PAS MATEMATIKA BLOG

Latihan Soal 9 Persiapan PAS Gasal Matematika Wajib Kelas XI

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

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

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

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

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

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

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

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

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

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

Latihan Soal 8 Persiapan PAS Gasal Matematika Wajib Kelas XI

$\begin{array}{ll}\\ 71.&(\textbf{SPMB 2003})\\ &\textrm{Diketahu matriks}\: \: \textrm{A}=\begin{pmatrix}a&b\\ c&d \end{pmatrix}.\\ &\textrm{Jika}\: \: \: \textrm{A}^{t}=\textrm{A}^{-1}\: \: \textrm{dengan}\: \: \textrm{A}^{t}\\ &\textrm{adalah transpose matriks A},\\ &\textrm{maka nilai}\: \: ad-bc=....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1\: \: \textrm{atau}\: \: -\sqrt{2}\\ \textrm{b}.&1\: \: \textrm{atau}\: \: \sqrt{2}\\ \textrm{c}.&-\sqrt{2}\: \: \textrm{atau}\: \: -\sqrt{2}\\ \color{red}\textrm{d}.&-1\: \: \textrm{atau}\: \: 1\\ \textrm{e}.&1\: \: \textrm{atau}\: \: -\sqrt{2}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{matriks}\: \: \textrm{A}=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{A}^{t}=\textrm{A}^{-1},\: \textrm{maka}\\ &\textrm{A}^{t}=\textrm{A}^{-1}\\ &\begin{pmatrix} a & b\\ c & d \end{pmatrix}^{t}=\displaystyle \frac{1}{ad-bc}\times \color{red}\textrm{Adjoin Matriks}\: \: \textrm{A}\\ &\begin{pmatrix} a & c\\ b & d \end{pmatrix}=\displaystyle \frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix},\\ & \color{red}\textrm{didapatkan hubungan}\\ &c=\displaystyle \frac{-b}{ad-bc}\quad ...............(1)\\ &b=\displaystyle \frac{-c}{ad-bc}\quad ...............(2)\\ &\textrm{Persamaan}\: \: (2)\: \: \textrm{disubstitusikan ke persamaan}\: \: (1)\\ &c=\displaystyle \displaystyle \frac{-\displaystyle \frac{-c}{ad-bc}}{ad-bc}\\ &1=(ad-bc)^{2}\\ &\color{red}(ad-bc)= -1\: \: \textrm{atau}\: \: 1 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 73.&(\textbf{UM UGM 2006})\\ &\textrm{Apabila}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{memenuhi}\\ &\textrm{persamaan matriks}\\ &\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -1\\ 2 \end{pmatrix},\\ &\textrm{maka}\: \: x+y=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \color{red}\textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \color{red}A.X&=B\\ \color{red}A^{-1}.A.X&=\color{red}A^{-1}.\color{blue}B\\ \color{red}A^{0}.X&=\color{red}A^{-1}.\color{blue}B\\ \color{red}X&=\color{red}A^{-1}.\color{blue}B\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}^{-1}\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{\begin{vmatrix} 1 & -2\\ -1 & 3 \end{vmatrix}}&\begin{pmatrix} 3 & 2\\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3-2}&\begin{pmatrix} 3.(-1)+2.2 \\ 1.(-1)+1.2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1\\ 1 \end{pmatrix}\\ x+y&=\color{red}1+1=2 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 75.&(\textbf{UM UGM 2005})\textrm{Jika}\\ &\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{pmatrix}=\begin{pmatrix} \sin A & \cos A \end{pmatrix}\\ &\textrm{dan}\: \: A\: \: \textrm{suatu konstanta, maka}\: \: x+y=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \color{red}\textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{pmatrix}=\begin{pmatrix} \sin A & \cos A \end{pmatrix}\\ &\begin{pmatrix} x\sin \alpha -y\cos \alpha & x\cos \alpha +y\sin \alpha \end{pmatrix}=\begin{pmatrix} \sin A & \cos A \end{pmatrix}\\ &\begin{cases} \sin A & =x\sin \alpha -y\cos \alpha =\sqrt{x^{2}+y^{2}}\cos \left ( \alpha -\tan ^{-1}\displaystyle \frac{x}{-y} \right ) \\ \cos A & =x\cos \alpha +y\sin \alpha =\sqrt{x^{2}+y^{2}}\cos \left (\alpha -\tan ^{-1}\displaystyle \frac{y}{x} \right ) \end{cases}\\ &\color{red}\textrm{Supaya}\: \: \color{blue}\cos A=\sqrt{x^{2}+y^{2}}\cos \left (\alpha -\tan ^{-1}\displaystyle \frac{y}{x} \right ),\: \: \color{red}\textrm{maka}\\ &\begin{cases} \sqrt{x^{2}+y^{2}} & =1 \\ \tan ^{-1}\displaystyle \frac{y}{x} & =0\Rightarrow \begin{cases} y & =0 \\ x & =1 \end{cases} \end{cases}\\ &\color{red}\textrm{Sehingga}\: \: \color{black}x+y=1+0=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 76.&(\textbf{UM UGM 2004})\\ &\textrm{Nilai-nilai}\: \: x\: \: \textrm{agar matriks}\\ &\qquad\quad\quad\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}\\ &\textrm{tidak memiliki invers adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&4\: \: \textrm{atau}\: \: 5\\ \color{red}\textrm{b}.&-2\: \: \textrm{atau}\: \: 2\\ \textrm{c}.&-4\: \: \textrm{atau}\: \: 5\\ \textrm{d}.&-6\: \: \textrm{atau}\: \: 4\\ \textrm{e}.&0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{supaya matriks}\: \: \color{black}\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}\\ &\textrm{tidak memiliki invers},\: \textrm{maka}\\ &\textrm{determinan matriks}\: \: \color{black}\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}=0\\ &\color{red}\textrm{Sehingga}\\ &\begin{vmatrix} 5x & 5\\ 4 & x \end{vmatrix}=0\\ &\Leftrightarrow 5x^{2}-20=0\\ &\Leftrightarrow x^{2}=\color{red}4\\ &\Leftrightarrow x=\color{red}\pm 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 77.&(\textbf{UM UGM 2005})\\ &\textrm{Matriks}\: \: \begin{pmatrix} x & 1\\ -2 & 1-x \end{pmatrix}\\ &\textrm{tidak memiliki invers untuk}\\ &\textrm{nilai}\: \: x=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{atau}\: \: -2\\ \textrm{b}.&-1\: \: \textrm{atau}\: \: 0\\ \textrm{c}.&-1\: \: \textrm{atau}\: \: 1\\ \color{red}\textrm{d}.&-1\: \: \textrm{atau}\: \: 2\\ \textrm{e}.&1\: \: \textrm{atau}\: \: 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\textrm{Mirip dengan pembahasan no. 26}\\ &\begin{aligned}&\textrm{Nilai}\: \: \color{black}\begin{vmatrix} x & 1\\ -2 & 1-x \end{vmatrix}=0\\ &\Leftrightarrow x-x^{2}-(-2)=0\\ &\Leftrightarrow 2+x-x^{2}=0\\ &\Leftrightarrow x^{2}-x-2=0\\ &\Leftrightarrow (x-2)(x+1)=0\\ &\Leftrightarrow \color{red}x=2\: \: \color{blue}\textrm{atau}\: \: \color{red}x=-1 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 79.&\textrm{Suatu translasi yang memetakan titik P(9,8) }\\ &\textrm{ke titik}\: \: \textrm{P}'(14,-2)\: \: \textrm{adalah}\: ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\begin{pmatrix} 5\\ -10 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 6\\ 6 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 5\\ 6 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} 23\\ -10 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} 5\\ 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\textrm{T}+\begin{pmatrix} x\\ y \end{pmatrix}\\ \textrm{T}&=\begin{pmatrix} x'-x\\ y'-y \end{pmatrix}\\ &=\begin{pmatrix} 14-9\\ -2-8 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ -10 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 80.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ &\textrm{Maka bayangan titik M}(2,-1)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (7,10)&&\textrm{d}.\quad (1,10)\\ \textrm{b}.\quad (10,7)&\textrm{c}.\quad \color{red}(1,4)&\textrm{e}.\quad (4,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ \left.\begin{matrix} x=2\\ y=-1 \end{matrix}\right\}&\Rightarrow \begin{cases} x' & =2(2)+3(-1)=4-3=1 \\ y' & =3(2)+2(-1)=6-2=4 \end{cases} \end{aligned} \end{array}$.

