CONTOH SOAL 9 MATRIKS DAN SPL

 $\begin{array}{l}\\ 11.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 4x+3y-11z & =-216 \\ 2x+5y+8z & =63\\ x+y+z & =0 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-6,\quad y=-9\quad \textrm{dan}\quad z=15\\ \textrm{b}.&x=-6,\quad y=9\quad \textrm{dan}\quad z=15\\ \textrm{c}.&x=6,\quad y=-9\quad \textrm{dan}\quad z=15\\ \textrm{d}.&x=-6,\quad y=-9\quad \textrm{dan}\quad z=-15\\ \textrm{e}.&x=-6,\quad y=9\quad \textrm{dan}\quad z=-15 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{ccc|c} 4&3&-11&-216\\ 2&5&8&63\\ 1&1&1&0 \end{array} \right]\quad\begin{matrix}\bullet R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-R_{\displaystyle 1}\\ \bullet R_{\displaystyle 3}\longleftarrow 4R_{\displaystyle 3}-R_{\displaystyle 1}\end{matrix}\\ &=2(2,5,8|63)-(4,3,-11|-216)=(0,7,27|342)\:\:\textrm{dan}\\ &=4(1,1,1|0)-(4,3,-11|-216)=(0,1,15|216)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 4&3&-11&-216\\ 0&7&27&342\\ 0&1&15&216 \end{array} \right]\quad \bullet R_{\displaystyle 3}\longleftarrow 7R_{\displaystyle 3}-R_{\displaystyle 2}\\ &=7(0,1,15|216)-(0,7,27|342)=(0,0,78|1170)\\ &\textrm{sehingga}\quad z=\displaystyle \frac{1170}{78}=15,\quad \textrm{maka}\quad y=-9,\:\: x=-6 \end{array}$

$\begin{array}{l}\\ 12.&\textrm{Bayu membeli sebuah buku tulis dan dua buah pensil}\\ &\textrm{Rp}10.000,00.\:\: \textrm{Di toko yang sama, Giri membeli dua }\\ &\textrm{buku tulis dan tiga buah pensil seharga Rp17.500,00}\\ &\textrm{Bentuk perkalian matriks dari narasi di atas adalah}....\\  &\begin{array}{llllllll} \textrm{a}.&\begin{pmatrix} 2 &  1\\ 2 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 10.000\\ 17.500 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 1 &  2\\ 3 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 10.000\\ 17.500 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 1 &  2\\ 2 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 17.500\\ 10.000 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} 1 &  2\\ 2 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 10.000\\ 17.500 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 2 &  1\\ 3 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 17.500\\ 10.000 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &x=\textrm{buku},\:\: y=\textrm{pensil}\\ & \end{array}$.

$\begin{array}{l}\\ 13.&\textrm{Penyelesaian dari sistem persamaan berikut}\\ &\begin{cases} 4x+3y+9& =0 \\ -x+y & =4 \end{cases}\\ &\textrm{Model matriks yang menyatakan bentuk di atas}....\\  &\begin{array}{llllllll} \textrm{a}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 &  -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ -4 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 &  3\\ -1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} x\\ y \end{pmatrix}=-\displaystyle \frac{1}{7}\begin{pmatrix} 1 &  -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 &  -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 &  -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} 9\\ 4 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{cases} 4x+3y & =-9 \\-x+y & =4 \end{cases}\Rightarrow \begin{pmatrix} 4 &  3\\ -1 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ &\Leftrightarrow \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{\begin{vmatrix}4 &  3\\ -1 & 1\end{vmatrix}}\begin{pmatrix} 1 &  -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ &\Leftrightarrow \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 &  -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix} \end{array}$.

$\begin{array}{l}\\ 14.&\textrm{Diberikan SPL berikut}\\ &\begin{cases} 4x+3y & =12 \\ 3x+2y & =7 \end{cases}\\ &\textrm{Nilai}\quad x+y\quad \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&-5\\ \textrm{b}.&5\\ \textrm{c}.&11\\ \textrm{d}.&-11\\ \textrm{e}.&10 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 4&3&12\\ 3&2&7 \end{array} \right]\quad\bullet R_{\displaystyle 1}\longleftarrow R_{\displaystyle 1}-R_{\displaystyle 2}\\ &=(4,3|12)-(3,2|7)=(1,1|5)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 1&1&5\\ 3&2&7 \end{array} \right]\\ &\textrm{sehingga}\quad x+y=5 \end{array}$.

$\begin{array}{l}\\ 15.&\textrm{Perpotongan dua garis yang tersaji sebagai}\\ &\textrm{persamaan matriks berikut}\\ &\begin{pmatrix} -1 &  3\\ 1 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 5\\ 5 \end{pmatrix}\\ & \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&(2,-1)\\ \textrm{b}.&(1,-2)\\ \textrm{c}.&(-1,2)\\ \textrm{d}.&(-1,-2)\\ \textrm{e}.&(1,2) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} -1&3&5\\ 1&2&5 \end{array} \right]\quad \begin{aligned}&\bullet R_{\displaystyle 2}\longleftarrow R_{\displaystyle 1}+R_{\displaystyle 2}\\ &\bullet R_{\displaystyle 1}\longleftarrow -R_{\displaystyle 1}\end{aligned}\\ &=(1,2|5)+(-1,3|5)=(0,5|10)\quad \textrm{dan}\\ &=(1,-3|-5)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 1&-3&-5\\ 0&5&10 \end{array} \right]\quad \begin{aligned}&\bullet R_{\displaystyle 1}\longleftarrow \displaystyle \frac{3}{5}R_{\displaystyle 1}+R_{\displaystyle 2}\\ &\bullet R_{\displaystyle 2}\longleftarrow \displaystyle \frac{1}{5}R_{\displaystyle 1}\end{aligned}\\ &\textrm{Selanjutnya diperoleh}\\&\left[ \begin{array}{cc|c} 1&0&1\\ 0&1&2 \end{array} \right]\\ &\textrm{Jadi, koordinat titik potongnya}: (1,2) \end{array}$.

