.&\textrm{Jika setiap titik pada grafik dengan}\\ &\textrm{dengan persamaan}\: \: y=\sqrt{x}\: \: \textrm{dicerminkan} \\ &\textrm{terhadap garis}\: \: y=x\: ,\: \textrm{maka persamaan}\\ &\textrm{grafik yang dihasilkan adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}y=x^{2}\: ,\: x\geq 0&&\\ \textrm{b}.\quad y=-\sqrt{x}\: ,\: x\geq 0&\\ \textrm{c}.\quad y=-x^{2}\: ,\: x\leq 0&\\ \textrm{d}.\quad y=\sqrt{-x}\: ,\: x\leq 0\\ \textrm{e}.\quad y=-\sqrt{-x}\: ,\: x\leq 0 \end{array}\\\\ &\quad\quad\qquad \textbf{UMB Tahun 2011 Kode 152}\\\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ y&=\sqrt{x},\: \: \textrm{atau}\: \: y^{2}=x\\ \textbf{Alt}&\textbf{ernatif 1}\\ \textrm{mak}&\textrm{a}\: \: \textrm{saat dicerminkan terhadap}\\ \textrm{gari}&\textrm{s}\: \: y=x,\: \textrm{adalah}\: \: \color{red}x^{2}=y\\ \textrm{atau}&\: \: \color{red}y=x^{2}.\\ \textbf{Alt}&\textbf{ernatif 2}\\ \textrm{Jika}\: &\textrm{ingin dikerjakan dengan rumus}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=M_{x=y}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} y\\ x \end{pmatrix}\\ \textrm{Sela}&\textrm{njutnya hasilnya disubstitusikan}\\ \textrm{ke p}&\textrm{ersamaan}\: \: y=\sqrt{x}\Rightarrow \color{red}x'=\sqrt{y'}\\ \sqrt{y'} &=x'\: \: \: \textrm{maka}\\ y'&=\left ( x' \right )^{2}\: \: \: \textrm{selanjutnya}\\ y&=x^{2} \end{aligned} \end{array}$.

Sebelum dicerminkan terhadap garis y=x

Gambar kurva/grafik setelah cerminkan terhadap garis y=x

$\begin{array}{ll}\\ 100.&\textrm{Transformasi}\: \: T\: \: \textrm{adalah pencerminan}\\ &\textrm{terhadap garis}\: \: y=\displaystyle \frac{x}{3}\: \: \textrm{dilanjutkan oleh} \\ &\textrm{pencerminan terhadap garis}\: \: y=-3x.\\ &\textrm{Matriks yang bersesuian dengan}\\ &\textrm{transformasi}\: \: T\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}&&\\ \textrm{b}.\quad \color{red}\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}&\\ \textrm{c}.\quad \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}&\\ \textrm{d}.\quad \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}\\ \textrm{e}.\quad \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \end{array}\\\\ &\quad\quad \textbf{SBMPTN Tahun 2013 Kode 433}\\\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \textrm{sebu}&\textrm{ah persamaan garis lurus}\\ \textrm{dapa}&\textrm{t dituliskan dengan}:\: y=\color{red}m\color{black}x\\ \textrm{Dike}&\textrm{tahui pula bahwa ada 2 garis}:\\ y_{1}&=\displaystyle \frac{1}{3}x\quad \textrm{dan}\: \: \: y_{2}=-3x\\ \textrm{seba}&\textrm{gai representasi transformasi}\: \: T.\\ \textrm{Kare}&\textrm{na}\: \: m_{1}\times m_{2}=\left ( \displaystyle \frac{1}{3} \right )(-3)=-1\\ \textrm{bera}&\textrm{rti 2 garis di atas saling tegak}\\ \textrm{luru}&\textrm{s dan hal ini seperti rotasi 2}\\ \textrm{kali}\: \: &90^{\circ}\: \: \textrm{atau}\: \: 180^{\circ}\\ \textrm{Jadi},&\: T=\color{purple}\begin{pmatrix} \cos 180^{\circ} & -\sin 180^{\circ}\\ \sin 180^{\circ} & \cos 180^{\circ} \end{pmatrix}\\ \Leftrightarrow &\: T=\color{red}\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix} \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.
3. Sharma,S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.

