Contoh Soal Matriks

\begin{array}{ll}\\ \fbox{1}.&\textrm{Diketahui matriks}\: \: \textrm{Z}=\begin{pmatrix} -5 & -4&-3&2\\ -5 & -6&-7&1\\ -5&4&-3&0\\ -5&6&-7&-1 \end{pmatrix}\\ &\textrm{Ordo dari matriks}\: \: \textrm{Z}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\times 2&&&\textrm{d}.&4\times 3\\ \textrm{b}.&3\times 3&\textrm{c}.&3\times 4&\textrm{e}.&4\times 4 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Cukup jelas karena matriks Z memiliki 4 baris dan 4 kolom} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Diketahui matriks}\: \: \textrm{A}=\begin{pmatrix} 12 & -4&3&20\\ 5 & 13&7&9\\ 11&14&3&-1\\ -15&6&17&18 \end{pmatrix}\\ &\textrm{Jika}\: \: \textrm{a}_{ij}\: \: \textrm{menunjukkan elemen yang terletak pada}\\ &\textrm{baris ke}-i\: \: \textrm{dan kolom ke}-j\: \: \textrm{pada matriks A di atas},\: \: \textrm{maka}\: \: a_{43}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3&&&\textrm{d}.&3\\ \textrm{b}.&9&\textrm{c}.&-1&\textrm{e}.&17 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Perh}&\textrm{atikan bahwa}\\ \textrm{A}_{4\times 4}&=\begin{pmatrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34}\\ a_{41} & a_{42} & a_{43} & a_{44} \end{pmatrix}=\begin{pmatrix} 12 & -4&3&20\\ 5 & 13&7&9\\ 11&14&3&-1\\ -15&6&17&18 \end{pmatrix}\\ \textrm{sehi}&\textrm{gga akan didapatkan}\: \: a_{43}=17 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui matriks}\: \: \textrm{A}\: \: \textrm{adalah matriks berordo}\: \: 3\times 3.\\ &\textrm{Jika}\: \: \textrm{a}_{ij}=4j-5i,\: \: \textrm{maka matriks A tersebut adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} -1 & 3 & 7\\ -6 & -2 & 2\\ -11 & -7 & -3 \end{pmatrix}&&&\textrm{b}.&\begin{pmatrix} -1 &-6 & -11\\ 3 & -2 & 2\\ 7 & 2 & -3 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} -1 & 7 & 3\\ -6 & 2 & -2\\ -7 & -11 & -3 \end{pmatrix}&&&\textrm{d}.&\begin{pmatrix} -1 & -2 & -3\\ 3 & -6 & -11\\ 7 & -7 & 2 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} -1 & -7 & -11\\ -6 & 7 & 3\\ -2 & 2 & -3 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\: \: a_{ij}=4j-5i,\: \: \textrm{maka}\\ \textrm{A}_{3\times 3}&=\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{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}\\ \fbox{4}.&\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{d}.&\textrm{X}^{t}=\begin{pmatrix} 4 & -7\\ -6 & 9\\ 15 & -1 \end{pmatrix}\\ \textrm{b}.&\textrm{X}^{t}=\begin{pmatrix} 15 & -6 & 4\\ -1 & 9 & -7 \end{pmatrix}&\textrm{c}.&\textrm{X}^{t}=\begin{pmatrix} -7 & 4\\ 9 & -6\\ -1 & 15 \end{pmatrix}&\textrm{e}.&\textrm{X}^{t}=\begin{pmatrix} 15 & -1\\ -6 & 9\\ 4 & -7 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\\ \textrm{X}_{2\times 3}&=\begin{pmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \end{pmatrix}=\begin{pmatrix} -7 & 9&-1\\ 4 & -6&15 \end{pmatrix}\\ \textrm{maka}&\\ \textrm{X}_{3\times 2}^{t}&=\begin{pmatrix} x_{11} & x_{21}\\ x_{12} & x_{22}\\ x_{13} & x_{23} \end{pmatrix}=\begin{pmatrix} -7 & 4\\ 9 & -6\\ -1 & 15 \end{pmatrix}\\ & \left ( x_{11},\: x_{12},\: x_{13},\: x_{21},...,\: x_{23}\: \: \: \textrm{hanya menunjukkan posisi perubahannya} \right )\\ \textrm{adal}&\textrm{ah sebuah matriks baru dengan ordo}\: \: 3\times 2 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\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{d}.&8\\ \textrm{b}.&3&\textrm{c}.&5&\textrm{e}.&10 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{P}&=2\textrm{Q}^{t}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=2\begin{pmatrix} 2c-3b & 2a+1\\ a & b+7 \end{pmatrix}^{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}\quad ......(\textrm{kesamaan 2 buah matriks})\\ \textrm{akibat}&\textrm{nya}\\ &\begin{cases} a &= 4c-6b \quad ..................\textcircled{1}\\ 4 &=2a \quad ...........................\textcircled{2}\\ 2b &=4a+2 \quad ......................\textcircled{3}\\ 3c &=2b+14 \quad ......................\textcircled{4} \end{cases}\\ \textrm{dari}&\: \textrm{persamaan}\: \: \textcircled{2}\: \: 2a=4\Rightarrow a=2\quad....\textcircled{5}\\ \textrm{pers}&\textrm{amaan}\: \: \textcircled{5}\: \: \textrm{hasilnya disbstitusikan ke persamaan}\: \: \textcircled{3},\: \textrm{yaitu}\\ 2b&=4a+2\Rightarrow 2b=4(2)+2=10\\ b&=5\quad.....................