Contoh Soal 3 Matriks

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

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

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

$\begin{array}{l}\\ 14.& \text{Diketahui matriks}\text{Z}=\left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)\\ & \text{dan}f(x)=x^2-x.\\ & \text{Jika}f(\text{Z})=\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right),\\ & \text{maka nilai}{p}^2-q^2=....\\ & \begin{array}{l}\\ \text{a}.& 5\\ \text{b}.& 7\\ \text{c}.& 9\\ \text{d}.& 12\\ \text{e}.& 15\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(\text{Z})& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ {\text{Z}}^2-\text{Z}& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ \left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)\times \left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)-\left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ \left(\begin{array}{cc}4-18& -12+30\\ 6-15& -18+25\end{array}\right)-\left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ \left(\begin{array}{cc}-12& 12\\ -6& 2\end{array}\right)& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\end{array}\\ & \begin{array}{rl}\text{Sehingga}& \\ -12& =-3p-8q.................(1)\\ -1& =p+q......................(2)\\ \text{persamaan}& (2)\text{ke persamaan}(1)\\ -12& =-3p-3q-5q=-3(p+q)-5q\\ -12& =-3(-1)-5q\\ -12& =3-5q\\ 5q& =3+12\\ q& =3........................(3)\\ \text{persamaan}& (3)\text{ke persamaan}(2)\\ p+q& =-1\\ p& =-1-q=-1-3=-4\\ p^2-q^2& =(-4{)}^2-3^2=16-9\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 15.& (\mathbf{\text{SBMPTN 2013}})\\ & \text{Jika}A=\left(\begin{array}{ccc}2& -1& 1\\ a& b& c\end{array}\right),\\ & B=\left(\begin{array}{cc}-2& 1\\ 1& -1\\ 0& 2\end{array}\right)\text{dan}\\ & AB=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \text{maka nilai}2c-a=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& AB=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \left(\begin{array}{ccc}2& -1& 1\\ a& b& c\end{array}\right)\left(\begin{array}{cc}-2& 1\\ 1& -1\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \left(\begin{array}{cc}-5& 5\\ -2a+b& a-b+2c\end{array}\right)=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \begin{array}{l}\\ -2a+b& =3& \\ a-b+2c& =-3& +\\ 2c-a& =0\end{array}\end{array}\end{array}$

DAFTAR PUSTAKA

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

Contoh Soal 2 Matriks

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

$\begin{array}{l}\\ 7.& \text{Hasil dari}\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right)\times \left(\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\end{array}\right)\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& \left(\begin{array}{cc}22& 28\\ 49& 64\end{array}\right)\\ \text{b}.& \left(\begin{array}{cc}22& 49\\ 28& 64\end{array}\right)\\ \text{c}.& \left(\begin{array}{cc}64& 28\\ 49& 22\end{array}\right)\\ \text{d}.& \left(\begin{array}{ccc}2& 8& 18\\ 4& 15& 30\end{array}\right)\\ \text{e}.& \left(\begin{array}{ccc}1& 4& 6\\ 4& 15& 30\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right)}_{2\times 3}\times {\left(\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\end{array}\right)}_{3\times 2}\\ & ={\left(\begin{array}{cc}1.1+2.3+3.5& 1.2+2.4+3.6\\ 4.1+5.3+6.5& 4.2+5.4+6.6\end{array}\right)}_{2\times 2}\\ & ={\left(\begin{array}{cc}1+6+15& 2+8+18\\ 4+15+30& 8+20+36\end{array}\right)}_{2\times 2}\\ & ={\left(\begin{array}{cc}22& 28\\ 49& 64\end{array}\right)}_{2\times 2}\end{array}\end{array}$

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

$\begin{array}{l}\\ 9.& (\mathbf{\text{SBMPTN Mat IPA 2014}})\\ & \text{Jika}\text{A}\text{adalah matriks yang berordo}\\ & 2\times 2\text{dan memenuhi}\\ & \left(\begin{array}{cc}x& 1\end{array}\right)\times \text{A}\times \left(\begin{array}{c}x\\ 1\end{array}\right)=x^2-5x+8,\\ & \text{maka matriks A yang mungkin adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left(\begin{array}{cc}1& -5\\ 8& 0\end{array}\right)\\ \text{b}.& \left(\begin{array}{cc}1& 5\\ 8& 0\end{array}\right)\\ \text{c}.& \left(\begin{array}{cc}1& 8\\ -5& 0\end{array}\right)\\ \text{d}.& \left(\begin{array}{cc}1& 3\\ -8& 8\end{array}\right)\\ \text{e}.& \left(\begin{array}{cc}1& -3\\ 8& -8\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{cc}x& 1\end{array}\right)\times \text{A}\times \left(\begin{array}{c}x\\ 1\end{array}\right)& =x^2-5x+8\\ \left(\begin{array}{cc}x& 1\end{array}\right)\times \left(\begin{array}{cc}p& q\\ r& s\end{array}\right)\times \left(\begin{array}{c}x\\ 1\end{array}\right)& =x^2-5x+8\\ \left(\begin{array}{cc}xp+r& xq+s\end{array}\right)\times \left(\begin{array}{c}x\\ 1\end{array}\right)& =x^2-5x+8\\ \left(\begin{array}{c}{x}^{2}p+xr+xq+s\end{array}\right)& =x^2-5x+8\\ p{x}^2+(q+r)x+s& =x^2-5x+8\end{array}\\ & \begin{array}{rl}& \{\begin{array}{ll}p& =1\\ q+r& =-5\\ s& =8\end{array}\Rightarrow \left(\begin{array}{cc}1& ...\\ ...& 8\end{array}\right)\\ & \text{Sehingga yang paling mungkin}\\ & \text{adalah}\left(\begin{array}{cc}1& 3\\ -8& 8\end{array}\right)\end{array}\end{array}$

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

Contoh Soal 1 Matriks

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

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

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

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

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

Lanjutan 2 Materi Matriks (Matematika Wajib Kelas XI)

 $\text{C. Tarnspose dan Kesamaan Dua Buah Matriks}$

$\begin{array}{|cl|}\hline 1.& \text{Transpose Matriks}\\ & \begin{array}{rl}& \text{Membentuk matriks baru dari matriks}\\ & \text{dengan cara mengubah baris matriks ke}-i\\ & \text{menjadi kolom ke}-i,\text{pada matriks baru}\\ & \text{dan demikian pula untuk kolomnya}.\text{Jika}\\ & \text{matriks pertama adalah A maka matriks}\\ & \text{transposenya adalah}{\text{A}}^{\prime }\text{atau}{\text{A}}^t\end{array}\\ 2.& \text{Kesamaan Duan Buah Matriks}\\ & \begin{array}{rl}& \text{Misalkan matriks}\text{A}=\left(a_{ij}\right)\text{dan}\text{B}=\left(b_{ij}\right)\\ & \text{adalah dua buah matriks berordo sama},\\ & \text{maka matriks A dikatakan sama dengan matriks B}\\ & \text{jika elemen-elemen yang seletak sama pada}\\ & \text{kedua matriks tersebut bernilai sama}\\ & \end{array}\\ \hline\end{array}$

$\begin{array}{|l|}\hline \text{Berikut contoh transpose}\\ \text{A}={\displaystyle \left(\begin{array}{cc}1& 2\\ -7& 0\\ 5& 4\end{array}\right)\Rightarrow {\text{A}}^t=\left(\begin{array}{ccc}1& -7& 5\\ 2& 0& 4\end{array}\right)}\\ \text{DAn berikut contoh kesamaan dua matriks}\\ \text{A}={\displaystyle \left(\begin{array}{cc}1& 2\\ -7& 0\\ 5& 4\end{array}\right),\text{D}={\displaystyle \left(\begin{array}{cc}1& 2\\ -7& 0\\ 5& 4\end{array}\right),\Rightarrow \text{A}=\text{D}}}\\ \hline\end{array}$