Latihan Soal 9 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{ll}\\ 81.&\textrm{Bayangan untuk titik A(1,3) oleh rotasi }\\ &\textrm{dengan pusat}\: \: \textit{O}(0,0)\textrm{sejauh}\: \: 90^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-1,3)&&\textrm{d}.\quad (1,-3)\\ \textrm{b}.\quad (-1,-3)&\textrm{c}.\quad \color{red}(-3,1)&\textrm{e}.\quad (3,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\: \: 90^{\circ},\\ \textrm{maka}\qquad\qquad\: \: &\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ}\\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga}\quad\qquad&\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -3\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 82.&\textrm{Suatu lingkaran dengan jari-jari 4 }\\ &\textrm{dengan pusat di O(0,0) dtranslasikan}\\ &\textrm{oleh}\: \: \textrm{T}=\begin{pmatrix} 2\\ -3 \end{pmatrix},\: \textrm{maka luas }\\ &\textrm{bayangan lingkaran tersebut adalah}\\ & ....\: \textrm{satuan luas}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \pi &&\textrm{d}.\quad 8\pi \\ \textrm{b}.\quad 2\pi &\textrm{c}.\quad 4\pi &\textrm{e}.\quad \color{red}16\pi \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran berpusat}\\ &\textrm{di O dengan}\: \: r=4.\: \textrm{Karena translasi adalah}\\ &\textrm{termasuk transformasi isometri(kongruen)}\\ &\textrm{maka jari-jari lingkaran bayangannya }\\ &\textrm{akan sama dengan bendanya. Sehingga}\\ &\textrm{ luas bayangan lingkarannya}\\ &=\pi r^{2}=\pi \times 4^{2}=\color{red}16\pi \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 83.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\textrm{Yang merupakan titik invarian (tidak berubah) }\\ &\textrm{adalah}\: ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (0,0)&&\textrm{d}.\quad (0,-1)\\ \textrm{b}.\quad \color{red}(1,-1)&\textrm{c}.\quad (1,0)&\textrm{e}.\quad (1,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}:\\ &\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\begin{array}{|c|c|c|c|}\hline \textrm{NO}&\textrm{Titik}&\begin{aligned}&\textrm{Disubstitusikan ke}\\ & \color{blue}\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases} \end{aligned}&\begin{aligned}&\textrm{Keterangan}\\ &\quad\textrm{Titik} \end{aligned}\\\hline \textrm{a}.&(0,0)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(0)-4=-4 \end{cases}&\textrm{Varian}\\\hline \textrm{b}&(1,-1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(-1)-4=-1 \end{cases}&\color{red}\textbf{Invarian}\\\hline \textrm{c}&(1,0)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(0)-4=-2 \end{cases}&\textrm{Varian}\\\hline \textrm{d}&(0,-1)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(-1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \textrm{e}&(1,1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \end{array} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 84.&\textrm{Bayangan untuk titik P(2,5) oleh rotasi }\\ &\textrm{dengan pusat}\: \textit{A}(1,3)\: \: \textrm{sejauh}\: \: 180^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (1,0)&&\textrm{d}.\quad (2,0)\\ \textrm{b}.\quad \color{red}(0,1)&\textrm{c}.\quad (0,2)&\textrm{e}.