PAS MATEMATIKA BLOG

LINGKARAN-SUDUT PUSAT DAN SUDUT KELILING-LANJUTAN

SELAMAT MEMPERINGATI MAULID NABI MUHAMMAD SAW TAHUN 2026

CONTOH SOAL 5 LINGKARAN-BUSUR-JURING-TEMBERENG

$\begin{aligned}21.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 154\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 77\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 44\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 38,5\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=\displaystyle \frac{1}{2}\:\textrm{luas lingkaran}_{\text{besar}}-\textrm{lingkaran}_{\text{kecil}}\\ &=\displaystyle \frac{1}{2}(\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}_{\textrm{besar}})-(\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}_{\textrm{kecil}})\\ &=\displaystyle \frac{1}{2}.\displaystyle \frac{22}{4.7}.14^{\displaystyle 2}-\displaystyle \frac{1}{4}.\displaystyle \frac{22}{7}.7^{\displaystyle 2}\\ &=\displaystyle \frac{22}{28}.(7.14-49)\\&=\displaystyle \frac{22}{28}.49\\ &=38,5\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.

$\begin{aligned}22.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 57\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 62,8\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 107\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 114\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=\textrm{Luas tembereng doubel}\\ &=2(\textrm{luas juring - luas segitiga siku-siku})\\ &=2\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r^{2}-\displaystyle \frac{1}{2}.r^{\displaystyle 2} \right)\\ &=2\left( \displaystyle \frac{1}{4}\pi.10^{\displaystyle 2}-\displaystyle \frac{1}{2}.10^{\displaystyle 2} \right)\\ &=2(78,5-50)=57\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.

$\begin{aligned}23.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 157\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 80\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 40\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 39,25\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=(\textrm{luas juring besar - luas juring kecil})\\ &=\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r_{\text{besar}}^{2}-\displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r_{\text{kecil}}^{2} \right)\\ &=\left( \displaystyle \frac{1}{4}\pi.10^{\displaystyle 2}-\displaystyle \frac{1}{4}\pi.7^{\displaystyle 2} \right)\\ &=\displaystyle \frac{1}{4}(3,14).(100-49)\approx 40\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.

$\begin{aligned}24.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 112\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 70\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 42\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 28\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=(\textrm{luas tembereng besar - 2 luas tembereng kecil})\\ &=\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r_{\text{besar}}^{2}-\displaystyle \frac{1}{2}\times r_{\text{besar}}^{2} \right)\\ &\quad-2\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times \pi.r_{ \textrm{kecil}}^{\displaystyle 2}-\displaystyle \frac{1}{2}\times r_{\text{kecil}}^{2} \right)\\ &=\displaystyle \frac{1}{4}.\displaystyle \frac{22}{7}.14^{\displaystyle 2}-\displaystyle \frac{1}{2}.14^{\displaystyle 2}-2\left( \displaystyle \frac{1}{4}.\frac{22}{7}.7^{\displaystyle 2}-\displaystyle \frac{1}{2}.7^{\displaystyle 2} \right)\\ &=154-98-(77-49)=28\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.

$\begin{aligned}25.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Jika}\:\: AB=5\:\: cm,\:\: AC=12\:\: cm,\:\: \textrm{dan}\:\: AB,\: AC\\ &\textrm{serta}\:\: BC\:\: \textrm{masing-masing adalah diamter lingkaran},\\ &\textrm{maka luas daerah terarsir adalah}=\: .... \\ &\text{a}.\quad 27\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 30\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 36\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 37,5\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=(2\:\textrm{luas setengah lingkaran - luas lingkaran besar})\\&\quad + \textrm{luas sebuah segitiga}\\ &=\displaystyle \frac{1}{2}.\frac{1}{4}\pi d_{2}^{\displaystyle 2}+\displaystyle \frac{1}{2}.\frac{1}{4}\pi d_{3}^{\displaystyle 2}-\displaystyle \frac{1}{2}.\frac{1}{4}\pi d_{1}^{\displaystyle 2}+\displaystyle \frac{1}{2}.d_{2} d_{3}\\ &=\displaystyle \frac{1}{8}(3,14) (d_{2}^{\displaystyle 2}+ d_{3}^{\displaystyle 2}- d_{1}^{\displaystyle 2})+\displaystyle \frac{1}{2}.d_{2} d_{3}\\ &=\displaystyle \frac{1}{8}(3,14) (12^{\displaystyle 2}+ 5^{\displaystyle 2}- 13^{\displaystyle 2})+\displaystyle \frac{1}{2}.12. 5\\ &=30\:\: \textrm{cm}^{\displaystyle 2} \end{aligned}\end{aligned}$.

CONTOH SOAL 4 LINGKARAN-BUSUR-JURING-TEMBERENG

$ \begin{aligned}16.\quad&\textrm{Diberikan pernyataan-pernyataan berikut}\\ &(i)\quad \pi(2r)\qquad\qquad\qquad (iii)\quad \displaystyle \frac{1}{2}\pi d\\ &(ii)\quad \pi.r^{\displaystyle 2}\qquad\qquad\qquad (iv)\quad \displaystyle \frac{1}{4}\pi d^{\displaystyle 2}\\&\textrm{jika r jari-jari lingkaran dan d adalah diameternya}\\&\textrm{Pernyataan di atas yang merupakan formula luas }\\ &\textrm{lingkaran adalah}\:....\\ &\text{a}.\quad (i)\quad \textrm{dan}\quad (ii)\\ &\text{b}.\quad (i)\quad \textrm{dan}\quad (iv)\\ &\text{c}.\quad (ii)\quad \textrm{dan}\quad (iii)\\ &\text{d}.\quad (ii)\quad \textrm{dan}\quad (iv)\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Cukup jelas}\end{aligned}$.

$ \begin{aligned}17.\quad&\textrm{Luas lingkaran yang berjari-jari}\quad 7\:\: \textrm{cm adalah}=\:....\\ &\text{a}.\quad 44\quad \textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 88\quad \textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 154\quad \textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 308\quad \textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &L=\pi r^{\displaystyle 2}=\displaystyle \frac{22}{7}.7^{\displaystyle 2}=154\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}$.

$ \begin{aligned}18.\quad&\textrm{Luas lingkaran yang bediameter}\quad 20\:\: \textrm{cm adalah}=\:....\\ &\text{a}.\quad 12,56\quad \textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 314\quad \textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 62,8\quad \textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 31,4\quad \textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &L=\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}=\displaystyle \frac{1}{4}.(3,14).20^{\displaystyle 2}=314\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}$.

$ \begin{aligned}19.\quad&\textrm{Jari-jari lingkaran dengan luasnya}\quad 36\pi\:\: cm^{\displaystyle 2}\:\textrm{adalah}=\:....\\ &\text{a}.\quad 6\quad \textrm{cm}\\ &\text{b}.\quad 9\quad \textrm{cm}\\ &\text{c}.\quad 12\quad \textrm{cm}\\ &\text{d}.\quad 6\pi\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &L=\pi r^{\displaystyle 2}\Leftrightarrow 36\pi=\pi r^{\displaystyle 2}\Leftrightarrow r=\sqrt{36}=6\:\: \textrm{cm}\end{aligned}$.

