Lanjutan Materi 2 Sistem Pertidaksamaan Dua Variabel

 3. Sistem Pertidaksamaan Dua Variabel Kuadrat dan Kuadrat (SPDVKK)

Bentuk Persamaan (SPDVKK)

$\begin{array}{rl}& \text{Bentuk Umum}:\\ & \{\begin{array}{l}a{x}^2+b{y}^2+cxy+dx+ey+f=0\\ p{x}^2+q{y}^2+rxy+sx+ty+u=0\end{array}\end{array}$.

Perhatikan tabel berikut

$\begin{array}{|cl|}\hline 1.& \text{Penyelesaian}\\ & y=y\\ 2.& \text{Proses}:\\ & \begin{array}{rl}& a{x}^2+bx+c=p{x}^2+qx+r\\ & \text{dengan}\text{syarat}\\ & a-p\ne 0\\ & \text{diubah kebentuk umum}\\ & A{x}^2+Bx+C=0\\ & \text{dengan}\\ & D=B^2-4AC\end{array}\\ 3.& \text{Sebagai Penjelasan}\\ & \begin{array}{rl}D& >0\\ & \text{SPDVKK mempunyai }\\ & \text{2 penyelesaian berbeda}\\ D& =0\\ & \text{SPDVKK mempunyai }\\ & \text{1 penyelesaian saja}\\ D& <0\\ & \text{SPDVKK tidak }\\ & \text{mempunyai penyelesaian}\end{array}\\ \hline\end{array}$.

Bentuk Pertidaksamaan (SPtDVKK)

$\begin{array}{rl}& \text{Bentuk Umum}:\\ & y\ge a{x}^2+bx+c\\ & y\le p{x}^2+qx+r\\ \\ & \underset{―}{\text{Catatan}}:\\ & \text{Tanda Ketaksamaan bisa diganti}\\ & \text{dengan tanda ketaksamaan yang}\\ & \text{lainnya}\end{array}$

Bentuk Pertidaksamaan dari SPtDVKK adalah wilayah atau daerah dengan menentukan penyelesaian seperti persamaan langkah awalnya kemudian dibuatkan wilayah yang menunjukkan pertidaksamaannya. Wilayah SPtDVKK ini adalah irisan antara wilayah penyelesaian dari pertidaksamaan kuadrat yang pertama dengan wilayah penyelesaian pertidaksamaan kuadrat yang kedua.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Carilah ada/tidaknya penyelesaian }\\ & \text{dari SPDVKK berikut ini}.\\ \\ & \{\begin{array}{l}y=-x^2\\ y=x^2+2x+1\end{array}\\ \\ & \text{dan sketsalah tafsiran geometrinya}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & \{\begin{array}{l}y=-x^2\\ y=x^2+2x+1\end{array}\\ & \text{substitusikan}y=-x^2\text{ke}y=x^2+2x+1\\ & \text{sehingga diperoleh}\\ & -x^2=x^2+2x+1\\ & 2x^2+2x+1=0\\ & \{\begin{array}{l}a=2\\ b=-2\\ c=1\end{array}\\ & \text{Karena, nilai}D=b^2-4ac\\ & =(-2{)}^2-4(2)(1)=4-8=-4<0\\ & \text{maka diperoleh keterangan }\\ & \text{bahwa kedua parabola tersebut}\\ & \mathbf{\text{tidak berpotongan dan tidak bersinggungan}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukan himpunan penyelesaian }\\ & \text{dari SPtDVLK berikut ini}\\ & \{\begin{array}{l}y\le 4-x^2\\ y\le x^2-2x-3\end{array}\\ \\ & \text{Jika ada, sketsalah tafsiran geometrinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ \mathbf{\text{a}}.& \text{Pembuat nol fungsi},\{\begin{array}{ll}y& =4-x^2\\ y& =x^2-2x-3\end{array},\\ & \text{yaitu}:\\ & \mathbf{\text{Untuk}}:y=f(x)=4-x^2\\ & 4-x^2=0\Rightarrow x^2=4\Rightarrow x=\pm 2\\ & \text{Jadi titik potongnya}:(-2,0)\text{dan}(2,0)\\ & \text{Titik puncak grafik fungsinya}(x_{ss},y),\\ & \text{yaitu}:\\ & x_{ss}=-{\displaystyle \frac{b}{a}\text{atau}{x}_{ss}={\displaystyle \frac{x_1+x_2}{2}}}\\ & \Rightarrow x_{ss}={\displaystyle \frac{2+(-2)}{2}=\frac{0}{2}=0,\text{maka}}\\ & y=f(0)=4-0^2=4\\ & \text{Jadi, titik puncaknya}:(0,4)\\ & \mathbf{\text{Dan untuk}}:y=f(x)=x^2-2x-3\\ & x^2-2x-3=0\Rightarrow (x+1)(x-3)=0\\ & x=-1\text{atau}x=3\\ & \text{Jadi titik potongnya}:(-1,0)\text{dan}(3,0)\\ & \text{Titik puncak grafik fungsinya}(x_{ss},y),\\ & \text{yaitu}:\\ & x_{ss}=-{\displaystyle \frac{b}{a}\text{atau}{x}_{ss}={\displaystyle \frac{x_1+x_2}{2}}}\\ & \Rightarrow x_{ss}={\displaystyle \frac{(-1)+3}{2}=\frac{2}{2}=1,\text{maka}}\\ & y=f(1)=1^2-2(1)-3=-4\\ & \text{Jadi, titik puncaknya}:(0,-4)\\ \mathbf{\text{b}}.& \text{Melakukan uji titik untuk menentukan}\\ & \text{wilayah pertidaksamaan, yaitu}:\\ & \text{Ambil titik bebas saja, misal kita pilih lagi}\\ & \text{titik}(0,0),\text{maka}\\ & \begin{array}{rl}(0,0)& \Rightarrow y\le 4-x^2\\ & 0\le 4-0^2\Rightarrow 0\le 4\mathbf{\text{benar}}\\ (0,0)& \Rightarrow y\le x^2-2x-3\\ & 0\le 0^2-2(0)-3\Rightarrow 0\le -3\mathbf{\text{salah}}\end{array}\\ & \text{ini berarti wilayah yang di dalamnya}\\ & \text{terdapat titik uji}(0,0)\text{bukan}\\ & \text{wilayah penyelesaian kedua pertidaksamaan}\\ & \text{tersebut}\\ \mathbf{\text{c}}.& \text{Menggambar wilayah pertidaksamaan}\\ & \text{berikut ilustrasi gambarnya}\end{array}\end{array}$.

atau

DAFTAR PUSTAKA

1. Kanginan, M., Akhmad, G., Nurdiansyah, H. 2014. Matematika untuk SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Yuana, R.A., Indriyastuti. 2017. Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Lanjutan Materi 1 Sistem Pertidaksamaan Dua Variabel

 2. Sistem Pertidaksamaan Dua Variabel Linear Kuadrat (SPtDVLK)

Bentuk Persamaan (SPDVLK)

$\begin{array}{rl}& \text{Bentuk Umum}:\\ & \{\begin{array}{l}ax+by+c=0\\ p{x}^2+q{y}^2+rxy+sx+ty+u=0\end{array}\end{array}$.

Perhatikanlah tabel berikut

$\begin{array}{|cl|}\hline 1.& \text{Penyelesaian}\\ & y=y\\ 2.& \text{Proses}:\\ & \begin{array}{rl}& ax+b=p{x}^2+qx+r\\ & \text{diubah ke bentuk umum}\\ & A{x}^2+Bx+C=0\\ & \text{dengan}\\ & D=B^2-4AC\end{array}\\ 3.& \text{Sebagai Penjelasan}\\ & \begin{array}{rl}D& >0\\ & \text{SPDVLK mempunyai }\\ & \text{2 penyelesaian berbeda}\\ D& =0\\ & \text{SPDVLK mempunyai }\\ & \text{1 penyelesaian saja}\\ D& <0\\ & \text{SPDVLK tidak }\\ & \text{mempunyai penyelesaian}\end{array}\\ \hline\end{array}$.

Bentuk Pertidaksamaan (SPtDVLK)

$\begin{array}{rl}& \text{Bentuk Umum}:\\ & y\ge a{x}^2+bx+c\\ & y\le px+q\\ \\ & \underset{―}{\text{Catatan}}:\\ & \text{Tanda Ketaksamaan bisa diganti}\\ & \text{dengan tanda ketaksamaan yang}\\ & \text{lainnya}\end{array}$

Bentuk Pertidaksamaan dari SPtDVLK adalah wilayah atau daerah dengan menentukan penyelesaian seperti persamaan langkah awalnya kemudian dibuatkan wilayah yang menunjukkan pertidaksamaannya. Wilayah SPtDVLK ini adalah irisan antara wilayah penyelesaian dari pertidaksamaan linear dan wilayah penyelesaian pertidaksamaan kuadrat.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Carilah ada/tidaknya penyelesaian }\\ & \text{dari SPDVLK berikut ini}\\ & \{\begin{array}{l}y=-x+2\\ y=x^2-3x+2\end{array}\\ \\ & \text{Jika ada, sketsalah tafsiran geometrinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & \{\begin{array}{l}y=-x+2\\ y=x^2-3x+2\end{array}\\ & \text{maka}\\ & y=y\\ & x^2-3x+2=-x+2\\ & x^2-2x=0\\ & x(x-2)=0\\ & x=0\text{atau}x=2\\ & \text{selanjutnya}\mathbf{\text{hasil ini dapat kita}}\\ & \text{substitusikan ke}y=-x+2\\ & \text{untuk mendapatkan nilai}y\text{yaitu}:\\ & \{\begin{array}{ll}x=0& \Rightarrow y=-0+2=2,\text{diperoleh titik}(0,2)\\ x=2& \Rightarrow y=-2+2=0,\text{diperoleh juga titik}(2,0)\end{array}\\ & \\ & \text{sehingga HP-nya}=\{(0,2),(2,0)\}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukan himpunan penyelesaian }\\ & \text{dari SPtDVLK berikut ini}\\ & \{\begin{array}{l}x+y<4\\ y\ge x^2-9\end{array}\\ \\ & \text{Jika ada, sketsalah tafsiran geometrinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ \mathbf{\text{a}}.& \text{Pembuat nol fungsi},y=f(x)=x^2-9,\\ & \text{yaitu}:\\ & x^2-9=0\Rightarrow x^2=9\Rightarrow x=\pm 3\\ & \text{Jadi titik potongnya}:(-3,0)\text{dan}(3,0)\\ & \text{Titik puncak grafik fungsinya}(x_{ss},y),\\ & \text{yaitu}:\\ & x_{ss}=-{\displaystyle \frac{b}{a}\text{atau}{x}_{ss}={\displaystyle \frac{x_1+x_2}{2}}}\\ & \Rightarrow x_{ss}={\displaystyle \frac{3+(-3)}{2}=\frac{0}{2}=0,\text{maka}}\\ & y=f(0)=0^2-9=-9\\ & \text{Jadi, titik puncaknya}:(0,-9)\\ \mathbf{\text{b}}.& \text{Melakukan uji titik untuk menentukan}\\ & \text{wilayah pertidaksamaan, yaitu}:\\ & \text{Ambil titik bebas saja, misal kita pilih}\\ & \text{titik}(0,0),\text{maka}\\ & \begin{array}{rl}(0,0)& \Rightarrow y\ge x^2-9\\ & 0\ge 0^2-9\Rightarrow 0\ge -9\mathbf{\text{benar}}\\ (0,0)& \Rightarrow x+y<4\text{juga}\mathbf{\text{benar}}\\ & 0+0<4\Rightarrow 0<4\end{array}\\ & \text{ini berarti wilayah yang di dalamnya}\\ & \text{terdapat titik uji}(0,0)\text{merupakan}\\ & \text{wilayah penyelesaian kedua pertidaksamaan}\\ & \text{tersebut}\\ \mathbf{\text{c}}.& \text{Menggambar wilayah pertidaksamaan}\\ & \text{berikut ilustrasi gambarnya}\end{array}\end{array}$.