\quad (1,2) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Karena rotasi de}&\textrm{ngan pusat A sebesar}\: \: 180^{\circ},\\ \textrm{maka}\qquad\qquad\: \: \: \: &\\ R\left ( A(1,3),180^{\circ} \right )&=\begin{pmatrix} \cos 180^{\circ} & -\sin 180^{\circ}\\ \sin 180^{\circ} & \cos 180^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\\ \textrm{sehingga bayang}&\textrm{an titik P(2,5)-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\color{blue}\begin{pmatrix} x-a\\ y-b \end{pmatrix}\color{black}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 2-1\\ 5-3 \end{pmatrix}+\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -1\\ -2 \end{pmatrix}+\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 0\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 85.&\textrm{Bayangan kurva}\: \: xy=6\: \: \textrm{oleh rotasi sebesar}\\ & \displaystyle \frac{\pi }{2}\: \: \textrm{dengan pusat}\: \: O(0,0)\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}xy=-6&&\textrm{d}.\quad x(y-x)=6\\ \textrm{b}.\quad xy=6&&\textrm{e}.\quad x(x+y)=-6\\ \textrm{c}.\quad x(x-y)=6&\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\\ \displaystyle \frac{\pi }{2}=90^{\circ},\: \: \textrm{maka}&\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga bayan}&\textrm{gan semua titik yang }\\ \textrm{terletak pada k}& \textrm{urva adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =-x' \end{cases}\\ \textrm{Selanjunya}\: \: \: \textrm{unt}&\textrm{uk bayangan kurvanya }\\ \textrm{adalah}:\qquad\quad&\\ xy&=6\\ y'.(-x')&=6\\ x'y'&=-6\\ \textrm{Jadi , persamaa}&\textrm{n kurva bayangannya}\\ \textrm{adalah}\: &\: \color{red}xy=-6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 86.&\textrm{Sebuah lingkaran yang berpusat di (3,4) }\\ &\textrm{dan menyinggung sumbu-X dicerminkan}\\ &\textrm{terhadap garis}\: \: y=x\: \textrm{, maka persamaan }\\ &\textrm{akhir lingkaran yang terjadi adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}x^{2}+y^{2}-8x-6y+9=0&&\\ \textrm{b}.\quad x^{2}+y^{2}+8x+6y+9=0&\\ \textrm{c}.\quad x^{2}+y^{2}+6x+8y+9=0&\\ \textrm{d}.\quad x^{2}+y^{2}-8x-6y+16=0\\ \textrm{e}.\quad x^{2}+y^{2}+8x+6y+16=0\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Refleksi l}&\textrm{ingkaran yang berpusat di (3,4) }\\ \textrm{dan men}&\textrm{yinggung sumbu-X, }\\ \textrm{dengan}\: \: r&=(y)=4,\\ \textrm{maka}\: \textrm{per}&\textrm{samaan lingkarannya adalah}:\\ (x-3)^{2}+&(y-4)^{2}=4^{2}.\: \textrm{Karena}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =x' \end{cases}\\ \textrm{selanjutn}&\textrm{ya untuk persamaan bayangan }\\ \textrm{lingkaran} &\textrm{nya adalah}:\\ &(y'-3)^{2}+(x'-4)^{2}=4^{2},\\ & \textbf{menjadi}\\ &(y-3)^{2}+(x-4)^{2}=4^{2},\quad \textrm{atau}:\\ &\color{red}x^{2}+y^{2}-8x-6y+9=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 87.&\textrm{Jika}\: \: M_{x}\: \: \textrm{adalah pencerminan terhadap sumbu-X }\\ &\textrm{dan}\: \: M_{y=x}\: \: \textrm{adalah pencerminan terhadap garis}\\ & y=x\: ,\: \textrm{maka matriks transformasi tunggal }\\ &\textrm{yang mewakili}\: \: M_{x}\circ M_{y=x}=\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\\ \textrm{b}.\quad \color{red}\begin{pmatrix} 0 &1 \\ -1 & 0 \end{pmatrix}&&\textrm{e}.\quad \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\\ \textrm{c}.\quad \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}\: &\textrm{bahwa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ M_{y=x} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \end{cases}\\ M_{x}\circ M_{y=x}&=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ &=\begin{pmatrix} 0+0 & 1+0\\ 0-1 & 0+0 \end{pmatrix}\\ &=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 88.