 $\text{D. Operasi Matriks}$

$\begin{array}{|clll|}\hline \text{No}& \text{Operasi}& \text{Ketentuan}& \text{Contoh}\\ & \text{Matriks}& & \\ 1& \text{Penjumlahan}\mathrm{\&}& \text{ordo sama}& A=\left(\begin{array}{c}1\\ 2\end{array}\right),B=\left(\begin{array}{c}8\\ 9\end{array}\right),\text{maka}\\ 2& \text{Pengurangan}& \text{ordo sama}& A+B=\left(\begin{array}{c}1+8\\ 2+9\end{array}\right)=\left(\begin{array}{c}9\\ 11\end{array}\right)\\ 3& \text{Perkalian}& \text{Dengan}& k\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)=\left(\begin{array}{cc}kp& kq\\ kr& ks\end{array}\right)\\ & \text{Skalar}& \text{mengalikan}& \\ & & \text{ke setiap elemen}& \\ 4& \text{Perkalian}& \begin{array}{rl}& \text{Dua matriks }\\ & \text{dapat dikalikan }\\ & \text{jika}\\ & \text{banyaknya kolom}\\ & \text{matriks pertama}\\ & \text{sama dengan}\\ & \text{banyaknya baris}\\ & \text{matriks kedua}\end{array}& \begin{array}{rl}E=& \left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right),F=\left(\begin{array}{c}5\\ 0\end{array}\right),\\ & \text{maka}\\ E& \times F\\ & ={\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)}_{2\times 2}\times {\left(\begin{array}{c}5\\ 0\end{array}\right)}_{2\times 1}\\ & \text{syarat memenuhi yaitu}:\\ & \text{kolom matriks 1}\\ & =\text{baris matriks 2}\\ & \text{dan hasilnya adalah }\\ & \text{matriks baru}\\ & \text{dengan ordo }\\ & \text{banyak baris matriks 1}\\ & \text{kali banyak}\\ & \text{kolom matriks 2}\\ & \text{Dan aturan perkaliannya }\\ & \text{adalah}\\ & \text{elemen baris matriks 1 kali}\\ & \text{elemen kolom matriks 2}\\ & \text{sehingga}\\ & ={\left(\begin{array}{c}1(5)+2(0)\\ 3(5)+-1(0)\end{array}\right)}_{2\times 1}\\ & =\left(\begin{array}{c}5+0\\ 15-0\end{array}\right)={\left(\begin{array}{c}5\\ 15\end{array}\right)}_{2\times 1}\end{array}\\ \hline\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ \bullet & \text{Penjumlahan}\\ & \begin{array}{rl}& \left(\begin{array}{cc}1& 2\\ 3& -4\end{array}\right)+\left(\begin{array}{cc}5& 6\\ 7& -8\end{array}\right)\\ & =\left(\begin{array}{cc}1+5& 2+6\\ 3+7& (-4)+(-8)\end{array}\right)\\ & =\left(\begin{array}{cc}6& 8\\ 10& -12\end{array}\right)\end{array}\\ \bullet & \text{Lawan suatu matriks}\\ & \begin{array}{rl}& \text{Jika}A=\left(\begin{array}{cc}1& 2\\ 3& -4\end{array}\right),\\ & \text{maka lawan matriks A adalah -A,}\\ & \text{Sehingga}-A=\left(\begin{array}{cc}-1& -2\\ -3& 4\end{array}\right)\end{array}\\ \bullet & \text{Pengurangan}\\ & \begin{array}{rl}& \left(\begin{array}{cc}1& 2\\ 3& -4\end{array}\right)-\left(\begin{array}{cc}5& 6\\ 7& -8\end{array}\right)\\ & =\left(\begin{array}{cc}1-5& 2-6\\ 3-7& (-4)-(-8)\end{array}\right)\\ & =\left(\begin{array}{cc}-4& -4\\ -4& 4\end{array}\right)\end{array}\\ \bullet & \text{Perkalian}\\ & \begin{array}{rl}& (1)\text{Perkalian suatu matriks dengan skalar}k\\ & 2\times \left(\begin{array}{cc}1& 2\\ 3& -4\end{array}\right)=\left(\begin{array}{cc}2\times 1& 2\times 2\\ 2\times 3& 2\times (-4)\end{array}\right)\\ & =\left(\begin{array}{cc}2& 4\\ 6& -8\end{array}\right)\end{array}\\ & \begin{array}{rl}& (2)\text{Perkalian antara dua buah matriks}\\ & \left(\begin{array}{cc}1& 2\\ 3& -4\end{array}\right)\times \left(\begin{array}{cc}5& 6\\ 7& -8\end{array}\right)\\ & \text{perhatikan syarat memenuhi}\\ & =\left(\begin{array}{cc}1\times 5+2\times 7& 1\times 6+2\times (-8)\\ 3\times 5+(-4)\times 7& 3\times 6+(-4)\times (-8)\end{array}\right)\\ & =\left(\begin{array}{cc}5+14& 6+(-16)\\ 15+(-28)& 18+32\end{array}\right)\\ & =\left(\begin{array}{cc}19& -10\\ -13& 50\end{array}\right)\end{array}\end{array}$

Lanjutan 1 Materi Matriks (Matematika Wajib Kelas XI)

$\text{B. Jenis-Jenis Matriks}$

$\begin{array}{|lllll|}\hline \text{No}& \text{Jenis}& \text{Keterangan}& \text{Contoh}& \text{Ordo}\\ 1.& \text{Matriks}& \text{Matriks yang elemen}& \left(\begin{array}{ccc}1& 3& -5\end{array}\right)& 1\times 3\\ & \text{Baris}& \text{penyusunnya satu baris saja}& & \\ 2.& \text{Matriks}& \text{Matriks yang elemen}& \left(\begin{array}{c}5\\ -5\\ 2\end{array}\right)& 3\times 1\\ & \text{Kolom}& \text{penyusunnya tepat satu kolom saja}& & \\ 3.& \text{Matriks}& \text{Matriks yang semua elemennya}& \left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\end{array}\right)& 2\times 3\\ & \text{Nol}& \text{adalah bilangan nol}& & \\ 4.& \text{Matriks}& \text{Matriks yang jumlah}& \left(\begin{array}{cc}2& 8\\ 6& 1\end{array}\right)& 2\times 2\\ & \text{Persegi}& \text{baris dan kolomnya sama}& & \\ \hline\end{array}$

$\begin{array}{|lllll|}\hline \text{No}& \text{Jenis}& \text{Keterangan}& \text{Contoh}& \text{Ordo}\\ 5.& \text{Matriks}& \text{Matriks Persegi yang semua}& \left(\begin{array}{cc}1& 0\\ 0& 7\end{array}\right)& 2\times 2\\ & \text{Diagonal}& \text{elemennya nol kecuali pada }& \left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& 6\end{array}\right)& 3\times 3\\ & & \text{diagonal utama}& & \\ 6.& \text{Matriks}& \text{Matriks yang elemen semuanya}& \left(\begin{array}{c}5\\ -5\\ 2\end{array}\right)& 3\times 1\\ & \text{Identitas}& \text{nol kecuali pada diagonal}& & \\ & & \text{utama berupa angka 1}& \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)& 2\times 2\\ \hline\end{array}$

$\begin{array}{|lllll|}\hline \text{No}& \text{Jenis}& \text{Keterangan}& \text{Contoh}& \text{Ordo}\\ 7.& \text{Matriks}& \text{matriks persegi yang semua elemen}& \left(\begin{array}{cc}5& 8\\ 0& 7\end{array}\right)& 2\times 3\\ & \text{Segitiga}& \text{di bawah diagonal utama berupa }& \left(\begin{array}{ccc}1& 2& 3\\ 0& 3& 5\\ 0& 0& 6\end{array}\right)& 3\times 3\\ & \text{atas}& \text{bilangan nol}& & \\ 8.& \text{Matriks}& \text{Matriks persegi yang semua elemen}& \left(\begin{array}{ccc}1& 0& 0\\ 4& 2& 0\\ 5& 6& 3\end{array}\right)& 3\times 3\\ & \text{segitiga}& \text{di bawah diagonal utama berupa}& & \\ & \text{bawah}& \text{angka nol}& \left(\begin{array}{cc}1& 0\\ 6& 2\end{array}\right)& 2\times 2\\ \hline\end{array}$

