$\begin{array}{l}\\ 51.&\textrm{Diketahui matriks}\\ &\textrm{A}=\begin{pmatrix} 2020 & -4&-3&2\\ 2020 & -6&-7&1\\ 2020&4&-3&0\\ 2020&6&-7&8 \end{pmatrix}\\ &\textrm{Ordo dari matriks}\: \: \textrm{A}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\times 2&&&\\ \textrm{b}.&3\times 3\\ \textrm{c}.&3\times 4\\ \textrm{d}.&4\times 3\\ \color{red}\textrm{e}.&4\times 4 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\textrm{Cukup jelas}\\ &\color{blue}\textrm{Karena matriknya mengandung}\\ &\color{red}\textrm{4 baris}\color{blue}\times \color{red}\textrm{4 kolom} \end{array}$
$\begin{array}{ll}\\ 52.&\textrm{Diketahui matriks}\\ &\textrm{B}=\begin{pmatrix} 1 & 2&3&2020\\ 5 & 13&7&2019\\ 11&14&3&-2018\\ -15&6&17&2017 \end{pmatrix}\\ &\textrm{Jika}\: \: \textrm{b}_{ij}\: \: \textrm{menunjukkan elemen}\\ &\textrm{yang terletak pada baris ke}-i\\ &\textrm{dan kolom ke}-j\: \: \textrm{pada matriks B}\\ &\textrm{ di atas, maka}\: \: b_{43}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\\ \textrm{b}.&9\\ \textrm{c}.&-1\\ \textrm{d}.&3\\ \color{red}\textrm{e}.&17 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\textrm{Perh}&\textrm{atikan bahwa}\\ \color{black}\textrm{B}_{4\times 4}&=\begin{pmatrix} b_{11} & b_{12} & b_{13} & b_{14}\\ b_{21} & b_{22} & b_{23} & b_{24}\\ b_{31} & b_{32} & b_{33} & b_{34}\\ b_{41} & b_{42} & \color{red}b_{43} & b_{44} \end{pmatrix}\\ &=\begin{pmatrix} 1 & 2&3&2020\\ 5 & 13&7&2019\\ 11&14&3&-2018\\ -15&6&\color{red}17&2017 \end{pmatrix}\\ \color{black}\textrm{sehi}&\color{black}\textrm{ngga entri}\: \: \color{red}b_{43}=17 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 53.&\textrm{Diketahui matriks}\: \: \textrm{C}\: \: \textrm{adalah matriks}\\ &\textrm{berordo}\: \: 3\times 3.\: \: \textrm{Jika}\: \: \textrm{c}_{ij}=4j-5i,\\ &\textrm{maka matriks C tersebut adalah}....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&\begin{pmatrix} -1 & 3 & 7\\ -6 & -2 & 2\\ -11 & -7 & -3 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} -1 & 7 & 3\\ -6 & 2 & -2\\ -7 & -11 & -3 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} -1 & -7 & -11\\ -6 & 7 & 3\\ -2 & 2 & -3 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} -1 &-6 & -11\\ 3 & -2 & 2\\ 7 & 2 & -3 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} -1 & -2 & -3\\ 3 & -6 & -11\\ 7 & -7 & 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\: \: \color{red}c_{ij}=4j-5i,\\ \textrm{mak}&\textrm{a}\\ \textrm{C}_{3\times 3}&=\begin{pmatrix} c_{\color{red}11} & c_{\color{red}12} & c_{\color{red}13} \\ c_{\color{red}21} & c_{\color{red}22} & c_{\color{red}23} \\ c_{\color{red}31} & c_{\color{red}32} & c_{\color{red}33} \end{pmatrix}\\ &=\begin{pmatrix} 4.1-5.1 & 4.2-5.1&4.3-5.1\\ 4.1-5.2 & 4.2-5.2&4.3-5.2\\ 4.1-5.3&4.2-5.3&4.3-5.3 \end{pmatrix}\\ &=\begin{pmatrix} 4-5 & 8-5 & 12-5\\ 4-10 & 8-10 & 12-10\\ 4-15 & 8-15 & 12-15 \end{pmatrix}\\ &=\begin{pmatrix} -1 & 3 & 7\\ -6 & -2 & 2\\ -11 & -7 & -3 \end{pmatrix} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 54.&\textrm{Jika diketahui matriks}\\ &\textrm{X}=\begin{pmatrix} -7 & 9&-1\\ 4 & -6&15 \end{pmatrix}.