Latihan Soal 7 Persiapan PAS Gasal Matematika Wajib Kelas XI

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

$\begin{array}{ll}\\ 62.&\textrm{Jika}\: \: \begin{pmatrix} -4x & 2y\\ y & x \end{pmatrix}\begin{pmatrix} 2\\ -3 \end{pmatrix}=\begin{pmatrix} 2\\ -12 \end{pmatrix},\\ & \textrm{maka nilai}\: \: xy=....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&-6\\ \textrm{b}.&-3\\ \textrm{c}.&2\\ \textrm{d}.&3\\ \textrm{e}.&6 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\begin{pmatrix} -4x & 2y\\ y & x \end{pmatrix}\begin{pmatrix} 2\\ -3 \end{pmatrix}&=\begin{pmatrix} 2\\ -12 \end{pmatrix}\\ \begin{pmatrix} -4x.2+2y.-3\\ y.2+x.-3 \end{pmatrix}&=\begin{pmatrix} 2\\ -12 \end{pmatrix}\\ \begin{pmatrix} -8x-6y\\ -3x+2y \end{pmatrix}&=\begin{pmatrix} 2\\ -12 \end{pmatrix}\\\\ \textbf{SPLDV}& \end{aligned} &\color{red}\begin{aligned}-8x-6y&=2\: \qquad (\times 1)\\ -3x+2y&=-12\quad (\times 3)\\ \textrm{menjadi}&\\ -8x-6y&=2\\ -9x+6y&=-36\quad _{+}\\ ----&---\\ -17x&=-34\\ x&=2 \end{aligned}\\\hline \color{black}\begin{aligned}-8x-6y&=2\\ -8(2)-6y&=2\\ -16-6y&=2\\ -6y&=2+16\\ -6y&=18\\ y&=-3\\ \textrm{sehingga}&\\ xy&=2.(-3)=-6 \end{aligned}&\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 63.&\textrm{Diketahui}\: \: \textrm{N}=\begin{pmatrix} -2&3\\ -1 & 4 \end{pmatrix}\\ & \textrm{dan}\: \: \textrm{M}=\begin{pmatrix} -1&3\\ -1&5 \end{pmatrix}.\\ &\textrm{Jika}\: \: \textrm{N}^{2}=p\textrm{N}-q\textrm{M},\\ &: \textrm{maka nilai}\: \: p-q=....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \textrm{d}.&5\\ \textrm{e}.&6 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{N}^{2}=p\textrm{N}-q\textrm{M}\\ &\begin{pmatrix} -2&3\\ -1 & 4 \end{pmatrix}\times \begin{pmatrix} -2&3\\ -1 & 4 \end{pmatrix}=p\begin{pmatrix} -2&3\\ -1 & 4 \end{pmatrix}-q\begin{pmatrix} -1&3\\ -1&5 \end{pmatrix}\\ &\begin{pmatrix} -2.-2+3.-1 & -2.3+3.4\\ -1.-2+4.-1 & -1.3+4.4 \end{pmatrix}=\begin{pmatrix} -2p+q & 3p-3q\\ -p+q & 4p-5q \end{pmatrix}\\ &\begin{pmatrix} 4-3 & -6+12\\ 2-4 & -3+16 \end{pmatrix}=\begin{pmatrix} -2p+q & 3p-3q\\ -p+q & 4p-5q \end{pmatrix}\\ &\begin{pmatrix} 1 & 6\\ -2 & 13 \end{pmatrix}=\begin{pmatrix} -2p+q & 3p-3q\\ -p+q & 4p-5q \end{pmatrix}\\\\ &\begin{array}{|c|c|}\hline \color{purple}\begin{aligned}-2p+q&=1\\ -p+q&=-2\quad _{-}\\ ----&---\\ -p\qquad&=3\\ p&=-3\\ &\\ & \end{aligned}&\color{red}\begin{aligned}-p+q&=-2\\ -(-3)+q&=-2\\ q&=-2-3\\ q&=-5\\ \textrm{sehingga}&\: \textrm{didapatkan}\\ p-q&=-3-(-5)\\ &=2 \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 64.&\textrm{Diketahui matriks}\: \: \textrm{Z}=\begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}\\ & \textrm{dan}\: \: f(x)=x^{2}-x.\\ &\textrm{Jika}\: \: f(\textrm{Z})=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix},\\ & \textrm{maka nilai}\: \: p^{2}-q^{2}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&5\\ \color{red}\textrm{b}.&7\\ \textrm{c}.&9\\ \textrm{d}.&12\\ \textrm{e}.&15 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}f(\color{red}\textrm{Z})&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \color{red}\textrm{Z}^{2}-\textrm{Z}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}\times \begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}-\begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \begin{pmatrix} 4-18 & -12+30\\ 6-15 & -18+25 \end{pmatrix}-\begin{pmatrix} -2 & 6\\ -3 & 5 \end{pmatrix}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \begin{pmatrix} -12 & 12\\ -6 & 2 \end{pmatrix}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \end{aligned} \\ &\begin{aligned} \color{black}\textrm{Sehingga}&\\ -12&=-3p-8q\quad.................(1)\\ -1&=p+q\quad......................(2)\\ \textrm{persamaan}&\: (2)\: \: \textrm{ke persamaan}\: \: (1)\\ -12&=-3p-3q-5q=-3(p+q)-5q\\ -12&=-3(-1)-5q\\ -12&=3-5q\\ 5q&=3+12\\ q&=3\quad........................(3)\\ \textrm{persamaan}&\: \: (3)\: \: \textrm{ke persamaan}\: \: (2)\\ \color{red}p+q&=-1\\ p&=-1-q=-1-3=-4\\ \color{red}p^{2}-q^{2}&=(-4)^{2}-3^{2}=16-9\\ &=7 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 65.&(\textbf{SBMPTN 2013})\\ &\textrm{Jika}\: \: A=\begin{pmatrix} 2 & -1 & 1\\ a & b & c \end{pmatrix},\\ &B=\begin{pmatrix} -2 & 1\\ 1 & -1\\ 0 & 2 \end{pmatrix}\: \: \textrm{dan}\\ &AB=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\textrm{maka nilai}\: \: 2c-a=\: ....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&5\\ \textrm{e}.&6 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&AB=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\begin{pmatrix} 2 & -1 & 1\\ a & b & c \end{pmatrix}\begin{pmatrix} -2 & 1\\ 1 & -1\\ 0 & 2 \end{pmatrix}=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\begin{pmatrix} -5 & 5\\ \color{black}-2a+b&\color{black}a-b+2c \end{pmatrix}=\begin{pmatrix} -5 & 5\\ \color{red}3 & \color{red}-3 \end{pmatrix}\\ &\begin{array}{lllll}\\ -2a+b&=3&\\ a-b+2c&=-3&+\\\hline \qquad \color{red}2c-a&=0 \end{array} \end{aligned} \end{array}$.

DAFTAR PUSTAKA

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

$\begin{array}{ll}\\ 66.&\textrm{Determinan untuk matriks}\: \: \begin{pmatrix} 2 & -5\\ 3 & -1 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-17\\ \textrm{b}.&-13\\ \textrm{c}.&11\\ \color{red}\textrm{d}.&13\\ \textrm{e}.&17 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Determinan}&\: \textrm{dari matriks}\: \: \begin{pmatrix} 2 & -5\\ 3 & -1 \end{pmatrix}\\ &=\begin{vmatrix} 2 & -5\\ 3 & -1 \end{vmatrix}=2(-1)-3(-5)\\ &=-2+15\\ &=13 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 67.&\textrm{Determinan untuk matriks}\\ &\begin{pmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&10\\ \textrm{b}.&18\\ \textrm{c}.&22\\ \textrm{d}.&30\\ \textrm{e}.&36 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Determinan}\: \: \textrm{dari matriks}\\ &\begin{pmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{pmatrix}\\ &=\begin{vmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{vmatrix}\\ &=+(2.4.3)+(-1.-1.1)+(-1.1.-2)\\ &\quad -(1.4.-1)-(-2.-1.2)-(3.1.-1)\\ &=24+1+2+4-24+3\\ &=10 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 68.&\textrm{Jika diketahu matriks}\\ &\textrm{A}=\begin{pmatrix} x+3&-2\\ -16&2x-6 \end{pmatrix},\\ &\textrm{maka nilai dari}\: \: \: x\: \: \textrm{supaya matriks}\\ &\textrm{A tidak memiliki invers adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \color{red}\textrm{e}.&5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\textrm{Invers}&\: \textrm{dari matriks A adalah}\: \: \: \textrm{A}^{-1}.\\ \textrm{A}^{-1}&=\displaystyle \frac{1}{det\: \textrm{A}}\times \color{red}Adjoin\: \textrm{A}.\\ \textrm{Karen}&\textrm{a}\: \textrm{tidak memiliki invers},\\ \textrm{maka}\: \, & det\: \textrm{A}=0,\: \textrm{sehingga}\\ det\: \textrm{A}&=\begin{vmatrix} x+3 & -2\\ -16 & 2x-6 \end{vmatrix}=0\\ &\Leftrightarrow (x+3)(2x-6)-(-16.-2)=0\\ &(\color{red}\textrm{masing-masing ruas dibagi 2})\\ &\Leftrightarrow (x+3)(x-3)-16=0\\ &\Leftrightarrow x^{2}-9-16=0\\ &\Leftrightarrow x^{2}-25=0\\ &\Leftrightarrow (x+5)(x-5)=0\\ &\Leftrightarrow x+5=0\quad \textrm{atau}\quad x-5=0\\ &\Leftrightarrow \: \: \: \, \, \color{red}x=-5\quad \textrm{atau}\quad x=5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 69.&\textrm{Jika}\: \: \begin{vmatrix} 5^{2x} & -5\\ 1 & 1 \end{vmatrix}=6.5^{x}\\ &\textrm{maka}\: \: 5^{2x}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&625\: \: \textrm{atau}\: \: 1\\ \color{red}\textrm{b}.&25\: \: \textrm{atau}\: \: 1\\ \textrm{c}.&25\: \: \textrm{atau}\: \: 0\\ \textrm{d}.&5\: \: \textrm{atau}\: \: 1\\ \textrm{e}.&5\: \: \textrm{atau}\: \: 0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&5^{2x}+5=6.5^{x}\\ &5^{2x}-6.5^{x}+5=0\\ &\left ( 5^{x}-1 \right )\left ( 5^{x}-5 \right )=0\\ &5^{x}-1=0\: \: \textrm{atau}\: \: 5^{x}-5=0\\ &5^{x}=1\: \: \textrm{atau}\: \: 5^{x}=5\\ &5^{x}=5^{0}\: \: \textrm{atau}\: \: 5^{x}=5^{1}\\ &x=0\: \: \textrm{atau}\: \: x=1\\ &\color{red}\textrm{maka}\\ &5^{2x}=\begin{cases} 5^{2.1} &=5^{2}=25 \\ 5^{2.0} &=5^{0}=1 \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 70.&\textrm{Diketahu determinan suatu}\\ &\textrm{matriks adalah}\: \: \begin{vmatrix} x & 1 & 2\\ x & 1 & x\\ 5 & -3 & 7 \end{vmatrix}=0.\\ &\textrm{Jika}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{adalah akar-akar}\\ &\textrm{yang memenuhi persamaan tersebut}\\ &\textrm{maka nilai dari}\: \: \: p+q\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-3\\ \textrm{b}.&-\displaystyle \frac{1}{3}\\ \textrm{c}.&-1\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{3}\\ \textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Diketahui ba}&\textrm{hwa}:\\ \begin{vmatrix} x & 1 & 2\\ x & 1 & x\\ 5 & -3 & 7 \end{vmatrix}&=0\\ +(x.1.7)+&(1.x.5)+(2.x.-3)\\ -(5.1.2)&-(-3.x.x)-(7.x.1)=0\\ 7x+5x-6x&-10+3x^{2}-7x=0\\ 3x^{2}-x-10&=0\begin{cases} p & \textrm{salah satu akar} \\ q & \textrm{salah satu akar yang lain}, \end{cases}\\ \color{red}\textrm{dengan}\: \: \: &\begin{cases} a &=3 \\ b &=-1 \\ c &=-10 \end{cases}.\\ \textrm{maka}\: \: \: p+q\: \: &=-\displaystyle \frac{b}{a}=-\displaystyle \frac{-1}{3}\\ &=\color{red}\displaystyle \frac{1}{3} \end{aligned} \end{array}$