$ \begin{aligned}20.\quad&\textrm{Dua lingkaran dengan jari-jari masing-masing}\\ &8\:\:\textrm{cm dan 10 cm. Perbandingan luas untuk}\\ &\textrm{kedua lingkaran tersebut adalah}=\:....\\ &\text{a}.\quad 4:5\\ &\text{b}.\quad 8:25\\ &\text{c}.\quad 16:25\\ &\text{d}.\quad 16:125\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\displaystyle \frac{L_{\displaystyle 8}}{L_{\displaystyle 10}}=\displaystyle \frac{\pi.8^{\displaystyle 2}}{\pi.10^{\displaystyle 2}}=\frac{16}{25}\end{aligned}$.

CONTOH SOAL 3 LINGKARAN-BUSUR-JURING-TEMBERENG

$\begin{aligned}11.\quad&\textrm{Sebuah lintasan lari berbentuk lingkaran yang}\\ &\textrm{berdiameter 56 meter. Banyak putaran yang harus}\\ &\textrm{dilakukan pelari jika ia ingin menempuh jarak}\\ &\textrm{902 meter adalah}\:....\\ &\text{a}.\quad 5\displaystyle \frac{3}{4}\quad \textrm{putaran}\\ &\text{b}.\quad 5\displaystyle \frac{1}{2}\quad \textrm{putaran}\\ &\text{c}.\quad 5\displaystyle \frac{1}{4}\quad \textrm{putaran}\\ &\text{d}.\quad 5\displaystyle \frac{1}{8}\quad \textrm{putaran}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: K=\pi d\:\: \textrm{meter} \\ &\textrm{maka, K}=\displaystyle \frac{22}{7}.56\:\: \textrm{meter}=176\:\: \textrm{meter}\\ &\textrm{Sehingga banyak putaran yang perlu dilakukan adalah}\\ &=\displaystyle \frac{902}{176}=5\displaystyle \frac{1}{8}\:\: \textrm{putaran}\end{aligned}$.

$\begin{aligned}12.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Berdasarkan gambar di atas, keliling daerah arsiran}=\:....\\ &\text{a}.\quad 44\:\:\textrm{cm}\\ &\text{b}.\quad 66\:\:\textrm{cm}\\ &\text{c}.\quad 88\:\:\textrm{cm}\\ &\text{d}.\quad 132\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}K&=\displaystyle \frac{1}{2}\textrm{lingkaran besar}+\textrm{1 lingkaran kecil penuh}\\ &=\displaystyle \frac{1}{2}\pi d_{\textrm{besar}}+\pi d_{\textrm{kecil}}\\ &=\displaystyle \frac{1}{2}\frac{22}{7}.28+\displaystyle \frac{22}{7}.14\\ &=44+44\\ &=88\:\: \textrm{cm}\end{aligned}\end{aligned}$.

$\begin{aligned}13.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Berdasarkan gambar di atas, keliling daerah arsiran}=\:....\\ &\text{a}.\quad 31,4\:\:\textrm{cm}\\ &\text{b}.\quad 62,8\:\:\textrm{cm}\\ &\text{c}.\quad 94,2\:\:\textrm{cm}\\ &\text{d}.\quad 125,6\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}K&=2\:\textrm{lingkaran besar}\\ &=2\pi d\\ &=2(3,14).20\\ &=125,6\:\: \textrm{cm}\end{aligned}\end{aligned}$.

$\begin{aligned}14.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Berdasarkan gambar di atas, besar keliling daerah}\\ &\textrm{jika diketahui jari-jari lingkarannyanya 10 cm }\\ &\textrm{adalah}\: .... \\ &\text{a}.\quad 82,8\:\:\textrm{cm}\\ &\text{b}.\quad 67,1\:\:\textrm{cm}\\ &\text{c}.\quad 61,1\:\:\textrm{cm}\\ &\text{d}.\quad 47,1\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}K&=\displaystyle \frac{3}{4}\:\textrm{lingkaran}+2r\\ &=\displaystyle \frac{3}{4}(2\pi r)+2r\\ &=\displaystyle \frac{3}{4}.2.(3,14).10+20\\ &=47,1+20\\ &=67,1\:\: \textrm{cm}\end{aligned}\end{aligned}$.

$\begin{aligned}15.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Berdasarkan gambar di atas, keliling daerah arsiran}=\: .... \\ &\text{a}.\quad 55\:\:\textrm{cm}\\ &\text{b}.\quad 62\:\:\textrm{cm}\\ &\text{c}.\quad 69\:\:\textrm{cm}\\ &\text{d}.\quad 83\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}K&=\displaystyle \frac{3}{4}\:\textrm{lingkaran}_{\text{kecil}}+\displaystyle \frac{1}{4}\:\textrm{lingkaran}_{\text{besar}}+7+7\\ &=\displaystyle \frac{3}{4}(2\pi r_{\textrm{kecil}})+\displaystyle \frac{1}{4}(2\pi r_{\textrm{besar}})+14\\ &=\displaystyle \frac{3}{4}.2.\displaystyle \frac{22}{7}.7+\displaystyle \frac{1}{4}.2.\displaystyle \frac{22}{7}.14\\ &=\displaystyle \frac{5}{4}.2.\displaystyle \frac{22}{7}.7\\ &=55\:\: \textrm{cm}\end{aligned}\end{aligned}$.

CONTOH SOAL 2 LINGKARAN-BUSUR-JURING-TEMBERENG

$\begin{aligned}6.\quad&\textrm{Keliling lingkaran yang berdiameter}\:\: d\:\: \textrm{adalah}\:....\\ &\text{a}.\quad 2\pi d\quad \textrm{satuan panjang}\\ &\text{b}.\quad \pi d\quad \textrm{satuan panjang}\\ &\text{c}.\quad \displaystyle \frac{1}{2}\pi d\quad \textrm{satuan panjang}\\ &\text{d}.\quad \displaystyle \frac{1}{4}\pi d\quad \textrm{satuan panjang}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Cukup jelas}\end{aligned}$.

$\begin{aligned}7.\quad&\textrm{Keliling lingkaran yang berdiameter}\:\: 10,5\:\: \textrm{cm}=\:....\\ &\text{a}.\quad 66\quad \textrm{cm}\\ &\text{b}.\quad 44\quad \textrm{cm}\\ &\text{c}.\quad 33\quad \textrm{cm}\\ &\text{d}.\quad 22\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: d=10,5\:\: \textrm{cm} \\ &\textrm{maka, K}=\pi d=\displaystyle \frac{22}{7}(10,5\quad \textrm{cm})=\displaystyle \frac{22}{7}.\frac{21}{2}=33\:\: \textrm{cm}\end{aligned}$.