$\begin{array}{|lllll|}\hline \text{No}& \text{Jenis}& \text{Keterangan}& \text{Contoh}& \text{Ordo}\\ 9.& \text{Matriks}& \text{Suatu matriks disebut sebagai matriks}& \left(\begin{array}{cc}5& 0\\ 8& 7\end{array}\right)& 2\times 2\\ & \text{Simetris}& \text{simetris jika dan hanya jika elemen-elemen}& & \\ & \text{utama}& \text{yang letaknya simetris terhadap diagonal }& & \\ & & \text{atau bernilai sama}& & \\ \hline\end{array}$

Matriks (Matematika Wajib Kelas XI)

$\text{A. Pendahuluan}$

Dalam kehidupan sehari-hari kita sering menjumpai atau mendapatkan informasi yang tersaji dalam bentuk daftar atau tabel. Sebagai misal ketika seorang wali kelas merekap data absensi kelas waliannya selama satu semester sebagaimana diperlihatkan dalam tabel berikut ini

$\begin{array}{l}\\ \begin{array}{|lccc|}\hline \text{Nama}& \text{Sakit}& \text{Izin}& \text{Tanpa}\\ \text{Siswa}& & & \text{Keterangan}\\ \text{Andi}& 2& 1& 3\\ \text{Budi}& 1& 4& 2\\ \text{Carli}& 1& 1& 5\\ \text{Dodi}& 2& 2& 1\\ \hline\end{array}& \begin{array}{rl}& \Leftarrow \text{Judul baris}\\ & \\ & \\ & \\ & \end{array}\\ \begin{array}{rl}& \Uparrow \\ & \text{Judul kolom}\end{array}& \end{array}$

Dari tabel di atas, jika kita tuliskan bilangannya saja maka akan kita dapatkan bilangan yang seolah-olah tersusun berbentuk persegi atau persegi panjang dan oleh kareanya sebagaimana ilsutrasi tabel di atas bilangan terbut juga tersusun dalam baris dan kolom sebagaimana berikut ini

$\begin{array}{l}\\ & 2& 1& 3\\ \\ & 1& 4& 2\\ \\ & 1& 1& 5\\ \\ & 2& 2& 1\end{array}$

Selanjutnya kaitanya dengan matriks apa bila susunan bilangan-bilangan di atas, diberikan tanda kurung tertentu jadil bentuk matriks. Dan dari paparan tersebut kita dapat mengakatakan matriks adalah susunan bilangan dalam bentuk persegi atau persegi panjang yang di atur menurut baris dan kolom dalam tanda kurung tertentu.

Selanjutnya bilangan yang diatus menurut baris dan kolom disebut unsur atau elemen atau entri dari suatu matri

Perhatikanlah ilustrasi berikut

Sebagai tambahan nama sebuah matrik adalah sebuah huruf besar dan memiliki ukuran sebuah matrik(selanjutnya dapat isebut sebagai ordo) = baris x kolom.

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Diketahui}M=\left[\begin{array}{ccccc}-2& 3& 0& 0& -3\\ -1& 2& 1& 1& 2\\ 0& 1& -3& 2& -2\\ 1& 0& -5& 3& 1\end{array}\right],\text{tentukanlah}\\ \\ & \text{a}.\text{ordo dari matrik tersebut}\\ & \text{b}.\text{elemen penyusun kolom pertama}\\ & \text{c}.\text{elemen penyusun baris pertama}\\ & \text{d}.\text{elemen penyusun kolom kedua}\\ & \text{e}.\text{elemen penyusun baris kedua}\\ & \text{f}.\text{elemen penyusun kolom ketiga}\\ & \text{g}.\text{elemen penyusun baris ketiga}\\ & \text{h}.\text{elemen penyusun kolom keempat}\\ & \text{i}.\text{elemen penyusun baris keempat}\\ & \text{j}.\text{elemen penyusun kolom kelima}\\ & \text{k}.\text{elemen penyusun baris kelima}\\ & \text{l}.\text{elemen yang terletak pada baris kelima dan kolom kelima}\\ & \text{m}.\text{elemen yang terletak pada baris pertama dan kolom kelima}\\ & \text{n}.\text{elemen yang terletak pada baris kedua dan kolom keempat}\\ & \text{o}.\text{elemen yang terletak pada baris ketiga dan kolom ketiga}\\ & \text{p}.\text{elemen yang terletak pada baris kedua dan kolom kedua}\\ & \text{q}.\text{elemen yang terletak pada baris pertama dan kolom pertama}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\\ & \text{a}.\text{ordo matriknya}4\times 5\\ & \begin{array}{|cccc|}\hline \text{b}.& -2,-1,0,1& \text{c}.& -2,3,0,0,-3\\ \text{d}.& 3,2,1,0& \text{e}.& -1,2,1,1,2\\ \text{f}.& 0,1,-3,-5& \text{g}.& 0,1,-3,2,-2\\ \text{h}.& 0,1,2,3& \text{i}.& 1,0,-5,3,1\\ \text{j}& -3,2,-2,1& \text{k}& \text{tidak ada}\\ \text{l}.& \text{tidak ada}& \text{m}& m_{15}=-3\\ \text{n}.& m_{24}=1& \text{o}.& m_{33}=-3\\ \text{p}.& m_{22}=2& \text{q}.& m_{11}=-2\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Diketahui}A=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],B=\left[\begin{array}{cc}3& 7\\ 4& 8\\ 5& 9\\ 6& 10\end{array}\right],\\ & C=\left[\begin{array}{ccc}11& 12& 13\\ 14& 15& 16\\ 17& 18& 19\\ 20& 21& 22\end{array}\right]\text{tentukanlah nilai}\\ \\ & \text{a}.a_{11},a_{31},b_{42},\text{dan}{c}_{43}\\ & \text{b}.b_{42}+c_{43}\\ & \text{c}.a_{41}-b_{31}+c_{21}-a_{11}\\ & \text{d}.a_{11}+b_{22}+c_{33}\\ & \text{e}.a_{11}^2+b_{22}^2+c_{33}^2\\ & \text{f}.{(a_{11}+b_{22})}^2-c_{33}^2\end{array}$

$.\begin{array}{rl}& \text{Jawab}\\ & \begin{array}{|clcl|}\hline \text{a}.& \begin{array}{rl}& \text{perhatikanlah}\\ & a_{11}=1,a_{31}=3\\ & b_{42}=10,c_{43}=22\end{array}& \text{b}.& \begin{array}{rl}& b_{42}+c_{43}\\ & =10+22\\ & =32\end{array}\\ \text{c}.& \begin{array}{rl}& a_{41}-b_{31}+c_{21}-a_{11}\\ & =4-5+14-1\\ & =12\end{array}& \text{d}.& \begin{array}{rl}& a_{11}+b_{22}+c_{33}\\ & =1+8+19\\ & =28\end{array}\\ \text{e}.& \begin{array}{rl}& a_{11}^2+b_{22}^2+c_{33}^2\\ & =1^2+8^2+{19}^2\\ & =1+64+361=426\end{array}& \text{f}.& \begin{array}{rl}& {(a_{11}+b_{22})}^2-c_{33}^2\\ & =(1+8{)}^2-{19}^2\\ & =81-361=-280\end{array}\\ \hline\end{array}\end{array}$