Latihan Soal 6 Persiapan PAS Gasal Matematika Wajib Kelas XI

$\begin{array}{l}\\ 51.&\textrm{Diketahui matriks}\\ &\textrm{A}=\begin{pmatrix} 2020 & -4&-3&2\\ 2020 & -6&-7&1\\ 2020&4&-3&0\\ 2020&6&-7&8 \end{pmatrix}\\ &\textrm{Ordo dari matriks}\: \: \textrm{A}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\times 2&&&\\ \textrm{b}.&3\times 3\\ \textrm{c}.&3\times 4\\ \textrm{d}.&4\times 3\\ \color{red}\textrm{e}.&4\times 4 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\textrm{Cukup jelas}\\ &\color{blue}\textrm{Karena matriknya mengandung}\\ &\color{red}\textrm{4 baris}\color{blue}\times \color{red}\textrm{4 kolom} \end{array}$

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

$\begin{array}{ll}\\ 53.&\textrm{Diketahui matriks}\: \: \textrm{C}\: \: \textrm{adalah matriks}\\ &\textrm{berordo}\: \: 3\times 3.\: \: \textrm{Jika}\: \: \textrm{c}_{ij}=4j-5i,\\ &\textrm{maka matriks C tersebut adalah}....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&\begin{pmatrix} -1 & 3 & 7\\ -6 & -2 & 2\\ -11 & -7 & -3 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} -1 & 7 & 3\\ -6 & 2 & -2\\ -7 & -11 & -3 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} -1 & -7 & -11\\ -6 & 7 & 3\\ -2 & 2 & -3 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} -1 &-6 & -11\\ 3 & -2 & 2\\ 7 & 2 & -3 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} -1 & -2 & -3\\ 3 & -6 & -11\\ 7 & -7 & 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\: \: \color{red}c_{ij}=4j-5i,\\ \textrm{mak}&\textrm{a}\\ \textrm{C}_{3\times 3}&=\begin{pmatrix} c_{\color{red}11} & c_{\color{red}12} & c_{\color{red}13} \\ c_{\color{red}21} & c_{\color{red}22} & c_{\color{red}23} \\ c_{\color{red}31} & c_{\color{red}32} & c_{\color{red}33} \end{pmatrix}\\ &=\begin{pmatrix} 4.1-5.1 & 4.2-5.1&4.3-5.1\\ 4.1-5.2 & 4.2-5.2&4.3-5.2\\ 4.1-5.3&4.2-5.3&4.3-5.3 \end{pmatrix}\\ &=\begin{pmatrix} 4-5 & 8-5 & 12-5\\ 4-10 & 8-10 & 12-10\\ 4-15 & 8-15 & 12-15 \end{pmatrix}\\ &=\begin{pmatrix} -1 & 3 & 7\\ -6 & -2 & 2\\ -11 & -7 & -3 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 54.&\textrm{Jika diketahui matriks}\\ &\textrm{X}=\begin{pmatrix} -7 & 9&-1\\ 4 & -6&15 \end{pmatrix}.\\ &\textrm{maka transpose matriks}\: \: \textrm{X}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\textrm{X}^{t}=\begin{pmatrix} 4 & -6 & 15\\ -7 & 9 & -1 \end{pmatrix}\\ \textrm{b}.&\textrm{X}^{t}=\begin{pmatrix} 15 & -6 & 4\\ -1 & 9 & -7 \end{pmatrix}\\ \color{red}\textrm{c}.&\textrm{X}^{t}=\begin{pmatrix} -7 & 4\\ 9 & -6\\ -1 & 15 \end{pmatrix}\\ \textrm{d}.&\textrm{X}^{t}=\begin{pmatrix} 4 & -7\\ -6 & 9\\ 15 & -1 \end{pmatrix}\\ \textrm{e}.&\textrm{X}^{t}=\begin{pmatrix} 15 & -1\\ -6 & 9\\ 4 & -7 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\\ \textrm{X}_{2\times 3}&=\begin{pmatrix} x_{\color{red}11} & x_{\color{red}12} & x_{\color{red}13} \\ x_{21} & x_{22} & x_{23} \end{pmatrix}\\ &=\begin{pmatrix} \color{red}-7 & \color{red}9&\color{red}-1\\ 4 & -6&15 \end{pmatrix}\\ \color{black}\textrm{maka}&\\ \textrm{X}_{3\times 2}^{t}&=\begin{pmatrix} x_{\color{red}11} & x_{21}\\ x_{\color{red}12} & x_{22}\\ x_{\color{red}13} & x_{23} \end{pmatrix}=\begin{pmatrix} \color{red}-7 & 4\\ \color{red}9 & -6\\ \color{red}-1 & 15 \end{pmatrix} \\ \textrm{adal}&\textrm{ah sebuah}\: \color{red}\textrm{matriks baru} \\ \textrm{deng}&\textrm{an ordo}\: \: \color{red}3\times 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 55.&\textrm{Diketahui matriks}\: \: \textrm{P}=\begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{Q}=\begin{pmatrix} 2c-3b & 2a+1\\ a & b+7 \end{pmatrix}.\\ &\textrm{Nilai}\: \: c\: \: \textrm{yang memenuhi jika}\\ & \textrm{P}=2\textrm{Q}^{t}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-2\\ \textrm{b}.&3\\ \textrm{c}.&5\\ \color{red}\textrm{d}.&8\\ \textrm{e}.&10 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{P}&=2\textrm{Q}^{\color{red}t}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=2\begin{pmatrix} 2c-3b & 2a+1\\ a & b+7 \end{pmatrix}^{\color{red}t}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=2\begin{pmatrix} 2c-3b & a\\ 2a+1 & b+7 \end{pmatrix}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=\begin{pmatrix} 4c-6b & 2a\\ 4a+2 & 2b+14 \end{pmatrix}\\ &(\color{red}\textrm{kesamaan 2 buah matriks})\\ \color{black}\textrm{akibat}&\color{black}\textrm{nya}\\ &\begin{cases} a &= 4c-6b \quad ....(1)\\ 4 &=2a \quad ..........(2)\\ 2b &=4a+2 \quad ........(3)\\ 3c &=2b+14 \quad ........(4) \end{cases}\\ \textrm{dari}&\: \textrm{persamaan}\: \: (2)\\ & 2a=4\Rightarrow a=2\quad....(5)\\ \textrm{pers}&\textrm{amaan}\: \: (5)\: \: \textrm{hasilnya}\\ \textrm{disu}&\textrm{bstitusikan ke persamaan}\: \: (3),\\ \color{blue}\textrm{yait}&\color{black}\textrm{u}\\ 2b&=4a+2\Rightarrow 2b=4(2)+2=10\\ b&=5\quad.....................(6)\\ \textrm{pers}&\textrm{amaan}\: \: (6)\: \: \textrm{hasilnya disbstitusikan}\\ \textrm{ke p}&\textrm{ersamaan}\\ (4),&\: \textrm{dan akan mendapatkan}\\ 3c&=2b+14\Rightarrow 3c=2(5)+14=24\\ \color{red}c&\color{red}=8 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 56.&\textrm{Diketahui matriks}\\ &\textrm{M}=\begin{pmatrix} -6 & 9&-15\\ 3 & -6&12 \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{N}=\begin{pmatrix} 2 & -3&5\\ -1 & 2&-4 \end{pmatrix}.\\ &\textrm{Nilai}\: \: k\: \: \textrm{yang memenuhi jika}\\ &\textrm{M}=k\textrm{N}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&-1\\ \color{red}\textrm{d}.&-3\\ \textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahu bahwa}\\ &\textrm{M}=k\textrm{N}\\ &(\color{red}\textrm{perkalian suatu matrik dengan skalar})\\ &\begin{pmatrix} -6 & 9&-15\\ 3 & -6&12 \end{pmatrix}\\ &=\begin{pmatrix} \color{red}-3.\color{black}2 & \color{red}-3.\color{black}-3 & \color{red}-3.\color{black}5\\ \color{red}-3.\color{black}-1 & \color{red}-3.\color{black}2 & \color{red}-3.\color{black}-4 \end{pmatrix}\\ &=\color{red}-3\color{black}\begin{pmatrix} 2 & -3 & 5\\ -1 & 2 & -4 \end{pmatrix}\\ &=k\begin{pmatrix} 2 & -3&5\\ -1 & 2&-4 \end{pmatrix}\\ &\textrm{sehingga dari kesamaan tersebut}\\ &\textrm{maka}\quad \color{red}k=-3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 57.&\textrm{Hasil dari}\: \: \begin{pmatrix} 1&2&3\\ 4&5&6 \end{pmatrix}\times \begin{pmatrix} 1 &2 \\ 3 &4 \\ 5 & 6 \end{pmatrix}\\ & \textrm{adalah}...\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&\begin{pmatrix} 22 & 28\\ 49&64 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 22&49\\ 28&64 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 64&28\\ 49&22 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} 2 & 8&18\\ 4&15 & 30 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 1&4&6\\ 4&15&30 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\begin{pmatrix} 1&2&3\\ 4&5&6 \end{pmatrix}_{\color{red}2\times \color{black}3}\times \begin{pmatrix} 1 &2 \\ 3 &4 \\ 5 & 6 \end{pmatrix}_{\color{black}3\times \color{red}2}\\ &=\begin{pmatrix} 1.1+2.3+3.5 & 1.2+2.4+3.6\\ 4.1+5.3+6.5 &4.2+5.4+6.6 \end{pmatrix}_{\color{red}2\times 2}\\ &=\begin{pmatrix} 1+6+15 & 2+8+18\\ 4+15+30 & 8+20+36 \end{pmatrix}_{\color{red}2\times 2}\\ &=\begin{pmatrix} 22 & 28\\ 49 & 64 \end{pmatrix}_{\color{red}2\times 2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 58.&\textrm{Jika diketahui matriks}\\ & \textrm{A}=\begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}.\\ &\textrm{maka hasil dari}\: \: \textrm{A}^{3}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} 5 & 8\\ 20&22 \end{pmatrix}\\ \color{red}\textrm{b}.&\begin{pmatrix} 6&7\\ 21&20 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 6&7\\ 20&22 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} 7 & 8\\ 20 & 23 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 7&9\\ 20&23 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\\ \textrm{A}&=\begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}\\ \textrm{mak}&\textrm{a}\\ \textrm{A}^{2}&=\textrm{A}\times \textrm{A}\\ &=\begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}\times \begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}\\ &=\begin{pmatrix} 0+3&0+2\\ 0+6&3+4 \end{pmatrix}\\ &=\color{purple}\begin{pmatrix} 3 & 2\\ 6 & 7 \end{pmatrix}\\ \textrm{A}^{3}&=\textrm{A}^{2}\times \textrm{A}\\ &=\begin{pmatrix} 3 & 2\\ 6 &7 \end{pmatrix}\times \begin{pmatrix} 0 & 1\\ 3 & 2 \end{pmatrix}\\ &=\begin{pmatrix} 0+6 & 3+4\\ 0+21 & 6+14 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 6 & 7\\ 21 & 20 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 59.&(\textbf{SBMPTN Mat IPA 2014})\\ &\textrm{Jika}\: \: \textrm{A}\: \: \textrm{adalah matriks yang berordo}\\ & 2\times 2\: \: \textrm{dan memenuhi}\\ &\: \: \begin{pmatrix} x & 1 \end{pmatrix}\times \textrm{A}\times \begin{pmatrix} x\\ 1 \end{pmatrix}=x^{2}-5x+8,\\ & \textrm{maka matriks A yang mungkin adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} 1 & -5\\ 8&0 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 1&5\\ 8&0 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 1&8\\ -5&0 \end{pmatrix}\\ \color{red}\textrm{d}.&\begin{pmatrix} 1 & 3\\ -8&8 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 1&-3\\ 8&-8 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\begin{pmatrix} x & 1 \end{pmatrix}\times \textrm{A}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=\color{red}x^{2}-5x+8\\ \begin{pmatrix} x & 1 \end{pmatrix}\times \begin{pmatrix} p & q\\ r & s \end{pmatrix}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=\color{red}x^{2}-5x+8\\ \begin{pmatrix} xp+r & xq+s \end{pmatrix}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=\color{red}x^{2}-5x+8\\ \begin{pmatrix} x^{2}p+xr+xq+s \end{pmatrix}&=\color{red}x^{2}-5x+8\\ px^{2}+(q+r)x+s&=\color{red}x^{2}-5x+8\\ \end{aligned}\\ &\color{blue}\begin{aligned}&\begin{cases} \color{red}p &=1 \\ q+r &=-5 \\ \color{red}s &=8 \end{cases}\quad\Rightarrow\quad \begin{pmatrix} 1 & ...\\ ... & 8 \end{pmatrix}\\ &\textrm{Sehingga yang paling mungkin}\\ & \textrm{adalah}\: \: \color{red}\begin{pmatrix} 1 & 3\\ -8 & 8 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 60.&\textrm{Diketahui}\\ &\begin{pmatrix} ^{x}\log a & \log (2a-6)\\ \log (b-2) & 1 \end{pmatrix}=\begin{pmatrix} \log b & 1\\ \log a & 1 \end{pmatrix}\\ &\textrm{maka nilai}\: \: x\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&6\\ \color{red}\textrm{e}.&8 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\begin{pmatrix} ^{x}\log a & \log (2a-6)\\ \log (b-2) & 1 \end{pmatrix}=\begin{pmatrix} \log b & 1\\ \log a & 1 \end{pmatrix}\\ &\color{black}\textrm{maka}\\ &\begin{cases} ^{x}\log a & =\log b \quad.........\color{red}(1)\\ \log (2a-6) &=1\quad..............\color{red}(2) \\ \log (b-2) &=\log a\quad.........\color{red}(3) \end{cases}\\ &\textrm{Sehingga}\: \textrm{dari persamaan}\: \: (2)\\ &\color{black}\textrm{akan didapatkan}\\ &\log (2a-6)=1=\log 10\\ &(2a-6)=10\\ &a=8\quad...........................(4)\\ &\textrm{persamaan}\: (4)\: \: \textrm{ke persamaan}\: \: (3),\\ & \color{black}\textrm{maka}\\ &\log (b-2) =\log a\\ &b-2=a=8\\ &b=10\quad.................................(5)\\ &\textrm{Selanjutnya dari persamaan}\: \: (5)\\ &\color{black}\textrm{akan diperoleh}\\ &^{x}\log a =\log b\\ &^{x}\log 8 =\log 10=1\\ &\qquad x^{1}=8\\ &\Leftrightarrow \: \: \color{red}x=8 \end{aligned} \end{array}$