$\begin{aligned}8.\quad&\textrm{Keliling lingkaran yang berjari-jari}\:\: 14\:\: \textrm{cm}=\:....\\ &\text{a}.\quad 88\quad \textrm{cm}\\ &\text{b}.\quad 132\quad \textrm{cm}\\ &\text{c}.\quad 154\quad \textrm{cm}\\ &\text{d}.\quad 308\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: r=14\:\: \textrm{cm} \\ &\textrm{maka, K}=2\pi r=2.\displaystyle \frac{22}{7}(14\quad \textrm{cm})=88\:\: \textrm{cm}\end{aligned}$.

$\begin{aligned}9.\quad&\textrm{Jari-jari lingkaran yang mempunyai keliling}\\ &132\:\: \textrm{cm adalah}\:....\\ &\text{a}.\quad 7\quad \textrm{cm}\\ &\text{b}.\quad 10,5\quad \textrm{cm}\\ &\text{c}.\quad 14\quad \textrm{cm}\\ &\text{d}.\quad 21\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: K=132\:\: \textrm{cm} \\ &\textrm{maka, r}=\displaystyle \frac{K}{2\pi}=\displaystyle \frac{132}{2.\displaystyle \frac{22}{7}}=21\:\: \textrm{cm}\end{aligned}$.

$\begin{aligned}10.\quad&\textrm{Diameter lingkaran yang mempunyai keliling}\\ &154\:\: \textrm{cm adalah}\:....\\ &\text{a}.\quad 14\quad \textrm{cm}\\ &\text{b}.\quad 21\quad \textrm{cm}\\ &\text{c}.\quad 24,5\quad \textrm{cm}\\ &\text{d}.\quad 49\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: K=154\:\: \textrm{cm} \\ &\textrm{maka, d}=\displaystyle \frac{K}{\pi}=\displaystyle \frac{154}{\displaystyle \frac{22}{7}}=49\:\: \textrm{cm}\end{aligned}$.

CONTOH SOAL 1 LINGKARAN-BUSUR-JURING-TEMBERENG

$\begin{aligned}1.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Daerah yang diarsir pada gambar di atas adalah}\:....\\ &\text{a}.\quad \textrm{juring}\\ &\text{b}.\quad \textrm{tembereng}\\ &\text{c}.\quad \textrm{apotema}\\ &\text{d}.\quad \textrm{tali busur}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Cukup jelas}\end{aligned}$.

Gunakan gambar berikut untuk menjawab soal nomor 2 sampai dengan nomor 5

$\begin{aligned}2.\quad&\textrm{Garis OD disebut}\:....\\ &\text{a}.\quad \textrm{apotema}\\ &\text{b}.\quad \textrm{tali busur}\\ &\text{c}.\quad \textrm{busur}\\ &\text{d}.\quad \textrm{jari-jari}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\textrm{Cukup jelas}\end{aligned}$.

$\begin{aligned}3.\quad&\textrm{Daerah yang diarsir adalah}\:....\\ &\text{a}.\quad \textrm{busur}\\ &\text{b}.\quad \textrm{tali busur}\\ &\text{c}.\quad \textrm{juring}\\ &\text{d}.\quad \textrm{tembereng}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas}\end{aligned}$.

$\begin{aligned}4.\quad&\textrm{Tali busur ditunjukkan oleh}\:....\\ &\text{a}.\quad AC\\ &\text{b}.\quad OA\\ &\text{c}.\quad OD\\ &\text{d}.\quad AB\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Cukup jelas}\end{aligned}$.

$\begin{aligned}5.\quad&\textrm{Yang merupakan diameter adalah}\:....\\ &\text{a}.\quad AB\\ &\text{b}.\quad AC\\ &\text{c}.\quad BC\\ &\text{d}.\quad OB\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas}\end{aligned}$.

LINGKARAN-BUSUR, TALI BUSUR, JURING, DAN TEMBERENG-LANJUTAN

C. Keliling Lingkaran

Perhatikan kembali gambar lingkaran berikut

Keliling lingkaran dapat dirumuskan sebagai berikut:

$\begin{aligned}&\bullet \: K=\pi d=2\pi r\\ &\textrm{Keterangan}:\\ &\qquad K:\textrm{keliling lingkarang}\\ &\qquad d:\textrm{diameter lingkarang}\\ &\qquad r:\textrm{jar-jari lingkarang}\\ &\qquad \pi:\textrm{pi}=\displaystyle \frac{22}{7}=3,14\\ \end{aligned}$.

D. Panjang Busur Lingkaran

$\begin{aligned}&\bullet \:\displaystyle \frac{\textrm{Besar sudut pusat}}{\textrm{Besar sudut satu putaran}}=\frac{\textrm{Panjang busur}}{\textrm{Keliling lingkaran}}\\ &\bullet \:\textrm{Panjang busur}=\displaystyle \frac{\textrm{Besar sudut pusat}}{360^{\displaystyle 0}}\times 2\pi r\end{aligned}$.

E. Luas Lingkaran

$\begin{aligned}&\bullet \:L=\pi r^{\displaystyle 2}=\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}\\ &\textrm{Keterangan}:\\ &\qquad L:\textrm{luas lingkarang}\\ &\qquad d:\textrm{diameter lingkarang}\\ &\qquad r:\textrm{jar-jari lingkarang}\\ &\qquad \pi:\textrm{pi}=\displaystyle \frac{22}{7}=3,14\\\end{aligned}$.

F. Luas Juring Lingkaran

$\begin{aligned}&\bullet \:\displaystyle \frac{\textrm{Besar sudut pusat}}{\textrm{Besar sudut satu putaran}}=\frac{\textrm{Luas juring}}{\textrm{Luas lingkaran}}\\ &\bullet \:\textrm{Luas juring}=\displaystyle \frac{\textrm{Besar sudut pusat}}{360^{\displaystyle 0}}\times \pi r^{\displaystyle 2}\end{aligned}$.

$\LARGE\fbox{CONTOH SOAL}$.

$\begin{aligned}1.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Tentukan panjang busur}\:\: \widehat{PQ}\\\\ &\textrm{Jawab}:\\ &\textrm{Panjang busur}\:\: \widehat{PQ}\\ &=\displaystyle \frac{\angle POQ}{360^{\displaystyle 0}}\text{Keliling lingkaran}_{r=OP}\\ &=\displaystyle \frac{144^{\displaystyle 0}}{360^{0}}.2\pi.r\\&=\displaystyle \frac{144^{\displaystyle 0}}{360^{0}}.2\pi.OP\\ &=\displaystyle \frac{2}{5}.2.\displaystyle \frac{22}{7}.7\\ &=\displaystyle \frac{2}{5}.44\\ &=\frac{88}{5}\\ &=17\displaystyle \frac{3}{5}\:\: cm\end{aligned}$.