Contoh Soal 8 Statistika

$\begin{array}{l}\\ 36.& \text{Diketahui nilai statistik lima serangkai}\\ & \text{dari empat kelompok data seperti terlihat}\\ & \text{dalam tabel berikut}\\ & \begin{array}{|lccccc|}\hline \text{Data}& \text{min}& Q_1& Q_2& Q_3& \text{mak}\\ \text{I}& 74& 80& 88& 92,5& 99\\ \text{II}& 66& 81,5& 86& 90,5& 96\\ \text{III}& 70& 77,5& 85& 92,5& 100\\ \text{IV}& 55& 80& 88& 90& 97,5\\ \hline\end{array}\\ & \text{Data yang memuat pencilan terdapat pada}\\ & \text{tabel}\\ & \begin{array}{l}\\ \text{a}.& \text{I dan II}\\ \text{b}.& \text{II dan III}\\ \text{c}.& \text{I dan III}\\ \text{d}.& \text{III dan IV}\\ \text{e}.& \text{II dan IV}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{Diketa}& \text{hui bahwa}\mathbf{\text{pencilan}}\text{adalah datum yang}\\ \text{bernil}& \text{ai kurang dari pagar dalam dan lebih besar}\\ \text{dari p}& \text{agar luar}\\ \text{Rumu}& \text{s pagar dalam}=Q_1-L\\ & =Q_1-{\displaystyle \frac{3}{2}(Q_3-Q_1)={\displaystyle \frac{5}{2}{Q}_1-\frac{3}{2}{Q}_3}}\\ \text{Rumu}& \text{s pagar luar}=Q_3+L\\ & =Q_3+{\displaystyle \frac{3}{2}(Q_3-Q_1)={\displaystyle \frac{5}{2}{Q}_3-\frac{3}{2}{Q}_1}}\\ \text{Data I}& \text{pagar dalamnya}={\displaystyle \frac{5}{2}(80)-\frac{3}{2}(92,5)}\\ & =200-138,75=61,25\\ \text{Data I}& \text{pagar luarnya}={\displaystyle \frac{5}{2}(92,5)-\frac{3}{2}(80)}\\ & =231,25-120=111,25\\ \text{Jadi, d}& \text{ata I tidak ada pencilan}\\ \text{Data I}& \text{I pagar dalamnya}={\displaystyle \frac{5}{2}(81,5)-\frac{3}{2}(90,5)}\\ & =203,75-135,75=68\\ \text{Data I}& \text{I pagar luarnya}={\displaystyle \frac{5}{2}(90,5)-\frac{3}{2}(81,5)}\\ & =226,25-122,25=104\\ \text{Jadi, d}& \text{ata II ada pencilan, yaitu}66<68\\ & \text{karena}66\text{adalah datum terkecil data II}\\ \text{Data I}& \text{II pagar dalamnya}={\displaystyle \frac{5}{2}(77,5)-\frac{3}{2}(92,5)}\\ & =193,75-138,75=55\\ \text{Data I}& \text{II pagar luarnya}={\displaystyle \frac{5}{2}(92,5)-\frac{3}{2}(77,5)}\\ & =231,25-116,25=115\\ \text{Jadi, d}& \text{ata III tidak ada pencilan}\\ \text{Data I}& \text{V pagar dalamnya}={\displaystyle \frac{5}{2}(80)-\frac{3}{2}(90)}\\ & =200-135=65\\ \text{Data I}& \text{V pagar luarnya}={\displaystyle \frac{5}{2}(90)-\frac{3}{2}(70)}\\ & =225-105=120\\ \text{Jadi, d}& \text{ata IV ada pencilan, yaitu}55<65\\ & \text{karena}55\text{adalah datum terkecil data IV}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastolan, & Sulasim. 2005. Kompetensi Matematika SMA Kelas 2 Semester 1 Program IPA Kurikulum Berbasis Kompetensi. Jakarta: YUDHISTIRA.
2. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI. Bandung: YRAMA WIDYA.
3. Sharma, dkk. 2017. Jelajah Matematika SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA.
4. Tim Supermat. 2007. Cara Mudah MenghadapiS SMBB TELKOM. Jakarta: LITERATUR MEDIA SUKSES.

Contoh Soal 7 Statistika

$\begin{array}{l}\\ 31.& (\mathbf{\text{SMBB TELKOM 06}})\\ & \text{Berikut tidak termasuk ukuran penyebaran}\\ & \text{data adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{rentang}\\ \text{b}.& \text{varians}\\ \text{c}.& \text{jarak antar kuartil}\\ \text{d}.& \text{kuartil}\\ \text{e}.& \text{simpangan baku}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Ingat bahwa mean, median, modus}\\ & \text{adalah bagian ukuran pemusatan data}\\ & \text{sedangkan kuartil adalah ukuran letak data}\end{array}\end{array}$.

$\begin{array}{l}\\ 32.& \text{Perhatikanlah barisan}1,-2,3,-4,5,-6,...\\ & \text{dengan suku ke}-n\text{adalah}(-1{)}^{n-1}\times n.\\ & \text{Rata-rata 200 suku pertama barisan tersebut}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& -0,5\\ \text{c}.& 0\\ \text{d}.& 0,5\\ \text{e}.& 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}:1,-2,3,-4,5,-6,...,199,-200\\ & {\bar{x}}_{200}={\displaystyle \frac{1-2+3-4+5-6+...+199-200}{200}}\\ & ={\displaystyle \frac{-100}{200}=-0,5}\end{array}\end{array}$

$\begin{array}{l}\\ 33.& (\mathbf{\text{STT TELKOM 2005}})\\ & \text{Seratus mahasiswa telah mengikuti ujian psikotes}\\ & \text{dan rata-rata skornya yang diperoleh adalah 100}.\\ & \text{Banyaknya mahasiswa junior yang mengikuti}\\ & \text{ujian psikotes}50\mathrm{\%}\text{lebih besar dari banyknya ma}-\\ & \text{hasiswa senior. Jika rata-rata skor dari mahasiswa}\\ & \text{senior}50\mathrm{\%}\text{lebih tinggi dari mahasiswa junior,}\\ & \text{maka rata-rata skor mahasiswa senior adalah}....\\ & \begin{array}{l}\\ \text{a}.& 110\\ \text{b}.& 115\\ \text{c}.& 120\\ \text{d}.& 125\\ \text{e}.& 150\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Perhatikanlah tabel berikut}\\ & \begin{array}{|ccc|}\hline \text{Mahasiswa}& \text{Mahasiswa}& \text{Total}\\ \text{senior}& \text{junior}& \\ n_{senior}& n_{junior}& 100\\ n_s& {\displaystyle \frac{3}{2}{n}_s}& 100\\ {\bar{x}}_{senior}& {\bar{x}}_{junior}& 100\\ {\bar{x}}_s& {\displaystyle \frac{2}{3}{\bar{x}}_s}& 100\\ \hline\end{array}\\ & n_s+{\displaystyle \frac{3}{2}{n}_s=100}\\ & n_s=40,\\ & \text{maka}{n}_j=60.\\ & \text{Selanjutnya kita tentukan nilai rata-rata}\\ & \text{mahasiswa senior, dengan}\\ & {\displaystyle \frac{40{\bar{x}}_s+(60){\displaystyle \frac{2}{3}{\bar{x}}_s}}{100}=100}\\ & 80{\bar{x}}_s=10000\\ & {\bar{x}}_s={\displaystyle \frac{10000}{80}}\\ & {\bar{x}}_s=125\end{array}\end{array}$

$\begin{array}{l}\\ 34.& (\mathbf{\text{SPMB 04}})\\ & \text{Median, rata-rata, dan modus dari data yang}\\ & \text{terdiri atas empat bilangan asli adalah 7. Jika}\\ & \text{selisih antara data terbesar dan terkecil adalah}\\ & \text{6, maka hasil kali keempat datum tersebut}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1.864\\ \text{b}.& 1.932\\ \text{c}.& 1.960\\ \text{d}.& 1.976\\ \text{e}.& 1.983\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}:x_1,x_2,x_3,\text{dan}{x}_4\text{adalah}\\ & \text{asli dengan}{x}_4-x_1=6........(1)\\ & \mathrm{\&}\text{modusnya adalah 7, maka data dapat dituliskan}\\ & :x_1,7,7,x_4.\text{Jawaban ini sesuai karena median}=7.\\ & \text{Karena mean}=7,\text{maka dapat dituliskan}\\ & {\displaystyle \frac{x_1+x_2+x_3+x_4}{4}=7\iff x_1+x_2+x_3+x_4=28}\\ & \iff x_1+14+x_4=28\iff x_4+x_1=14........(2)\\ & \text{Selanjutnya kita eliminasi}(1)\mathrm{\&}(2)\\ & \begin{array}{l}\\ x_4-x_1& =6& \\ x_4+x_1& =14+\\ 2x_4& =20& \\ x_4& =10.......(3)\\ \text{maka}{x}_1& =4.........(4)\end{array}\\ & \text{Jadi},x_1\times x_2\times x_3\times x_4=(4).(7).(7).(10)=1960\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Desil ke-8}\left(D_8\right)\text{dari data berikut adalah}....\\ & \begin{array}{|cc|}\hline \text{Nilai}& f\\ 41-45& 7\\ 46-50& 12\\ 51-55& 9\\ 56-60& 8\\ 61-65& 4\\ \hline\end{array}\\ & \begin{array}{l}\\ \text{a}.& 58\\ \text{b}.& 57,5\\ \text{c}.& 57\\ \text{d}.& 56,75\\ \text{e}.& 56,25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui}& \text{desil ke}-8=D_8,\text{dengan}n=\sum f=40\\ D_i& =t_b+p\left({\displaystyle \frac{{\displaystyle \frac{i\times n}{10}-F}}{f}}\right)\\ D_8& =\text{datum ke}-\left({\displaystyle \frac{8n}{10}}\right)=x_{\frac{8\times 40}{10}}=x_{32}\\ \text{Dan}{x}_{32}& \text{terletak di kelas interval}:56-60\\ D_8& =55,5+5\left({\displaystyle \frac{32-28}{8}}\right)\\ & =55,5+2,5\\ & =58\end{array}\end{array}$