Latihan Soal 5 Persiapan PAS Gasal Matematika Wajib Kelas XI

$\begin{array}{ll}\\ 41.&\textrm{Pada gambar berikut ini, pertidaksamaan}\\ &\textrm{yang memenuhi adalah}\\\\ \end{array}$

$.\: \: \: \quad\begin{array}{ll}\\ \textrm{a}.&2x+y-4\leq 0,\: 2x+3y-6\geq 0,\: x\geq 0,\: y\geq 0\\ \textrm{b}.&2x+y-4\geq 0,\: 2x+3y-6\leq 0,\: x\geq 0,\: y\geq 0\\ \textrm{c}.&2x+y-4\leq 0,\: 2x+3y-6\leq 0,\: x\geq 0,\: y\geq 0\\ \color{red}\textrm{d}.&\left (2x+y-4 \right )\left (2x+3y-6 \right )\leq 0,\: x\geq 0,\: y\geq 0\\ \textrm{e}.&\left (2x+y-4 \right )\left (2x+3y-6 \right )\geq 0,\: x\geq 0,\: y\geq 0\\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\textrm{Misalkan titik potong kedua garis adalah M}\\ &\color{magenta}\textrm{Persamaan garisnya sebelah kiri M}:\\ &(1)\: 4x+2y-8=0\\ &\: \: \: \quad\textrm{kendala}:2x+y-4\leq 0\\ &(2)\: 2x+3y=6\\ &\: \: \: \quad \textrm{kendala}:2x+3y-6\geq 0\\ &(3)\: y=0,\: \: \textrm{kendala}:y\geq 0 \\ &(4)\: x=0,\: \: \textrm{kendala}:x\geq 0\\ &\color{blue}\textrm{Persamaan garisnya sebelah kanan M}:\\ &(5)\: 4x+2y-8=0\\ &\: \: \: \quad\textrm{kendala}:2x+y-4\geq 0\\ &(6)\: 2x+3y=6\\ &\: \: \: \quad \textrm{kendala}:2x+3y-6\leq 0\\ &(7)\: y=0,\: \: \textrm{kendala}:y\geq 0 \\ &(8)\: x=0,\: \: \textrm{kendala}:x\geq 0 \end{aligned}$

$\begin{array}{ll}\\ 42.&\textrm{Pada gambar berikut ini, pertidaksamaan}\\ &\textrm{yang memenuhi adalah}\\\\ \end{array}$

$.\: \: \: \quad\begin{array}{ll}\\ \color{red}\textrm{a}.&2x-y-4\leq 0,\: x-y-3\leq 0,\: x\geq 0,\: y\geq 0\\ \textrm{b}.&2x-y-4\geq 0,\: x-y-3\geq 0,\: x\geq 0,\: y\leq 0\\ \textrm{c}.&2x-y-4\leq 0,\: x-y-3\geq 0,\: x\geq 0,\: y\leq 0\\ \textrm{d}.&\left (2x-y-4 \right )\left (x-y-3 \right )\geq 0,\: x\geq 0,\: y\leq 0\\ \textrm{e}.&\left (2x-y-4 \right )\left (x-y-3 \right )\leq 0,\: x\geq 0,\: y\leq 0\\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\textrm{Misalkan titik potong kedua garis adalah M}\\ &\color{magenta}\textrm{Persamaan garisnya di atas M dan model}\\ &\color{magenta}\textrm{matematikanya adalah sebagai berikut}:\\ &(1)\: -3x+3y=-9\\ &\: \: \: \quad\textrm{garisnya menjadi}:\: -3x+3y+9=0\\ &\: \: \: \quad\textrm{maka}:\: -x+y+3=0\\ &\: \: \: \quad\textrm{kendala}:-x+y+3\geq 0,\: \: \textrm{atau}\\ &\: \: \: \quad\color{blue}\textrm{kendala}:x-y-3\leq 0\\ &(2)\: -4x+2y=-8\\ &\: \: \: \quad\textrm{garisnya menjadi}:\: -4x+2y+8=0\\ &\: \: \: \quad\textrm{maka}:\: -2x+y+4=0\\ &\: \: \: \quad\textrm{kendala}:-2x+y+4\geq 0,\: \: \textrm{atau}\\ &\: \: \: \quad \quad\color{blue}\textrm{kendala}:2x-y-4\leq 0\\ &(3)\: y=0,\: \: \textrm{kendala}:y\geq 0 \\ &(4)\: x=0,\: \: \textrm{kendala}:x\geq 0\\ \end{aligned}$