&\textrm{Diketahui vektor}\: \: \vec{x}\: \: \textrm{dirotasikan terhadap titik asal}\\ & O\: \: \textrm{sebesar}\: \: \theta >0\: \: \textrm{searah jarum jam}.\\ &\textrm{Kemudian hasilnya dicerminkan terhadap garis}\: \: y=0\\ & \textrm{menghasilkan vektor}\: \: \vec{y}.\\ &\textrm{Jika}\: \: \vec{y}=A.\vec{x}\: ,\: \textrm{maka matriks}\: \: A-\textrm{nya adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}&&\\ \textrm{b}.\quad \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}&&\\ \textrm{c}.\quad \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}&&\\ \textrm{d}.\quad \color{red}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\\ \textrm{e}.\quad \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui bah}\textrm{wa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ R_{-\theta } & =\begin{pmatrix} \cos (-\theta ) & -\sin (-\theta )\\ \sin (-\theta ) & \cos (-\theta ) \end{pmatrix}=\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{cases}\\ &A=M_{x}\circ R_{-\theta }=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 89.&\textrm{Titik A(1,-2) dirotasikan sejauh}\: \: 15^{\circ}\\ & \textrm{kemudian dilanjutkan}\: \: 75^{\circ}\: \: \textrm{dengan pusat }\\ &O(0,0)\: \: \textrm{maka bayangan akhir titik A adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-2,1)&&\textrm{d}.\quad \color{red}(2,1)\\ \textrm{b}.\quad (-1,2)&\textrm{c}.\quad (1,2)&\textrm{e}.\quad (-2,-1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos (\theta _{1}+\theta _{2}) & -\sin (\theta _{1}+\theta _{2})\\ \sin (\theta _{1}+\theta _{2}) & \cos (\theta _{1}+\theta _{2}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos (75^{\circ}+15^{\circ})& -\sin (75^{\circ}+15^{\circ})\\ \sin (75^{\circ}+15^{\circ}) & \cos (75^{\circ}+15^{\circ}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 90^{\circ}&-\sin 90^{\circ}\\ \sin 90^{\circ}&\cos 90^{\circ} \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 90.&\textrm{Jika garis}\: \: 3x-2y+5=0\: \: \textrm{dicerminkan }\\ &\textrm{terhadap garis}\: \: y=-x\: \: \textrm{kemudian}\\ &\textrm{didilatasikan dengan pusat (1,-2) }\\ &\textrm{dengan faktor skala 2, maka persamaan}\\ & \textrm{bayangannya adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x-2y-10=0&&\\ \textrm{b}.\quad x+2y-10=0&\\ \textrm{c}.\quad x-6y+5=0&\\ \textrm{d}.\quad x+2y-12=0\\ \textrm{e}.\quad \color{red}2x-3y+18=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Proses}&\: \textrm{untuk refleksinya}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&-1\\ -1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ -x \end{pmatrix}\\ \textrm{proses}&\: \textrm{dilatasinya}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 2&0\\ 0&2 \end{pmatrix}\begin{pmatrix} x'-1\\ y'+2 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-2\\ 2y'+4 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-1\\ 2y'+2 \end{pmatrix}\\ &=\begin{pmatrix} 2(-y)-1\\ 2(-x)+2 \end{pmatrix}\\ &\begin{cases} x &=-\displaystyle \frac{1}{2}(y''-2) \\ y &=-\displaystyle \frac{1}{2}(x''+1) \end{cases} \end{aligned}\\ &\begin{aligned}\textrm{Sehingga persam}&\textrm{aan bayangan}\\ \textrm{garisnya adalah}:&\\ 3x&-2y+5=0\\ 3\left ( -\displaystyle \frac{1}{2}(y''-2) \right )&-2\left ( -\displaystyle \frac{1}{2}(x''+1) \right )+5=0\\ -\displaystyle \frac{3}{2}y''+3 &+(x''+1)+5=0\\ 2x&-3y+6+2+10=0\\ 2x&-3y+18=0 \end{aligned} \end{array}$.