atau

$\begin{array}{l}\\ 3.& \text{Tentukan himpunan penyelesaian }\\ & \text{dari SPtDVLK berikut ini}\\ & \{\begin{array}{l}y\le -x+2\\ y\ge x^2-3x+2\end{array}\\ \\ & \text{Jika ada, sketsalah tafsiran geometrinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dengan cara sama seperti no. 2}:\\ & \text{ditambah dengan}\\ & \{\begin{array}{l}y=-x+2\\ y=x^2-3x+2\end{array}\\ & \text{maka}\\ & y=y\\ & x^2-3x+2=-x+2\\ & x^2-2x=0\\ & x(x-2)=0\\ & x=0\text{atau}x=2\\ & \text{selanjutnya}\mathbf{\text{hasil ini dapat kita}}\\ & \text{substitusikan ke}y=-x+2\\ & \text{untuk mendapatkan nilai}y\text{yaitu}:\\ & \{\begin{array}{ll}x=0& \Rightarrow y=-0+2=2,\text{diperoleh titik}(0,2)\\ x=2& \Rightarrow y=-2+2=0,\text{diperoleh juga titik}(2,0)\end{array}\\ & \\ & \text{sehingga titik potongnya}=\{(0,2),(2,0)\}\\ & \text{Dan wilayah solusinya adalah irisan}\\ & \text{dari kedua pertidaksamaan tersebut}\\ & \text{Berikut Gambar wilayahnya}\end{array}\end{array}$.

atau

Sistem Pertidaksamaan Dua Variabel

Sebelumnya telah diketahui sistem persamaan linear dua variabel, silahkan lihat di sini.

1. Sistem Persamaan Dua Variabel Linear dan Kuadrat

Sebelumnya akan kita singgung dulu fungsi linear dan kuadrat sebagai mana tabel berikut:

Fungsi Linear

$\begin{array}{rl}& \\ \text{Fungsi}& \text{Linear adalah}:\\ & \text{fungsi di aman fungsi yang }\\ & \text{hanya memiliki satu variabel}\\ & \text{atau peubah dan berpangkat satu}.\\ \\ \text{Misal},& f:x\mapsto ax+b\\ & \end{array}$.

Dalam menentukan persamaan linear/garis lurus adalah:

$\begin{array}{|cc|}\hline \text{Melalui titik}(x_1,y_1)& \text{Melalui titik}(x_1,y_1)\\ \text{dan bergradien}m& \text{dan}(x_2,y_2)\\ y=m(x-x_1)+y_1& {\displaystyle \frac{y-y_1}{y_2-y_1}={\displaystyle \frac{x-x_1}{x_2-x_1}}}\\ & \begin{array}{rl}\text{dengan}:& \\ m& ={\displaystyle \frac{y_2-y_1}{x_2-x_1}}\end{array}\\ \text{Sejajar dengan}& \text{Tegak lurus dengan}\\ \text{garis bergradien}{m}_1& \text{garis bergradien}{m}_1\\ \text{Syarat dua garis}& \text{Syarat dua garis}\\ \text{Sejajar}{m}_1=m_2& \text{Tegak lurus}{m}_1\times m_2=-1\\ y=m_2(x-x_1)+y_1& y=-{\displaystyle \frac{1}{m_2}(x-x_1)+y_1}\\ \hline\end{array}$

Fungsi Kuadrat

$\begin{array}{|ll|}\hline \text{Pengertian}& \begin{array}{rl}& \text{Suatu fungsi yang berbentuk}\\ & f(x)=a{x}^2+bx+c\\ & a,b,c,\in \mathbb{R},a\ne 0\end{array}\\ \text{Grafik Fungsi}& \text{Keterangan}\\ \text{Titik potong sumbu x}& \text{Jika ada}\\ & \begin{array}{rl}& \text{untuk titik potong}\\ & \text{terhadap sumbu x }\\ & \text{Jika y = 0 maka }\\ & a{x}^2+bx+c=0\\ & \text{Selanjutnya tinggal}\\ & \text{menentukan nilai D}\\ & D=b^2-4ac\text{adalah}\\ & \text{nilai diskriminan}.\\ & \text{Jika}D>0\\ & \text{maka grafik}\\ & \text{memotong sumbu x}\\ & \text{di dua tempat berbeda}\\ & \text{yaitu di}(x_1,0)\text{dan}(x_2,0).\\ & \text{dan jika D = 0}\\ & \text{maka grafik}\\ & \text{ hanya menyinggung}\\ & \text{sumbu x di satu titik}\\ & \text{yaitu di }(x_1,0)\\ & \text{dan jika}D<0\\ & \text{maka grafik}\\ & \text{tidak memotong}\\ & \text{atau menyinggung sumbu x}\end{array}\\ \text{Titik potong sumbu y}& \begin{array}{rl}& \text{titik potong terhadap}\\ & \text{sumbu y, jika x = 0}\\ & y=f(x)=a{x}^2+bx+c\\ & y=f(0)=a(0{)}^2+b(0)+c\\ & y=c\end{array}\\ \text{Sumbu Simetri (SS)}& x={\displaystyle \frac{-b}{2a}}\\ \text{Titik Puncak}& \left({\displaystyle \frac{-b}{2a},{\displaystyle \frac{D}{-4a}}}\right)\\ \text{Posisi grafik}& \text{Jika}a>0\text{maka}\\ & \text{grafik terbuka ke atas}\\ & \text{Dan jika nilai}a<0\text{maka}\\ & \text{grafik terbuka ke bawah}\\ \hline\end{array}$.

Selanjutnya cara membuat grafik fungsi kudratnya adalah sebagai berikut:

$\begin{array}{|cc|}\hline \text{Jika memotong sumbu}-\text{X}& \text{Jika menyinggung sumbu}-\text{X}\\ \text{di titik}(x_1,0)\text{dan}(x_2,0)& \text{di titik}(x_1,0)\text{dan melalui}\\ \text{dan melalui sebuah titik lain}& \text{sebuah titik lain}\\ & \\ y=f(x)=a(x-x_1)(x-x_2)& y=f(x)=a{(x-x_1)}^2\\ & \\ \text{Jika grafik fungsi itu melalui}& \text{Jika grafik fungsi itu melalui}\\ \text{Titik puncak}P(x_p,y_p)\text{dan}& \text{tiga buah titik yaitu}(x_1,y_1)\\ \text{sebuah titik lain}& (x_2,y_2)\text{dan}(x_3,y_3)\\ & \\ y=f(x)=a{(x-x_p)}^2+y_p& y=f(x)=a{x}^2+bx+c\\ & \\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}f\text{adalah fungsi linear dengan}\\ & f(2)-f(-2)=8,\\ & \text{maka nilai dari}f(4)-f(-2)\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & f(x)=ax+b\\ & f(2)-f(-2)\\ & =(a(2)+b)-(a(-2)+b)=8\\ & 8=2a+2a\\ & 8=4a\\ & 2=a\\ & f(x)=2x+b,\text{dengan}b\text{konstan}\\ & \text{Sehingga nilai}\\ & f(4)-f(-2)=(2(4)+b)-(2(-2)+b)\\ & =8+b+4-b\\ & =12\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Ubahlah}8-6x-x^2\text{ke dalam bentuk}\\ & a-(x+b{)}^2,\text{selanjutnya tentukan}\\ & \text{daerah hasil dari}f(x)=8-6x-x^2\\ & \text{untuk}x\text{bilangan real}\\ & (\mathit{\text{NTU Entrance Examination AO-level}})\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cl|}\hline 1.& \text{Diketahui}\\ & \begin{array}{rl}\text{Misal}& \\ 8-6x-x^2& =f(x)\\ f(x)& =-x^2-6x+8\\ & =-(x^2+6x-8)\\ & =-(x^2+6x+9-17)\\ & =-((x+3{)}^2-17)\\ & =-(x+3{)}^2+17\\ & \end{array}\\ 2.& \text{Mencari koordinat}(x_{SS},y_{SS})\\ & \begin{array}{rl}f(x)& =-x^2-6x+8\{\begin{array}{c}a=-1\\ b=-6\\ c=8\end{array}\\ \text{Maka}& \\ x_{SS}& =\frac{-b}{2a}={\displaystyle \frac{-(-6)}{2(-1)}}\\ & =-3\\ y_{SS}& =f(-3)=-{(-3+3)}^2+17=17\\ \therefore & (x_{SS},y_{SS})=(-3,17)\end{array}\\ 3.& \text{Nilai fungsi}\\ & \begin{array}{rl}\text{Karena}& a=-1<0\\ \text{maka f}& \text{ungsi menghadap}\\ \mathbf{\text{ke ba}}& \mathbf{\text{wah}},\text{sehingga}\\ \text{daerah}& \text{hasilnya}\left(R_f\right)\\ \text{adalah}& :\\ & \{-\mathrm{\infty }<y\le 17\}\\ & \\ & \text{Berikut ilustrasinya}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}\alpha \text{dan}\beta \text{adalah akar-akar dari }\\ & \text{persamaan kuadrat}{x}^2+mx+m=0,\\ & \text{maka nilai}m\text{yang menyebabkan }\\ & \text{jumlah kuadrat akar-akar mencapai}\\ & \text{minimum adalah}....\\ & \mathbf{\text{(UM UNDIP 2014 Mat Das)}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}{x}^2+mx+m=0\\ & \mathbf{\text{persamaan kuadrat}}\text{dalam}x,\\ & \text{maka}\\ & x^2+mx+m=x^2-(\alpha +\beta )x+(\alpha \beta )=0\\ & \{\begin{array}{ll}\alpha +\beta & =-m\\ & \\ \alpha \beta & =m\end{array}\\ & \text{Selanjutnya}\\ & {\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta \\ & =(-m{)}^2-2m\text{dan dapat kita tuliskan sebagai}\\ & f(m)=m^2-2m\{\begin{array}{ll}a& =1\\ b& =-2\\ c& =0\end{array}\\ & \text{fungsi kuadrat dalam}m,\\ & \text{sehingga kita perlu mencari titik}(m_{SS},f\left(m_{SS}\right)),\\ & \text{tetapi yang kita perlukan}\\ & \text{cuma}m-\text{nya saja, yaitu}:m=m_{SS},\\ & \text{dengan}{m}_{SS}={\displaystyle \frac{-b}{2a}=\frac{-(-2)}{2.1}=1}\end{array}\end{array}$.