\\ &\textrm{maka transpose matriks}\: \: \textrm{X}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\textrm{X}^{t}=\begin{pmatrix} 4 & -6 & 15\\ -7 & 9 & -1 \end{pmatrix}\\ \textrm{b}.&\textrm{X}^{t}=\begin{pmatrix} 15 & -6 & 4\\ -1 & 9 & -7 \end{pmatrix}\\ \color{red}\textrm{c}.&\textrm{X}^{t}=\begin{pmatrix} -7 & 4\\ 9 & -6\\ -1 & 15 \end{pmatrix}\\ \textrm{d}.&\textrm{X}^{t}=\begin{pmatrix} 4 & -7\\ -6 & 9\\ 15 & -1 \end{pmatrix}\\ \textrm{e}.&\textrm{X}^{t}=\begin{pmatrix} 15 & -1\\ -6 & 9\\ 4 & -7 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\\ \textrm{X}_{2\times 3}&=\begin{pmatrix} x_{\color{red}11} & x_{\color{red}12} & x_{\color{red}13} \\ x_{21} & x_{22} & x_{23} \end{pmatrix}\\ &=\begin{pmatrix} \color{red}-7 & \color{red}9&\color{red}-1\\ 4 & -6&15 \end{pmatrix}\\ \color{black}\textrm{maka}&\\ \textrm{X}_{3\times 2}^{t}&=\begin{pmatrix} x_{\color{red}11} & x_{21}\\ x_{\color{red}12} & x_{22}\\ x_{\color{red}13} & x_{23} \end{pmatrix}=\begin{pmatrix} \color{red}-7 & 4\\ \color{red}9 & -6\\ \color{red}-1 & 15 \end{pmatrix} \\ \textrm{adal}&\textrm{ah sebuah}\: \color{red}\textrm{matriks baru} \\ \textrm{deng}&\textrm{an ordo}\: \: \color{red}3\times 2 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 55.&\textrm{Diketahui matriks}\: \: \textrm{P}=\begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{Q}=\begin{pmatrix} 2c-3b & 2a+1\\ a & b+7 \end{pmatrix}.\\ &\textrm{Nilai}\: \: c\: \: \textrm{yang memenuhi jika}\\ & \textrm{P}=2\textrm{Q}^{t}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-2\\ \textrm{b}.&3\\ \textrm{c}.&5\\ \color{red}\textrm{d}.&8\\ \textrm{e}.&10 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{P}&=2\textrm{Q}^{\color{red}t}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=2\begin{pmatrix} 2c-3b & 2a+1\\ a & b+7 \end{pmatrix}^{\color{red}t}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=2\begin{pmatrix} 2c-3b & a\\ 2a+1 & b+7 \end{pmatrix}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=\begin{pmatrix} 4c-6b & 2a\\ 4a+2 & 2b+14 \end{pmatrix}\\ &(\color{red}\textrm{kesamaan 2 buah matriks})\\ \color{black}\textrm{akibat}&\color{black}\textrm{nya}\\ &\begin{cases} a &= 4c-6b \quad ....(1)\\ 4 &=2a \quad ..........(2)\\ 2b &=4a+2 \quad ........(3)\\ 3c &=2b+14 \quad ........(4) \end{cases}\\ \textrm{dari}&\: \textrm{persamaan}\: \: (2)\\ & 2a=4\Rightarrow a=2\quad....(5)\\ \textrm{pers}&\textrm{amaan}\: \: (5)\: \: \textrm{hasilnya}\\ \textrm{disu}&\textrm{bstitusikan ke persamaan}\: \: (3),\\ \color{blue}\textrm{yait}&\color{black}\textrm{u}\\ 2b&=4a+2\Rightarrow 2b=4(2)+2=10\\ b&=5\quad.....................(6)\\ \textrm{pers}&\textrm{amaan}\: \: (6)\: \: \textrm{hasilnya disbstitusikan}\\ \textrm{ke p}&\textrm{ersamaan}\\ (4),&\: \textrm{dan akan mendapatkan}\\ 3c&=2b+14\Rightarrow 3c=2(5)+14=24\\ \color{red}c&\color{red}=8 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 56.&\textrm{Diketahui matriks}\\ &\textrm{M}=\begin{pmatrix} -6 & 9&-15\\ 3 & -6&12 \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{N}=\begin{pmatrix} 2 & -3&5\\ -1 & 2&-4 \end{pmatrix}.