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}$

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

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

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

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

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

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