$\begin{aligned}2.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Tentukan luas daerah yang diarsir}\\ &\textrm{jika}\quad OR=2\: cm\quad \textrm{dan}\:\:\quad PR=1\: cm\\\\ &\textrm{Jawab}:\\ &\textrm{Luas daerah arsiran}\\ &=\textrm{Luas juring POQ}-\textrm{Luas juring ROS}\\ &=\displaystyle \frac{\angle POQ}{360^{\displaystyle 0}}\text{Luas lingkaran}_{r=OP}-\displaystyle \frac{\angle ROS}{360^{\displaystyle 0}}\text{Luas lingkaran}_{r=OR}\\ &=\displaystyle \frac{72^{\displaystyle 0}}{360^{0}}.\pi.OP^{\displaystyle 2}-\displaystyle \frac{72^{\displaystyle 0}}{360^{0}}.\pi.OR^{\displaystyle 2}\\ &=\displaystyle \frac{1}{5}\pi.3^{\displaystyle 2}-\frac{1}{5}.\pi.2^{\displaystyle 2}\\ &=\displaystyle \frac{1}{5}\pi.9-\frac{1}{5}\pi.4\\ &=\frac{1}{5}\pi.(9-4)\\ &=\frac{1}{5}\pi.5\\ &=\pi\:\: cm^{\displaystyle 2}\qquad \textrm{atau}\\ &=\displaystyle \frac{22}{7}\:\: cm^{\displaystyle 2}\end{aligned}$.

$\begin{aligned}3.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Tentukan luas daerah yang diarsir (tembereng)}\\\\ &\textrm{Jawab}:\\ &\textrm{Luas daerah arsiran}\\ &=\textrm{Luas juring POQ}-\textrm{Luas segitiga POQ}\\ &=\displaystyle \frac{\angle POQ}{360^{\displaystyle 0}}\text{Luas lingkaran}_{r=OP}-\displaystyle \frac{1}{2}.\textrm{alas}\times \textrm{tinggi}\\ &=\displaystyle \frac{90^{\displaystyle 0}}{360^{0}}.\pi.OP^{\displaystyle 2}-\displaystyle \frac{1}{2}.r^{\displaystyle 2}\\ &=\displaystyle \frac{1}{4}(3,14).10^{\displaystyle 2}-\frac{1}{2}.10^{\displaystyle 2}\\ &=\frac{(3,14-2)}{4}.100\\ &=1,14\times 25\\ &=28,5\\ &=\displaystyle \frac{22}{7}\:\: cm^{\displaystyle 2}\end{aligned}$.

$\begin{aligned}4.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Tentukan luas daerah yang diarsir (4 tembereng)}\\\\ &\textrm{Jawab}:\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: d=21\: cm\\&\textrm{Luas daerah arsiran (4 tembereng)}\\ &Alternatif\:1\\&=\textrm{Luas lingkaran}-\textrm{Luas persegi ABCD}\\ &=\displaystyle \frac{1}{4}\pi.d^{\displaystyle 2}-\displaystyle \frac{1}{2}d^{\displaystyle 2}=\displaystyle \frac{1}{4}.\frac{22}{7}.21^{\displaystyle 2}-\displaystyle \frac{1}{2}21^{\displaystyle 2}\\&=\displaystyle \frac{1}{2}.11.63-\frac{1}{2}.441\\&=126\:\: cm^{\displaystyle 2}\\&Alternatif\:2\\&\textrm{diserahkan ke pembaca yang budiman}\end{aligned}$.

$\begin{aligned}5.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Jika garis AB adalah diameter dan luas daerah}\\ &\textrm{arsiran}\:\: 22,5\:\: cm^{\displaystyle 2}\:\: \textrm{dan luas juring BOC}\:\: 10\:\: cm^{\displaystyle 2}\\ &\textrm{maka besar sudut}\:\: \angle \,\textrm{BOC}\:\: \textrm{adalah}\:....\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\textrm{Besar sudut pusat}}{\textrm{Besar sudut satu putaran}}=\frac{\textrm{Luas juring}}{\textrm{Luas lingkaran}}\end{aligned}\\ &\Leftrightarrow \displaystyle \frac{\angle BOC}{180^{\displaystyle 0}}=\frac{10\:\: cm^{\displaystyle 2}}{22,5\:\:cm^{\displaystyle 2}}\\ &\Leftrightarrow \angle BOC=\frac{10\:\: cm^{\displaystyle 2}}{22,5\:\:cm^{\displaystyle 2}}\times 180^{\displaystyle 0}\\ &\Leftrightarrow \angle BOC=80^{\displaystyle 0}\end{aligned}$.

DAFTAR PUSTAKA

Kurniawan. 2008. Mandiri Matematika Mengasah Kemampuan Diri SMP Kelas VIII Jilid 2 KTSP 2006. Jakarta: ERLANGGA.
Santoso, N.E., Sksin, N. 2024. Matematika untuk SMA/MA/SMK/MAK Kelas XII Kurikulum Merdeka. Yogyakarta: INTAN PARIWARA EDUKASI.

LINGKARAN-BUSUR, TALI BUSUR, JURING, DAN TEMBERENG

A. Pendahuluan

Lingkaran adalah bangun datar yang dibatasi oleh sekumpulan titik-titik pada bidang yang semuanya meiliki jarak yang sama dari satu titik tetap. Titik tetap tersebut disebut pusat lingkaran, sedangkan jarak dari pusat ke setiap titik pada lingkaran disebut jari-jari.

B. Beberapa Unsur Penting Lingkaran

Pusat: titik tengah lingkaran
Jari-jari (r): jarak dari pusat ke tepi lingkaran
Diameter (d): garis yang melalui pusat dan menghubungkan dua titik pada lingkaran. Diameter = 2 x jari-jari.
Busur: bagian lengkung pada keliling lingkaran
Tali busur: garis lurus yang menghubungkan dua titik pada lingkaran
Juring: daerah yang dibatasi oleh dua jari-jari dan sebuah busur
Temberang: daerah yang dibatasi oleh tali busur dan busur
Apotema: jarak tegak lurus dari pusat lingkaran ke sebuah tali busur
Sudut pusat: sudut yang titik sudutnya berada di pusat lingkaran
Sudut keliling: sudut yang titik sudutnya berada pada lingkaran
Garis singgung: garis yang menyentuh lingkaran tepat di satu titik
Titik-titik singgung: Titik tempat garis singgung menyentuh lingkaran
Garis sekan: garis yang memotong lingkaran di dua titik
Sudut antara tali busur: sudut yang terbentuk oleh dua tali busur
Busur setengah lingkaran: busur berukuran $180^{0}$, dibatasi oleh diameter.
Busur minor: busur yang ukurannya kurang dari $180^{0}$
Busur mayor: busur yang ukurannya lebih dari $180^{0}$.

$\LARGE\fbox{CONTOH SOAL}$.