Latihan Soal 10 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 91.& \text{Titik A(4,-4) dicerminkan terhadap}\\ & \text{garis}y=x\tan {15}^{\circ }\text{menghasilkan}\\ & \text{bayangan}{A}^{\prime }(a,b)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.\sqrt{3}& & \text{d}.4\sqrt{3}\\ \text{b}.2\sqrt{3}& \text{c}.3\sqrt{3}& \text{e}.6\sqrt{3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{c}a\\ b\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {2.15}^{\circ }& \sin {2.15}^{\circ }\\ \sin {2.15}^{\circ }& -\cos {2.15}^{\circ }\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {30}^{\circ }& \sin {30}^{\circ }\\ \sin {30}^{\circ }& -\cos {30}^{\circ }\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{c}2\sqrt{3}-2\\ 2+2\sqrt{3}\end{array}\right)\\ & \{\begin{array}{ll}a& =2\sqrt{3}-2\\ b& =2+2\sqrt{3}\end{array}\\ \text{mak}& \text{a nilai dari}\\ a+b& =(2\sqrt{3}-2+2+2\sqrt{3})\\ & =4\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 92.& \text{Lingkaran}{x}^2+y^2-5x+8y+7=0\\ & \text{ditranslasikan oleh}T=\left(\begin{array}{c}m\\ n\end{array}\right)\text{menghasilkan}\\ & \text{bayangan}{x}^2+y^2-9x+2y+6=0.\\ & \text{Nilai}m+n=...\\ & \begin{array}{l}\\ \text{a}.2& & \text{d}.5\\ \text{b}.3& \text{c}.4& \text{e}.6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Dik}& \text{etahui sebuah lingkaran dengan persamaan}:\\ & x^2+y^2-5x+8y+7=0\\ \text{kar}& \text{ena akibat translasi, maka}\\ & \{\begin{array}{ll}x& =x^{\prime }-m\\ y& =y^{\prime }-n\end{array}\\ & x^2+y^2-5x+8y+7=0\\ \text{seh}& \text{ingga}\\ & \iff (x^{\prime }-m{)}^2+(y^{\prime }-n{)}^2-5(x^{\prime }-m)+8(y^{\prime }-n)+7=0\\ & \iff x^{\prime 2}+y^{\prime 2}-2m{x}^{\prime }-2n{y}^{\prime }+m^2+n^2-5x^{\prime }+5m+8y^{\prime }-8n+7=0\\ & \iff x^{\prime 2}+y^{\prime 2}-(2m+5)x^{\prime }+(8-2n)y^{\prime }+m^2+n^2+5m-8n+7=0\\ & \equiv x^{\prime 2}+y^{\prime 2}-9x^{\prime }+2y^{\prime }+6=0(\mathbf{\text{akhir bayangan}})\\ & \{\begin{array}{ll}9& =2m+5\Rightarrow m=2\\ 2& =8-2n\Rightarrow n=3\end{array}\\ \text{Jad}& \text{i , nilai}m+n=2+3=5\end{array}\end{array}$.

$\begin{array}{l}\\ 93.& \text{Jika titik A(-2,1) dicerminkan terhadap garis}\\ & y=-{\displaystyle \frac{1}{3}x\sqrt{3},\text{maka bayangan dari}}\\ & \text{titik textit{A} tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.A^{\prime }(1-{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})& & \\ \text{b}.A^{\prime }(-1-{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})& \\ \text{c}.A^{\prime }(-1-{\displaystyle \frac{1}{2}\sqrt{3},{\displaystyle \frac{1}{2}-\sqrt{3}}})& \\ \text{d}.A^{\prime }(1-{\displaystyle \frac{1}{2}\sqrt{3},{\displaystyle \frac{1}{2}-\sqrt{3}}})\\ \text{e}.A^{\prime }(-1+{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ y& =-{\displaystyle \frac{1}{3}x\sqrt{3}=(-{\displaystyle \frac{1}{3}\sqrt{3}})x}\\ & =(-\tan {30}^{\circ })x=\tan ({180}^{\circ }-{30}^{\circ })x\\ & =\tan {150}^{\circ }.x\\ \text{maka}\theta & ={150}^{\circ }\Rightarrow 2\theta ={300}^{\circ }\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {300}^{\circ }& \sin {300}^{\circ }\\ \sin {300}^{\circ }& -\cos {300}^{\circ }\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\sqrt{3}}\\ -{\displaystyle \frac{1}{2}\sqrt{3}}& -{\displaystyle \frac{1}{2}}\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)\\ & =\left(\begin{array}{c}-1-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \sqrt{3}-{\displaystyle \frac{1}{2}}\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 94.& \text{Bayangan titik A(2,4) dicerminkan }\\ & \text{terhadap garis}y-x=0\text{dilanjutkan}\\ & \text{ke garis}x\sqrt{3}-3y=0\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.A^{\prime }(2+\sqrt{3},-1+2\sqrt{3})& & \\ \text{b}.A^{\prime }(2+\sqrt{3},1-2\sqrt{3})& \\ \text{c}.A^{\prime }(1-\sqrt{3},-2+\sqrt{3})& \\ \text{d}.A^{\prime }(-2+\sqrt{3},1+2\sqrt{3})\\ \text{e}.A^{\prime }(2-\sqrt{3},1-2\sqrt{3})\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ & \{\begin{array}{ll}x\sqrt{3}-3y=0& \iff y={\displaystyle \frac{1}{3}\sqrt{3}x}\\ & \iff y=\tan {30}^{\circ }.x\\ \\ x-y=0& \iff y=x\end{array}\\ \\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {2.30}^{\circ }& \sin {2.30}^{\circ }\\ \sin {2.30}^{\circ }& -\cos {2.30}^{\circ }\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}\sqrt{3}}\\ {\displaystyle \frac{1}{2}\sqrt{3}}& -{\displaystyle \frac{1}{2}}\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}2\\ 4\end{array}\right)\\ & =\left(\begin{array}{c}\sqrt{3}+2\\ -1+2\sqrt{3}\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 95.& \text{Jika}{T}_1=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\text{dan}{T}_2=\left(\begin{array}{cc}-2& 5\\ -1& 3\end{array}\right)\\ & \text{maka bayangan garis}x+y+1=0\\ & \text{oleh}{T}_2\circ T_1\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.x-2y-1=0& & \\ \text{b}.x+2y-1=0& \\ \text{c}.x+2y+1=0& \\ \text{d}.x-2y+1=0\\ \text{e}.x+y-1=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T_2\circ T_1\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}-2& 5\\ -1& 3\end{array}\right)\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}-2+5& -4+5\\ -1+3& -2+3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}3& 1\\ 2& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}3x+y\\ 2x+y\end{array}\right)\\ \text{Dipe}& \text{roleh}\\ & \begin{array}{l}\\ x^{\prime }& =3x+y\\ y^{\prime }& =2x+y-\\ x^{\prime }-y^{\prime }& =x\\ \iff x& =x^{\prime }-y^{\prime }....(1)\\ \text{maka}\\ y& =x^{\prime }-3x\\ & =x^{\prime }-3(x^{\prime }-y^{\prime })\\ & =3y^{\prime }-2x^{\prime }....(2)\end{array}\\ \text{Sehin}& \text{gga}\\ x+y& +1=0\\ x^{\prime }-& y^{\prime }+3y^{\prime }-2x^{\prime }+1=0\\ -x^{\prime }+& 2y^{\prime }+1=0\\ x^{\prime }-& 2y^{\prime }-1=0\\ \text{maka}& \text{bayangan garisnya}\\ x-2& y-1=0\end{array}\end{array}$.

$\begin{array}{l}\\ 96.& \text{Garis}2x+y+4=0\text{ditranslasikan}\\ & \text{oleh}\left(\begin{array}{c}-2\\ 5\end{array}\right)\text{dilanjutkan transformasi}\\ & \text{oleh}\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\text{persamaan bayangannya}\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.2x+y+3=0& & \\ \text{b}.2x-3y+3=0& \\ \text{c}.2x+3y+3=0& \\ \text{d}.3x+2y+3=0\\ \text{e}.3x-2y+3=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}-2\\ 5\end{array}\right)=\left(\begin{array}{c}x-2\\ y+5\end{array}\right)\\ \left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)& =\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & =\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}x-2\\ y+5\end{array}\right)\\ & =\left(\begin{array}{c}x-2+2y+10\\ y+5\end{array}\right)\\ & =\left(\begin{array}{c}x+2y+8\\ y+5\end{array}\right)\\ \text{Diper}& \text{oleh}\\ & \begin{array}{l}\\ x^{″}& =x+2y+8\\ 2y^{″}& =2y+10-\\ x^{″}-2y^{″}& =x-2\\ \iff x& =x^{″}-2y^{″}+2....(1)\\ \text{maka}\\ y& =y^{″}-5....(2)\end{array}\\ \text{sehin}& \text{gga}\\ 2x+& y+4=0\\ 2(x^{″}& -2y^{″}+2)+(y^{″}-5)+4=0\\ 2x^{″}-& 3y^{″}+3=0\\ \text{maka}& \text{bayangan garisnya}\\ 2x-& 3y+3=0\end{array}\end{array}$.