CONTOH SOAL 8 MATRIKS DAN SPL

 $\begin{array}{l}\\ 6.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 2x-7y & =-19 \\ 3x-y & =0 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-3\quad \textrm{dan}\quad y=-1\\ \textrm{b}.&x=3\quad \textrm{dan}\quad y=1\\ \textrm{c}.&x=1\quad \textrm{dan}\quad y=3\\ \textrm{d}.&x=-3\quad \textrm{dan}\quad y=1\\ \textrm{e}.&x=1\quad \textrm{dan}\quad y=-3 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 2&-7&-19\\ 3&-1&0 \end{array} \right]\quad\bullet R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-3R_{\displaystyle 1}\\ &=2(3,-1|0)-3(2,-7|-19)=(0,19|57)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 2&-7&-19\\ 0&19&57 \end{array} \right]\\ &\textrm{sehingga}\quad y=\displaystyle \frac{57}{19}=3,\quad \textrm{maka}\quad x=1 \end{array}$.

$\begin{array}{l}\\ 7.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 4x-3y & =14 \\ -x+2y & =-1 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=2\quad \textrm{dan}\quad y=5\\ \textrm{b}.&x=5\quad \textrm{dan}\quad y=-2\\ \textrm{c}.&x=5\quad \textrm{dan}\quad y=2\\ \textrm{d}.&x=-5\quad \textrm{dan}\quad y=-2\\ \textrm{e}.&x=-2\quad \textrm{dan}\quad y=-5 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 4&-3&14\\ -1&2&-1 \end{array} \right]\quad\bullet R_{\displaystyle 2}\longleftarrow R_{\displaystyle 2}+\displaystyle \frac{1}{4}R_{\displaystyle 1}\\ &=(-1,2|-1)+\displaystyle \frac{1}{4}(4,-3|14)=(0,\displaystyle \frac{5}{4}|\displaystyle \frac{10}{4})\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 4&-3&14\\ 0&\displaystyle \frac{5}{4}&\displaystyle \frac{10}{4} \end{array} \right]\\ &\textrm{sehingga}\quad y=\displaystyle \frac{\displaystyle \frac{10}{4}}{\displaystyle \frac{5}{4}}=2,\quad \textrm{maka}\quad x=1 \end{array}$.

$\begin{array}{l}\\ 8.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 8x+3y & =37 \\ 4x+y & =15 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=2\quad \textrm{dan}\quad y=7\\ \textrm{b}.&x=2\quad \textrm{dan}\quad y=-7\\ \textrm{c}.&x=7\quad \textrm{dan}\quad y=-2\\ \textrm{d}.&x=7\quad \textrm{dan}\quad y=2\\ \textrm{e}.&x=-2\quad \textrm{dan}\quad y=-7 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 8&3&37\\ 4&1&15 \end{array} \right]\quad\bullet R_{\displaystyle 2}\longleftarrow R_{\displaystyle 2}-\displaystyle \frac{1}{2}R_{\displaystyle 1}\\ &=(4,1|15)-\displaystyle \frac{1}{2}(8,3|37)=(0,-\displaystyle \frac{1}{2}|-\displaystyle \frac{7}{2})\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 8&3&37\\ 0&-\displaystyle \frac{1}{2}&-\displaystyle \frac{7}{2} \end{array} \right]\\ &\textrm{sehingga}\quad y=\displaystyle \frac{-\displaystyle \frac{7}{2}}{-\displaystyle \frac{2}{2}}=7,\quad \textrm{maka}\quad x=2 \end{array}$.

$\begin{array}{l}\\ 9.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 4x+2y+3z & =0 \\ 2x+3y+5z & =9\\ 3x+y+7z & =9 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-2,\quad y=1\quad \textrm{dan}\quad z=2\\ \textrm{b}.&x=2,\quad y=1\quad \textrm{dan}\quad z=2\\ \textrm{c}.&x=-2,\quad y=-1\quad \textrm{dan}\quad z=-2\\ \textrm{d}.&x=-2,\quad y=1\quad \textrm{dan}\quad z=-2\\ \textrm{e}.&x=-2,\quad y=-1\quad \textrm{dan}\quad z=2 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{ccc|c} 4&2&3&0\\ 2&3&5&9\\ 3&1&7&9 \end{array} \right]\quad\begin{matrix} \bullet R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-R_{\displaystyle 1}\: .\\ \bullet R_{\displaystyle 3}\longleftarrow 4R_{\displaystyle 3}-3R_{\displaystyle 1} \end{matrix}\\ &=2(2,3,5|9)-(4,2,3|0)=(0,4,7|18)\quad \textrm{dan}\\ &=4(2,3,5|9)-3(4,2,3|0)=(0,-2,19|36)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 4&2&3&0\\ 0&4&7&18\\ 0&-2&19&36 \end{array} \right]\quad \bullet R_{\displaystyle 3}\longleftarrow 2R_{\displaystyle 3}+R_{\displaystyle 2}\\ &=2(0,-2,19|36)+(0,4,7|18)=(0,0,45|90)\\ &\textrm{sehingga}\quad z=\displaystyle \frac{90}{45}=2,\quad \textrm{maka}\quad y=1,\:\: x=-2 \end{array}$.