Contoh Soal 6 Statistika

$\begin{array}{l}\\ 26.& \text{Jika rata-rata dari}{x}_1,x_2,x_3,x_4,...,x_{10}\\ & \text{adalah}{x}_0,\text{maka rata-rata dari data}\\ & (x_1-1),(x_2+2),(x_3-3),(x_4+4),..\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x_0+5,5\\ \text{b}.& x_0+25\\ \text{c}.& x_0+0,5\\ \text{d}.& x_0-0,5\\ \text{e}.& x_0-2,5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Rata}-& \text{ratanya adalah}:\\ \bar{x}=x_0& ={\displaystyle \frac{x_1+x_2+x_3+...+x_{10}}{10}}\\ \iff 10x_0& =x_1+x_2+x_3+...+x_{10}\\ \text{Selanju}& \text{tnya penghitungan rata-rata yang data baru},\\ {\bar{x}}_{baru}& ={\displaystyle \frac{(x_1-1)+(x_2+2)+(x_3-3)+...+(x_{10}+10)}{10}}\\ & ={\displaystyle \frac{x_1+x_2+...+x_{10}+(2-1+4-3+..+10-9)}{10}}\\ & ={\displaystyle \frac{10x_0+5}{10}}\\ & =x_0+0,5\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Nilai rata-rata dari 20 bilangan adalah}14,2\\ & \text{Jika rata-rata dari 12 bilangan pertama adalah}\\ & \text{12,6 dan rata-rata dari 6 bilangan berikutnya}\\ & \text{adalah 18,2, rata-rata 2 bilangan tersisa}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 10,4\\ \text{b}.& 11,8\\ \text{c}.& 12,2\\ \text{d}.& 12,8\\ \text{e}.& 13,8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\bar{x}}_{total}& ={\displaystyle \frac{{\bar{x}}_{12}\times (12)+{\bar{x}}_6\times (6)+{\bar{x}}_2\times (2)}{20}}\\ 14,2& ={\displaystyle \frac{(12,6\times 12)+(18,2\times 6)+{\bar{x}}_2\times (2)}{20}}\\ 284& =151,2+109,2+2{\bar{x}}_2\\ 2{\bar{x}}_2& =284-(151,2+109,2)=23,6\\ {\bar{x}}_2& ={\displaystyle \frac{23,6}{2}}\\ & =11,8\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Dari 3 bilangan yang terkecil adalah 39}\\ & \text{dan terbesarnya adalah 75, maka rata-rata}\\ & \text{hitung ketiga bilangan tersebut tidak}\\ & \text{mungkin sama dengan}....\\ & \begin{array}{l}\\ \text{a}.& 49\\ \text{b}.& 52\\ \text{c}.& 53\\ \text{d}.& 59\\ \text{e}.& 60\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\bar{x}}_3& ={\displaystyle \frac{39+A+75}{3}}\\ \text{Se}& \text{lanjutnya rentang nilai}A\text{akan berada di}\\ :& 39\le A\le 75\\ \text{Se}& \text{hingga},\\ \text{un}& \text{tuk}A=39,\text{maka}\\ {\bar{x}}_3& ={\displaystyle \frac{39+39+75}{3}=\frac{153}{3}=51,\text{dan}}\\ \text{un}& \text{tuk}A=75,\text{maka}\\ {\bar{x}}_3& ={\displaystyle \frac{39+75+75}{3}=\frac{189}{3}=63}\end{array}\end{array}$

$\begin{array}{l}\\ 29.& (\mathbf{\text{SPMB 04}})\\ & \text{Nilai rata-rata tes Matematika dari kelompok}\\ & \text{siswa dan kelompok siswi di suatu kelas berturut-}\\ & \text{turut adalah 5 dan 7. Jika nilai rata-rata di kelas}\\ & \text{tersebut adalah 6,2, maka perbandingan banyaknya}\\ & \text{siswa dan siswi adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2:3\\ \text{b}.& 3:4\\ \text{c}.& 2:5\\ \text{d}.& 3:5\\ \text{e}.& 4:5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\bar{x}}_{gabungan}& ={\displaystyle \frac{n_1{\bar{x}}_1+n_2{\bar{x}}_2}{n_1+n_2}}\\ 6,2& ={\displaystyle \frac{n_1(5)+n_2(7)}{n_1+n_2}}\\ 6,2(n_1+n_2)& =5n_1+7n_2\\ 6,2n_1-5n_1& =7n_2-6,2n_2\\ 1,2n_1& =0,8n_2\\ {\displaystyle \frac{n_1}{n_2}}& ={\displaystyle \frac{0,8}{1,2}=\frac{2}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 30.& (\mathbf{\text{SPMB 05}})\\ & \text{Nilai rata-rata ulangan kelas A adalah}{\bar{x}}_A\text{dan}\\ & \text{kelas B adalah}{\bar{x}}_B.\text{Setelah kedua kelas digabungkan}\\ & \text{nilai rata-ratanya adalah}\bar{x}.\text{Perbandingan nilai}\\ & \text{kelas A dan B adalah}10:9.\text{Jika perbandingan nilai}\\ & \text{rata-rata kedua kelas dan kelas B adalah}85:81,\\ & \text{maka perbandinganbanyaknya siswa kelas A dan B}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 8:9\\ \text{b}.& 4:5\\ \text{c}.& 3:4\\ \text{d}.& 3:5\\ \text{e}.& 9:10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\bar{x}}_A:{\bar{x}}_B& =10:9=90:81\\ \bar{x}:{\bar{x}}_B& =85:81,\text{maka}\\ \bar{x}:{\bar{x}}_A:{\bar{x}}_B& =85:90:81\\ (n_A+n_B)\bar{x}& =n_A\times {\bar{x}}_A+n_B\times {\bar{x}}_B\\ {\displaystyle \frac{n_A}{n_B}}& ={\displaystyle \frac{\bar{x}-{\bar{x}}_B}{{\bar{x}}_A-\bar{x}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{85}{81}{\bar{x}}_B-{\bar{x}}_B}}{{\displaystyle \frac{10}{9}{\bar{x}}_B-{\displaystyle \frac{85}{81}{\bar{x}}_B}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{4}{81}{\bar{x}}_B}}{{\displaystyle \frac{5}{81}{\bar{x}}_B}}}\\ & ={\displaystyle \frac{4}{5}}\end{array}\end{array}$