$\begin{array}{l}\\ 97.& \text{Diketahui}M\text{adalah pencerminan terhadap}\\ & \text{garis}y=-x\text{dan}T\text{adalah transformasi}\\ & \text{yang dinyatakan oleh matriks}\left(\begin{array}{cc}2& 3\\ 0& -1\end{array}\right)\\ & \text{Koordinat bayangan titik}A(2,-8)\text{oleh}\\ & \text{transformasi}M\text{dilanjutkan oleh}T\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(-10,2)& & \\ \text{b}.(-2,-10)& \\ \text{c}.(10,2)& \\ \text{d}.(-10,-2)\\ \text{e}.(2,10)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T\circ M\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}2& 3\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{cc}0-3& -2+0\\ 0+1& 0+0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{cc}-3& -2\\ 1& 0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{c}-6+16\\ 2+0\end{array}\right)\\ & =\left(\begin{array}{c}10\\ 2\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 98.& \text{Jika}W\text{adalah transformasi oleh}\\ & \text{matriks}\left(\begin{array}{cc}1& 0\\ 3& 1\end{array}\right),\text{maka titik mula}\\ & \text{dari}{W}^{\prime }(-2,5)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(-11,-2)& & \\ \text{b}.(11,-2)& \\ \text{c}.(-2,11)& \\ \text{d}.(2,11)\\ \text{e}.(12,11)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dimi}& \text{salkan}:\\ A& =\left(\begin{array}{c}-2\\ 5\end{array}\right),\text{dan}\\ W& =\left(\begin{array}{cc}1& 0\\ 3& 1\end{array}\right),\text{serta}X=\left(\begin{array}{c}x\\ y\end{array}\right)\\ \text{mak}& \text{a}\\ & \begin{array}{|c|}\hline \begin{array}{rl}A& =BX\\ B^{-1}A& =B^{-1}BX\\ B^{-1}A& =I.X\\ B^{-1}A& =X\\ X& =B^{-1}A\end{array}\\ \hline\end{array}\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\displaystyle \frac{1}{\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|}\left(\begin{array}{cc}1& 0\\ -3& 1\end{array}\right)\left(\begin{array}{c}-2\\ 5\end{array}\right)}\\ & =1.\left(\begin{array}{c}-2+0\\ 6+5\end{array}\right)\\ & =\left(\begin{array}{c}-2\\ 11\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 99.& \text{Jika setiap titik pada grafik dengan}\\ & \text{dengan persamaan}y=\sqrt{x}\text{dicerminkan}\\ & \text{terhadap garis}y=x,\text{maka persamaan}\\ & \text{grafik yang dihasilkan adalah}...\\ & \begin{array}{l}\\ \text{a}.y=x^2,x\ge 0& & \\ \text{b}.y=-\sqrt{x},x\ge 0& \\ \text{c}.y=-x^2,x\le 0& \\ \text{d}.y=\sqrt{-x},x\le 0\\ \text{e}.y=-\sqrt{-x},x\le 0\end{array}\\ \\ & \mathbf{\text{UMB Tahun 2011 Kode 152}}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ y& =\sqrt{x},\text{atau}{y}^2=x\\ \mathbf{\text{Alt}}& \mathbf{\text{ernatif 1}}\\ \text{mak}& \text{a}\text{saat dicerminkan terhadap}\\ \text{gari}& \text{s}y=x,\text{adalah}{x}^2=y\\ \text{atau}& y=x^2.\\ \mathbf{\text{Alt}}& \mathbf{\text{ernatif 2}}\\ \text{Jika}& \text{ingin dikerjakan dengan rumus}\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =M_{x=y}\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}y\\ x\end{array}\right)\\ \text{Sela}& \text{njutnya hasilnya disubstitusikan}\\ \text{ke p}& \text{ersamaan}y=\sqrt{x}\Rightarrow x^{\prime }=\sqrt{y^{\prime }}\\ \sqrt{y^{\prime }}& =x^{\prime }\text{maka}\\ y^{\prime }& ={\left(x^{\prime }\right)}^2\text{selanjutnya}\\ y& =x^2\end{array}\end{array}$.

Sebelum dicerminkan terhadap garis y=x

Gambar kurva/grafik setelah cerminkan terhadap garis y=x

$\begin{array}{l}\\ 100.& \text{Transformasi}T\text{adalah pencerminan}\\ & \text{terhadap garis}y={\displaystyle \frac{x}{3}\text{dilanjutkan oleh}}\\ & \text{pencerminan terhadap garis}y=-3x.\\ & \text{Matriks yang bersesuian dengan}\\ & \text{transformasi}T\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)& & \\ \text{b}.\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)& \\ \text{c}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)& \\ \text{d}.\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ \text{e}.\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\end{array}\\ \\ & \mathbf{\text{SBMPTN Tahun 2013 Kode 433}}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \text{sebu}& \text{ah persamaan garis lurus}\\ \text{dapa}& \text{t dituliskan dengan}:y=mx\\ \text{Dike}& \text{tahui pula bahwa ada 2 garis}:\\ y_1& ={\displaystyle \frac{1}{3}x\text{dan}{y}_2=-3x}\\ \text{seba}& \text{gai representasi transformasi}T.\\ \text{Kare}& \text{na}{m}_1\times m_2=\left({\displaystyle \frac{1}{3}}\right)(-3)=-1\\ \text{bera}& \text{rti 2 garis di atas saling tegak}\\ \text{luru}& \text{s dan hal ini seperti rotasi 2}\\ \text{kali}& {90}^{\circ }\text{atau}{180}^{\circ }\\ \text{Jadi},& T=\left(\begin{array}{cc}\cos {180}^{\circ }& -\sin {180}^{\circ }\\ \sin {180}^{\circ }& \cos {180}^{\circ }\end{array}\right)\\ \iff & T=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\end{array}\end{array}$.

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

 $\begin{array}{l}\\ 81.& \text{Bayangan untuk titik A(1,3) oleh rotasi }\\ & \text{dengan pusat}\mathit{\text{O}}(0,0)\text{sejauh}{90}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.(-1,3)& & \text{d}.(1,-3)\\ \text{b}.(-1,-3)& \text{c}.(-3,1)& \text{e}.(3,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Karena rotasi d}& \text{engan pusat O sebesar}{90}^{\circ },\\ \text{maka}& \\ R(O(0,0),{90}^{\circ })& =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\\ \text{sehingga}& \\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-3\\ 1\end{array}\right)\end{array}\end{array}$.

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

$\begin{array}{l}\\ 83.& \text{Sebuah transformasi yang didefiniskan oleh}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \text{Yang merupakan titik invarian (tidak berubah) }\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(0,0)& & \text{d}.(0,-1)\\ \text{b}.(1,-1)& \text{c}.(1,0)& \text{e}.(1,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{D}& \text{iketahui bahwa}:\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \begin{array}{|cccc|}\hline \text{NO}& \text{Titik}& \begin{array}{rl}& \text{Disubstitusikan ke}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\end{array}& \begin{array}{rl}& \text{Keterangan}\\ & \text{Titik}\end{array}\\ \text{a}.& (0,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(0)-4=-4\end{array}& \text{Varian}\\ \text{b}& (1,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(-1)-4=-1\end{array}& \mathbf{\text{Invarian}}\\ \text{c}& (1,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(0)-4=-2\end{array}& \text{Varian}\\ \text{d}& (0,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(-1)-4=-3\end{array}& \text{Varian}\\ \text{e}& (1,1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(1)-4=-3\end{array}& \text{Varian}\\ \hline\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 84.& \text{Bayangan untuk titik P(2,5) oleh rotasi }\\ & \text{dengan pusat}\mathit{\text{A}}(1,3)\text{sejauh}{180}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.(1,0)& & \text{d}.(2,0)\\ \text{b}.(0,1)& \text{c}.(0,2)& \text{e}.(1,2)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Karena rotasi de}& \text{ngan pusat A sebesar}{180}^{\circ },\\ \text{maka}& \\ R(A(1,3),{180}^{\circ })& =\left(\begin{array}{cc}\cos {180}^{\circ }& -\sin {180}^{\circ }\\ \sin {180}^{\circ }& \cos {180}^{\circ }\end{array}\right)\\ & =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ \text{sehingga bayang}& \text{an titik P(2,5)-nya adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}2-1\\ 5-3\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-1\\ -2\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}\end{array}$.

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

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

$\begin{array}{l}\\ 87.& \text{Jika}{M}_x\text{adalah pencerminan terhadap sumbu-X }\\ & \text{dan}{M}_{y=x}\text{adalah pencerminan terhadap garis}\\ & y=x,\text{maka matriks transformasi tunggal }\\ & \text{yang mewakili}{M}_x\circ M_{y=x}=....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)& & \text{d}.\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ \text{b}.\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)& & \text{e}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\\ \text{c}.\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ & \{\begin{array}{ll}{M}_x& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ M_{y=x}& =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\end{array}\\ M_x\circ M_{y=x}& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ & =\left(\begin{array}{cc}0+0& 1+0\\ 0-1& 0+0\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 88.& \text{Diketahui vektor}\overrightarrow{x}\text{dirotasikan terhadap titik asal}\\ & O\text{sebesar}\theta >0\text{searah jarum jam}.\\ & \text{Kemudian hasilnya dicerminkan terhadap garis}y=0\\ & \text{menghasilkan vektor}\overrightarrow{y}.\\ & \text{Jika}\overrightarrow{y}=A.\overrightarrow{x},\text{maka matriks}A-\text{nya adalah}....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)& & \\ \text{b}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)& & \\ \text{c}.\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)& & \\ \text{d}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\\ \text{e}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui bah}\text{wa}:\\ & \{\begin{array}{ll}{M}_x& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ R_{-\theta }& =\left(\begin{array}{cc}\cos (-\theta )& -\sin (-\theta )\\ \sin (-\theta )& \cos (-\theta )\end{array}\right)=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\end{array}\\ & A=M_x\circ R_{-\theta }=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 89.& \text{Titik A(1,-2) dirotasikan sejauh}{15}^{\circ }\\ & \text{kemudian dilanjutkan}{75}^{\circ }\text{dengan pusat }\\ & O(0,0)\text{maka bayangan akhir titik A adalah}...\\ & \begin{array}{l}\\ \text{a}.(-2,1)& & \text{d}.(2,1)\\ \text{b}.(-1,2)& \text{c}.(1,2)& \text{e}.(-2,-1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos ({\theta }_1+{\theta }_2)& -\sin ({\theta }_1+{\theta }_2)\\ \sin ({\theta }_1+{\theta }_2)& \cos ({\theta }_1+{\theta }_2)\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos ({75}^{\circ }+{15}^{\circ })& -\sin ({75}^{\circ }+{15}^{\circ })\\ \sin ({75}^{\circ }+{15}^{\circ })& \cos ({75}^{\circ }+{15}^{\circ })\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2\\ 1\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 90.& \text{Jika garis}3x-2y+5=0\text{dicerminkan }\\ & \text{terhadap garis}y=-x\text{kemudian}\\ & \text{didilatasikan dengan pusat (1,-2) }\\ & \text{dengan faktor skala 2, maka persamaan}\\ & \text{bayangannya adalah}....\\ & \begin{array}{l}\\ \text{a}.x-2y-10=0& & \\ \text{b}.x+2y-10=0& \\ \text{c}.x-6y+5=0& \\ \text{d}.x+2y-12=0\\ \text{e}.2x-3y+18=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{Proses}& \text{untuk refleksinya}\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-y\\ -x\end{array}\right)\\ \text{proses}& \text{dilatasinya}\\ \left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)& =\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\left(\begin{array}{c}{x}^{\prime }-1\\ y^{\prime }+2\end{array}\right)+\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2x^{\prime }-2\\ 2y^{\prime }+4\end{array}\right)+\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2x^{\prime }-1\\ 2y^{\prime }+2\end{array}\right)\\ & =\left(\begin{array}{c}2(-y)-1\\ 2(-x)+2\end{array}\right)\\ & \{\begin{array}{ll}x& =-{\displaystyle \frac{1}{2}(y^{″}-2)}\\ y& =-{\displaystyle \frac{1}{2}(x^{″}+1)}\end{array}\end{array}\\ & \begin{array}{rl}\text{Sehingga persam}& \text{aan bayangan}\\ \text{garisnya adalah}:& \\ 3x& -2y+5=0\\ 3(-{\displaystyle \frac{1}{2}(y^{″}-2)})& -2(-{\displaystyle \frac{1}{2}(x^{″}+1)})+5=0\\ -{\displaystyle \frac{3}{2}{y}^{″}+3}& +(x^{″}+1)+5=0\\ 2x& -3y+6+2+10=0\\ 2x& -3y+18=0\end{array}\end{array}$.