$\begin{array}{l}\\ 10.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 7x+2y+z & =-40 \\ 2x+7y+4z & =45\\ 5x-4y+6z & =8 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-8,\quad y=-3\quad \textrm{dan}\quad z=10\\ \textrm{b}.&x=-8,\quad y=3\quad \textrm{dan}\quad z=10\\ \textrm{c}.&x=-8,\quad y=3\quad \textrm{dan}\quad z=-10\\ \textrm{d}.&x=8,\quad y=3\quad \textrm{dan}\quad z=10\\ \textrm{e}.&x=8,\quad y=-3\quad \textrm{dan}\quad z=10 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{ccc|c} 7&2&1&-40\\ 2&7&4&45\\ 5&-4&6&8 \end{array} \right]\quad\begin{matrix}\bullet R_{\displaystyle 2}\longleftarrow 7R_{\displaystyle 2}-2R_{\displaystyle 1}\\ \bullet R_{\displaystyle 3}\longleftarrow 7R_{\displaystyle 3}-5R_{\displaystyle 1}\end{matrix}\\ &=7(2,7,4|45)-2(7,2,1|-40)=(0,45,26|395)\:\:\textrm{dan}\\ &=7(5,-4,6|8)-5(7,2,1|-40)=(0,-38,37|256)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 7&2&1&-40\\ 0&45&26&395\\ 0&-38&37&256 \end{array} \right]\quad \bullet R_{\displaystyle 3}\longleftarrow \displaystyle \frac{1}{45}R_{\displaystyle 3}+\displaystyle \frac{1}{38}R_{\displaystyle 2}\\ &=\displaystyle \frac{1}{45}(0,-38,37|256)+\displaystyle \frac{1}{38}(0,45,26|395)=(0,0,\displaystyle \frac{2653}{1710}|\displaystyle \frac{26530}{1710})\\ &\textrm{sehingga}\quad z=\displaystyle \frac{\displaystyle \frac{226530}{1710}}{\displaystyle \frac{2653}{1710}}=10,\quad \textrm{maka}\quad y=3,\:\: x=-8 \end{array}$.

CONTOH SOAL 7 MATRIKS DAN SPL

 $\begin{array}{ll}\\ 1.&\textrm{Untuk sistem persamaan linear dua variabel}\\ &\textrm{berikut, jika dinyatakan dalam perkalian}\\ &\textrm{matriks adalah}....\\ &\begin{cases} 2x+3y & =5 \\ 3x+y & =1 \end{cases}\\ &\begin{array}{llll}\\ \textrm{a}.&\left( \begin{matrix} 2 &  3\\ 3 & 1 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 5 \\ 1 \end{matrix} \right)\\ \textrm{b}.&\left( \begin{matrix} 2 &  1\\ 3 & 3 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 5 \\ 1 \end{matrix} \right)\\ \textrm{c}.&\left( \begin{matrix} 3 &  2\\ 1 & 3 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 5 \\ 1 \end{matrix} \right)\\ \textrm{d}.&\left( \begin{matrix} 2 &  3\\ 3 & 1 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 1 \\ 5 \end{matrix} \right)\\ \textrm{e}.&\left( \begin{matrix} 3 &  2\\ 1 & 3 \end{matrix} \right)\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} 1 \\ 5 \end{matrix} \right)\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\textrm{Cukup jelas} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui persamaan}\begin{cases} x-y & =2 \\ kx+y & =3 \end{cases}\\ &\textrm{memiliki solusi}\: \: (x,y)\: \: \textrm{di kuadran I}\\ &\textrm{Jika dan hanya jika nilai}\: \: k\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle k=-1\\ \textrm{b}.&k>-1\\ \textrm{c}.&k<\displaystyle \frac{3}{2}\\ \textrm{d}.&0<k<\displaystyle \frac{3}{2}\\ \textrm{e}.&-1<k<\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x-y=2\: \: \: \quad....(1)\\ kx+y=3\quad\: ....(2)\end{matrix}\right.\\ &\textrm{Dengan metode matriks didapatkan}\\ &x=\displaystyle \frac{\begin{vmatrix} 2 & -1\\ 3& 1 \end{vmatrix}}{\begin{vmatrix} 1 & -1\\ k & 1 \end{vmatrix}}=\displaystyle \frac{2-(-3)}{1+k}=\frac{5}{k+1}\\ &\textrm{Dengan cara yang sama pula}\\ &y=\displaystyle \frac{\begin{vmatrix} 1 & 2\\ k & 3 \end{vmatrix}}{\begin{vmatrix} 1 & -1\\ k & 1 \end{vmatrix}}=\displaystyle \frac{3-2k}{k+1}\\ &\textrm{Supaya memiliki solusi di kwadran I},\\ &\textrm{maka baik}\: \: x\: \: \textrm{maupun}\: \: y\\ &\textrm{haruslah positif, akibatnya}:\\ & k+1>0\Rightarrow k>-1\\ &\textrm{Sebagai akibat yang lain adalah}:\\ &3-2k>0\Rightarrow k<\displaystyle \frac{3}{2}\\ &\textrm{Jadi},\: \: -1<k<\displaystyle \frac{3}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Untuk sistem persamaan linear dua variabel}\\ &\textrm{berikut, jika dinyatakan dalam perkalian}\\ &\textrm{matriks adalah}....\\ &\begin{cases} x+y+z & =5 \\ 2x-5y+7z & =4\\ x-y+4z&=9 \end{cases}\\ &\begin{array}{llll}\\ \textrm{a}.&\left( \begin{matrix} 1 &  1&1\\ 2 & 5&7\\ 1&-1&4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right)\\ \textrm{b}.&\left( \begin{matrix} 1 &  1&1\\ 2 & -5&7\\ 1&1&4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right)\\ \textrm{c}.&\left( \begin{matrix} 1 &  1&1\\ 2 & -5&7\\ 1&-1&4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right)\\ \textrm{d}.&\left( \begin{matrix} 1 &  1&1\\ 2 & -5&7\\ 1&-1&-4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right)\\ \textrm{e}.&\left( \begin{matrix} -1 &  1&1\\ 2 & -5&7\\ 1&-1&4 \end{matrix} \right)\left( \begin{matrix} x \\ y\\ z \end{matrix} \right)=\left( \begin{matrix} 5 \\ 4\\ 9 \end{matrix} \right) \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Cukup jelas} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Perhatikanlah sistem persamaan berikut}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{agar sistem persamaan ini tidak}\\ &\textrm{memiliki penyelesaian, maka nilai}\: \: k=....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Agar sistem persamaan}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{tidak berpenyelesaian, maka}\\ &\textrm{ingat penyelesaian metode matrik}\\ &\textrm{buatlah penyebutnya}=0,\: \: \textrm{yaitu}:\\ &\begin{vmatrix} 3 & 2 & -5\\ 2 & -6 & k\\ 5 & -4 & -1 \end{vmatrix}=0\\ &\textrm{Selanjutnya}\\ &3\begin{vmatrix} -6 & k\\ -4 & -1 \end{vmatrix}-2\begin{vmatrix} 2 & k\\ 5 & -1 \end{vmatrix}-5\begin{vmatrix} 2 & -6\\ 5 & -4 \end{vmatrix}=0\\ &3(6+4k)-2(-2-5k)-5(-8+30)=0\\ &18+12k+4+10k+40-150=0\\ &22x=88\\ &\quad x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Himpunan penyelesaian dari}\\ &\left\{\begin{matrix} x+y+4z=15\quad\\ x-y+z=2\qquad\\ x+2y-3z=-4 \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ (-1,1,3) \right \}\\ \textrm{b}.&\left \{ (1,2,3) \right \}\\ \textrm{c}.&\left \{ (-2,1,1) \right \}\\ \textrm{d}.&\left \{ (3,2,-1) \right \}\\ \textrm{e}.&\left \{ (1,-2,3) \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Semunya dikerjakan dengan metode}\\ &\textrm{matriks}\: (\textbf{Cara Cramer})\\ &\begin{aligned} x&=\displaystyle \frac{\begin{vmatrix} 15 & 1 & 4\\ 2& -1 & 1\\ -4& 2 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{15\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}+4\begin{vmatrix} 2 & -1\\ -4 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{15(3-2)-1(-6+4)+4(4-4)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{15(1)-1(-2)+4(0)}{1(1)-1(-4)+4(3)}=\frac{17}{17}=1 \\ y&=\displaystyle \frac{\begin{vmatrix} 1 & 15 & 4\\ 1& 2 & 1\\ 1& -4 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}-15\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(-6+4)-15(-3-1)+4(-4-2)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(-2)-15(-4)+4(-6)}{1(1)-1(-4)+4(3)}=\frac{34}{17}=2\\ z&=\displaystyle \frac{\begin{vmatrix} 1 & 1 & 15\\ 1& -1 & 2\\ 1& 2 & -4 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} -1 & 2\\ 2 & -4 \end{vmatrix}-1\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}+15\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(4-4)-1(-4-2)+15(2+1)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(0)-1(-6)+15(3)}{1(1)-1(-4)+4(3)}=\frac{51}{17}=3 \end{aligned} \end{array}$