Perhatikan gambar berikut

Yang tampak pada gambar di atas adalah:

Pusat: O
Jari-jari (r): OA, OB, OC, OD
Diameter (d): AC, BD
Busur: $\widehat{AF},\widehat{AB}, \widehat{AC},\widehat{AD}$, dan lain-lain
Tali busur: $\overline{\textrm{AC}},\overline{\textrm{BD}},\overline{\textrm{CF}},\overline{\textrm{CD}},\overline{\textrm{DF}}$
Juring: juring AOB, BOC, COD, AOD, AOC, dan lain-lain
Temberang: tembereng CD, tembereng CF, dan tembereng DF
Apotema: OE
Sudut pusat: $\angle \textrm{AOB}, \angle \textrm{BOC}, \angle \textrm{COD}, \angle \textrm{AOD},\angle \textrm{AOC}$, dan lain-lain
Sudut keliling: $\angle \textrm{FCD}, \angle \textrm{CDF}, \angle \textrm{CFD}$.
Garis singgung: tak tampak pada gambar
Titik-titik singgung: tak tampak pada gambar
Garis sekan: tak tampak pada gambar dan bedakan antara tali busur dengan garis sekan. Misalkan CD adalah tali busur, tetapi jika garis CD diperpanjang dapat menjadi garis sekan
Sudut antara tali busur: $\angle \textrm{AGF},\angle \textrm{CGD}$. Bedakan dengan sudut keliling. Jika sudut keliling melibatkan dua tai busur dan titik sudutnya pada lingkaran, tetapi untuk sudut antara tali busur adalah yang ada dalam lingkaran.
Busur setengah lingkaran: $\widehat{\textrm{AC}},\widehat{\textrm{BD}}$.
Busur minor: $\widehat{AF},\widehat{AB}, \widehat{AC},\widehat{AD}$
Busur mayor: $\widehat{ACF},\widehat{ACB},\widehat{ACD}$, dan lain-lain

CONTOH SOAL 10 MATRIKS DAN SPL

$\begin{array}{l}\\ 16.&\textrm{Jika diketahui}\quad x_{0}\quad \textrm{dan}\quad y_{0}\quad \textrm{adalah selesaian}\\ &\textrm{dari sistem persamaan}\quad \begin{cases} 2x+3y & =60 \\ 3x+2y & =60 \end{cases}\\ &\textrm{dengan}\quad x_{0}=\displaystyle \frac{a}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}\quad \textrm{dan}\quad y_{0}=\displaystyle \frac{b}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}\:, \: \textrm{maka}\\ &\textrm{nilai}\quad a+b\quad\textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&-240\\ \textrm{b}.&-180\\ \textrm{c}.&-120\\ \textrm{d}.&-60\\ \textrm{e}.&0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Dengan aturan Cramer didapatkan}\\ &x_{0}=\displaystyle \frac{a}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}=\displaystyle \frac{\begin{vmatrix} 60 & 3 \\ 60 & 2 \end{vmatrix}}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}=\displaystyle \frac{120-180}{4-9}=\displaystyle \frac{-60}{-5}\\ &y_{0}=\displaystyle \frac{b}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}=\displaystyle \frac{\begin{vmatrix} 2 & 60 \\ 3 & 60 \end{vmatrix}}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}=\displaystyle \frac{120-180}{4-9}=\displaystyle \frac{-60}{-5}\\ &\textrm{Jadi, nilai}\quad a+b=(-60)+(-60)=-120 \end{array}$.

$\begin{array}{l}\\ 17.&\textrm{Sebuah garis}\quad k\quad \textrm{dinayatkan dalam bentuk}\\ &\begin{vmatrix} 1 & x & y\\ a & 1 & 1\\ 1 & 2 & 3 \end{vmatrix}=0.\quad \textrm{Nilai}\quad a\quad \textrm{saat garis tersebut}\\ &\textrm{melalui titik}\quad (1,1)\quad \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&5\\ \textrm{b}.&4\\ \textrm{c}.&3\\ \textrm{d}.&2\\ \textrm{e}.&1 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\textrm{Dengan substitusi langsung ke garis, maka}\\ &\begin{vmatrix} 1 & x & y\\ a & 1 & 1\\ 1 & 2 & 3 \end{vmatrix}=\begin{vmatrix} 1 & 1 & 1\\ a & 1 & 1\\ 1 & 2 & 3 \end{vmatrix}=3+1+2a-1-3a-1=0\\ &\Leftrightarrow 4+2a-3a-2=0\Leftrightarrow 2-a=0\Leftrightarrow a=2 \end{array}$.

$\begin{array}{l}\\ 18.&\textrm{Jika garis yang dinyatakan dengan bentuk}\\ &\begin{vmatrix} x & y & 1\\ 2 & 1 & 1\\ 4 & 5 & 1 \end{vmatrix}=0\quad \textrm{sejajar dengan garis yang }\\ &\textrm{dinyatakan pula dalam bentuk}\quad \begin{vmatrix} x & y & 1\\ 4 & a & 1\\ 3 & 6 & 1 \end{vmatrix}=0\\ &\textrm{maka nilai}\quad a\quad \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&5\\ \textrm{d}.&6\\ \textrm{e}.&8 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Perhatikan bahwa untuk kedua garis tersebut adalah}\\ &\bullet \quad\begin{vmatrix} x & y & 1\\ 2 & 1 & 1\\ 4 & 5 & 1 \end{vmatrix}=0\Leftrightarrow x+4y+10-4-2y-5x=0\\ &\Leftrightarrow -4x+2y+6=0\Leftrightarrow y=2x-3,\quad \textrm{maka},\:m_{1}=2\\ &\bullet \quad \begin{vmatrix} x & y & 1\\ 4 & a & 1\\ 3 & 6 & 1 \end{vmatrix}=0\Leftrightarrow ax+3y+24-3a-6x-4y=0\\ &\Leftrightarrow (a-6)x-y+24-3a=0\Leftrightarrow y=(a-6)x+24-3a\\ &\textrm{maka}\: m_{2}=a-6\\ &\textrm{Syarat dua garis lurus sejajar adalah}:m_{1}=m_{2}\\ &\Leftrightarrow 2=a-6 \Leftrightarrow a=8 \end{array}$.