\\ &\textrm{Nilai}\: \: k\: \: \textrm{yang memenuhi jika}\\ &\textrm{M}=k\textrm{N}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&-1\\ \color{red}\textrm{d}.&-3\\ \textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahu bahwa}\\ &\textrm{M}=k\textrm{N}\\ &(\color{red}\textrm{perkalian suatu matrik dengan skalar})\\ &\begin{pmatrix} -6 & 9&-15\\ 3 & -6&12 \end{pmatrix}\\ &=\begin{pmatrix} \color{red}-3.\color{black}2 & \color{red}-3.\color{black}-3 & \color{red}-3.\color{black}5\\ \color{red}-3.\color{black}-1 & \color{red}-3.\color{black}2 & \color{red}-3.\color{black}-4 \end{pmatrix}\\ &=\color{red}-3\color{black}\begin{pmatrix} 2 & -3 & 5\\ -1 & 2 & -4 \end{pmatrix}\\ &=k\begin{pmatrix} 2 & -3&5\\ -1 & 2&-4 \end{pmatrix}\\ &\textrm{sehingga dari kesamaan tersebut}\\ &\textrm{maka}\quad \color{red}k=-3 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 57.&\textrm{Hasil dari}\: \: \begin{pmatrix} 1&2&3\\ 4&5&6 \end{pmatrix}\times \begin{pmatrix} 1 &2 \\ 3 &4 \\ 5 & 6 \end{pmatrix}\\ & \textrm{adalah}...\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&\begin{pmatrix} 22 & 28\\ 49&64 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 22&49\\ 28&64 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 64&28\\ 49&22 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} 2 & 8&18\\ 4&15 & 30 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 1&4&6\\ 4&15&30 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\begin{pmatrix} 1&2&3\\ 4&5&6 \end{pmatrix}_{\color{red}2\times \color{black}3}\times \begin{pmatrix} 1 &2 \\ 3 &4 \\ 5 & 6 \end{pmatrix}_{\color{black}3\times \color{red}2}\\ &=\begin{pmatrix} 1.1+2.3+3.5 & 1.2+2.4+3.6\\ 4.1+5.3+6.5 &4.2+5.4+6.6 \end{pmatrix}_{\color{red}2\times 2}\\ &=\begin{pmatrix} 1+6+15 & 2+8+18\\ 4+15+30 & 8+20+36 \end{pmatrix}_{\color{red}2\times 2}\\ &=\begin{pmatrix} 22 & 28\\ 49 & 64 \end{pmatrix}_{\color{red}2\times 2} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 58.&\textrm{Jika diketahui matriks}\\ & \textrm{A}=\begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}.\\ &\textrm{maka hasil dari}\: \: \textrm{A}^{3}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} 5 & 8\\ 20&22 \end{pmatrix}\\ \color{red}\textrm{b}.&\begin{pmatrix} 6&7\\ 21&20 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 6&7\\ 20&22 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} 7 & 8\\ 20 & 23 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 7&9\\ 20&23 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\\ \textrm{A}&=\begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}\\ \textrm{mak}&\textrm{a}\\ \textrm{A}^{2}&=\textrm{A}\times \textrm{A}\\ &=\begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}\times \begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}\\ &=\begin{pmatrix} 0+3&0+2\\ 0+6&3+4 \end{pmatrix}\\ &=\color{purple}\begin{pmatrix} 3 & 2\\ 6 & 7 \end{pmatrix}\\ \textrm{A}^{3}&=\textrm{A}^{2}\times \textrm{A}\\ &=\begin{pmatrix} 3 & 2\\ 6 &7 \end{pmatrix}\times \begin{pmatrix} 0 & 1\\ 3 & 2 \end{pmatrix}\\ &=\begin{pmatrix} 0+6 & 3+4\\ 0+21 & 6+14 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 6 & 7\\ 21 & 20 \end{pmatrix} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 59.