$.\quad\quad \textrm{Cara di atas}$  full matriks-Cramer

MATRIKS-PENERAPAN MATRIKS

 H. Penggunaan Matriks dalam Sistem Persamaan Linear (SPL)

H.1 Tokoh yang Sangat Terkenal Terkait Penggunaan Matriks dalam SPL

Dalam mempelajari determinan dan SPL, kedua tokoh berikut akan selalu dikenang akan sumbangsihnya yang sangat luar biasa. Pertama Gabriel Cramer seorang matematikawan Swiss (1704-1752) dengan aturan Cramer dalam penggunaan determinan untuk mencari nilai variabel. Sedangkan yang kedua adalah Carl Friedrich Gauss seorang matematikawan Jerman (1777-1855) dengan eliminasi Gaussnya dalam penggunaan operasi baris pada matriks untuk menyelesaikan SPL.

H.2 Aturan Cramer untuk SPL dua variabel 

Misalkan diberikan

$\begin{aligned}&\begin{cases} a_{1}x &+&b_{1}y=c_{1} \\ a_{2}x &+&b_{2}y=c_{2} \end{cases} \end{aligned}$.

maka koefisiennya adalah:

$\begin{pmatrix} a_{1} &  b_{1}\\ a_{2} & b_{2} \end{pmatrix}$.

Determinan utama:

$D=\begin{vmatrix} a_{1} &  b_{1}\\ a_{2} & b_{2} \end{vmatrix}=a_{1}b_{2}-a_{2}b_{1}$.

syaratnya $D\neq 0$.

Kemudian kolom x diganti dengan konstanta:

$D_{x}=\begin{vmatrix} c_{1} &  b_{1}\\ c_{2} & b_{2} \end{vmatrix}=c_{1}b_{2}-c_{2}b_{1}$.

dan kolom y diganti dengan konstanta:

$D_{y}=\begin{vmatrix} a_{1} &  c_{1}\\ a_{2} & c_{2} \end{vmatrix}=a_{1}c_{2}-a_{2}c_{1}$,

maka solusi SPLDV adalah: $x=\displaystyle \frac{D_{\displaystyle x}}{D},\quad y=\displaystyle \frac{D_{\displaystyle y}}{D}$.