$\begin{array}{l}\\ 19.&\textrm{Diketahui suatu sistem persamaan berupa}\\ &\begin{cases} x+y+2z & =4 \\ 2x-y-2z & =-1\\ 3x-2y-z&=3 \end{cases}\quad \textrm{memiliki selesaian}\\ &(x_{0},y_{0},z_{0}).\quad \textrm{Jika}\quad x_{0}=\displaystyle \frac{a}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}\\&\textrm{dan}\quad z_{0}=\displaystyle \frac{b}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}\\ &\textrm{maka nilai}\quad a+b\quad \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&27\\ \textrm{b}.&25\\ \textrm{c}.&0\\ \textrm{d}.&-25\\ \textrm{e}.&-27 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Dengan aturan Cramer kita akan mendapatkan}\\ &x_{0}=\displaystyle \frac{a}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}=\displaystyle \frac{\begin{vmatrix} 4 & 1 & 2\\ -1 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}\\ &\quad=\displaystyle \frac{4+(-6)+4-(-6)-16-1}{1+(-6)+(-8)-(-6)-4-(-2)}\\ &\quad =\displaystyle \frac{-9}{-9}\\ &z_{0}=\displaystyle \frac{b}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}=\displaystyle \frac{\begin{vmatrix} 1 & 1 & 4\\ 2 & -1 & -1\\ 3 & -2 & 3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}\\ &\quad=\displaystyle \frac{(-3)+(-3)+(-16)-(-12)-2-6}{1+(-6)+(-8)-(-6)-4-(-2)}\\ &\quad =\displaystyle \frac{-18}{-9}\\ &\textrm{maka nilai}\quad a+b=(-9)+(-18)=-27 \end{array}$.

$\begin{array}{l}\\ 20.&\textrm{Diketahui matriks transformasi berikut}\\ &\begin{pmatrix}\displaystyle \frac{1}{2}\sqrt{2} & \displaystyle \frac{1}{2}\sqrt{2}\\-\displaystyle \frac{1}{2}\sqrt{2} & \displaystyle \frac{1}{2}\sqrt{2}\end{pmatrix}\\ &\textrm{Peta berupa}\quad (x',y')\quad \textrm{merupakan}\:...\textrm{terhadap}\\&\textrm{titik}\quad (x,y)\\&\begin{array}{llllllll}\\ \textrm{a}.&\textrm{rotasi sejauh}\quad 30^{0}\\ \textrm{b}.&\textrm{rotasi sejauh}\quad 45^{0}\\ \textrm{c}.&\textrm{rotasi sejauh}\quad -45^{0}\\ \textrm{d}.&\textrm{rotasi sejauh}\quad 60^{0}\\ \textrm{e}.&\textrm{rotasi sejauh}\quad -60^{0} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Ingat matriks rotasi dengan bentuk}\\ &\begin{pmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{pmatrix}\\ &\textrm{dengan memilih}\quad \alpha=-45^{0},\quad \textrm{maka akan sama}\\ &\textrm{dengan yang diketahui pada soal di atas} \end{array}$.

CONTOH SOAL 9 MATRIKS DAN SPL

$\begin{array}{l}\\ 11.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 4x+3y-11z & =-216 \\ 2x+5y+8z & =63\\ x+y+z & =0 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-6,\quad y=-9\quad \textrm{dan}\quad z=15\\ \textrm{b}.&x=-6,\quad y=9\quad \textrm{dan}\quad z=15\\ \textrm{c}.&x=6,\quad y=-9\quad \textrm{dan}\quad z=15\\ \textrm{d}.&x=-6,\quad y=-9\quad \textrm{dan}\quad z=-15\\ \textrm{e}.&x=-6,\quad y=9\quad \textrm{dan}\quad z=-15 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{ccc|c} 4&3&-11&-216\\ 2&5&8&63\\ 1&1&1&0 \end{array} \right]\quad\begin{matrix}\bullet R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-R_{\displaystyle 1}\\ \bullet R_{\displaystyle 3}\longleftarrow 4R_{\displaystyle 3}-R_{\displaystyle 1}\end{matrix}\\ &=2(2,5,8|63)-(4,3,-11|-216)=(0,7,27|342)\:\:\textrm{dan}\\ &=4(1,1,1|0)-(4,3,-11|-216)=(0,1,15|216)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 4&3&-11&-216\\ 0&7&27&342\\ 0&1&15&216 \end{array} \right]\quad \bullet R_{\displaystyle 3}\longleftarrow 7R_{\displaystyle 3}-R_{\displaystyle 2}\\ &=7(0,1,15|216)-(0,7,27|342)=(0,0,78|1170)\\ &\textrm{sehingga}\quad z=\displaystyle \frac{1170}{78}=15,\quad \textrm{maka}\quad y=-9,\:\: x=-6 \end{array}$

$\begin{array}{l}\\ 12.&\textrm{Bayu membeli sebuah buku tulis dan dua buah pensil}\\ &\textrm{Rp}10.000,00.\:\: \textrm{Di toko yang sama, Giri membeli dua }\\ &\textrm{buku tulis dan tiga buah pensil seharga Rp17.500,00}\\ &\textrm{Bentuk perkalian matriks dari narasi di atas adalah}....\\ &\begin{array}{llllllll} \textrm{a}.&\begin{pmatrix} 2 & 1\\ 2 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 10.000\\ 17.500 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 1 & 2\\ 3 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 10.000\\ 17.500 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 1 & 2\\ 2 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 17.500\\ 10.000 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} 1 & 2\\ 2 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 10.000\\ 17.500 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 2 & 1\\ 3 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 17.500\\ 10.000 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &x=\textrm{buku},\:\: y=\textrm{pensil}\\ & \end{array}$.

$\begin{array}{l}\\ 13.&\textrm{Penyelesaian dari sistem persamaan berikut}\\ &\begin{cases} 4x+3y+9& =0 \\ -x+y & =4 \end{cases}\\ &\textrm{Model matriks yang menyatakan bentuk di atas}....\\ &\begin{array}{llllllll} \textrm{a}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ -4 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & 3\\ -1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} x\\ y \end{pmatrix}=-\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} 9\\ 4 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{cases} 4x+3y & =-9 \\-x+y & =4 \end{cases}\Rightarrow \begin{pmatrix} 4 & 3\\ -1 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ &\Leftrightarrow \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{\begin{vmatrix}4 & 3\\ -1 & 1\end{vmatrix}}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ &\Leftrightarrow \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix} \end{array}$.

$\begin{array}{l}\\ 14.&\textrm{Diberikan SPL berikut}\\ &\begin{cases} 4x+3y & =12 \\ 3x+2y & =7 \end{cases}\\ &\textrm{Nilai}\quad x+y\quad \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&-5\\ \textrm{b}.&5\\ \textrm{c}.&11\\ \textrm{d}.&-11\\ \textrm{e}.&10 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 4&3&12\\ 3&2&7 \end{array} \right]\quad\bullet R_{\displaystyle 1}\longleftarrow R_{\displaystyle 1}-R_{\displaystyle 2}\\ &=(4,3|12)-(3,2|7)=(1,1|5)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 1&1&5\\ 3&2&7 \end{array} \right]\\ &\textrm{sehingga}\quad x+y=5 \end{array}$.