Contoh Soal 5 Statistika

$\begin{array}{l}\\ 21.& \text{Simpangan kuartil dari data}\\ & 71,70,68,40,45,48,52,53,53,67,62\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 8\\ \text{b}.& 10\\ \text{c}.& 15\\ \text{d}.& 18\\ \text{e}.& 20\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}(\text{degan total datum ganjil})\\ :& 71,70,68,40,45,48,52,53,53,67,62\\ \text{Sete}& \text{lah data diurutkan menjadi}\\ :& 40,45,48,52,53,53,62,67,68,70,71\\ \text{Dike}& \text{tahui}n=11\mathbf{\text{ganjil}}\\ Q_1& =x_{\frac{1}{4}(n+1)}=x_{\frac{1}{4}.12}=x_3=48\\ Q_2& =x_{\frac{2}{4}(n+1)}=x_{\frac{2}{4}.12}=x_6=53\\ Q_3& =x_{\frac{3}{4}(n+1)}=x_{\frac{3}{4}.12}=x_9=68\\ \text{Sela}& \text{njutnya data dapat dituliskan}\\ & 40,45,\underset{\begin{array}{c}\Downarrow \\ Q_1\end{array}}{\underbrace{48}},52,53,53,62,67,\underset{\begin{array}{c}\Downarrow \\ Q_3\end{array}}{\underbrace{68}},70,71\\ \text{Simp}& \text{angan kuartil data tunggal adalah}:\\ & ={\displaystyle \frac{1}{2}(Q_3-Q_1)}\\ & ={\displaystyle \frac{1}{2}(68-48)}\\ & ={\displaystyle \frac{1}{2}.20=10}\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Data penjualan suatu barang setiap bulan}\\ & \text{di sebuah toko pada tahun 2019 adalah}:\\ & 20,3,9,11,4,12,1,9,9,12,8,10.\\ & \text{Median, kuartil bawah, dan kuartil atasnya}\\ & \text{berturut-turut adalah}....\\ & \begin{array}{l}\\ \text{a}.& 6{\displaystyle \frac{1}{2},3\frac{1}{2},\text{dan}9\frac{1}{2}}\\ \text{b}.& 9,6,\text{dan}11{\displaystyle \frac{1}{2}}\\ \text{c}.& 6{\displaystyle \frac{1}{2},9,\text{dan}12}\\ \text{d}.& 9,4,\text{dan}12\\ \text{e}.& 9,3{\displaystyle \frac{1}{2},\text{dan}12}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}\\ :& 20,3,9,11,4,12,1,9,9,12,8,10\\ \text{Sete}& \text{lah data diurutkan}\\ :& 1,3,4,8,9,9,9,10,11,12,12,20\\ \text{Dike}& \text{tahui}n=12\mathbf{\text{genap}}\\ Q_1& =x_{\frac{1}{4}n+\frac{1}{2}}=x_{\frac{1}{4}.12+\frac{1}{2}}=x_{3,5}=6\\ Q_2& =x_{\frac{2}{4}n+\frac{1}{2}}=x_{\frac{2}{4}.12+\frac{1}{2}}=x_{6,5}=9=M_e\\ Q_3& =x_{\frac{3}{4}n+\frac{1}{2}}=x_{\frac{3}{4}.12+\frac{1}{2}}=x_{9,5}=11{\displaystyle \frac{1}{2}}\\ \text{Sela}& \text{njutnya data dapat dituliskan}\\ & 1,3,\underset{\begin{array}{c}\Downarrow \\ Q_1\end{array}}{\underbrace{4,8}},9,\underset{\begin{array}{c}\Downarrow \\ Q_2=M_e\end{array}}{\underbrace{9,9}},10,\underset{\begin{array}{c}\Downarrow \\ Q_3\end{array}}{\underbrace{11,12}},12,20\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Ragam(varians) dari data}\\ & 6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 1{\displaystyle \frac{3}{8}}\\ \text{c}.& 1{\displaystyle \frac{1}{8}}\\ \text{d}.& {\displaystyle \frac{7}{8}}\\ \text{e}.& {\displaystyle \frac{5}{8}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}\\ :& 6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \text{Sete}& \text{lah data diurutkan(untuk memudahkan)}\\ :& 5,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9\\ \text{Dike}& \text{tahui}n=16.\text{Selanjutnya kita cari}\\ \bar{x}& ={\displaystyle \frac{5.1+6.4+7.6+8.4+9.1}{16}}\\ & =\frac{112}{16}=7\\ \text{Dan}& \text{rumus untuk menghitung ragam adalah}:\\ S^2& ={\displaystyle \frac{{\displaystyle \sum _{i=1}^{16}{(x_i-\bar{x})}^2}}{n}}\\ & ={\displaystyle \frac{(5-7{)}^2+4(6-7{)}^2+6(7-7{)}^2+4(8-7{)}^2+(9-7{)}^2}{16}}\\ & ={\displaystyle \frac{4+4.1+6.0+4.1+4}{16}}\\ & ={\displaystyle \frac{16}{16}}\\ & =1\end{array}\end{array}$

$\begin{array}{l}\\ 24.& \text{Diketahui}{x}_1=2,x_2=3,5,x_3=5,x_4=7,\\ & \text{dan}{x}_5=7,5.\text{Deviasi rata-rata data di atas}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& 1,8\\ \text{d}.& 2,6\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}\\ :& 2,3{\displaystyle \frac{1}{2},5,7,7{\displaystyle \frac{1}{2}}}\\ \text{Dike}& \text{tahui}n=5.\text{Selanjutnya kita cari}\\ \bar{x}& ={\displaystyle \frac{2+3,5+5+7+7,5}{5}=\frac{25}{5}=5}\\ \text{Dan}& \text{rumus simpangan rata-rata adalah}:\\ SR& ={\displaystyle \frac{{\displaystyle \sum _{i=1}^5|x_i-\bar{x}|}}{n}}\\ & ={\displaystyle \frac{|2-5|+|3,5-5|+|5-5|+|7-5|+|7,5-5|}{5}}\\ & ={\displaystyle \frac{|-3|+|-1,5|+\left|0\right|+\left|2\right|+|2,5|}{5}}\\ & ={\displaystyle \frac{3+1,5+0+2+2,5}{5}=\frac{9}{5}}\\ & =1,8\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Jumlah rataan dan median dari}\\ & (x-6),(x+5),(x+4),(x-7),\\ & (x+9),\text{dan}(x-2)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2x-1,5\\ \text{b}.& 2x-0,5\\ \text{c}.& 2x+1,5\\ \text{d}.& 2x+2,5\\ \text{e}.& 2x+3,5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}\\ :& (x-6),(x+5),(x+4),(x-7),(x+9),(x-2)\\ \text{Dike}& \text{tahui}n=6.\text{Selanjutnya kita urutkan datanya}\\ & (x-7),(x-6),(x-2),(x+4),(x+5),(x+9)\\ \text{Rata}& \text{annya}:\\ \bar{x}& ={\displaystyle \frac{6x+3}{6}=x+0,5}\\ \text{Medi}& \text{annya}:\\ M_e& =x_{3,4}={\displaystyle \frac{x_3+x_4}{2}=\frac{(x-2)+(x+4)}{2}}\\ & ={\displaystyle \frac{2x+2}{2}=x+1}\\ \text{Rata}& \text{an+median}=x+0,5+x+1=2x+1,5\end{array}\end{array}$

Contoh Soal 4 Statistika

$\begin{array}{l}\\ 16.& (\mathbf{\text{UN IPA 2014}})\\ & \text{Kuartil atas dari data pada tabel berikut}\\ & \text{adalah}....\\ & \begin{array}{|cc|}\hline \text{Data}& f\\ 20-25& 4\\ 26-31& 6\\ 32-37& 6\\ 38-43& 10\\ 44-49& 12\\ 50-55& 8\\ 56-61& 4\\ \hline\end{array}\\ & \begin{array}{l}\\ \text{a}.& 49,25\\ \text{b}.& 48,75\\ \text{c}.& 48,25\\ \text{d}.& 47,75\\ \text{e}.& 47,25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Kuartil atas}=Q_3,\text{dengan}n=\sum f=50\\ & \text{Kita sertakan lagi tabel di atas berikut}\\ & \begin{array}{|cc|}\hline \text{Data}& f\\ 20-25& 4\\ 26-31& 6\\ 32-37& 6\\ 38-43& 10\\ \text{44-49}& \text{12}\\ 50-55& 8\\ 56-61& 4\\ \hline\end{array}\\ & Q_3=\text{Datum ke}-\left({\displaystyle \frac{3n}{4}}\right)=x_{\frac{3.50}{4}}=x_{37,5}\\ & \text{dan}{x}_{37,5}\text{terletak di kelas interval}44-49\\ & Q_3=t_b+p\left({\displaystyle \frac{{\displaystyle \frac{3n}{4}-F}}{f}}\right)\\ & =43,5+6\left({\displaystyle \frac{37,5-26}{12}}\right)\\ & =43,5+{\displaystyle \frac{11,5}{2}}\\ & =49,5+5,75\\ & =49,25\end{array}\end{array}$