H.3 Aturan Cramer untuk SPL tiga variabel 

Misalkan diketahui sistem persamaan berikut:

$\begin{aligned}&\begin{cases} a_{1}x &+&b_{1}y&+&c_{1}z&=&d_{1} \\ a_{2}x &+&b_{2}y&+&c_{2}z&=&d_{2}\\ a_{3}x &+&b_{3}y&+&c_{3}z&=&d_{3} \end{cases} \end{aligned}$.

Dengan cara yang sama seperti pada penyelesaian di bagian H.2 di atas dalam penyelesaian SPL tersebut dengan determinan, yaitu: $D=\begin{vmatrix} a_{1} & b_{1} &  c_{1}\\ a_{2} & b_{2} &  c_{2}\\ a_{3} & b_{3} & c_{3} \end{vmatrix}\neq 0$,  $D_{\displaystyle x}=\begin{vmatrix} d_{1} & b_{1} &  c_{1}\\ d_{2} & b_{2} &  c_{2}\\ d_{3} & b_{3} & c_{3} \end{vmatrix}$, $D_{\displaystyle y}=\begin{vmatrix} a_{1} & d_{1} &  c_{1}\\ a_{2} & d_{2} &  c_{2}\\ a_{3} & d_{3} & c_{3} \end{vmatrix}$, dan $D_{\displaystyle z}=\begin{vmatrix} a_{1} & b_{1} &  d_{1}\\ a_{2} & b_{2} &  d_{2}\\ a_{3} & b_{3} & d_{3} \end{vmatrix}$.

Selanjutnya untuk mendapatkan nilai variabel x, y, dan z adalah: $x=\displaystyle \frac{D_{\displaystyle x}}{D},\quad y=\displaystyle \frac{D_{\displaystyle y}}{D},\quad \textrm{dan}\quad z=\displaystyle \frac{D_{\displaystyle z}}{D}$.

Sebagai catatan bahwa untuk mencarai nilai dterminan ada banyak cara, baik dari cara Sarrus, maupun ekspansi baris atau ekspansi kolom 

H.4 Metode Eliminasi Gauss untuk SPL tiga variabel 

Langkah-langkah untuk metode eliminasi Gauss adalah sebagai berikut:

• Ubahlah SPLTV menjadi matriks augmented (matriks berupa koefisien variabel yang diperluas dengan kolom konstanta
• Mengenolkan elemen di bawah pivot (elemen utama) pertama dan kedua dan usahakan dapat membentuk sperti matriks identitas ordo 3x3
• Lakukan sedikit/bebrapa langkah yang perlu agar minimal menjadi bentuk eselon baris, yaitu bentuk matriks yang susunan elemennya sudah dibuat seperti tangga, sehingga lebih mudah digunakan untuk mencari solusi SPL dengan eliminasi Gauss 

Terkait model matriks bentuk eselon baris, perhatikan 3 hal mendasar berikut

1. Semua baris yang seluruh elemennya nol berada di bagian paling bawah
2. Pada setipa baris yang tidak seluruhnya nol, elemen pertama yang bukan nol disebut pivot, posisi pivot pada baris berikutnya harus semakin ke kanan
3. Semua elemen di bawah setiap pivot harus nol.