Contoh Soal Persiapan Menghadapi Semester Gasal Tahun 2022 untuk Kelas XI MA/SMA/MAK Matematika Wajib (K13 Revisi)

 MATEMATIKA WAJIB

Kelas XI

1. Induksi matematika dan pola bilangan

2. Program linear

3. Matrik dan operasi matriks

4. Transformasi geometri

Contoh soal dan pembahasan materi terkait di atas ada sebanyak 100 contoh soal diserati dengan pembahasannya, berikut link soal yang dimaksud

no. 1 sampai dengan 10

no. 11 sampai dengan 20

no. 21 sampai dengan 30

no. 31 sampai dengan 40

no. 41 sampai dengan 50

no. 51 sampai dengan 60

no. 61 sampai dengan 70

no. 71 sampai dengan 80

no. 81 sampai dengan 90

no. 91 sampai dengan 100

Tambahan contoh soal khusus pola bilangan dan barisan serta deret aritmetika dan geometri

no. 1 sampai dengan 5

no. 6 sampai dengan 10

no. 11 sampai dengan 15

no. 16 sampai dengan 20

no. 21 sampai dengan 25

no. 26 sampai dengan 30

no. 31 sampai dengan 35

Tambahan contoh soal khusus transformasi geometri

no. 1 sampai dengan 5

no. 6 sampai dengan 10

no. 11 sampai dengan 15

no. 16 sampai dengan 20

no. 20 sampai dengan 22

Contoh Soal 2 Transformasi Geometri

$\begin{array}{ll}\\ 6.&\textrm{Bayangan untuk titik P(2,5) oleh rotasi }\\ &\textrm{dengan pusat}\: \textit{A}(1,3)\: \: \textrm{sejauh}\: \: 180^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (1,0)&&\textrm{d}.\quad (2,0)\\ \textrm{b}.\quad \color{red}(0,1)&\textrm{c}.\quad (0,2)&\textrm{e}.\quad (1,2) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Karena rotasi de}&\textrm{ngan pusat A sebesar}\: \: 180^{\circ},\\ \textrm{maka}\qquad\qquad\: \: \: \: &\\ R\left ( A(1,3),180^{\circ} \right )&=\begin{pmatrix} \cos 180^{\circ} & -\sin 180^{\circ}\\ \sin 180^{\circ} & \cos 180^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\\ \textrm{sehingga bayang}&\textrm{an titik P(2,5)-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\color{blue}\begin{pmatrix} x-a\\ y-b \end{pmatrix}\color{black}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 2-1\\ 5-3 \end{pmatrix}+\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -1\\ -2 \end{pmatrix}+\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 0\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Bayangan kurva}\: \: xy=6\: \: \textrm{oleh rotasi sebesar}\\ & \displaystyle \frac{\pi }{2}\: \: \textrm{dengan pusat}\: \: O(0,0)\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}xy=-6&&\textrm{d}.\quad x(y-x)=6\\ \textrm{b}.\quad xy=6&&\textrm{e}.\quad x(x+y)=-6\\ \textrm{c}.\quad x(x-y)=6&\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\\ \displaystyle \frac{\pi }{2}=90^{\circ},\: \: \textrm{maka}&\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga bayan}&\textrm{gan semua titik yang }\\ \textrm{terletak pada k}& \textrm{urva adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =-x' \end{cases}\\ \textrm{Selanjunya}\: \: \: \textrm{unt}&\textrm{uk bayangan kurvanya }\\ \textrm{adalah}:\qquad\quad&\\ xy&=6\\ y'.