Latihan Soal 8 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 71.& (\mathbf{\text{SPMB 2003}})\\ & \text{Diketahu matriks}\text{A}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right).\\ & \text{Jika}{\text{A}}^t={\text{A}}^{-1}\text{dengan}{\text{A}}^t\\ & \text{adalah transpose matriks A},\\ & \text{maka nilai}ad-bc=....\\ & \begin{array}{l}\\ \text{a}.& -1\text{atau}-\sqrt{2}\\ \text{b}.& 1\text{atau}\sqrt{2}\\ \text{c}.& -\sqrt{2}\text{atau}-\sqrt{2}\\ \text{d}.& -1\text{atau}1\\ \text{e}.& 1\text{atau}-\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\text{matriks}\text{A}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ & \text{dan}{\text{A}}^t={\text{A}}^{-1},\text{maka}\\ & {\text{A}}^t={\text{A}}^{-1}\\ & {\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)}^t={\displaystyle \frac{1}{ad-bc}\times \text{Adjoin Matriks}\text{A}}\\ & \left(\begin{array}{cc}a& c\\ b& d\end{array}\right)={\displaystyle \frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right),}\\ & \text{didapatkan hubungan}\\ & c={\displaystyle \frac{-b}{ad-bc}...............(1)}\\ & b={\displaystyle \frac{-c}{ad-bc}...............(2)}\\ & \text{Persamaan}(2)\text{disubstitusikan ke persamaan}(1)\\ & c={\displaystyle {\displaystyle \frac{-{\displaystyle \frac{-c}{ad-bc}}}{ad-bc}}}\\ & 1=(ad-bc{)}^2\\ & (ad-bc)=-1\text{atau}1\end{array}\end{array}$

$\begin{array}{l}\\ 72.& \text{Diketahu matriks}\text{H}\\ & \text{yang memenuhi persamaan}\\ & \text{H}\left(\begin{array}{cc}3& 2\\ 1& 4\end{array}\right)=\left(\begin{array}{cc}7& 8\\ 4& 6\end{array}\right),\\ & \text{maka nilai dari}det\text{H}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -3\\ \text{b}.& -2\\ \text{c}.& -1\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{|c|}\hline \mathbf{\text{Alternatif 1}}\\ \begin{array}{rl}\text{H.A}& =\text{B}\\ {\text{H.A.A}}^{-1}& ={\text{B.A}}^{-1}\\ \text{H}& ={\text{B.A}}^{-1}\\ & =\left(\begin{array}{cc}7& 8\\ 4& 6\end{array}\right).{\displaystyle \frac{1}{\left|\begin{array}{cc}3& 2\\ 1& 4\end{array}\right|}\left(\begin{array}{cc}4& -2\\ -1& 3\end{array}\right)}\\ & =\left(\begin{array}{cc}7& 8\\ 4& 6\end{array}\right).{\displaystyle \frac{1}{12-2}\left(\begin{array}{cc}4& -2\\ -1& 3\end{array}\right)}\\ & ={\displaystyle \frac{1}{10}\left(\begin{array}{cc}28+(-8)& (-14)+24\\ 16+(-6)& (-8)+18\end{array}\right)}\\ \text{H}& ={\displaystyle \frac{1}{10}\left(\begin{array}{cc}20& 10\\ 10& 10\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)}\\ det\text{H}& =\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|=2.1-1.1=2-1=1\end{array}\\ \mathbf{\text{Alternatif 2}}\\ \begin{array}{rl}\text{H.A}& =\text{B}\{\begin{array}{ll}det\text{H}& =\left|\text{H}\right|\\ det\text{A}& =\left|\text{A}\right|=\left|\begin{array}{cc}3& 4\\ 2& 1\end{array}\right|\\ & =12-2=10\\ det\text{B}& =\left|\text{B}\right|=\left|\begin{array}{cc}7& 8\\ 4& 6\end{array}\right|\\ & =42-32=10\end{array}\\ \left|\text{H}\right|.\left|\text{A}\right|& =\left|\text{B}\right|\\ \left|\text{H}\right|& ={\displaystyle \frac{\left|\text{B}\right|}{\left|\text{A}\right|}}\\ & ={\displaystyle \frac{10}{10}}\\ & =1\end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 73.& (\mathbf{\text{UM UGM 2006}})\\ & \text{Apabila}x\text{dan}y\text{memenuhi}\\ & \text{persamaan matriks}\\ & \left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-1\\ 2\end{array}\right),\\ & \text{maka}x+y=....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ A.X& =B\\ A^{-1}.A.X& =A^{-1}.B\\ A^0.X& =A^{-1}.B\\ X& =A^{-1}.B\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)}^{-1}\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{\left|\begin{array}{cc}1& -2\\ -1& 3\end{array}\right|}}& \left(\begin{array}{cc}3& 2\\ 1& 1\end{array}\right)\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3-2}}& \left(\begin{array}{c}3.(-1)+2.2\\ 1.(-1)+1.2\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}1\\ 1\end{array}\right)\\ x+y& =1+1=2\end{array}\end{array}$

$\begin{array}{l}\\ 74.& (\mathbf{\text{KSM Matematika Kabupten 2019}})\\ & \text{Matriks}A\text{dengan entri bulat dan}\\ & \text{berukuran 2x2},\text{dikalikan dengan matriks}\\ & \left(\begin{array}{cc}1& 2\\ 2& 2\end{array}\right)\text{dari kanan menghasilkan matriks}\\ & \text{yang semua entrinya bilangan prima}.\\ & \text{Jika determinan dari matriks}A\text{juga}\\ & \text{bilangan prima, maka nilai minimum dari}\\ & detA\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 5\\ \text{d}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left(\begin{array}{cc}1& 2\\ 2& 2\end{array}\right)& \times A_{2\times 2}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\\ \left|\begin{array}{cc}1& 2\\ 2& 2\end{array}\right|& \times \left|A_{2\times 2}\right|=\left|\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right|\\ & \left|A_{2\times 2}\right|={\displaystyle \frac{\left|\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right|}{\left|\begin{array}{cc}1& 2\\ 2& 2\end{array}\right|}}\\ & \left|A_{2\times 2}\right|={\displaystyle \frac{(\alpha \delta -\beta \gamma )}{-2}}\\ & \left|A_{2\times 2}\right|={\displaystyle \frac{(\beta \gamma -\alpha \delta )}{2}}\\ \text{Karena}& \left|A_{2\times 2}\right|\text{bilangan prima}\\ \text{akan m}& \text{engakibatkan}(\beta \gamma -\alpha \delta )\\ \text{harus h}& \text{abis dibagi}2,\text{oleh karenanya}\\ \text{menyeb}& \text{abkan}(\beta \gamma -\alpha \delta )\text{berupa bilangan}\\ \text{genap.}& \text{Dan karena}(\beta \gamma -\alpha \delta )\text{genap},\\ \text{maka p}& \text{astilah}\left|A_{2\times 2}\right|\text{juga bernilai genap}\\ \text{sehingg}& \text{a nilai}\left|A_{2\times 2}\right|\text{pastilah 2}\end{array}\end{array}$

$\begin{array}{l}\\ 75.& (\mathbf{\text{UM UGM 2005}})\text{Jika}\\ & \left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right)=\left(\begin{array}{cc}\sin A& \cos A\end{array}\right)\\ & \text{dan}A\text{suatu konstanta, maka}x+y=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right)=\left(\begin{array}{cc}\sin A& \cos A\end{array}\right)\\ & \left(\begin{array}{cc}x\sin \alpha -y\cos \alpha & x\cos \alpha +y\sin \alpha \end{array}\right)=\left(\begin{array}{cc}\sin A& \cos A\end{array}\right)\\ & \{\begin{array}{ll}\sin A& =x\sin \alpha -y\cos \alpha =\sqrt{x^2+y^2}\cos (\alpha -{\tan }^{-1}{\displaystyle \frac{x}{-y}})\\ \cos A& =x\cos \alpha +y\sin \alpha =\sqrt{x^2+y^2}\cos (\alpha -{\tan }^{-1}{\displaystyle \frac{y}{x}})\end{array}\\ & \text{Supaya}\cos A=\sqrt{x^2+y^2}\cos (\alpha -{\tan }^{-1}{\displaystyle \frac{y}{x}}),\text{maka}\\ & \{\begin{array}{ll}\sqrt{x^2+y^2}& =1\\ {\tan }^{-1}{\displaystyle \frac{y}{x}}& =0\Rightarrow \{\begin{array}{ll}y& =0\\ x& =1\end{array}\end{array}\\ & \text{Sehingga}x+y=1+0=1\end{array}\end{array}$.

$\begin{array}{l}\\ 76.& (\mathbf{\text{UM UGM 2004}})\\ & \text{Nilai-nilai}x\text{agar matriks}\\ & \left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)\\ & \text{tidak memiliki invers adalah}....\\ & \begin{array}{l}\\ \text{a}.& 4\text{atau}5\\ \text{b}.& -2\text{atau}2\\ \text{c}.& -4\text{atau}5\\ \text{d}.& -6\text{atau}4\\ \text{e}.& 0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{supaya matriks}\left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)\\ & \text{tidak memiliki invers},\text{maka}\\ & \text{determinan matriks}\left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)=0\\ & \text{Sehingga}\\ & \left|\begin{array}{cc}5x& 5\\ 4& x\end{array}\right|=0\\ & \iff 5x^2-20=0\\ & \iff x^2=4\\ & \iff x=\pm 2\end{array}\end{array}$

$\begin{array}{l}\\ 77.& (\mathbf{\text{UM UGM 2005}})\\ & \text{Matriks}\left(\begin{array}{cc}x& 1\\ -2& 1-x\end{array}\right)\\ & \text{tidak memiliki invers untuk}\\ & \text{nilai}x=....\\ & \begin{array}{l}\\ \text{a}.& -1\text{atau}-2\\ \text{b}.& -1\text{atau}0\\ \text{c}.& -1\text{atau}1\\ \text{d}.& -1\text{atau}2\\ \text{e}.& 1\text{atau}2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \text{Mirip dengan pembahasan no. 26}\\ & \begin{array}{rl}& \text{Nilai}\left|\begin{array}{cc}x& 1\\ -2& 1-x\end{array}\right|=0\\ & \iff x-x^2-(-2)=0\\ & \iff 2+x-x^2=0\\ & \iff x^2-x-2=0\\ & \iff (x-2)(x+1)=0\\ & \iff x=2\text{atau}x=-1\end{array}\end{array}$