I. Penggunaan Matriks dalam Transformasi Geometri

Lihat materi tentang transformasi ini di sini

$\LARGE\fbox{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah dengan metode matriks}\\ &\textrm{(cara Cramer) SPLDV berikut}:\\ &\begin{cases} 2x-y & =7 \\ x-y & =-1 \end{cases}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}x&=\displaystyle \frac{D_{\displaystyle x}}{D}=\displaystyle \frac{\begin{vmatrix} 7 & -1\\ -1 & -1 \end{vmatrix}}{\begin{vmatrix} 2 & -1\\ 1 & -1 \end{vmatrix}}=\frac{7(-1)-(-1).(-1)}{2.(-1)-(-1).1}\\ &=\displaystyle \frac{-7-1}{-2+1}=\frac{-8}{-1}=8\\ y&=\displaystyle \frac{\begin{vmatrix} 2 & 7\\ 1 & -1 \end{vmatrix}}{\begin{vmatrix} 2 & -1\\ 1 & -1 \end{vmatrix}}=\frac{2(-1)-(7).1}{2.(-1)-(-1).1}\\ &=\displaystyle \frac{-2-7}{-2+1}=\frac{-9}{-1}=9\\ \textrm{J}&\textrm{adi}\: \: (x,y)=(8,9) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah dengan metode matriks}\\ &\textrm{(cara Cramer) SPLTV berikut}:\\ &\begin{cases} 2x+2y+3z & =14 \\ 3x-y+4z & =8\\ 5x+y+2z&=12 \end{cases}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}x&=\displaystyle \frac{D_{\displaystyle x}}{D}=\displaystyle \frac{\begin{vmatrix} 14 & 2&3\\ 8 & -1&4\\ 12&1&2 \end{vmatrix}}{\begin{vmatrix} 2 & 2&3\\ 3 & -1&4\\ 5&1&2 \end{vmatrix}}\\ x&=\displaystyle \frac{14\begin{vmatrix} -1&4\\ 1&2 \end{vmatrix}-2\begin{vmatrix} 8 & 4\\ 12 & 2 \end{vmatrix}+3\begin{vmatrix} 8 & -1\\ 12 & 1 \end{vmatrix}}{2\begin{vmatrix} -1 & 4\\ 1 & 2 \end{vmatrix}-2\begin{vmatrix} 3 & 4\\ 5 & 2 \end{vmatrix}+3\begin{vmatrix} 3 & -1\\ 5 & 1 \end{vmatrix}} \\ &=\displaystyle \frac{14(-2-4)-2(16-48)+3(8+12)}{2(-2-4)-2(6-20)+3(3+5)}\\ &=\displaystyle \frac{14(-6)-2(-32)+3(20)}{2(-6)-2(-14)+3(8)}=\displaystyle \frac{-84+64+60}{-12+28+24}\\ &=\frac{40}{40}=1\\  \end{aligned}\\ &\begin{aligned}y&=\displaystyle \frac{D_{\displaystyle y}}{D}=\displaystyle \frac{\begin{vmatrix} 2 & 14&3\\ 3 & 8&4\\ 5&12&2 \end{vmatrix}}{\begin{vmatrix} 2 & 2&3\\ 3 & -1&4\\ 5&1&2 \end{vmatrix}}\\ y&=\displaystyle \frac{2\begin{vmatrix} 8&4\\ 12&2 \end{vmatrix}-14\begin{vmatrix} 3 & 4\\ 5 & 2 \end{vmatrix}+3\begin{vmatrix} 3 & 8\\ 5 & 12 \end{vmatrix}}{2\begin{vmatrix} -1 & 4\\ 1 & 2 \end{vmatrix}-2\begin{vmatrix} 3 & 4\\ 5 & 2 \end{vmatrix}+3\begin{vmatrix} 3 & -1\\ 5 & 1 \end{vmatrix}} \\ &=\displaystyle \frac{2(16-48)-14(6-20)+3(36-40)}{2(-2-4)-2(6-20)+3(3+5)}\\ &=\displaystyle \frac{2(-32)-14(-14)+3(-4)}{2(-6)-2(-14)+3(8)}=\displaystyle \frac{-64+196-12}{-12+28+24}\\ &=\frac{120}{40}=3\\  \end{aligned}\\ &\begin{aligned}z&=\displaystyle \frac{D_{\displaystyle z}}{D}=\displaystyle \frac{\begin{vmatrix} 2 & 2&14\\ 3 & -1&8\\ 5&1&12 \end{vmatrix}}{\begin{vmatrix} 2 & 2&3\\ 3 & -1&4\\ 5&1&2 \end{vmatrix}}\\ z&=\displaystyle \frac{2\begin{vmatrix} -1&8\\ 1&12 \end{vmatrix}-2\begin{vmatrix} 3 & 8\\ 5 & 12 \end{vmatrix}+14\begin{vmatrix} 3 & -1\\ 5 & 1 \end{vmatrix}}{2\begin{vmatrix} -1 & 4\\ 1 & 2 \end{vmatrix}-2\begin{vmatrix} 3 & 4\\ 5 & 2 \end{vmatrix}+3\begin{vmatrix} 3 & -1\\ 5 & 1 \end{vmatrix}} \\ &=\displaystyle \frac{2(-12-8)-2(36-40)+14(3+5)}{2(-2-4)-2(6-20)+3(3+5)}\\ &=\displaystyle \frac{2(-20)-2(-4)+14(8)}{2(-6)-2(-14)+3(8)}=\displaystyle \frac{-40+8+112}{-12+28+24}\\ &=\frac{80}{40}=2\\  \textrm{J}&\textrm{adi, nilai }\: \: (x,y,y)=\left ( \displaystyle 1,3,2 \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah dengan metode matriks}\\ &\textrm{(metode eliminasi Gauss) SPLTV berikut}:\\ &\begin{cases} 2x+2y+3z & =14 \\ 3x-y+4z & =8\\ 5x+y+2z&=12 \end{cases}\\\\ &\textrm{Jawab}:\\ &\textrm{Matriks augmentednya adalah}:\\ &\left[ \begin{array}{ccc|c} 2&2&3&14\\ 3&-1&4&8\\ 5&1&2&12 \end{array} \right]\\ &\textrm{Jadikan elemen di bawah pivot pertama menjadi nol}\\ &(\textrm{gunakan baris pertama sebagai pivot})\\ &\bullet \quad R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-3R_{\displaystyle 1}\\ &\bullet \quad R_{\displaystyle 3}\longleftarrow 2R_{\displaystyle 3}-5R_{\displaystyle 1}\\ &\textrm{maka}\\ &R_{\displaystyle 2}=2(3,-1,4|8)-3(2,2,3|14)=(0,-8,-1|-26)\\ &R_{\displaystyle 3}=2(5,1,2|12)-5(2,2,3|14)=(0,-8,-11|-46)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 2&2&3&14\\ 0&-8&-1&-26\\ 0&-8&-11&-46 \end{array} \right]\\ &\textrm{Buat elemen di bawah pivot kedua menjadi nol}\\ &\bullet \quad R_{\displaystyle 3}\longleftarrow R_{\displaystyle 3}-R_{\displaystyle 2}\\ &\textrm{maka}\\ &R_{\displaystyle 3}=(0,-8,-11|-46)-(0,-8,-1|-26)\\ &\:\:\:\quad=(0,0,-10|-20)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 2&2&3&14\\ 0&-8&-1&-26\\ 0&0&-10&-20 \end{array} \right]\\ &\textrm{Bentuk di atas sudah}\,\: \textbf{membentuk eselon baris}\\ &\textrm{Langkah berikutnya adalah substitusi, yaitu}\\ &\textrm{Dari baris ke tiga didapatkan}:-10z=-20\Leftrightarrow z=2\\  &\textrm{Pada baris ke dua}:-8y-z=-26\Leftrightarrow -8y-2=-26\\ &\Leftrightarrow -8y=-24\Leftrightarrow y=3\\ &\textrm{Dari baris pertama }:2x+2y+3z=14\\ &\Leftrightarrow 2x+2(3)+3(2)=14 \Leftrightarrow x=1\\ &\textrm{Jadi, penyelesaiansaiannya adalah}\\ &x=1,\quad, y=3,\quad \textrm{dan} \quad z=2\end{array}$.