$\begin{array}{ll}\\ 43.&\textrm{Seorang penjual hewan kurban memiliki}\\ &\textrm{15 kandang ternak untuk memelihara sapi}\\ &\textrm{dan kambing. Setiap kandang hanya berisi}\\ &\textrm{kambing saja atau sapi saja. Setiap kandang}\\ &\textrm{dapat menampung sapi sebanyak 20 ekor}\\ &\textrm{atau kambing sebanyak 38 ekor. Penjual}\\ &\textrm{hewan kurban tersebut menaksir biaya}\\ &\textrm{perawatan yang dikeluarkan untuk setiap}\\ &\textrm{kandang sapi setiap bulannya sebesar}\\ &Rp500.000,00\: \: \textrm{dan kambing}\: \: Rp300.000,00.\\ &\textrm{Sementara itu, jumlah hewan yang}\\ &\textrm{direncanakan tidak lebih dari 300 ekor}.\\ &\textrm{Jika banyak kandang yang berisi sapi}\\ &\textrm{disebut}\: \: x\: \: \textrm{dan banyak kandang yang berisi}\\ &\textrm{kambing disebut}\: \: y,\: \: \textrm{sistem pertidaksamaan}\\ &\textrm{yang harus dipenuhi oleh}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{serta}\\ &\textrm{fungsi objektif untuk meminimumkan biaya}\\ &\textrm{perawatan hewan kurban adalah}.... \end{array}$

$.\: \: \: \quad\begin{array}{ll}\\ \textrm{a}.&x\geq 0,\: y\geq 0,\: 20x+38y\leq 15,\: x+y\leq 300\\ &\textrm{minimum}\: \: f(x,y)=500.000x+300.000y\\ \textrm{b}.&x\geq 0,\: y\geq 0,\: 38x+20y\leq 15,\:x+y\leq 300 \\ &\textrm{minimum}\: \: f(x,y)=500.000x+300.000y\\ \textrm{c}.&x\geq 0,\: y\geq 0,\: 28x+30y\leq 300,\:x+y\leq 15 \\ &\textrm{minimum}\: \: f(x,y)=500.000x+300.000y\\ \textrm{d}.&x\geq 0,\: y\geq 0,\: 38x+20y\leq 300,\:x+y\leq 15 \\ &\textrm{minimum}\: \: f(x,y)=500.000x+300.000y\\ \color{red}\textrm{e}.&x\geq 0,\: y\geq 0,\: 20x+38y\leq 300,\:x+y\leq 15 \\ &\textrm{minimum}\: \: f(x,y)=500.000x+300.000y\\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\textrm{Misalkan titik potong kedua garis adalah M}\\ &\color{magenta}\textrm{Misalkan sapi}=x,\: \: \textrm{kambing}=y,\: \: \textrm{maka}\\ &(1)\: \textrm{Sapi}+\textrm{Kambing}= 15\: \: \textrm{ekor}\\ &\: \: \: \quad \textrm{persamaan garisnya}:\\ &\: \: \: \quad x+y= 15,\: \: \textrm{dan}\\ &\: \: \: \quad\textrm{kendalanya}:x+y\leq 15\\ &(2)\: \textrm{Daya tampung kandang}\\ &\: \: \: \quad \textrm{persamaan garisnya}:\\ &\: \: \: \quad20x+38y=300\\ &\: \: \: \quad \textrm{kendala}:20x+38y\leq 300\\ &(3)\: y=0,\: \: \textrm{kendala}:y\geq 0 \\ &(4)\: x=0,\: \: \textrm{kendala}:x\geq 0\\ &(5)\: \color{red}\textrm{Fungsi optimumnya adalah}:\\ &\: \: \: \color{blue}\quad f(x,y)=500.000x+300.000y \end{aligned}$

$\begin{array}{ll}\\ 44.&\textrm{Suatu perusahaan bangunan merencanakan}\\ &\textrm{pembangunan paling banyak 150 unit rumah}\\ &\textrm{untuk disewakan kepada 500 orang. Ada dua}\\ &\textrm{jenis rumah, yaitu rumah jenis A dengan}\\ &\textrm{kapasitas 4 orang yang akan disewakan dengan}\\ &\textrm{harga}\: \: Rp2.000.000,00\: \: \textrm{per tahun dan rumah}\\ &\textrm{jenis B dengan kapasitas 6 orang yang disewakan}\\ &Rp2.500.000,00\: \: \textrm{per tahun. Jika rumah jenis}\\ &\textrm{A dibuat sebanyak}\: \: x\: \: \textrm{unit dan jenis B sebanyak}\\ &y\: \: \textrm{unit},\: \color{blue}\textbf{model matematika}\: \color{black}\textrm{dari masalah tersebut}\\ &\textrm{adalah}\:.... \end{array}$

$.\: \: \: \quad\begin{array}{ll}\\ \textrm{a}.&x\geq 0,\: y\geq 0,\: x+y\leq 100,\: 4x+6y\leq 500\\ \color{red}\textrm{b}.&x\geq 0,\: y\geq 0,\: x+y\leq 150,\:4x+6y\leq 500 \\ \textrm{c}.&x\geq 0,\: y\geq 0,\: x+y\leq 200,\:4x+6y\leq 250 \\ \textrm{d}.&x\geq 0,\: y\geq 0,\: x+y\leq 200,\:6x+4y\leq 250 \\ \textrm{e}.&x\geq 0,\: y\geq 0,\: x+y\leq 500,\:6x+4y\leq 250 \\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\textrm{Model matematikanya adalah}:\\ &\color{magenta}\textrm{Misalkan rumah jenis A}=x,\: \: \textrm{jenis B}=y,\: \: \textrm{maka}\\ &(1)\: \textrm{Jenis A}+\textrm{Jenis B}= 150\: \: \textrm{unit}\\ &\: \: \: \quad \textrm{persamaan garisnya}:\\ &\: \: \: \quad x+y= 150,\: \: \textrm{dan}\\ &\: \: \: \quad\textrm{kendalanya}:x+y\leq 150\\ &(2)\: \textrm{Kapasitas atau daya tampung}\\ &\: \: \: \quad \textrm{Rumah jenis A muat 4 orang dan jenis B}\\ &\: \: \: \quad \textrm{6 orang sedangkan targetnya 500 orang, maka}\\ &\: \: \: \quad \textrm{persamaan garisnya}:\\ &\: \: \: \quad 4x+6y=500\\ &\: \: \: \quad \textrm{kendala}:4x+6y\leq 500\\ &(3)\: y=0,\: \: \textrm{kendala}:y\geq 0 \\ &(4)\: x=0,\: \: \textrm{kendala}:x\geq 0\\ &(5)\: \color{red}\textrm{Fungsi optimumnya adalah}:\\ &\: \: \: \color{blue}\quad f(x,y)=2.000.000x+2.500.000y \end{aligned}$

$\begin{array}{ll}\\ 45.&\textrm{Pedagang teh mempunyai lemari yang hanya}\\ &\textrm{cukup ditempati 40 boks teh. Teh A dibeli}\\ &\textrm{dengan harga}\: \: Rp6.000,00\: \: \textrm{setiap boks dan teh B}\\ &\textrm{dibeli dengan harga}\: \: Rp8.000,00\: \: \textrm{setiap boks}\\ &\textrm{Jika pedang tersebut mempunyai modal sebesar}\\ &Rp300.000,00\: \: \textrm{untuk membeli}\: \: x\: \: \textrm{boks teh A dan}\\ &y\: \: \textrm{boks teh B, maka sistem pertidaksamaan dari}\\ &\textrm{permasalahan tersebut adalah}\: .... \end{array}$

$.\: \: \: \quad\begin{array}{ll}\\ \textrm{a}.&3x+4y\geq 150,\: x+y\geq 40,\: x\geq 0,\: y\geq 0\\ \color{red}\textrm{b}.&3x+4y\leq 150,\: x+y\leq 40,\:x\geq 0,y\geq 0 \\ \textrm{c}.&3x+4y\geq 150,\: x+y\leq 40,\:x\geq 0,\: y\geq 0 \\ \textrm{d}.&6x+8y\leq 300,\: x+y\geq 40,\:x\geq 0,\: y\geq 0 \\ \textrm{e}.&8x+4y\leq 300,\: x+y\leq 40,\:x\geq 0,\: y\geq 0 \\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\textrm{Pembahasan diserahkan kepada pembaca} \end{aligned}$.

$\begin{array}{ll}\\ 46.&\textrm{Pada pertidaksamaan}\\ & 2y\geq x\: ;\: y\leq 2x\: ;\: 2y+x\leq 20\: ;\: x+y\geq 9\\ &\textrm{Nilai maksimum untuk}\: \: \color{red}3y-x\: \: \color{black}\textrm{dicapai saat}\: ....\\\\ \end{array}$

$.\: \: \: \quad\begin{array}{ll}\\ \textrm{a}.&\textrm{P}\\ \textrm{b}.&\textrm{Q}\\ \color{red}\textrm{c}.&\textrm{R}\\ \textrm{d}.&\textrm{S}\\ \textrm{e}.&\textrm{T}\\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\textrm{Dengan membuat garis(selidik)}:f(x,y)=3y-x\\ &\color{blue}\textrm{digeser dari bawah ke atas, maka akan didapatkan}\\ &\color{blue}\textrm{titik sudut(verteks) yang diinginkan}\\ & \end{aligned}$

$\begin{array}{ll}\\ 47.&\textrm{Nilai minimum dari}\: \: -2x+4y+6\\ &\textrm{untuk}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\\ &\begin{cases} 2x+y-20 &\leq 0 \\ 2x-y+10&\geq 0 \\ x+y-5 &\geq 0 \\ x-2y-5 &\leq 0 \\ x\geq 0 & \\ y\geq 0 & \end{cases}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll}\\ \textrm{a}.&-14\\ \textrm{b}.&-11\\ \textrm{c}.&-9\\ \textrm{d}.&-6\\ \color{red}\textrm{e}.&-4\\ \end{array}\\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\textrm{Diketahui fungsi objektif}:f(x,y)=-2x+4y+6\\ &\color{blue}\textrm{dan kendala-kendalanya}\\ &\begin{cases} 2x+y-20 &\leq 0 \\ 2x-y+10&\geq 0 \\ x+y-5 &\geq 0 \\ x-2y-5 &\leq 0 \\ x\geq 0 & \\ y\geq 0 & \end{cases}\\ &\color{blue}\textrm{maka daerah penyelesaiannya adalah}: \end{aligned}$