$\begin{array}{l}\\ 15.&\textrm{Perpotongan dua garis yang tersaji sebagai}\\ &\textrm{persamaan matriks berikut}\\ &\begin{pmatrix} -1 & 3\\ 1 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 5\\ 5 \end{pmatrix}\\ & \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&(2,-1)\\ \textrm{b}.&(1,-2)\\ \textrm{c}.&(-1,2)\\ \textrm{d}.&(-1,-2)\\ \textrm{e}.&(1,2) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} -1&3&5\\ 1&2&5 \end{array} \right]\quad \begin{aligned}&\bullet R_{\displaystyle 2}\longleftarrow R_{\displaystyle 1}+R_{\displaystyle 2}\\ &\bullet R_{\displaystyle 1}\longleftarrow -R_{\displaystyle 1}\end{aligned}\\ &=(1,2|5)+(-1,3|5)=(0,5|10)\quad \textrm{dan}\\ &=(1,-3|-5)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 1&-3&-5\\ 0&5&10 \end{array} \right]\quad \begin{aligned}&\bullet R_{\displaystyle 1}\longleftarrow \displaystyle \frac{3}{5}R_{\displaystyle 1}+R_{\displaystyle 2}\\ &\bullet R_{\displaystyle 2}\longleftarrow \displaystyle \frac{1}{5}R_{\displaystyle 1}\end{aligned}\\ &\textrm{Selanjutnya diperoleh}\\&\left[ \begin{array}{cc|c} 1&0&1\\ 0&1&2 \end{array} \right]\\ &\textrm{Jadi, koordinat titik potongnya}: (1,2) \end{array}$.

CONTOH SOAL 8 MATRIKS DAN SPL

$\begin{array}{l}\\ 6.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 2x-7y & =-19 \\ 3x-y & =0 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-3\quad \textrm{dan}\quad y=-1\\ \textrm{b}.&x=3\quad \textrm{dan}\quad y=1\\ \textrm{c}.&x=1\quad \textrm{dan}\quad y=3\\ \textrm{d}.&x=-3\quad \textrm{dan}\quad y=1\\ \textrm{e}.&x=1\quad \textrm{dan}\quad y=-3 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 2&-7&-19\\ 3&-1&0 \end{array} \right]\quad\bullet R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-3R_{\displaystyle 1}\\ &=2(3,-1|0)-3(2,-7|-19)=(0,19|57)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 2&-7&-19\\ 0&19&57 \end{array} \right]\\ &\textrm{sehingga}\quad y=\displaystyle \frac{57}{19}=3,\quad \textrm{maka}\quad x=1 \end{array}$.

$\begin{array}{l}\\ 7.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 4x-3y & =14 \\ -x+2y & =-1 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=2\quad \textrm{dan}\quad y=5\\ \textrm{b}.&x=5\quad \textrm{dan}\quad y=-2\\ \textrm{c}.&x=5\quad \textrm{dan}\quad y=2\\ \textrm{d}.&x=-5\quad \textrm{dan}\quad y=-2\\ \textrm{e}.&x=-2\quad \textrm{dan}\quad y=-5 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 4&-3&14\\ -1&2&-1 \end{array} \right]\quad\bullet R_{\displaystyle 2}\longleftarrow R_{\displaystyle 2}+\displaystyle \frac{1}{4}R_{\displaystyle 1}\\ &=(-1,2|-1)+\displaystyle \frac{1}{4}(4,-3|14)=(0,\displaystyle \frac{5}{4}|\displaystyle \frac{10}{4})\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 4&-3&14\\ 0&\displaystyle \frac{5}{4}&\displaystyle \frac{10}{4} \end{array} \right]\\ &\textrm{sehingga}\quad y=\displaystyle \frac{\displaystyle \frac{10}{4}}{\displaystyle \frac{5}{4}}=2,\quad \textrm{maka}\quad x=1 \end{array}$.

$\begin{array}{l}\\ 8.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 8x+3y & =37 \\ 4x+y & =15 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=2\quad \textrm{dan}\quad y=7\\ \textrm{b}.&x=2\quad \textrm{dan}\quad y=-7\\ \textrm{c}.&x=7\quad \textrm{dan}\quad y=-2\\ \textrm{d}.&x=7\quad \textrm{dan}\quad y=2\\ \textrm{e}.&x=-2\quad \textrm{dan}\quad y=-7 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 8&3&37\\ 4&1&15 \end{array} \right]\quad\bullet R_{\displaystyle 2}\longleftarrow R_{\displaystyle 2}-\displaystyle \frac{1}{2}R_{\displaystyle 1}\\ &=(4,1|15)-\displaystyle \frac{1}{2}(8,3|37)=(0,-\displaystyle \frac{1}{2}|-\displaystyle \frac{7}{2})\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 8&3&37\\ 0&-\displaystyle \frac{1}{2}&-\displaystyle \frac{7}{2} \end{array} \right]\\ &\textrm{sehingga}\quad y=\displaystyle \frac{-\displaystyle \frac{7}{2}}{-\displaystyle \frac{2}{2}}=7,\quad \textrm{maka}\quad x=2 \end{array}$.

$\begin{array}{l}\\ 9.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 4x+2y+3z & =0 \\ 2x+3y+5z & =9\\ 3x+y+7z & =9 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-2,\quad y=1\quad \textrm{dan}\quad z=2\\ \textrm{b}.&x=2,\quad y=1\quad \textrm{dan}\quad z=2\\ \textrm{c}.&x=-2,\quad y=-1\quad \textrm{dan}\quad z=-2\\ \textrm{d}.&x=-2,\quad y=1\quad \textrm{dan}\quad z=-2\\ \textrm{e}.&x=-2,\quad y=-1\quad \textrm{dan}\quad z=2 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{ccc|c} 4&2&3&0\\ 2&3&5&9\\ 3&1&7&9 \end{array} \right]\quad\begin{matrix} \bullet R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-R_{\displaystyle 1}\: .\\ \bullet R_{\displaystyle 3}\longleftarrow 4R_{\displaystyle 3}-3R_{\displaystyle 1} \end{matrix}\\ &=2(2,3,5|9)-(4,2,3|0)=(0,4,7|18)\quad \textrm{dan}\\ &=4(2,3,5|9)-3(4,2,3|0)=(0,-2,19|36)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 4&2&3&0\\ 0&4&7&18\\ 0&-2&19&36 \end{array} \right]\quad \bullet R_{\displaystyle 3}\longleftarrow 2R_{\displaystyle 3}+R_{\displaystyle 2}\\ &=2(0,-2,19|36)+(0,4,7|18)=(0,0,45|90)\\ &\textrm{sehingga}\quad z=\displaystyle \frac{90}{45}=2,\quad \textrm{maka}\quad y=1,\:\: x=-2 \end{array}$.