DAFTAR PUSTAKA

1. Aprilianto, B. ..... Buku Ajar Fokus Matematika (Program IPA) untuk SMA/MA Kelas XII. Solo: CV. SINDUNATA.
2. Johanes, Kastola & Sulasim. 2006. Kompetensi Matematika 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
3. Kusnandar, Muharman, I., Saputra, E.S. 2017. Pendalaman Buku Teks Matematika 2A SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.

DIRGAHAYU REPUBLIK INDONESIA KE-81

Dirgahayu Republik Indonesia Ke-81

Indonesia Berdaulaut, Adil, dan Makmur

Kemerdekaan adalah momentum untuk merawat persatuan, bersemangat gotong royong, serta mengisi hari jadi bangsa melalui karya nyata dan kontribusi terbaik bagi negeri

MATRIKS-INVERS MATRIKS ORDO 3x3

 G. Invers Matriks Ordo 3x3

Invers matriks ordo 3x3 adalah suatu matriks berordo 3x3 yang apabila dikalikan dengan matriks asalnya akan menghasilkan matris identitas. Salah satu metode untuk memperolehnya metode adjoin. Dan untuk menentukan invers matriks ordo 3x3 dengan metode adjoin ini, kita perlu memahami beberapa istilah seperti minor dan kofaktor, determinan matriks ordo 3x3, serta transpose matriks kofaktor sehingga menjadi adjoin untuknantinya kita bagi dengan determinan sehingga mengasilkan invers matriks yang kita inginkan.

G. 1 Matrik Identitas Ordo 3x3

Matriks identitas ordo 3x3 dinyatakan sebagai berikut:

$\textrm{I}=\begin{pmatrix} 1 & 0 &  0\\ 0 & 1 &  0\\ 0 & 0 & 1 \end{pmatrix}$.

G.2 Minor 

Jika unsur-unsur pada baris ke-i kolom ke-j pada sebuah matriks ordo 3x3 dihapus, maka akan diperoleh matriks persegi ordo 2x2. Determinan matriks ordo 2x2 setelah penghapusan tersebut dinamakan minor $a_{ij}$ dan dinotasikan dengan $\left| M_{ij} \right|$.

G.3 Kofaktor

Setelah kita mendapatkan $\left| M_{ij} \right|$ sebagai minor $a_{ij}$ dari matriks A, selanjutnya kita bisa mendapatkan kofaktor $a_{ij}$ denga formula $\alpha_{\displaystyle ij}=(-1)^{\displaystyle i+j}.\left| M_{\displaystyle ij} \right|$.

G.4 Determinan

Determinan dibahas di sini adalah determinan matriks yang tidak bernilai nol atau disebut juga matriks non singular, sedangkan matriks yang nilai determinannya nol adalah matriks singular. Untuk mendapatkan invers sebuah matriks ordo 3x3, maka matriks tersebut haruslah matriks non singular.

G.5 Adjoin

Jika sebuah matriks $\textrm{A}=\begin{pmatrix} a_{\displaystyle 11} & a_{\displaystyle 12} & a_{\displaystyle 13} \\ a_{\displaystyle 21} & a_{\displaystyle 22} & a_{\displaystyle 23} \\ a_{\displaystyle 31} & a_{\displaystyle 32} & a_{\displaystyle 33} \end{pmatrix}$ dengan $\alpha_{\displaystyle ij}$ sebagai kofaktor $a_{\displaystyle ij}$, maka formulasi untuk adjoin matriks A adalah:

$\textrm{adj A}=\begin{pmatrix} \alpha_{\displaystyle 11} & \alpha_{\displaystyle 21} & \alpha_{\displaystyle 31} \\ \alpha_{\displaystyle 12} & \alpha_{\displaystyle 22} & \alpha_{\displaystyle 32} \\ \alpha_{\displaystyle 13} & \alpha_{\displaystyle 23} & \alpha_{\displaystyle 33} \end{pmatrix}$.

G.6 Invers Matriks Ordo 3x3

Perhatikan kembali matriks persegi yaitu matriks A di atas, formulasi invers yang dimaksud adalah:

$\textrm{A}^{\displaystyle -1}=\displaystyle \frac{1}{\textrm{det A}}.\textrm{Adj A}$.

$\LARGE\fbox{CONTOH SOAL}$.