(-x')&=6\\ x'y'&=-6\\ \textrm{Jadi , persamaa}&\textrm{n kurva bayangannya}\\ \textrm{adalah}\: &\: \color{red}xy=-6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Sebuah lingkaran yang berpusat di (3,4) }\\ &\textrm{dan menyinggung sumbu-X dicerminkan}\\ &\textrm{terhadap garis}\: \: y=x\: \textrm{, maka persamaan }\\ &\textrm{akhir lingkaran yang terjadi adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}x^{2}+y^{2}-8x-6y+9=0&&\\ \textrm{b}.\quad x^{2}+y^{2}+8x+6y+9=0&\\ \textrm{c}.\quad x^{2}+y^{2}+6x+8y+9=0&\\ \textrm{d}.\quad x^{2}+y^{2}-8x-6y+16=0\\ \textrm{e}.\quad x^{2}+y^{2}+8x+6y+16=0\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Refleksi l}&\textrm{ingkaran yang berpusat di (3,4) }\\ \textrm{dan men}&\textrm{yinggung sumbu-X, }\\ \textrm{dengan}\: \: r&=(y)=4,\\ \textrm{maka}\: \textrm{per}&\textrm{samaan lingkarannya adalah}:\\ (x-3)^{2}+&(y-4)^{2}=4^{2}.\: \textrm{Karena}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =x' \end{cases}\\ \textrm{selanjutn}&\textrm{ya untuk persamaan bayangan }\\ \textrm{lingkaran} &\textrm{nya adalah}:\\ &(y'-3)^{2}+(x'-4)^{2}=4^{2},\\ & \textbf{menjadi}\\ &(y-3)^{2}+(x-4)^{2}=4^{2},\quad \textrm{atau}:\\ &\color{red}x^{2}+y^{2}-8x-6y+9=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: M_{x}\: \: \textrm{adalah pencerminan terhadap sumbu-X }\\ &\textrm{dan}\: \: M_{y=x}\: \: \textrm{adalah pencerminan terhadap garis}\\ & y=x\: ,\: \textrm{maka matriks transformasi tunggal }\\ &\textrm{yang mewakili}\: \: M_{x}\circ M_{y=x}=\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\\ \textrm{b}.\quad \color{red}\begin{pmatrix} 0 &1 \\ -1 & 0 \end{pmatrix}&&\textrm{e}.\quad \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\\ \textrm{c}.\quad \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}\: &\textrm{bahwa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ M_{y=x} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \end{cases}\\ M_{x}\circ M_{y=x}&=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ &=\begin{pmatrix} 0+0 & 1+0\\ 0-1 & 0+0 \end{pmatrix}\\ &=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Diketahui vektor}\: \: \vec{x}\: \: \textrm{dirotasikan terhadap titik asal}\\ & O\: \: \textrm{sebesar}\: \: \theta >0\: \: \textrm{searah jarum jam}.\\ &\textrm{Kemudian hasilnya dicerminkan terhadap garis}\: \: y=0\\ & \textrm{menghasilkan vektor}\: \: \vec{y}.\\ &\textrm{Jika}\: \: \vec{y}=A.\vec{x}\: ,\: \textrm{maka matriks}\: \: A-\textrm{nya adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}&&\\ \textrm{b}.\quad \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}&&\\ \textrm{c}.\quad \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}&&\\ \textrm{d}.\quad \color{red}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\\ \textrm{e}.\quad \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui bah}\textrm{wa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ R_{-\theta } & =\begin{pmatrix} \cos (-\theta ) & -\sin (-\theta )\\ \sin (-\theta ) & \cos (-\theta ) \end{pmatrix}=\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{cases}\\ &A=M_{x}\circ R_{-\theta }=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{aligned} \end{array}$.