$\begin{array}{l}\\ 10.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 7x+2y+z & =-40 \\ 2x+7y+4z & =45\\ 5x-4y+6z & =8 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-8,\quad y=-3\quad \textrm{dan}\quad z=10\\ \textrm{b}.&x=-8,\quad y=3\quad \textrm{dan}\quad z=10\\ \textrm{c}.&x=-8,\quad y=3\quad \textrm{dan}\quad z=-10\\ \textrm{d}.&x=8,\quad y=3\quad \textrm{dan}\quad z=10\\ \textrm{e}.&x=8,\quad y=-3\quad \textrm{dan}\quad z=10 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{ccc|c} 7&2&1&-40\\ 2&7&4&45\\ 5&-4&6&8 \end{array} \right]\quad\begin{matrix}\bullet R_{\displaystyle 2}\longleftarrow 7R_{\displaystyle 2}-2R_{\displaystyle 1}\\ \bullet R_{\displaystyle 3}\longleftarrow 7R_{\displaystyle 3}-5R_{\displaystyle 1}\end{matrix}\\ &=7(2,7,4|45)-2(7,2,1|-40)=(0,45,26|395)\:\:\textrm{dan}\\ &=7(5,-4,6|8)-5(7,2,1|-40)=(0,-38,37|256)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 7&2&1&-40\\ 0&45&26&395\\ 0&-38&37&256 \end{array} \right]\quad \bullet R_{\displaystyle 3}\longleftarrow \displaystyle \frac{1}{45}R_{\displaystyle 3}+\displaystyle \frac{1}{38}R_{\displaystyle 2}\\ &=\displaystyle \frac{1}{45}(0,-38,37|256)+\displaystyle \frac{1}{38}(0,45,26|395)=(0,0,\displaystyle \frac{2653}{1710}|\displaystyle \frac{26530}{1710})\\ &\textrm{sehingga}\quad z=\displaystyle \frac{\displaystyle \frac{226530}{1710}}{\displaystyle \frac{2653}{1710}}=10,\quad \textrm{maka}\quad y=3,\:\: x=-8 \end{array}$.

CONTOH SOAL 7 MATRIKS DAN SPL

$\begin{array}{ll}\\ 1.&\textrm{Untuk sistem persamaan linear dua variabel}\\ &\textrm{berikut, jika dinyatakan dalam perkalian}\\ &\textrm{matriks adalah}....\\ &\begin{cases} 2x+3y & =5 \\ 3x+y & =1 \end{cases}\\ &\begin{array}{llll}\\ \textrm{a}.&\left( \begin{matrix} 2 & 3\\ 3 & 1 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 5 \\ 1 \end{matrix} \right)\\ \textrm{b}.&\left( \begin{matrix} 2 & 1\\ 3 & 3 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 5 \\ 1 \end{matrix} \right)\\ \textrm{c}.&\left( \begin{matrix} 3 & 2\\ 1 & 3 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 5 \\ 1 \end{matrix} \right)\\ \textrm{d}.&\left( \begin{matrix} 2 & 3\\ 3 & 1 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 1 \\ 5 \end{matrix} \right)\\ \textrm{e}.&\left( \begin{matrix} 3 & 2\\ 1 & 3 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 1 \\ 5 \end{matrix} \right)\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\textrm{Cukup jelas} \end{aligned} \end{array}$.

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

$\begin{array}{ll}\\ 3.&\textrm{Untuk sistem persamaan linear dua variabel}\\ &\textrm{berikut, jika dinyatakan dalam perkalian}\\ &\textrm{matriks adalah}....\\ &\begin{cases} x+y+z & =5 \\ 2x-5y+7z & =4\\ x-y+4z&=9 \end{cases}\\ &\begin{array}{llll}\\ \textrm{a}.&\left( \begin{matrix} 1 & 1&1\\ 2 & 5&7\\ 1&-1&4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right)\\ \textrm{b}.&\left( \begin{matrix} 1 & 1&1\\ 2 & -5&7\\ 1&1&4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right)\\ \textrm{c}.&\left( \begin{matrix} 1 & 1&1\\ 2 & -5&7\\ 1&-1&4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right)\\ \textrm{d}.&\left( \begin{matrix} 1 & 1&1\\ 2 & -5&7\\ 1&-1&-4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right)\\ \textrm{e}.&\left( \begin{matrix} -1 & 1&1\\ 2 & -5&7\\ 1&-1&4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Cukup jelas} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Perhatikanlah sistem persamaan berikut}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{agar sistem persamaan ini tidak}\\ &\textrm{memiliki penyelesaian, maka nilai}\: \: k=....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Agar sistem persamaan}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{tidak berpenyelesaian, maka}\\ &\textrm{ingat penyelesaian metode matrik}\\ &\textrm{buatlah penyebutnya}=0,\: \: \textrm{yaitu}:\\ &\begin{vmatrix} 3 & 2 & -5\\ 2 & -6 & k\\ 5 & -4 & -1 \end{vmatrix}=0\\ &\textrm{Selanjutnya}\\ &3\begin{vmatrix} -6 & k\\ -4 & -1 \end{vmatrix}-2\begin{vmatrix} 2 & k\\ 5 & -1 \end{vmatrix}-5\begin{vmatrix} 2 & -6\\ 5 & -4 \end{vmatrix}=0\\ &3(6+4k)-2(-2-5k)-5(-8+30)=0\\ &18+12k+4+10k+40-150=0\\ &22x=88\\ &\quad x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Himpunan penyelesaian dari}\\ &\left\{\begin{matrix} x+y+4z=15\quad\\ x-y+z=2\qquad\\ x+2y-3z=-4 \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ (-1,1,3) \right \}\\ \textrm{b}.&\left \{ (1,2,3) \right \}\\ \textrm{c}.&\left \{ (-2,1,1) \right \}\\ \textrm{d}.&\left \{ (3,2,-1) \right \}\\ \textrm{e}.&\left \{ (1,-2,3) \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Semunya dikerjakan dengan metode}\\ &\textrm{matriks}\: (\textbf{Cara Cramer})\\ &\begin{aligned} x&=\displaystyle \frac{\begin{vmatrix} 15 & 1 & 4\\ 2& -1 & 1\\ -4& 2 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{15\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}+4\begin{vmatrix} 2 & -1\\ -4 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{15(3-2)-1(-6+4)+4(4-4)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{15(1)-1(-2)+4(0)}{1(1)-1(-4)+4(3)}=\frac{17}{17}=1 \\ y&=\displaystyle \frac{\begin{vmatrix} 1 & 15 & 4\\ 1& 2 & 1\\ 1& -4 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}-15\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(-6+4)-15(-3-1)+4(-4-2)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(-2)-15(-4)+4(-6)}{1(1)-1(-4)+4(3)}=\frac{34}{17}=2\\ z&=\displaystyle \frac{\begin{vmatrix} 1 & 1 & 15\\ 1& -1 & 2\\ 1& 2 & -4 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} -1 & 2\\ 2 & -4 \end{vmatrix}-1\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}+15\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(4-4)-1(-4-2)+15(2+1)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(0)-1(-6)+15(3)}{1(1)-1(-4)+4(3)}=\frac{51}{17}=3 \end{aligned} \end{array}$

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