$.\: \: \: \quad\begin{aligned}&\textrm{Dan persamaan garisnya adalah}\\ &\begin{cases} \textrm{L}_{1}\equiv &2x+y=20 \\ \textrm{L}_{2}\equiv &2x-y=-10\\ \textrm{L}_{3}\equiv &x+y=5 \\\textrm{L}_{4} \equiv &x-2y=5 \end{cases}\\ &\color{blue}\textrm{Perpotongan untuk garis}\: \: \textrm{L}_{1}\&\textrm{L}_{2}\\ &\textrm{akan didapatkan titik C}\: \left ( \displaystyle \frac{5}{2},15 \right )\\ &\color{blue}\textrm{Perpotongan untuk garis}\: \: \textrm{L}_{1}\&\textrm{L}_{4}\\ &\textrm{akan didapatkan titik B}\: \left ( \displaystyle 9,2 \right )\\ \end{aligned}$

$.\: \: \: \quad\begin{array}{|l|}\hline \begin{aligned}\color{red}\textrm{untuk}\: &\color{red}\textrm{mendapatkan titik potong}\\ \textrm{garis}:\: &\textrm{L}_{1}\&\textrm{L}_{2}\: \: \color{black}\textrm{adalah sebagai berikut}:\\ &\begin{cases} 2x+y & =20 \\ 2x-y & =-10 \end{cases}\\ &-------\: \: .^{-}\\ &\: \: \: \quad\quad\quad2y=30\\ &\: \qquad\qquad y=15\Rightarrow x=\displaystyle \frac{5}{2}\\ \textrm{Sehing}&\textrm{ga didapatkan titik}\: \: \left ( \displaystyle \frac{5}{2},15 \right ) \end{aligned} \\\hline \begin{aligned}\color{magenta}\textrm{untuk}\: &\color{red}\textrm{mendapatkan titik potong}\\ \textrm{garis}:\: &\textrm{L}_{1}\&\textrm{L}_{4}\: \: \color{black}\textrm{adalah sebagai berikut}:\\ &\begin{cases} 2x+y & =20 \\ x-2y & =5 \end{cases}\\ &\begin{cases} 4x+2y & =40 \\ x-2y & =5 \end{cases}\\ &-------\: \: .^{+}\\ &\qquad\quad\quad5x=45\\ &\: \: \qquad\qquad x=9\Rightarrow y=2\\ \textrm{Sehing}&\textrm{ga didapatkan titik}\: \: \left ( 9,2 \right ) \end{aligned}\\\hline \end{array}$

$.\: \: \: \quad\textrm{Selanjutnya}$

$.\: \: \: \quad\begin{array}{|c|c|c|}\hline \textrm{Verteks}&\color{magenta}\textrm{Nilai}:& \textrm{Keterangan}\\ &-2x+4y+6&\\\hline \textrm{A}(5,0)&-2(5)+4.0+6=-4&\color{blue}\textrm{Minimum}\\\hline \textrm{B}(9,2)&-2(9)+4.2+6=-4&\color{blue}\textrm{Minimum}\\\hline \textrm{C}\left ( \displaystyle \frac{5}{2},15 \right )&-2\left ( \displaystyle \frac{5}{2} \right )+4.15+6=61&\color{red}\textrm{Maksimum}\\\hline \textrm{D}(0,10)&-2.0+4.10+6=46&\\\hline \textrm{E}(0,5)&-2.0+4.5+6=26&\\\hline \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Nilai minimum}\: \: f(x,y)=3+4x-5y\\ &\textrm{untuk}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\\ &\begin{cases} -x+y &\leq 1 \\ x+2y&\geq 5 \\ 2x+y &\leq 10 \\ \end{cases}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll}\\ \textrm{a}.&-19\\ \textrm{b}.&-6\\ \color{red}\textrm{c}.&-5\\ \textrm{d}.&-3\\ \textrm{e}.&23\\ \end{array} \\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\textrm{Diketahui fungsi objektif}:f(x,y)=3+4x-5y\\ &\color{blue}\textrm{dan kendala-kendalanya}\\ &\begin{cases} -x+y &\leq 1 \\ x+2y&\geq 5 \\ 2x+y &\leq 10 \\ \end{cases}\\ &\color{blue}\textrm{maka daerah penyelesaiannya adalah}: \end{aligned}$

$.\: \: \: \quad\begin{array}{|c|c|c|}\hline \textrm{Verteks}&\color{magenta}\textrm{Nilai}:& \textrm{Keterangan}\\ &3+4x-5y&\\\hline \textrm{A}(1,2)&3+4.1-5.2=-3&\\\hline \textrm{B}(3,4)&3+4.3-5.4=-5&\color{blue}\textrm{Minimum}\\\hline \textrm{C}(5,0)&3+4.5-5.0=23&\color{red}\textrm{Maksimum}\\\hline \end{array}$

$\begin{array}{ll}\\ 49.&\textrm{Fungsi}\: \: f(x,y)=10x+15y\\ &\textrm{untuk}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\\ &\begin{cases} x &\geq 0 \\ y&\geq 0 \\ x &\leq 800 \\ x&+\: \: y\: \: \leq 1000\\ \end{cases}\\ &\textrm{mempunyai nilai maksimum}\: ....\\ &\begin{array}{ll}\\ \textrm{a}.&9.000\\ \textrm{b}.&11.000\\ \color{red}\textrm{c}.&13.000\\ \textrm{d}.&15.000\\ \textrm{e}.&16.000\\ \end{array} \\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\textrm{Diketahui fungsi objektif}:f(x,y)=10x+15y\\ &\color{blue}\textrm{dan kendala-kendalanya}\\ &\begin{cases} x &\geq 0 \\ y&\geq 0 \\ x &\leq 800 \\ x&+\: \: y\: \: \leq 1000\\ \end{cases}\\ &\color{blue}\textrm{maka daerah penyelesaiannya adalah}: \end{aligned}$

$.\: \: \: \quad\begin{array}{|c|c|c|}\hline \textrm{Verteks}&\color{magenta}\textrm{Nilai}:& \textrm{Keterangan}\\ &f(x,y)=10x+15y&\\\hline \textrm{A}(800,0)&800.10+0=8000&\color{red}\textrm{Minimum}\\\hline \textrm{B}(800,200)&800.10+15.200=11.000&\\\hline \textrm{C}(400,600)&10.400+15.600=13.000&\color{blue}\textrm{Maksimum}\\\hline \textrm{D}(0,600)&3+0+15.600=9.000&\\\hline \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Nilai maksimum fungsi sasaran}\\ & f(x,y)=4x+5y\\ &\textrm{untuk}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\\ &\begin{cases} x&\geq 0 \\ y&\geq 0 \\ (&2x+y-4\: )(\: 2x+3y-6\: )\: \: \leq 0\\ \end{cases}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll}\\ \textrm{a}.&11\\ \textrm{b}.&12\\ \textrm{c}.&16\\ \color{red}\textrm{d}.&20\\ \textrm{e}.&24\\ \end{array} \\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\textrm{Diketahui fungsi objektif}:f(x,y)=4x+5y\\ &\color{blue}\textrm{dan kendala-kendalanya}\\ &\begin{cases} x&\geq 0 \\ y&\geq 0 \\ (&2x+y-4\: )(\: 2x+3y-6\: )\: \: \leq 0\\ \end{cases}\\ &\color{blue}\textrm{maka daerah penyelesaiannya adalah}: \end{aligned}$

$.\: \: \: \quad\begin{aligned}\textrm{untuk}\: &\textrm{mendapatkan titik potongnya}\\ &\begin{cases} 2x+y & =4 \\ 2x+3y & =6 \end{cases}\\ &-------\: \: .^{-}\\ &\: \, \qquad-2y=-2\\ &\qquad\qquad y=1\Rightarrow x=\displaystyle \frac{3}{2}\\ \textrm{sehing}&\textrm{ga akan didapatkan}\\ \color{blue}\textrm{titik p}&\color{blue}\textrm{otongnya adalah}:\: \: \left ( \displaystyle \frac{3}{2},1 \right )\\ \textrm{Selanj}&\textrm{utnya, kita dapat menentukan}\\ \textrm{nilai}\: \: \: \: &\textrm{maksimunya dengan bantuan tabel berikut} \end{aligned}$

$.\: \: \: \quad\begin{array}{|c|c|c|}\hline \textrm{Verteks}&\color{magenta}\textrm{Nilai}:& \textrm{Keterangan}\\ &f(x,y)=4x+5y&\\\hline (2,0)&4.2+0=8&\color{red}\textrm{Minimum}\\\hline (3,0)&4.3+0=12&\\\hline \left ( \displaystyle \frac{3}{2},1 \right )&4.\left ( \displaystyle \frac{3}{2} \right )+5.1=11&\\\hline (0,2)&0+52=10&\\\hline (0,4)&0+5.4=20&\color{blue}\textrm{Maksimum}\\\hline \end{array}$

Latihan Soal 4 Persiapan PAS Gasal Matematika Wajib Kelas XI

$\begin{array}{ll}\\ 31.&\textrm{Daerah yang diarsir berikut adalah himpunan}\\ &\textrm{penyelesaian pertidaksamaan dari}....\\ \end{array}$

$.\: \: \: \begin{array}{ll}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&y\geq 0,\: x-2y\geq -2,\: 3x+4y\leq 12\\ \textrm{b}.&y\geq 0,\: x-2y\geq -2,\: 3x+4y\geq 12\\ \textrm{c}.&y\geq 0,\: -2x+y\geq -2,\: 4x+3y\leq 12\\ \textrm{d}.&x\geq 0,\: -2x+y\leq -2,\: 4x+3y\geq 12\\ \textrm{e}.&x\geq 0,\: x-2y\leq -2,\: 3x+4y\leq 12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\textrm{Cukup jelas. Anda bisa mengecek dengan titik uji} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Daerah yang diarsir berikut adalah himpunan}\\ &\textrm{penyelesaian pertidaksamaan dari}....\\ \end{array}$