$\begin{array}{l}\\ 78.& (\mathbf{\text{Mat Das SIMAK UI 2014}})\\ & \text{Jika matriks}\text{A}\text{adalah invers}\\ & \text{dari matriks}{\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)\text{dan}}\\ & \text{A}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right)\text{maka nilai}2x+y\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{10}{3}}\\ \text{b}.& -{\displaystyle \frac{1}{3}}\\ \text{c}.& 1\\ \text{d}.& {\displaystyle \frac{9}{7}}\\ \text{e}.& {\displaystyle \frac{20}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Misalkan diketahui matriks}\\ & \text{B}={\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right),}\\ & \text{maka}\text{A}={\left({\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)}\right)}^{-1}\\ & \text{selanjutnya}\text{A}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & \left(\begin{array}{c}x\\ y\end{array}\right)=A^{-1}\left(\begin{array}{c}1\\ 3\end{array}\right),\\ & \text{ingat bahwa}{\left({\mathbf{\text{A}}}^{-1}\right)}^{-1}=\mathbf{\text{A}}\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\left({\left({\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)}\right)}^{-1}\right)}^{-1}\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)\left(\begin{array}{c}1\\ 3\end{array}\right)}\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3}\left(\begin{array}{c}-1-9\\ 4+15\end{array}\right)=\left(\begin{array}{c}{\displaystyle -\frac{10}{3}}\\ {\displaystyle \frac{19}{3}}\end{array}\right)}\\ & 2x+y=2(-{\displaystyle \frac{10}{3}})+\frac{19}{3}\\ & ={\displaystyle \frac{-20+19}{3}=-{\displaystyle \frac{1}{3}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 79.& \text{Suatu translasi yang memetakan titik P(9,8) }\\ & \text{ke titik}{\text{P}}^{\prime }(14,-2)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{c}5\\ -10\end{array}\right)& & \text{d}.\left(\begin{array}{c}6\\ 6\end{array}\right)\\ \text{b}.\left(\begin{array}{c}5\\ 6\end{array}\right)& \text{c}.\left(\begin{array}{c}23\\ -10\end{array}\right)& \text{e}.\left(\begin{array}{c}5\\ 2\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\text{T}+\left(\begin{array}{c}x\\ y\end{array}\right)\\ \text{T}& =\left(\begin{array}{c}{x}^{\prime }-x\\ y^{\prime }-y\end{array}\right)\\ & =\left(\begin{array}{c}14-9\\ -2-8\end{array}\right)\\ & =\left(\begin{array}{c}5\\ -10\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 80.& \text{Sebuah transformasi yang didefiniskan oleh}\\ & \{\begin{array}{ll}{x}^{\prime }& =2x+3y\\ y^{\prime }& =3x+2y\end{array}\\ & \text{Maka bayangan titik M}(2,-1)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(7,10)& & \text{d}.(1,10)\\ \text{b}.(10,7)& \text{c}.(1,4)& \text{e}.(4,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ & \{\begin{array}{ll}{x}^{\prime }& =2x+3y\\ y^{\prime }& =3x+2y\end{array}\\ \begin{array}{c}x=2\\ y=-1\end{array}\}& \Rightarrow \{\begin{array}{ll}{x}^{\prime }& =2(2)+3(-1)=4-3=1\\ y^{\prime }& =3(2)+2(-1)=6-2=4\end{array}\end{array}\end{array}$.

Latihan Soal 7 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 61.& \text{Diketahui matriks}\\ & \text{A}=\left(\begin{array}{cc}{}^a\log 6& 2\\ 1& a+3b\end{array}\right),\\ & \text{B}=\left(\begin{array}{cc}0& -5\\ -6& 3a-5b\end{array}\right),\text{dan}\\ & \text{C}=\left(\begin{array}{cc}{}^a\log 2& -{\displaystyle \frac{1}{2}}\\ -2(b+c)& 3\end{array}\right),\\ & \text{serta}\text{I}\text{adalah matriks identitas}.\\ & \text{Jika}2\text{A}+\text{B}-2\text{C}=2\text{I},\\ & \text{maka nilai}4a+b+c\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 5\\ \text{c}.& 7\\ \text{d}.& 11\\ \text{e}.& 13\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 2\text{A}+\text{B}-2\text{C}=2\text{I}\\ & 2\left(\begin{array}{cc}{}^a\log 6& 2\\ 1& a+3b\end{array}\right)+\left(\begin{array}{cc}0& -5\\ -6& 3a-5b\end{array}\right)\\ & -2\left(\begin{array}{cc}{}^a\log 2& -{\displaystyle \frac{1}{2}}\\ -2(b+c)& 3\end{array}\right)=2\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ & \left(\begin{array}{cc}2.{}^a\log 6-2.{}^2\log 2& 2.2-5-2(-{\displaystyle \frac{1}{2}})\\ 2.1-6-2(-2(b+c))& 2(a+3b)+3a-5b-2.3\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\\ & \{\begin{array}{ll}2& =2.{}^a\log 6-2.{}^2\log 2\\ 0& =2.1-6-2(-2(b+c))\\ 2& =2(a+3b)+3a-5b-2.3\end{array}\\ & \begin{array}{|cc|}\hline \text{dari persamaan}(1)& \text{dari persamaan}(2)\\ \begin{array}{rl}2.{}^a\log 6-2.{}^2\log 2& =2\\ {}^a\log 6^2-{}^2\log 2^2& =2\\ {}^a\log {\displaystyle \frac{6^2}{2^2}}& =2\\ {}^a\log 9& =2\\ 9& =a^2\\ 3& =a\\ 12& =4a\end{array}& \begin{array}{rl}2.1-6-2(-2(b+c))& =0\\ 2-6+4(b+c)& =0\\ 4(b+c)& =4\\ b+c& =1\\ & \\ \text{sehingga diperoleh}& ,\\ 4a+b+c=12+1& \\ =13& \end{array}\\ \hline\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 62.& \text{Jika}\left(\begin{array}{cc}-4x& 2y\\ y& x\end{array}\right)\left(\begin{array}{c}2\\ -3\end{array}\right)=\left(\begin{array}{c}2\\ -12\end{array}\right),\\ & \text{maka nilai}xy=....\\ & \begin{array}{l}\\ \text{a}.& -6\\ \text{b}.& -3\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\left(\begin{array}{cc}-4x& 2y\\ y& x\end{array}\right)\left(\begin{array}{c}2\\ -3\end{array}\right)& =\left(\begin{array}{c}2\\ -12\end{array}\right)\\ \left(\begin{array}{c}-4x.2+2y.-3\\ y.2+x.-3\end{array}\right)& =\left(\begin{array}{c}2\\ -12\end{array}\right)\\ \left(\begin{array}{c}-8x-6y\\ -3x+2y\end{array}\right)& =\left(\begin{array}{c}2\\ -12\end{array}\right)\\ \\ \mathbf{\text{SPLDV}}& \end{array}& \begin{array}{rl}-8x-6y& =2(\times 1)\\ -3x+2y& =-12(\times 3)\\ \text{menjadi}& \\ -8x-6y& =2\\ -9x+6y& =-36{}_{+}\\ ----& ---\\ -17x& =-34\\ x& =2\end{array}\\ \begin{array}{rl}-8x-6y& =2\\ -8(2)-6y& =2\\ -16-6y& =2\\ -6y& =2+16\\ -6y& =18\\ y& =-3\\ \text{sehingga}& \\ xy& =2.(-3)=-6\end{array}& \\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 63.& \text{Diketahui}\text{N}=\left(\begin{array}{cc}-2& 3\\ -1& 4\end{array}\right)\\ & \text{dan}\text{M}=\left(\begin{array}{cc}-1& 3\\ -1& 5\end{array}\right).\\ & \text{Jika}{\text{N}}^2=p\text{N}-q\text{M},\\ & :\text{maka nilai}p-q=....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\text{N}}^2=p\text{N}-q\text{M}\\ & \left(\begin{array}{cc}-2& 3\\ -1& 4\end{array}\right)\times \left(\begin{array}{cc}-2& 3\\ -1& 4\end{array}\right)=p\left(\begin{array}{cc}-2& 3\\ -1& 4\end{array}\right)-q\left(\begin{array}{cc}-1& 3\\ -1& 5\end{array}\right)\\ & \left(\begin{array}{cc}-2.-2+3.-1& -2.3+3.4\\ -1.-2+4.-1& -1.3+4.4\end{array}\right)=\left(\begin{array}{cc}-2p+q& 3p-3q\\ -p+q& 4p-5q\end{array}\right)\\ & \left(\begin{array}{cc}4-3& -6+12\\ 2-4& -3+16\end{array}\right)=\left(\begin{array}{cc}-2p+q& 3p-3q\\ -p+q& 4p-5q\end{array}\right)\\ & \left(\begin{array}{cc}1& 6\\ -2& 13\end{array}\right)=\left(\begin{array}{cc}-2p+q& 3p-3q\\ -p+q& 4p-5q\end{array}\right)\\ \\ & \begin{array}{|cc|}\hline \begin{array}{rl}-2p+q& =1\\ -p+q& =-2{}_{-}\\ ----& ---\\ -p& =3\\ p& =-3\\ & \\ & \end{array}& \begin{array}{rl}-p+q& =-2\\ -(-3)+q& =-2\\ q& =-2-3\\ q& =-5\\ \text{sehingga}& \text{didapatkan}\\ p-q& =-3-(-5)\\ & =2\end{array}\\ \hline\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 64.& \text{Diketahui matriks}\text{Z}=\left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)\\ & \text{dan}f(x)=x^2-x.\\ & \text{Jika}f(\text{Z})=\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right),\\ & \text{maka nilai}{p}^2-q^2=....\\ & \begin{array}{l}\\ \text{a}.& 5\\ \text{b}.& 7\\ \text{c}.& 9\\ \text{d}.& 12\\ \text{e}.& 15\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(\text{Z})& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ {\text{Z}}^2-\text{Z}& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ \left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)\times \left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)-\left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ \left(\begin{array}{cc}4-18& -12+30\\ 6-15& -18+25\end{array}\right)-\left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ \left(\begin{array}{cc}-12& 12\\ -6& 2\end{array}\right)& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\end{array}\\ & \begin{array}{rl}\text{Sehingga}& \\ -12& =-3p-8q.................(1)\\ -1& =p+q......................(2)\\ \text{persamaan}& (2)\text{ke persamaan}(1)\\ -12& =-3p-3q-5q=-3(p+q)-5q\\ -12& =-3(-1)-5q\\ -12& =3-5q\\ 5q& =3+12\\ q& =3........................(3)\\ \text{persamaan}& (3)\text{ke persamaan}(2)\\ p+q& =-1\\ p& =-1-q=-1-3=-4\\ p^2-q^2& =(-4{)}^2-3^2=16-9\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 65.& (\mathbf{\text{SBMPTN 2013}})\\ & \text{Jika}A=\left(\begin{array}{ccc}2& -1& 1\\ a& b& c\end{array}\right),\\ & B=\left(\begin{array}{cc}-2& 1\\ 1& -1\\ 0& 2\end{array}\right)\text{dan}\\ & AB=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \text{maka nilai}2c-a=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& AB=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \left(\begin{array}{ccc}2& -1& 1\\ a& b& c\end{array}\right)\left(\begin{array}{cc}-2& 1\\ 1& -1\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \left(\begin{array}{cc}-5& 5\\ -2a+b& a-b+2c\end{array}\right)=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \begin{array}{l}\\ -2a+b& =3& \\ a-b+2c& =-3& +\\ 2c-a& =0\end{array}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Budhi, W.S. 2018. Bupena Matematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: ERLANGGA
2. Kanginan, M., Terzalgi, Z. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: SEWU.
3. Sharma, S. N. 2017. Jelajah Matematika 2 SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
4. Suparmin, S. Malau, A. 2014. Mainstream Matematika Dasar & Matematika IPA untuk Siswa SMA/MA Kelompok IPA. Bandung: YRAMA WIDYA.
5. 
   