&(\textbf{SBMPTN Mat IPA 2014})\\ &\textrm{Jika}\: \: \textrm{A}\: \: \textrm{adalah matriks yang berordo}\\ & 2\times 2\: \: \textrm{dan memenuhi}\\ &\: \: \begin{pmatrix} x & 1 \end{pmatrix}\times \textrm{A}\times \begin{pmatrix} x\\ 1 \end{pmatrix}=x^{2}-5x+8,\\ & \textrm{maka matriks A yang mungkin adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} 1 & -5\\ 8&0 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 1&5\\ 8&0 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 1&8\\ -5&0 \end{pmatrix}\\ \color{red}\textrm{d}.&\begin{pmatrix} 1 & 3\\ -8&8 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 1&-3\\ 8&-8 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\begin{pmatrix} x & 1 \end{pmatrix}\times \textrm{A}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=\color{red}x^{2}-5x+8\\ \begin{pmatrix} x & 1 \end{pmatrix}\times \begin{pmatrix} p & q\\ r & s \end{pmatrix}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=\color{red}x^{2}-5x+8\\ \begin{pmatrix} xp+r & xq+s \end{pmatrix}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=\color{red}x^{2}-5x+8\\ \begin{pmatrix} x^{2}p+xr+xq+s \end{pmatrix}&=\color{red}x^{2}-5x+8\\ px^{2}+(q+r)x+s&=\color{red}x^{2}-5x+8\\ \end{aligned}\\ &\color{blue}\begin{aligned}&\begin{cases} \color{red}p &=1 \\ q+r &=-5 \\ \color{red}s &=8 \end{cases}\quad\Rightarrow\quad \begin{pmatrix} 1 & ...\\ ... & 8 \end{pmatrix}\\ &\textrm{Sehingga yang paling mungkin}\\ & \textrm{adalah}\: \: \color{red}\begin{pmatrix} 1 & 3\\ -8 & 8 \end{pmatrix} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 60.&\textrm{Diketahui}\\ &\begin{pmatrix} ^{x}\log a & \log (2a-6)\\ \log (b-2) & 1 \end{pmatrix}=\begin{pmatrix} \log b & 1\\ \log a & 1 \end{pmatrix}\\ &\textrm{maka nilai}\: \: x\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&6\\ \color{red}\textrm{e}.&8 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\begin{pmatrix} ^{x}\log a & \log (2a-6)\\ \log (b-2) & 1 \end{pmatrix}=\begin{pmatrix} \log b & 1\\ \log a & 1 \end{pmatrix}\\ &\color{black}\textrm{maka}\\ &\begin{cases} ^{x}\log a & =\log b \quad.........\color{red}(1)\\ \log (2a-6) &=1\quad..............\color{red}(2) \\ \log (b-2) &=\log a\quad.........\color{red}(3) \end{cases}\\ &\textrm{Sehingga}\: \textrm{dari persamaan}\: \: (2)\\ &\color{black}\textrm{akan didapatkan}\\ &\log (2a-6)=1=\log 10\\ &(2a-6)=10\\ &a=8\quad...........................(4)\\ &\textrm{persamaan}\: (4)\: \: \textrm{ke persamaan}\: \: (3),\\ & \color{black}\textrm{maka}\\ &\log (b-2) =\log a\\ &b-2=a=8\\ &b=10\quad.................................(5)\\ &\textrm{Selanjutnya dari persamaan}\: \: (5)\\ &\color{black}\textrm{akan diperoleh}\\ &^{x}\log a =\log b\\ &^{x}\log 8 =\log 10=1\\ &\qquad x^{1}=8\\ &\Leftrightarrow \: \: \color{red}x=8 \end{aligned} \end{array}$
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