$.\: \: \: \begin{array}{ll}\\ &\begin{array}{llll}\\ \textrm{a}.&x\geq 0,\: 4x+y\geq 4,\: x+y\leq 2\\ \color{red}\textrm{b}.&x\geq 0,\: 4x+y\leq 4,\: x+y\geq 2\\ \textrm{c}.&x\geq 0,\: 4x+y> 4,\: x+y< 2\\ \textrm{d}.&x\geq 0,\: x+4y> 4,\: x+y< 2\\ \textrm{e}.&x\geq 0,\: x+4y\leq 4,\: x+y\geq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\textrm{Cukup jelas} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Pada gambar berikut ini, yang merupakan}\\ &\textrm{himpunan penyelesaian sistem pertidaksamaan}\\\\ &\color{magenta}\begin{cases} 2x+y & \leq 24 \\ x+2y &\geq 12 \\ x-y & \geq -2 \end{cases}\\\\ &\textrm{adalah}\: ....\\ & \end{array}$

$.\: \: \: \begin{array}{ll}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{I}\\ \textrm{b}.&\textrm{II}\\ \color{red}\textrm{c}.&\textrm{III}\\ \textrm{d}.&\textrm{IV}\\ \textrm{e}.&\textrm{V} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\textrm{Pembahasan diserahkan kepada pembaca} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Pada gambar berikut ini, yang merupakan}\\ &\textrm{himpunan penyelesaian sistem pertidaksamaan}\\\\ &\color{magenta}\begin{cases} x+2y & \geq 6 \\ 4x+5y &\leq 20 \\ 2x+y & \geq 6 \end{cases}\\\\ &\textrm{adalah}\: ....\\ \end{array}$

$.\: \: \: \begin{array}{ll}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{I}\\ \color{red}\textrm{b}.&\textrm{II}\\ \textrm{c}.&\textrm{III}\\ \textrm{d}.&\textrm{IV}\\ \textrm{e}.&\textrm{V} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\textrm{Pembahasan diserahkan kepada pembaca} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Pada gambar berikut ini, yang merupakan}\\ &\textrm{himpunan penyelesaian sistem pertidaksamaan}\\\\ &\color{magenta}\begin{cases} 2x+y & \geq 4 \\ x+y &\geq 3 \\ x-4y & \geq 4 \end{cases}\\\\ &\textrm{adalah}\: ....\\ \end{array}$

$\begin{array}{ll}\\ 36.&\textrm{Pada gambar berikut ini, yang merupakan}\\ &\textrm{himpunan penyelesaian sistem pertidaksamaan}\\\\ &\color{magenta}\begin{cases} -x+2y & \leq 2 \\ 4x+3y &\leq 12 \\ x \geq 0&\\ y\geq 0& \end{cases}\\\\ &\textrm{adalah}\: ....\\ \end{array}$

$.\: \: \: \begin{array}{ll}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\textrm{I}\\ \textrm{b}.&\textrm{II}\\ \textrm{c}.&\textrm{III}\\ \textrm{d}.&\textrm{I}\: \: \textrm{dan}\: \: \textrm{IV}\\ \textrm{e}.&\textrm{II}\: \: \textrm{dan}\: \: \textrm{II} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\textrm{Pembahasan diserahkan kepada pembaca} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Pada gambar berikut ini, yang merupakan}\\ &\textrm{himpunan penyelesaian sistem pertidaksamaan}\\\\ &\color{magenta}\begin{cases} 5x+y & \geq 10\\ 2x+y &\leq 8 \\ y\geq 2& \end{cases}\\\\ &\textrm{adalah}\: ....\\ \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Pada gambar berikut ini, yang merupakan}\\ &\textrm{himpunan penyelesaian sistem pertidaksamaan}\\\\ &\color{magenta}\begin{cases} x+y & \geq 4\\ x+2y &\leq 6 \\ y\geq 1& \end{cases}\\\\ &\textrm{adalah}\: ....\\ \end{array}$

$\begin{array}{ll}\\ 39.&(\textrm{SPMB 2003})\\ &\textrm{Daerah yang diarsir pada ilustrasi gambar berikut}\\ &\textrm{adalah himpunan semua}\: \: (x,y)\: \: \textrm{yang memenuhi} \end{array}$

$.\: \: \: \quad\begin{array}{ll}\\ \color{red}\textrm{a}.&2x+y\leq 30,\: 3x+4y\leq 60,\: x\geq 0,\: y\geq 0\\ \textrm{b}.&2x+y\geq 30,\: 3x+4y\geq 60,\: x\geq 0,\: y\geq 0\\ \textrm{c}.&x+2y\geq 30,\: 3x+4y\geq 60,\: x\geq 0,\: y\geq 0\\ \textrm{d}.&x+2y\leq 30,\: 4x+3y\leq 60,\: x\geq 0,\: y\geq 0\\ \textrm{e}.&2x+y\geq 30,\: 4x+3y\leq 60,\: x\geq 0,\: y\geq 0 \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\textrm{Persamaan garisnya adalah}:\\ &(1)\: 15x+20y=300\\ &\: \: \: \quad\textrm{kendala}:3x+4y\leq 60\\ &(2)\: 30x+15y=450\\ &\: \: \: \quad \textrm{kendala}:2x+y\leq 30\\ &(3)\: y=0,\: \: \textrm{kendala}:y\geq 0 \\ &(4)\: x=0,\: \: \textrm{kendala}:x\geq 0 \end{aligned}$

$\begin{array}{ll}\\ 40.&(\textrm{SPMB 2004})\\ &\textrm{Daerah yang diarsir pada ilustrasi gambar berikut}\\ &\textrm{adalah himpunan penyelesaian yang dipenuhi oleh} \end{array}$

$.\: \: \: \quad\begin{array}{ll}\\ \textrm{a}.&6x+5y-30\leq 0,\: x+6y-6\leq 0,\: x-y\leq 0\\ \textrm{b}.&6x+5y-30\geq 0,\: x+6y-6\leq 0,\: x-y\leq 0\\ \color{red}\textrm{c}.&6+5y-30\leq 0,\: x+6y-6\geq 0,\: x-y\geq 0\\ \textrm{d}.&6x+5y-30\leq 0,\: x+6y-6\geq 0,\: x-y\leq 0\\ \textrm{e}.&6x+5y-30\geq 0,\: x+6y-6\geq 0,x-y\geq 0 \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\textrm{Perhatikan bahwa kendala-kendalanya}:\\ &\begin{array}{|c|c|c|}\hline \begin{aligned}6x+5y&=6\times 5\\ 6x+5y&=30 \end{aligned}&\begin{aligned}x+6y&=1\times 6\\ x+6y&=6 \end{aligned}&\begin{aligned}x&=y\\ \end{aligned}\\\hline \textbf{Sebelah kiri}&\textbf{Sebelah kanan}&\\\hline \begin{aligned}6x+5y&\leq 30\\ 6x+5y-30&\leq 0 \end{aligned}&\begin{aligned}x+6y&\geq 6\\ x+6y-6&\geq 0 \end{aligned}&\begin{aligned}x&\geq y\\ x-y&\geq 0 \end{aligned}\\\hline \end{array} \end{aligned}$