Lanjutan Contoh Soal 3 Distribusi Normal

$\begin{array}{ll}\\ 11.&\textrm{Perhatikan ilustrasi berikut}\:   ....\\    \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Jika luas yang diarsir adalah 0,90 dan}\: \: n=21,\\ &\textrm{maka nilai}\: \: t\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad1,32\\ &\textrm{b}.\quad \color{red}1,72\\&\textrm{c}.\quad 2,08\\ &\textrm{d}.\quad 2,09\\ &\textrm{e}.\quad 2,53\\\\ &\textbf{Jawab}:\\ &\textrm{Luas arsiran (distribusi student) adalah}:\\ &=1-0,90=0,10,\: \textrm{karena luas kedua ujung}\\ &\textrm{sama, mulai dari}\: t\: \textrm{ke kanan}=\displaystyle \frac{0,10}{2}=0,05\\ &\textrm{Sehingga luas dari}\: t\: \textrm{ke kiri}=1-0,05=0,95\\ &\textrm{Dari ini diketahui luas arsirannya}=p=\color{red}0,95\\ &\textrm{Dengan}\: dk=n-1=21-1=20,\: \textrm{maka}\\ &t=\color{red}\pm 1,72   \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Misalkan luas dari}\: t\: \textrm{ke kiri}=0,05,\\ &\textrm{dengan}\: \: dk=15.\: \textrm{Nilai}\: \: t\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad -1,76\\ &\textrm{b}.\quad \color{red}-1,75\\&\textrm{c}.\quad -1,74\\ &\textrm{d}.\quad 1,75\\ &\textrm{e}.\quad 1,76\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui dari soal di atas}\\ &\textrm{dari}\: \: t\: \: \textrm{ke kiri}=0,05,\: \textrm{maka}\\ &\textrm{luas arsiran}=p=1-0,05=0,95\\ &t=\color{red}-1,75 \: \: (\textrm{lihat tabel})\end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Transformasi Z}-score\: \: \textrm{dikaitan dengan}\\ &\textrm{data dimaksudkan agar data}\\ &\textrm{terdistribusi secara}\: ....\\ &\textrm{a}.\quad \textrm{tak normal}\\ &\textrm{b}.\quad \textrm{binomial}\\&\textrm{c}.\quad Chi-square\\ &\textrm{d}.\quad \textrm{Poison}\\ &\textrm{e}.\quad \color{red}\textrm{normal}\\\\ &\textrm{Jawab}:\\ &\begin{aligned} &\color{red}\textrm{Cukup jelas} \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Perhatikan ilustrasi berikut}\:   ....\\    \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Jika kurva normal di atas memiliki simpangan}\\ &\textrm{bakunya}\: \: 12,\: \textrm{maka luar daerah arsiran adalah}\: ....\\ &\textrm{a}.\quad 0,4821\\ &\textrm{b}.\quad 0,4966\\&\textrm{c}.\quad 0,4999\\ &\textrm{d}.\quad \color{red}0,5934\\ &\textrm{e}.\quad 0,6921\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui},\: \mu =60,\sigma =12,\textrm{x}_{1}=50,\textrm{x}_{2}=70\\ &\textrm{dengan nilai}\: \: z=\displaystyle \frac{\textrm{x}-\mu }{\sigma },\: \:\textrm{ maka}\\ &\begin{aligned}\textrm{z}_{1}&=\displaystyle \frac{\textrm{x}_{1}-\mu }{\sigma }=\frac{50-60}{12}=\frac{-10}{12}=-0.83\\ \textrm{z}_{2}&=\displaystyle \frac{\textrm{x}_{2}-\mu }{\sigma }=\frac{70-60}{12}=\frac{10}{12}=0.83 \end{aligned}\\ &\textrm{Luas}=P(-z\leq Z\leq z)=2\times P(0\leq Z\leq z)\\ &\qquad\, \:  =2\times (0,2967)=\color{red}0,5934   \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Perhatikan ilustrasi berikut}\:   ....\\    \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Jika kurva normal di atas memiliki simpangan}\\ &\textrm{bakunya}\: \: 5,\: \textrm{maka luar daerah arsiran adalah}\: ....\\ &\textrm{a}.\quad 0,1585\\ &\textrm{b}.\quad 0,1815\\&\textrm{c}.\quad 0,3413\\ &\textrm{d}.\quad 0,8172\\ &\textrm{e}.\quad \color{red}0,8185\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui},\: \mu =40,\sigma =5,\textrm{x}_{1}=30,\textrm{x}_{2}=45\\ &\textrm{dengan nilai}\: \: z=\displaystyle \frac{\textrm{x}-\mu }{\sigma },\: \:\textrm{ maka}\\ &\begin{aligned}\textrm{z}_{1}&=\displaystyle \frac{\textrm{x}_{1}-\mu }{\sigma }=\frac{30-40}{5}=\frac{-10}{5}=-2\\ \textrm{z}_{2}&=\displaystyle \frac{\textrm{x}_{2}-\mu }{\sigma }=\frac{45-40}{5}=\frac{5}{5}=1 \end{aligned}\\ &\textrm{Luas}=P(z_{1}\leq Z\leq z_{2})= P(0\leq Z\leq z_{1})\\ &\qquad\: \: +P(0\leq Z\leq z_{2})\\ &\qquad\, \:  =P(0\leq Z\leq 2)+P(0\leq Z\leq 1)\\ &\qquad\, \:  =0,3413+0,4772=\color{red}0,8185   \end{array}$.