$\begin{array}{l}\\ 66.& \text{Determinan untuk matriks}\left(\begin{array}{cc}2& -5\\ 3& -1\end{array}\right)=....\\ & \begin{array}{l}\\ \text{a}.& -17\\ \text{b}.& -13\\ \text{c}.& 11\\ \text{d}.& 13\\ \text{e}.& 17\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Determinan}& \text{dari matriks}\left(\begin{array}{cc}2& -5\\ 3& -1\end{array}\right)\\ & =\left|\begin{array}{cc}2& -5\\ 3& -1\end{array}\right|=2(-1)-3(-5)\\ & =-2+15\\ & =13\end{array}\end{array}$

$\begin{array}{l}\\ 67.& \text{Determinan untuk matriks}\\ & \left(\begin{array}{ccc}2& -1& -1\\ 1& 4& -1\\ 1& -2& 3\end{array}\right)=....\\ & \begin{array}{l}\\ \text{a}.& 10\\ \text{b}.& 18\\ \text{c}.& 22\\ \text{d}.& 30\\ \text{e}.& 36\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Determinan}\text{dari matriks}\\ & \left(\begin{array}{ccc}2& -1& -1\\ 1& 4& -1\\ 1& -2& 3\end{array}\right)\\ & =\left|\begin{array}{ccc}2& -1& -1\\ 1& 4& -1\\ 1& -2& 3\end{array}\right|\\ & =+(2.4.3)+(-1.-1.1)+(-1.1.-2)\\ & -(1.4.-1)-(-2.-1.2)-(3.1.-1)\\ & =24+1+2+4-24+3\\ & =10\end{array}\end{array}$

$\begin{array}{l}\\ 68.& \text{Jika diketahu matriks}\\ & \text{A}=\left(\begin{array}{cc}x+3& -2\\ -16& 2x-6\end{array}\right),\\ & \text{maka nilai dari}x\text{supaya matriks}\\ & \text{A tidak memiliki invers adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{Invers}& \text{dari matriks A adalah}{\text{A}}^{-1}.\\ {\text{A}}^{-1}& ={\displaystyle \frac{1}{det\text{A}}\times Adjoin\text{A}.}\\ \text{Karen}& \text{a}\text{tidak memiliki invers},\\ \text{maka}& det\text{A}=0,\text{sehingga}\\ det\text{A}& =\left|\begin{array}{cc}x+3& -2\\ -16& 2x-6\end{array}\right|=0\\ & \iff (x+3)(2x-6)-(-16.-2)=0\\ & (\text{masing-masing ruas dibagi 2})\\ & \iff (x+3)(x-3)-16=0\\ & \iff x^2-9-16=0\\ & \iff x^2-25=0\\ & \iff (x+5)(x-5)=0\\ & \iff x+5=0\text{atau}x-5=0\\ & \iff x=-5\text{atau}x=5\end{array}\end{array}$

$\begin{array}{l}\\ 69.& \text{Jika}\left|\begin{array}{cc}{5}^{2x}& -5\\ 1& 1\end{array}\right|={6.5}^x\\ & \text{maka}{5}^{2x}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 625\text{atau}1\\ \text{b}.& 25\text{atau}1\\ \text{c}.& 25\text{atau}0\\ \text{d}.& 5\text{atau}1\\ \text{e}.& 5\text{atau}0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& 5^{2x}+5={6.5}^x\\ & 5^{2x}-{6.5}^x+5=0\\ & (5^x-1)(5^x-5)=0\\ & 5^x-1=0\text{atau}{5}^x-5=0\\ & 5^x=1\text{atau}{5}^x=5\\ & 5^x=5^0\text{atau}{5}^x=5^1\\ & x=0\text{atau}x=1\\ & \text{maka}\\ & 5^{2x}=\{\begin{array}{ll}{5}^{2.1}& =5^2=25\\ 5^{2.0}& =5^0=1\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 70.& \text{Diketahu determinan suatu}\\ & \text{matriks adalah}\left|\begin{array}{ccc}x& 1& 2\\ x& 1& x\\ 5& -3& 7\end{array}\right|=0.\\ & \text{Jika}p\text{dan}q\text{adalah akar-akar}\\ & \text{yang memenuhi persamaan tersebut}\\ & \text{maka nilai dari}p+q\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -3\\ \text{b}.& -{\displaystyle \frac{1}{3}}\\ \text{c}.& -1\\ \text{d}.& {\displaystyle \frac{1}{3}}\\ \text{e}.& 3\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Diketahui ba}& \text{hwa}:\\ \left|\begin{array}{ccc}x& 1& 2\\ x& 1& x\\ 5& -3& 7\end{array}\right|& =0\\ +(x.1.7)+& (1.x.5)+(2.x.-3)\\ -(5.1.2)& -(-3.x.x)-(7.x.1)=0\\ 7x+5x-6x& -10+3x^2-7x=0\\ 3x^2-x-10& =0\{\begin{array}{ll}p& \text{salah satu akar}\\ q& \text{salah satu akar yang lain},\end{array}\\ \text{dengan}& \{\begin{array}{ll}a& =3\\ b& =-1\\ c& =-10\end{array}.\\ \text{maka}p+q& =-{\displaystyle \frac{b}{a}=-{\displaystyle \frac{-1}{3}}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$

Latihan Soal 6 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 51.& \text{Diketahui matriks}\\ & \text{A}=\left(\begin{array}{cccc}2020& -4& -3& 2\\ 2020& -6& -7& 1\\ 2020& 4& -3& 0\\ 2020& 6& -7& 8\end{array}\right)\\ & \text{Ordo dari matriks}\text{A}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3\times 2& & & \\ \text{b}.& 3\times 3\\ \text{c}.& 3\times 4\\ \text{d}.& 4\times 3\\ \text{e}.& 4\times 4\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \text{Cukup jelas}\\ & \text{Karena matriknya mengandung}\\ & \text{4 baris}\times \text{4 kolom}\end{array}$

$\begin{array}{l}\\ 52.& \text{Diketahui matriks}\\ & \text{B}=\left(\begin{array}{cccc}1& 2& 3& 2020\\ 5& 13& 7& 2019\\ 11& 14& 3& -2018\\ -15& 6& 17& 2017\end{array}\right)\\ & \text{Jika}{\text{b}}_{ij}\text{menunjukkan elemen}\\ & \text{yang terletak pada baris ke}-i\\ & \text{dan kolom ke}-j\text{pada matriks B}\\ & \text{ di atas, maka}{b}_{43}=....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 9\\ \text{c}.& -1\\ \text{d}.& 3\\ \text{e}.& 17\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{Perh}& \text{atikan bahwa}\\ {\text{B}}_{4\times 4}& =\left(\begin{array}{cccc}{b}_{11}& b_{12}& b_{13}& b_{14}\\ b_{21}& b_{22}& b_{23}& b_{24}\\ b_{31}& b_{32}& b_{33}& b_{34}\\ b_{41}& b_{42}& b_{43}& b_{44}\end{array}\right)\\ & =\left(\begin{array}{cccc}1& 2& 3& 2020\\ 5& 13& 7& 2019\\ 11& 14& 3& -2018\\ -15& 6& 17& 2017\end{array}\right)\\ \text{sehi}& \text{ngga entri}{b}_{43}=17\end{array}\end{array}$

$\begin{array}{l}\\ 53.& \text{Diketahui matriks}\text{C}\text{adalah matriks}\\ & \text{berordo}3\times 3.\text{Jika}{\text{c}}_{ij}=4j-5i,\\ & \text{maka matriks C tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left(\begin{array}{ccc}-1& 3& 7\\ -6& -2& 2\\ -11& -7& -3\end{array}\right)\\ \text{b}.& \left(\begin{array}{ccc}-1& 7& 3\\ -6& 2& -2\\ -7& -11& -3\end{array}\right)\\ \text{c}.& \left(\begin{array}{ccc}-1& -7& -11\\ -6& 7& 3\\ -2& 2& -3\end{array}\right)\\ \text{d}.& \left(\begin{array}{ccc}-1& -6& -11\\ 3& -2& 2\\ 7& 2& -3\end{array}\right)\\ \text{e}.& \left(\begin{array}{ccc}-1& -2& -3\\ 3& -6& -11\\ 7& -7& 2\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}{c}_{ij}=4j-5i,\\ \text{mak}& \text{a}\\ {\text{C}}_{3\times 3}& =\left(\begin{array}{ccc}{c}_{11}& c_{12}& c_{13}\\ c_{21}& c_{22}& c_{23}\\ c_{31}& c_{32}& c_{33}\end{array}\right)\\ & =\left(\begin{array}{ccc}4.1-5.1& 4.2-5.1& 4.3-5.1\\ 4.1-5.2& 4.2-5.2& 4.3-5.2\\ 4.1-5.3& 4.2-5.3& 4.3-5.3\end{array}\right)\\ & =\left(\begin{array}{ccc}4-5& 8-5& 12-5\\ 4-10& 8-10& 12-10\\ 4-15& 8-15& 12-15\end{array}\right)\\ & =\left(\begin{array}{ccc}-1& 3& 7\\ -6& -2& 2\\ -11& -7& -3\end{array}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 54.& \text{Jika diketahui matriks}\\ & \text{X}=\left(\begin{array}{ccc}-7& 9& -1\\ 4& -6& 15\end{array}\right).\\ & \text{maka transpose matriks}\text{X}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\text{X}}^t=\left(\begin{array}{ccc}4& -6& 15\\ -7& 9& -1\end{array}\right)\\ \text{b}.& {\text{X}}^t=\left(\begin{array}{ccc}15& -6& 4\\ -1& 9& -7\end{array}\right)\\ \text{c}.& {\text{X}}^t=\left(\begin{array}{cc}-7& 4\\ 9& -6\\ -1& 15\end{array}\right)\\ \text{d}.& {\text{X}}^t=\left(\begin{array}{cc}4& -7\\ -6& 9\\ 15& -1\end{array}\right)\\ \text{e}.& {\text{X}}^t=\left(\begin{array}{cc}15& -1\\ -6& 9\\ 4& -7\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}\\ {\text{X}}_{2\times 3}& =\left(\begin{array}{ccc}{x}_{11}& x_{12}& x_{13}\\ x_{21}& x_{22}& x_{23}\end{array}\right)\\ & =\left(\begin{array}{ccc}-7& 9& -1\\ 4& -6& 15\end{array}\right)\\ \text{maka}& \\ {\text{X}}_{3\times 2}^t& =\left(\begin{array}{cc}{x}_{11}& x_{21}\\ x_{12}& x_{22}\\ x_{13}& x_{23}\end{array}\right)=\left(\begin{array}{cc}-7& 4\\ 9& -6\\ -1& 15\end{array}\right)\\ \text{adal}& \text{ah sebuah}\text{matriks baru}\\ \text{deng}& \text{an ordo}3\times 2\end{array}\end{array}$