Latihan Soal 3 Persiapan PAS Gasal Matematika Wajib Kelas XI

$\begin{array}{ll}\\ 21.&\textrm{Diketahui bahwa}\: \: S(n)\: \: \textrm{adalah formula dari}\\ &3+6+12+24+...+\left ( 3.2^{n-1} \right )=3.\left ( 2^{n}-1 \right )\\ &\textrm{Jika}\: \: S(n)\: \: \textrm{benar, untuk}\: \: n=k+1,\: \: \textrm{maka}\\ &\textrm{ruas kiri persamaan tersebut dapat dituliskan}\\ &\textrm{dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&3+6+12+24+...+ 3.2^{k+1} \\ \textrm{b}.&3+6+12+24+...+ 3.2^{k-1} \\ \color{red}\textrm{c}.&3+6+12+24+...+ 3.2^{k-1}+3.2^{k} \\ \textrm{d}.&3+6+12+24+...+ 3.2^{k-1}+3.2^{k+1} \\ \textrm{e}.&3+6+12+24+...+ 3.2^{k}+3.2^{k+1} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&3+6+12+24+...+ 3.2^{n-1} =3.\left ( 2^{n}-1 \right )\\ &\color{red}3+6+12+24+...+ 3.2^{k-1}+3.2^{k}\color{black}=\color{blue}3.\left ( 2^{k+1}-1 \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 22.&\textbf{(EBTANAS 1999)}\\ &\textrm{Nilai dari}\: \: \displaystyle \sum_{k=1}^{100}5k-\sum_{k=1}^{100}(2k-1)\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 30.900\\ &\textrm{b}.\quad 30.500\\ &\textrm{c}.\quad 16.250\\ &\textrm{d}.\quad 15.450\\ &\textrm{e}.\quad \color{red}15.250\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\textrm{Diketahi}\\ &\begin{aligned}\displaystyle \sum_{k=1}^{100}5k-\sum_{k=1}^{100}(2k-1)&=\sum_{k=1}^{100}(5k-2k+1)\\ &=\displaystyle \sum_{k=1}^{100}(3k+1)\\ &=3\displaystyle \sum_{k=1}^{100}k+1.100\\ &=3\left ( \displaystyle \frac{100}{2}(1+100) \right )+100\\ &=3.(5.050)+100\\ &=15.150+100\\ &=\color{red}15.250\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textbf{(EBTANAS 2000)}\\ &\textrm{Diketahui}\: \: \displaystyle \sum_{k=5}^{25}(2-pk)=0, \textrm{maka nilai}\\ & \displaystyle \sum_{k=5}^{25}pk= ... .\\ &\textrm{a}.\quad 20\\ &\textrm{b}.\quad 28\\ & \textrm{c}.\quad 30\\ &\textrm{d}.\quad \color{red}42\\ & \textrm{e}.\quad 112\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\displaystyle \sum_{k=5}^{25}(2-pk)=\displaystyle \sum_{k=5}^{25}2-\sum_{k=5}^{25}pk&=0\\ \displaystyle \sum_{k=5-4}^{25-4}2-\sum_{k=5}^{25}pk&=0\\ \displaystyle \sum_{k=5}^{25}pk&=\displaystyle \sum_{k=1}^{21}2\\ &=21.2\\ &=\color{red}42 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textbf{(EBTANAS 2000)}\\ &\textrm{Nilai dari}\: \: \displaystyle \sum_{k=1}^{7}\left ( \displaystyle \frac{1}{2} \right )^{k+1}=... .\\ &\textrm{a}.\quad \displaystyle \frac{127}{1024}\\\\ &\textrm{b}.\quad \displaystyle \frac{127}{256}\\\\ & \textrm{c}.\quad \displaystyle \frac{255}{512}\\\\ &\textrm{d}.\quad \displaystyle \frac{127}{128}\\\\ & \textrm{e}.\quad \displaystyle \frac{255}{256}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}&\displaystyle \sum_{k=1}^{7}\left ( \displaystyle \frac{1}{2} \right )^{k+1}\\ &=\left ( \displaystyle \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{2} \right )^{4}+\left ( \frac{1}{2} \right )^{5}+\left ( \frac{1}{2} \right )^{6}+\left ( \frac{1}{2} \right )^{7} +\left ( \displaystyle \frac{1}{2} \right )^{8}\\ &=\displaystyle \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\\ &=\displaystyle \frac{64+32+16+8+4+2+1}{256}\\ &=\color{red}\displaystyle \frac{127}{256} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textbf{EBTANAS 1999}\\ &\textrm{Diketahui jumlah n suku pertama }\\ &\textrm{deret aritmetika dinyatakan sebagai}\\ &S_{n}=n^{2}+2n.\: \textrm{Beda dari deret tersebut }\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad 3 \\ &\textrm{b}.\quad \color{red}2\\ & \textrm{c}.\quad 1\\ &\textrm{d}.\quad -2\\ & \textrm{e}.\quad -3\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Diketahui bahwa}\: \: S_{n}=n^{2}+2n,\\ &\textrm{dengan}\: \begin{cases} S_{1} & =U_{1}=a \\ S_{2} & =U_{1}+U_{2} \\ S_{3} & =U_{1}+U_{2}+U_{3} \\ &\vdots \\ S_{n} & =U_{1}+U_{2}+U_{3}+\cdots +U_{n} \end{cases}\\ &\begin{aligned}\textrm{Beda}=b&=U_{2}-U_{1}\\ &=(S_{2}-S_{1})-S_{1}\\ &=S_{2}-2S_{1}\\ &=\left ( 2^{2}+2.(2) \right )-2\left ( 1^{2}+2.(1) \right )\\ &=\left ( 4+4 \right )-2\left ( 1+2 \right )=8-6\\ &=\color{red}2\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 26.&\textbf{(UMPTN 1994)}\\ &\textrm{Diketahui jumlah n suku pertama suatu }\\ &\textrm{deret dinyatakan sebagai}\quad S_{n}=12n-n^{2}.\\ & \textrm{Suku kelima dari deret tersebut adalah}\: ....\\ &\textrm{a}.\quad -1 \\ &\textrm{b}.\quad 1\\ & \textrm{c}.\quad -3\\ &\textrm{d}.\quad \color{red}3\\ & \textrm{e}.\quad 0\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\textrm{Diketahui bahwa}\\ &S_{n}=12n-n^{2}\\ &\begin{aligned}U_{5}&=S_{5}-S_{4}\\ &=\left ( 12.(5)-(5)^{2} \right )-\left ( 12.(4)-(4)^{2} \right )\\ &=\left ( 60-25 \right )-\left ( 48-16 \right )\\ &=\color{red}3\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 27.&\textbf{(UMPTN 1997)}\\ &\textrm{Diketahui}\: \: U_{n}\: \: \textrm{adalah suku ke - n }\\ &\textrm{deret aritmetika dengan}\\ &U_{1}+U_{2}+U_{3}=-9\: \: \textrm{dan}\\ & \: U_{3}+U_{4}+U_{5}=15.\\ & \textrm{Maka jumlah lima suku pertama}\\ &\textrm{deret aritmetika tersebut adalah}\: ....\\ &\textrm{a}.\quad 4\\ &\textrm{b}.\quad \color{red}5\\ &\textrm{c}.\quad 6\\ &\textrm{d}.\quad 15\\ &\textrm{e}.\quad 24\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&U_{1}+U_{2}+U_{3}=-9,\\ &\Leftrightarrow a+(a+b)+(a+2b)=-9\\ &\Leftrightarrow \color{blue}3a+3b=-9\\ &U_{3}+U_{4}+U_{5}=15,\\ &\Leftrightarrow (a+2b)+(a+3b)+(a+4b)=15\\ &\Leftrightarrow \color{blue}3a+9b=15\quad _{-}\\ & -----------------\\ &\, \qquad\qquad\qquad -6b=-24\Rightarrow b=\color{red}4\\ &\, \qquad \textrm{sehingga akan diperoleh}\: \: a=\color{red}-7\\ &\textrm{Selanjutnya}\\ &S_{5}=\displaystyle \frac{5}{2}\left ( U_{1}+U_{5} \right )\\ &=\displaystyle \frac{5}{2}\left ( a+a+(5-1)b \right )\\ &=\displaystyle \frac{5}{2}\left ( -7-7+4.4 \right )\\ &=\displaystyle \frac{5}{2}(2)\\ &=\color{red}5\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 28.&\textbf{(UN 2013)}\\ &\textrm{Diketahui suatu barisan aritmetika dengan }\\ &\textrm{suku ketiga adalah 4 dan suku ketujuhnya }\\ &\textrm{adalah 16. Jumlah 10 suku pertama dari }\\ &\textrm{deret tersebut adalah}\: ...\: .\\ &\textrm{a}.\quad \color{red}115\\ &\textrm{b}.\quad 125\\ & \textrm{c}.\quad 130\\ &\textrm{d}.\quad 135\\ &\textrm{e}.\quad \displaystyle 140\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}U_{3}=a+2b&=4\\ U_{7}=a+6b&=16\quad _{-}\\ ------&--\\ -4b&=-12\\ b&=3\\ \textrm{Sehingga}&\: \textrm{didapatkan}\\ \textrm{nilai}\: \: a=&4-2b\\ =&4-2.3\\ =&-2 \end{aligned}\\ &\begin{aligned}&\textrm{Maka}\: \textrm{jumlah 10 suku pertama }\\ &\textrm{deret tersebut adalah}\\ &S_{n}=\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ &S_{10}=\displaystyle \frac{10}{2}\left ( 2.(-2)+(10-1).3 \right )\\ &\: \: \quad=5\left ( -4+27 \right )\\ &\: \: \quad=5(23)\\ &\: \: \quad=\color{red}115 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 29.&\textbf{(UN 2014)}\\ &\textrm{Diketahui tempat duduk gedung pertunjukan }\\ &\textrm{film diatur mulai dari baris depan ke belakang }\\ &\textrm{dengan banyak banyak baris dibelakang lebih }\\ &\textrm{4 kursi dari baris di depannya.}\\ &\textrm{Jika dalam gedung pertunjukan terdapat 15}\\ &\textrm{baris kursi dan baris terdepan ada 20 kursi, }\\ &\textrm{maka kapasitas gedung pertunjukan tersebut }\\ &\textrm{adalah}\: ...\: .\: \textrm{kursi}\\ &\textrm{a}.\quad 1200\\ &\textrm{b}.\quad 800\\ & \textrm{c}.\quad \color{red}720\\ &\textrm{d}.\quad 600\\ &\textrm{e}.\quad 300\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diketahui}&\:\begin{cases} a &=U_{1}=20 \\ b & =4 \\ n & =15 \\ S_{n} & =\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right ) \end{cases}\\ &\\ & \end{aligned}\\ &\begin{aligned}S_{n}&=\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ &=\displaystyle \frac{15}{2}\left ( 2(20)+(15-1).4 \right )\\ &=15(20+28)\\ &=15(48)\\ &=\color{red}750 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textbf{(UN 2015)}\\ &\textrm{Diketahui suatu barisan aritmetika dengan }\\ &\textrm{suku ke-3 adalah 2 dan suku ke-8 adalah -13}.\\ &\textrm{Jumlah 20 suku pertama dari deret tersebut }\\ &\textrm{adalah}\: ...\: .\\ &\textrm{a}.\quad -580\\ &\textrm{b}.\quad -490\\ &\textrm{c}.\quad -440\\ &\textrm{d}.\quad \color{red}-410\\ &\textrm{e}.\quad -380\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}U_{3}=a+2b&=2\\ U_{8}=a+7b&=-13\quad _{-}\\ ------&---\\ -5b&=15\\ b&=-3\\ \textrm{Sehingga}&\: \textrm{didapatkan}\\ \textrm{nilai}\: \: a=&2-2b\\ =&2-2.(-3)\\ =&8 \end{aligned}\\ &\begin{aligned}&\textrm{Maka}\: \textrm{jumlah 20 suku pertama }\\ &\textrm{deret tersebut adalah}\\ &S_{n}=\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ &S_{20}=\displaystyle \frac{20}{2}\left ( 2.(8)+(20-1).(-3) \right )\\ &\: \: \quad=10\left ( 16-57 \right )\\ &\: \: \quad=10(-41)\\ &\: \: \quad=\color{red}-410\\ &\\ & \end{aligned} \end{array}$

DAFTAR PUSTAKA

Budhi, W.S. 2018. Bupena Matematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: ERLANGGA.

DAFTAR PUSTAKA WEB

Thohir, A. https://ahmadthohir1089.wordpress.com/2016/01/11/insyaallah-44/