$\begin{array}{l}\\ 17.& (\mathbf{\text{UN IPA 2014}})\\ & \text{Perhatikanlah histrogram berikut}\\ & \end{array}$

$.\begin{array}{l}\\ & \text{Modus dari data pada histogram adalah}....\\ & \begin{array}{l}\\ \text{a}.& 23,25\\ \text{b}.& 23,75\\ \text{c}.& 24,00\\ \text{d}.& 25,75\\ \text{e}.& 26,25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui dari data histogram di atas adalah}:\\ & \begin{array}{|cc|}\hline \text{Data}& f\\ 3-7& 4\\ 8-12& 6\\ 13-17& 8\\ 18-22& 10\\ \text{23-27}& \text{12}\\ 28-32& 6\\ 33-37& 4\\ 38-42& 2\\ \hline\end{array}\\ & \text{Saat menentukan batas interval kurang lebih sama}\\ & \text{seperti menentukan panjang interval kelas}\\ & \text{Modus dari histogram di atas}\\ & M_0=t_b+p\left({\displaystyle \frac{\mathrm{△}{f}_1}{\mathrm{△}{f}_1+\mathrm{△}{f}_2}}\right)\\ & =22,5+5\left({\displaystyle \frac{(12-10)}{(12-10)+(12-6)}}\right)\\ & =22,5+\left({\displaystyle \frac{2}{2+6}}\right)=22,5+{\displaystyle \frac{10}{8}}\\ & =22,5+1,25=23,75\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Median dari data}\\ & 3,4,7,5,6,9,9,7,6,5,8\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 5\\ \text{b}.& 6\\ \text{c}.& 7\\ \text{d}.& 8\\ \text{e}.& 9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketah}& \text{ui data adalah ganjil}\\ \text{Median}& \text{ (datum tengah) data tunggal}:\\ \text{Data}& :3,4,7,5,6,9,9,7,6,5,8\\ \text{setel}& \text{ah diurutkan}\\ \text{Data}& :3,4,5,5,6,6,7,7,8,9,9\\ \text{Data}& :3,4,5,5,6,6,7,7,8,9,9\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Dari data berikut yang memiliki}\\ & \mathbf{\text{mean}}7{\displaystyle \frac{1}{2}\text{dan}\mathbf{\text{median}}7\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 2,5,6,9,7,8,5,14,8,11\\ \text{b}.& 6,3,7,8,6,4,11,8,9,8\\ \text{c}.& 3,7,10,7,9,5,10,2,14,11\\ \text{d}.& 4,1,6,12,8,11,4,5,8,2\\ \text{e}.& 2,3,4,3,10,8,12,6,15,12\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{|ll|}\hline \text{Mean}& \bar{x}={\displaystyle \frac{75}{10}=7,5}\\ \text{Median}& 2,5,6,9,7,8,5,14,8,11\\ \text{a}& 2,5,5,6,7,8,8,9,11,14\\ \text{Mean}& \bar{x}={\displaystyle \frac{70}{10}=7}\\ \text{Median}& 6,3,7,8,6,4,11,8,9,8\\ \text{b}& 3,4,6,6,7,8,8,8,9,11\\ \text{Mean}& \bar{x}={\displaystyle \frac{78}{10}=7,8}\\ \text{Median}& 3,7,10,7,9,5,10,2,14,11\\ \text{c}& 2,3,5,7,7,9,10,10,11,14\\ \text{Mean}& \bar{x}={\displaystyle \frac{61}{10}=6,1}\\ \text{Median}& 4,1,6,12,8,11,4,5,8,2\\ \text{d}& 1,2,4,4,5,6,8,8,11,12\\ \text{Mean}& \bar{x}={\displaystyle \frac{75}{10}=7,5}\\ \text{Median}& 2,3,4,3,10,8,12,6,15,12\\ \text{e}& 2,3,3,4,6,8,10,12,12,15\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Berikut adalah daftar nilai matematika}\\ & \text{kelas XII IA1}\\ & \begin{array}{|lcccccccc|}\hline \text{Nilai}& 3& 4& 5& 6& 7& 8& 9& 10\\ \text{Frekuensi}& 3& 5& 5& 9& 8& 6& 2& 2\\ \hline\end{array}\\ & \text{Jika siswa yang nilainya di atas rata-rata}\\ & \text{akan diikutsertakan dalam seleksi}\\ & \text{olimpiade matematika, maka banyak}\\ & \text{siswa yang mengikuti seleksi olimpiade}\\ & \text{adalah}...\text{siswa}\\ & \begin{array}{l}\\ \text{a}.& 6\\ \text{b}.& 8\\ \text{c}.& 10\\ \text{d}.& 18\\ \text{e}.& 27\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \begin{array}{|lcccccccc|}\hline x_i& 3& 4& 5& 6& 7& 8& 9& 10\\ f_i& 3& 5& 5& 9& 8& 6& 2& 2\\ x_i{f}_i& 9& 20& 25& 54& 56& 48& 18& 20\\ \hline\end{array}\\ & \text{Rata-rata nilai matematikanya}\\ & \bar{x}={\displaystyle \frac{\sum x_i{f}_i}{\sum f_i}}\\ & ={\displaystyle \frac{250}{40}=6,25}\\ & \text{Jadi, nilai rata-ratanya}6,25\\ & \text{Sehingga yang bisa ikut selesksi adalah}\\ & \text{nilai di atas rata-rata yaitu}:7,8,9,10\\ & \text{dan totalnya yang mendapatkan nilai}\\ & \text{itu sebanyak}:8+6+2+2=18\text{siswa}\end{array}\end{array}$

Contoh Soal 3 Statistika

$\begin{array}{l}\\ 11.& \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut}\\ & \begin{array}{|cc|}\hline \text{Interval}& f\\ 2-6& 2\\ 7-11& 3\\ 12-16& 3\\ 17-21& 6\\ 22-26& 6\\ \hline\end{array}\\ & \text{Rata-ratanya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 17,5\\ \text{b}.& 17\\ \text{c}.& 16,75\\ \text{d}.& 16,5\\ \text{e}.& 15,5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \begin{array}{|cccc|}\hline \text{Interval}& x_i& f_i& x_i.f_i\\ 2-6& 4& 2& 8\\ 7-11& 9& 3& 27\\ 12-16& 14& 3& 42\\ 17-21& 19& 6& 114\\ 22-26& 24& 6& 144\\ & \sum _{i=1}^5& 20& 335\\ \hline\end{array}\\ & \text{maka rata-ratanya}\\ & \bar{x}={\displaystyle \frac{\sum x_i.f_i}{\sum f_i}=\frac{335}{20}=16,75}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut}\\ & \begin{array}{|cc|}\hline \text{Interval}& f\\ 51-60& 5\\ 61-70& 7\\ 71-80& 14\\ 81-90& 8\\ 91-100& 6\\ \hline\end{array}\\ & \text{Rata-ratanya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 75,50\\ \text{b}.& 76,25\\ \text{c}.& 76,50\\ \text{d}.& 78,25\\ \text{e}.& 80,50\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \begin{array}{|cccc|}\hline \text{Interval}& x_i& f_i& x_i.f_i\\ 51-60& 55,5& 5& 277,5\\ 61-70& 65,5& 7& 458,5\\ 71-80& 75,5& 14& 1057\\ 81-90& 85,5& 8& 684\\ 91-100& 95,5& 6& 573\\ & \sum _{i=1}^5& 40& 3050\\ \hline\end{array}\\ & \text{maka rata-ratanya}\\ & \bar{x}={\displaystyle \frac{\sum x_i.f_i}{\sum f_i}=\frac{3050}{40}=76,25}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut(data sama dengan no.12 di atas)}\\ & \begin{array}{|cc|}\hline \text{Interval}& f\\ 51-60& 5\\ 61-70& 7\\ 71-80& 14\\ 81-90& 8\\ 91-100& 6\\ \hline\end{array}\\ & \text{Mediannya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 75,50\\ \text{b}.& 76,20\\ \text{c}.& 76,21\\ \text{d}.& 77,22\\ \text{e}.& 78,23\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui},n=\sum f=40,\text{Perhatikan tabel}\\ & \text{berikut ini}\\ & \begin{array}{|cc|}\hline \text{Interval}& f_i\\ 51-60& 5\\ 61-70& 7\\ 71-80& 14\\ 81-90& 8\\ 91-100& 6\\ & 40\\ \hline\end{array}\\ & Q_k=\text{Datum ke}-\left({\displaystyle \frac{kn}{4}}\right)\\ & Median=Q_2=\text{Datum ke}-\left({\displaystyle \frac{2.40}{4}}\right)=x_{20}\\ & x_{20}\text{terletak pada kelas interval}:71-80\\ & \text{dengan}f=14,F\text{sebelum}{Q}_2=12,\\ & t_b=70,5,\text{serta}p=10\\ & \text{maka mediannya}\\ & Q_2=t_b+p\left({\displaystyle \frac{{\displaystyle \frac{2n}{4}-F}}{f}}\right)\\ & =70,5+10\left({\displaystyle \frac{20-12}{14}}\right)\\ & =70,5+5,714=76,21\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut}\\ & \begin{array}{|cc|}\hline \text{Skor}& f\\ 40-49& 8\\ 50-59& 9\\ 60-69& 22\\ 70-79& 15\\ 80-89& 6\\ \hline\end{array}\\ & \text{Modusnya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 65,50\\ \text{b}.& 66,00\\ \text{c}.& 66,50\\ \text{d}.& 67,00\\ \text{e}.& 85,50\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui}n=\sum f=60,modusnya\\ & \text{terdapat pada kelas dengan frekuensi terbanyak}\\ & \text{yaitu}:60-69,\text{dengan}p=10\\ & \{\begin{array}{ll}\mathrm{△}{f}_i& =f-f_1=22-9=13\\ \mathrm{△}{f}_{ii}& =f-f_2=22-15=7\end{array}\\ & \text{Sehingga}\\ & M_0=t_b+p\left({\displaystyle \frac{\mathrm{△}{f}_1}{\mathrm{△}{f}_1+\mathrm{△}{f}_2}}\right)\\ & M_0=59,5+10\left({\displaystyle \frac{22-9}{(22-9)+(22-15)}}\right)\\ & =59,5+{\displaystyle \frac{10.13}{13+7}}\\ & =59,5+6,5=66,0\end{array}\end{array}$