$\begin{array}{l}\\ 55.& \text{Diketahui matriks}\text{P}=\left(\begin{array}{cc}a& 4\\ 2b& 3c\end{array}\right)\\ & \text{dan}\text{Q}=\left(\begin{array}{cc}2c-3b& 2a+1\\ a& b+7\end{array}\right).\\ & \text{Nilai}c\text{yang memenuhi jika}\\ & \text{P}=2{\text{Q}}^t\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& 3\\ \text{c}.& 5\\ \text{d}.& 8\\ \text{e}.& 10\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{P}& =2{\text{Q}}^t\\ \left(\begin{array}{cc}a& 4\\ 2b& 3c\end{array}\right)& =2{\left(\begin{array}{cc}2c-3b& 2a+1\\ a& b+7\end{array}\right)}^t\\ \left(\begin{array}{cc}a& 4\\ 2b& 3c\end{array}\right)& =2\left(\begin{array}{cc}2c-3b& a\\ 2a+1& b+7\end{array}\right)\\ \left(\begin{array}{cc}a& 4\\ 2b& 3c\end{array}\right)& =\left(\begin{array}{cc}4c-6b& 2a\\ 4a+2& 2b+14\end{array}\right)\\ & (\text{kesamaan 2 buah matriks})\\ \text{akibat}& \text{nya}\\ & \{\begin{array}{ll}a& =4c-6b....(1)\\ 4& =2a..........(2)\\ 2b& =4a+2........(3)\\ 3c& =2b+14........(4)\end{array}\\ \text{dari}& \text{persamaan}(2)\\ & 2a=4\Rightarrow a=2....(5)\\ \text{pers}& \text{amaan}(5)\text{hasilnya}\\ \text{disu}& \text{bstitusikan ke persamaan}(3),\\ \text{yait}& \text{u}\\ 2b& =4a+2\Rightarrow 2b=4(2)+2=10\\ b& =5.....................(6)\\ \text{pers}& \text{amaan}(6)\text{hasilnya disbstitusikan}\\ \text{ke p}& \text{ersamaan}\\ (4),& \text{dan akan mendapatkan}\\ 3c& =2b+14\Rightarrow 3c=2(5)+14=24\\ c& =8\end{array}\end{array}$.

$\begin{array}{l}\\ 56.& \text{Diketahui matriks}\\ & \text{M}=\left(\begin{array}{ccc}-6& 9& -15\\ 3& -6& 12\end{array}\right)\\ & \text{dan}\text{N}=\left(\begin{array}{ccc}2& -3& 5\\ -1& 2& -4\end{array}\right).\\ & \text{Nilai}k\text{yang memenuhi jika}\\ & \text{M}=k\text{N}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{3}}\\ \text{b}.& {\displaystyle \frac{1}{3}}\\ \text{c}.& -1\\ \text{d}.& -3\\ \text{e}.& 3\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahu bahwa}\\ & \text{M}=k\text{N}\\ & (\text{perkalian suatu matrik dengan skalar})\\ & \left(\begin{array}{ccc}-6& 9& -15\\ 3& -6& 12\end{array}\right)\\ & =\left(\begin{array}{ccc}-3.2& -3.-3& -3.5\\ -3.-1& -3.2& -3.-4\end{array}\right)\\ & =-3\left(\begin{array}{ccc}2& -3& 5\\ -1& 2& -4\end{array}\right)\\ & =k\left(\begin{array}{ccc}2& -3& 5\\ -1& 2& -4\end{array}\right)\\ & \text{sehingga dari kesamaan tersebut}\\ & \text{maka}k=-3\end{array}\end{array}$

$\begin{array}{l}\\ 57.& \text{Hasil dari}\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right)\times \left(\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\end{array}\right)\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& \left(\begin{array}{cc}22& 28\\ 49& 64\end{array}\right)\\ \text{b}.& \left(\begin{array}{cc}22& 49\\ 28& 64\end{array}\right)\\ \text{c}.& \left(\begin{array}{cc}64& 28\\ 49& 22\end{array}\right)\\ \text{d}.& \left(\begin{array}{ccc}2& 8& 18\\ 4& 15& 30\end{array}\right)\\ \text{e}.& \left(\begin{array}{ccc}1& 4& 6\\ 4& 15& 30\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right)}_{2\times 3}\times {\left(\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\end{array}\right)}_{3\times 2}\\ & ={\left(\begin{array}{cc}1.1+2.3+3.5& 1.2+2.4+3.6\\ 4.1+5.3+6.5& 4.2+5.4+6.6\end{array}\right)}_{2\times 2}\\ & ={\left(\begin{array}{cc}1+6+15& 2+8+18\\ 4+15+30& 8+20+36\end{array}\right)}_{2\times 2}\\ & ={\left(\begin{array}{cc}22& 28\\ 49& 64\end{array}\right)}_{2\times 2}\end{array}\end{array}$

$\begin{array}{l}\\ 58.& \text{Jika diketahui matriks}\\ & \text{A}=\left(\begin{array}{cc}0& 1\\ 3& 2\end{array}\right).\\ & \text{maka hasil dari}{\text{A}}^3\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left(\begin{array}{cc}5& 8\\ 20& 22\end{array}\right)\\ \text{b}.& \left(\begin{array}{cc}6& 7\\ 21& 20\end{array}\right)\\ \text{c}.& \left(\begin{array}{cc}6& 7\\ 20& 22\end{array}\right)\\ \text{d}.& \left(\begin{array}{cc}7& 8\\ 20& 23\end{array}\right)\\ \text{e}.& \left(\begin{array}{cc}7& 9\\ 20& 23\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}\\ \text{A}& =\left(\begin{array}{cc}0& 1\\ 3& 2\end{array}\right)\\ \text{mak}& \text{a}\\ {\text{A}}^2& =\text{A}\times \text{A}\\ & =\left(\begin{array}{cc}0& 1\\ 3& 2\end{array}\right)\times \left(\begin{array}{cc}0& 1\\ 3& 2\end{array}\right)\\ & =\left(\begin{array}{cc}0+3& 0+2\\ 0+6& 3+4\end{array}\right)\\ & =\left(\begin{array}{cc}3& 2\\ 6& 7\end{array}\right)\\ {\text{A}}^3& ={\text{A}}^2\times \text{A}\\ & =\left(\begin{array}{cc}3& 2\\ 6& 7\end{array}\right)\times \left(\begin{array}{cc}0& 1\\ 3& 2\end{array}\right)\\ & =\left(\begin{array}{cc}0+6& 3+4\\ 0+21& 6+14\end{array}\right)\\ & =\left(\begin{array}{cc}6& 7\\ 21& 20\end{array}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 59.& (\mathbf{\text{SBMPTN Mat IPA 2014}})\\ & \text{Jika}\text{A}\text{adalah matriks yang berordo}\\ & 2\times 2\text{dan memenuhi}\\ & \left(\begin{array}{cc}x& 1\end{array}\right)\times \text{A}\times \left(\begin{array}{c}x\\ 1\end{array}\right)=x^2-5x+8,\\ & \text{maka matriks A yang mungkin adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left(\begin{array}{cc}1& -5\\ 8& 0\end{array}\right)\\ \text{b}.& \left(\begin{array}{cc}1& 5\\ 8& 0\end{array}\right)\\ \text{c}.& \left(\begin{array}{cc}1& 8\\ -5& 0\end{array}\right)\\ \text{d}.& \left(\begin{array}{cc}1& 3\\ -8& 8\end{array}\right)\\ \text{e}.& \left(\begin{array}{cc}1& -3\\ 8& -8\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{cc}x& 1\end{array}\right)\times \text{A}\times \left(\begin{array}{c}x\\ 1\end{array}\right)& =x^2-5x+8\\ \left(\begin{array}{cc}x& 1\end{array}\right)\times \left(\begin{array}{cc}p& q\\ r& s\end{array}\right)\times \left(\begin{array}{c}x\\ 1\end{array}\right)& =x^2-5x+8\\ \left(\begin{array}{cc}xp+r& xq+s\end{array}\right)\times \left(\begin{array}{c}x\\ 1\end{array}\right)& =x^2-5x+8\\ \left(\begin{array}{c}{x}^{2}p+xr+xq+s\end{array}\right)& =x^2-5x+8\\ p{x}^2+(q+r)x+s& =x^2-5x+8\end{array}\\ & \begin{array}{rl}& \{\begin{array}{ll}p& =1\\ q+r& =-5\\ s& =8\end{array}\Rightarrow \left(\begin{array}{cc}1& ...\\ ...& 8\end{array}\right)\\ & \text{Sehingga yang paling mungkin}\\ & \text{adalah}\left(\begin{array}{cc}1& 3\\ -8& 8\end{array}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 60.& \text{Diketahui}\\ & \left(\begin{array}{cc}{}^x\log a& \log (2a-6)\\ \log (b-2)& 1\end{array}\right)=\left(\begin{array}{cc}\log b& 1\\ \log a& 1\end{array}\right)\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 6\\ \text{e}.& 8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \left(\begin{array}{cc}{}^x\log a& \log (2a-6)\\ \log (b-2)& 1\end{array}\right)=\left(\begin{array}{cc}\log b& 1\\ \log a& 1\end{array}\right)\\ & \text{maka}\\ & \{\begin{array}{ll}{}^x\log a& =\log b.........(1)\\ \log (2a-6)& =1..............(2)\\ \log (b-2)& =\log a.........(3)\end{array}\\ & \text{Sehingga}\text{dari persamaan}(2)\\ & \text{akan didapatkan}\\ & \log (2a-6)=1=\log 10\\ & (2a-6)=10\\ & a=8...........................(4)\\ & \text{persamaan}(4)\text{ke persamaan}(3),\\ & \text{maka}\\ & \log (b-2)=\log a\\ & b-2=a=8\\ & b=10.................................(5)\\ & \text{Selanjutnya dari persamaan}(5)\\ & \text{akan diperoleh}\\ & {}^x\log a=\log b\\ & {}^x\log 8=\log 10=1\\ & x^1=8\\ & \iff x=8\end{array}\end{array}$