\textcircled{6}\\ \textrm{pers}&\textrm{amaan}\: \: \textcircled{6}\: \: \textrm{hasilnya disbstitusikan ke persamaan}\: \: \textcircled{4},\: \textrm{dan akan mendapatkan}\\ 3c&=2b+14\Rightarrow 3c=2(5)+14=24\\ c&=8 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\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{d}.&-3\\ \textrm{b}.&\displaystyle \frac{1}{3}&\textrm{c}.&-1&\textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Diketahu bahwa}\:\:\: &\\ \textrm{M}&=k\textrm{N}.........(\textrm{perkalian suatu matrik dengan skalar})\\ \begin{pmatrix} -6 & 9&-15\\ 3 & -6&12 \end{pmatrix}&=\begin{pmatrix} -3.2 & -3.-3 & -3.5\\ -3.-1 & -3.2 & -3.-4 \end{pmatrix}\\ &=-3\begin{pmatrix} 2 & -3 & 5\\ -1 & 2 & -4 \end{pmatrix}\\ &=k\begin{pmatrix} 2 & -3&5\\ -1 & 2&-4 \end{pmatrix}\\ \textrm{sehingga dari}&\: \textrm{persamaan di atas diketahui bahwa nilai skalar}\quad k=-3 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{7}.&\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}\\ \textrm{a}.&\begin{pmatrix} 22 & 28\\ 49&64 \end{pmatrix}&&&\textrm{d}.&\begin{pmatrix} 2 & 8&18\\ 4&15 & 30 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 22&49\\ 28&64 \end{pmatrix}&\textrm{c}.&\begin{pmatrix} 64&28\\ 49&22 \end{pmatrix}&\textrm{e}.&\begin{pmatrix} 1&4&6\\ 4&15&30 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\begin{pmatrix} 1&2&3\\ 4&5&6 \end{pmatrix}_{2\times 3}\times \begin{pmatrix} 1 &2 \\ 3 &4 \\ 5 & 6 \end{pmatrix}_{3\times 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}_{2\times 2}\\ &=\begin{pmatrix} 1+6+15 & 2+8+18\\ 4+15+30 & 8+20+36 \end{pmatrix}_{2\times 2}\\ &=\begin{pmatrix} 22 & 28\\ 49 & 64 \end{pmatrix}_{2\times 2} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{8}.&\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}&&&\textrm{d}.&\begin{pmatrix} 7 & 8\\ 20 & 23 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 6&7\\ 21&20 \end{pmatrix}&\textrm{c}.&\begin{pmatrix} 6&7\\ 20&22 \end{pmatrix}&\textrm{e}.&\begin{pmatrix} 7&9\\ 20&23 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \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}=\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}\\ &=\begin{pmatrix} 6 & 7\\ 21 & 20 \end{pmatrix} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{9}.&(\textbf{SBMPTN Mat IPA 2014})\\ &\textrm{Jika}\: \: \textrm{A}\: \: \textrm{adalah sebuah 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{d}.&\begin{pmatrix} 1 & 3\\ -8&8 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 1&5\\ 8&0 \end{pmatrix}&\textrm{c}.&\begin{pmatrix} 1&8\\ -5&0 \end{pmatrix}&\textrm{e}.&\begin{pmatrix} 1&-3\\ 8&-8 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} x & 1 \end{pmatrix}\times \textrm{A}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=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}&=x^{2}-5x+8\\ \begin{pmatrix} xp+r & xq+s \end{pmatrix}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=x^{2}-5x+8\\ \begin{pmatrix} x^{2}p+xr+xq+s \end{pmatrix}&=x^{2}-5x+8\\ px^{2}+(q+r)x+s&=x^{2}-5x+8\\ &\begin{cases} p &=1 \\ q+r &=-5 \\ s &=8 \end{cases}\quad\Rightarrow\quad \begin{pmatrix} 1 & ...\\ ... & 8 \end{pmatrix}\\ \textrm{Sehingga}&\: \textrm{yang paling mungkin adalah}\: \: \begin{pmatrix} 1 & 3\\ -8 & 8 \end{pmatrix} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{10}.&\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{d}.&6\\ \textrm{b}.&2&\textrm{c}.&4&\textrm{e}.&8 \end{array}\\\\ &\textbf{Jawab}:\quad \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}\\ \textrm{maka}&\\ &\begin{cases} ^{x}\log a & =\log b \quad.........\textcircled{1}\\ \log (2a-6) &=1\quad..............\textcircled{2} \\ \log (b-2) &=\log a\quad.........\textcircled{3} \end{cases}\\ \textrm{Sehingga}&\: \textrm{dari persamaan}\: \: \textcircled{2}\: \: \textrm{akan didapatkan}\\ \log (2a-6)&=1=\log 10\\ (2a-6)&=10\\ a&=8\quad.......................................\textcircled{4}\\ \textrm{persamaan}&\: \textcircled{4}\: \: \textrm{ke persamaan}\: \: \textcircled{3},\: \textrm{maka}\\ \log (b-2) &=\log a\\ b-2&=a=8\\ b&=10\quad.......................................\textcircled{5}\\ \textrm{Selanjutnya}&\: \: \textrm{dari persamaan}\: \: \textcircled{5}\: \: \textrm{akan diperoleh}\\ ^{x}\log a & =\log b\\ ^{x}\log 8 & =\log 10=1\\ x^{1}&=8\\ \Leftrightarrow \: \: x&=8 \end{aligned} \end{array}.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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