$\begin{array}{l}\\ 15.& (\mathbf{\text{UN 2013}})\\ & \text{Perhatikan tabel distribusi frekuensi}\\ & \text{berikut}\\ & \begin{array}{|cc|}\hline \text{Berat Badan (Kg)}& f\\ 45-49& 3\\ 50-54& 6\\ 55-59& 10\\ 60-64& 12\\ 65-69& 15\\ 70-74& 6\\ 75-79& 4\\ \hline\end{array}\\ & \text{Kuartil atasnya adalah}....\\ & \begin{array}{l}\\ \text{a}.& 66{\displaystyle \frac{5}{6}}\\ \text{b}.& 67{\displaystyle \frac{1}{6}}\\ \text{c}.& 67{\displaystyle \frac{5}{6}}\\ \text{d}.& 68{\displaystyle \frac{1}{6}}\\ \text{e}.& 68{\displaystyle \frac{4}{6}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Kuartil atas}=Q_3,\text{dengan}n=\sum f=56\\ & Q_3=\text{Datum ke}-\left({\displaystyle \frac{3n}{4}}\right)=x_{\frac{3.56}{4}}=x_{42}\\ & \text{dan}{x}_{42}\text{terletak di kelas interval}65-69\\ & Q_3=t_b+p\left({\displaystyle \frac{{\displaystyle \frac{3n}{4}-F}}{f}}\right)\\ & =64,5+5\left({\displaystyle \frac{42-(3+6+10+12)}{15}}\right)\\ & =64{\displaystyle \frac{1}{2}+{\displaystyle \frac{11}{3}}}\\ & =64{\displaystyle \frac{3}{6}+3{\displaystyle \frac{4}{6}=68{\displaystyle \frac{1}{6}}}}\end{array}\end{array}$

Contoh Soal 2 Statistika

$\begin{array}{l}\\ 6.& \text{Tabel berikut adalah nilai tes matematika}\\ & \begin{array}{|cc|}\hline \text{Nilai}& f\\ 21-30& 1\\ 31-40& 2\\ 41-50& 5\\ 51-60& 7\\ 61-70& 8\\ 71-80& 8\\ 81-90& 6\\ 91-100& 1\\ \hline\end{array}\\ & \text{Banyak siswa yang mendapatkan nilai 71}\\ & \text{atau lebih adalah}....\\ & \begin{array}{l}\\ \text{a}.& 16\\ \text{b}.& 15\\ \text{c}.& 12\\ \text{d}.& 12\\ \text{e}.& 10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Perhatikan kembali tabelnya}\\ & \begin{array}{|cc|}\hline \text{Nilai}& f\\ 21-30& 1\\ 31-40& 2\\ 41-50& 5\\ 51-60& 7\\ 61-70& 8\\ 71-80& 8\\ 81-90& 6\\ 91-100& 1\\ \hline\end{array}\\ & \text{Nilai yang lebih dari 71 adalah}:\\ & 8+6+1=17\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Jangkauan dari tabel distribusi}\\ & \text{frekuensi pada no.6 di atas adalah}....\\ & \begin{array}{l}\\ \text{a}.& 60\\ \text{b}.& 70\\ \text{c}.& 79\\ \text{d}.& 89\\ \text{e}.& 100\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Karena data berkelompok, maka}\\ & \text{Jangkauan}=\left(\text{Nilai tengah kelas tertinggi}\right)\\ & -\left(\text{Nilai tengah kelas pertama}\right)\\ & ={\displaystyle \frac{1}{2}((100+91)-(30+21))=70}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}n=\text{banyak data},k=\text{banyak interval}\\ & \text{kelas, maka menurut aturan Sturges, rumus}\\ & \text{untuk menentukan nilai}k\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& k=\log \left(10.n^{3,3}\right)\\ \text{b}.& k=1+3,3\log n\\ \text{c}.& k=1-3,3\log (n-1)\\ \text{d}.& k=\log ({10}^{3,3}.n)\\ \text{e}.& k=\log n^{3,3}+2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Cukup jelas}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Rata-rata data soal no.6}\\ & \text{di atas adalah}....\\ & \begin{array}{l}\\ \text{a}& {\displaystyle 64,45}\\ \text{b}& {\displaystyle 64,55}\\ \text{c}& {\displaystyle 65,45}\\ \text{d}& {\displaystyle 65,55}\\ \text{e}& {\displaystyle 66}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \begin{array}{|cccc|}\hline \text{Nilai}& x_i& f_i& x_i.f_i\\ 21-30& 25,5& 1& 25,5\\ 31-40& 35,5& 2& 71\\ 41-50& 45,5& 5& 227,5\\ 51-60& 55,5& 7& 388,5\\ 61-70& 65,5& 8& 524\\ 71-80& 75,5& 8& 604\\ 81-90& 85,5& 6& 513\\ 91-100& 95,5& 1& 95,5\\ & \sum _{i=1}^8& 38& 2449\\ \hline\end{array}\\ & \text{Sehingga, rata-rata data nilai di atas adalah}\\ & \bar{x}={\displaystyle \frac{\sum x_i{f}_i}{\sum f_i}=\frac{2449}{38}=64,447368421\approx 64,45}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Rumus untuk menentukan}\mathrm{\%}{f}_{\text{rel}}=....\\ & \begin{array}{l}\\ \text{a}& {\displaystyle \frac{f_i+1}{\sum f}\times 100\mathrm{\%}}\\ \text{b}& {\displaystyle \frac{f_i-1}{\sum f}\times 100\mathrm{\%}}\\ \text{c}& {\displaystyle \frac{f_i}{\sum f}\times 100\mathrm{\%}}\\ \text{d}& {\displaystyle \frac{f_i}{\sum f+1}\times 100\mathrm{\%}}\\ \text{e}& {\displaystyle \frac{f_i}{\sum f-1}\times 100\mathrm{\%}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Cukup jelas}\end{array}\end{array}$