$\begin{aligned}1.\quad&\textrm{Tentukan invers dari matriks A berikut}\\&A=\begin{pmatrix} 2 & 1 &  1\\ 1 & 2 &  1\\ 1 & 1 & 2 \end{pmatrix}\\\\ &\textrm{Jawab}:\\ &\textrm{Kita tentukan adjoinya, yaitu}\\ &\alpha_{\displaystyle 11}=(-1)^{\displaystyle 1+1}\left| M_{\displaystyle 11} \right|=1.\begin{vmatrix} 2 &  1\\ 1 & 2 \end{vmatrix}\\ &\qquad=1.(4-1)=3\\&\alpha_{\displaystyle 12}=(-1)^{\displaystyle 1+2}\left| M_{\displaystyle 12} \right|=(-1).\begin{vmatrix} 1 &  1\\ 1 & 2 \end{vmatrix}\\ &\qquad=(-1).(2-1)=-1\\&\alpha_{\displaystyle 13}=(-1)^{\displaystyle 1+3}\left| M_{\displaystyle 13} \right|=1.\begin{vmatrix} 1 &  2\\ 1 & 1 \end{vmatrix}\\ &\qquad=1.(1-2)=-1\\&\alpha_{\displaystyle 21}=(-1)^{\displaystyle 2+1}\left| M_{\displaystyle 21} \right|=(-1).\begin{vmatrix} 1 &  1\\ 1 & 2 \end{vmatrix}\\ &\qquad=(-1).(2-1)=-1\\&\alpha_{\displaystyle 22}=(-1)^{\displaystyle 2+2}\left| M_{\displaystyle 22} \right|=1.\begin{vmatrix} 2 &  1\\ 1 & 2 \end{vmatrix}\\ &\qquad=1.(4-1)=3\\&\alpha_{\displaystyle 23}=(-1)^{\displaystyle 2+3}\left| M_{\displaystyle 23} \right|=(-1).\begin{vmatrix} 2 &  1\\ 1 & 1 \end{vmatrix}\\ &\qquad=(-1).(2-1)=-1\\&\alpha_{\displaystyle 31}=(-1)^{\displaystyle 3+1}\left| M_{\displaystyle 31} \right|=1.\begin{vmatrix} 1 &  1\\ 2 & 1 \end{vmatrix}\\ &\qquad=1.(1-2)=-1\\&\alpha_{\displaystyle 32}=(-1)^{\displaystyle 3+2}\left| M_{\displaystyle 32} \right|=(-1).\begin{vmatrix} 2 &  1\\ 1 & 1 \end{vmatrix}\\ &\qquad=(-1).(2-1)=-1\\&\alpha_{\displaystyle 33}=(-1)^{\displaystyle 3+3}\left| M_{\displaystyle 33} \right|=1.\begin{vmatrix} 2 &  1\\ 1 & 2 \end{vmatrix}\\ &\qquad=1.(4-1)=3\\&\textrm{Dengan menggunakan data kofaktor-kofaktor}\\ &\textrm{di atas dari matriks A, diperoleh}\\ &\textrm{adj A}=\begin{pmatrix} \alpha_{\displaystyle 11} & \alpha_{\displaystyle 21} & \alpha_{\displaystyle 31} \\ \alpha_{\displaystyle 12} & \alpha_{\displaystyle 22} & \alpha_{\displaystyle 32} \\ \alpha_{\displaystyle 13} & \alpha_{\displaystyle 23} & \alpha_{\displaystyle 33} \end{pmatrix}=\begin{pmatrix} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{pmatrix}\\ &\textrm{Sedangkan nilai det A atau}\:\left| A \right|\\&\begin{aligned}\quad &=\begin{vmatrix} 2 & 1&1\\ 1 & 2&1\\ 1&1&2 \end{vmatrix}\\ &=(2)(2)(2)+\\ &\: \: \quad (1)(1)(1)+\\ &\: \: \quad (1)(1)(1)+\\ &\: \: \quad -(1)(2)(1)\\ &\: \: \quad -(1)(1)(2)\\ &\: \: \quad -(2)(1)(1)\\ &=8+1+1\\ &\: \: \: -2-2-2\\ &=4 \\\end{aligned}\\ &\textrm{Jadi},\: \textrm{A}^{\displaystyle -1}=\displaystyle \frac{1}{\left| A \right|}.\textrm{Adj A}=\displaystyle \frac{1}{4}\begin{pmatrix} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{pmatrix} \end{aligned}$.

$\begin{aligned}2.\quad&\textrm{Tunjukkan bahwa untuk matriks A pada soal no.1}\\&\textrm{di atas dan matriks inversnya yaitu}\quad A^{\displaystyle -1}\\ &\textrm{jika dikalikan akan mengahsilkan matriks identitas}\\\\ &\textrm{Bukti}:\\ &\textrm{Akan dibuktikan secara langsung bahwa}:A.A^{\displaystyle -1}=I\\ &\Leftrightarrow \begin{pmatrix} 2 & 1 &  1\\ 1 & 2 &  1\\ 1 & 1 & 2 \end{pmatrix}.\displaystyle \frac{1}{4}\begin{pmatrix} 3 & -1 &  -1\\ -1 & 3 &  -1\\ -1 & -1 & 3 \end{pmatrix}\\ &=\displaystyle \frac{1}{4}\begin{pmatrix} 6-1-1 & 3-2-1 &  3-1-2\\ -2+3-1 & -1+6-1 &  -1+3-2\\ -2-1+3 & -1-2+3 & -1-1+6 \end{pmatrix}\\ &=\displaystyle \frac{1}{4}\begin{pmatrix} 4 & 0 &  0\\ 0 & 4 &  0\\ 0 & 0 & 4 \end{pmatrix}=\begin{pmatrix} 1 & 0 &  0\\ 0 & 1 &  0\\ 0 & 0 & 1 \end{pmatrix}\quad \textbf{terbukti} \end{aligned}$.

$\begin{aligned}3.\quad&\textrm{Tentukan invers dari matriks B berikut}\\&B=\begin{pmatrix} 1 & 2 &  3\\ 4 & 5 &  6\\ 1 & 2 & 0 \end{pmatrix}\\\\ &\textrm{Jawab}:\\ &\textrm{Kita tentukan adjoinya, yaitu}\\ &\alpha_{\displaystyle 11}=(-1)^{\displaystyle 1+1}\left| M_{\displaystyle 11} \right|=1.\begin{vmatrix} 5 &  6\\ 2 & 0 \end{vmatrix}\\ &\qquad=1.(0-12)=-12\\&\alpha_{\displaystyle 12}=(-1)^{\displaystyle 1+2}\left| M_{\displaystyle 12} \right|=(-1).\begin{vmatrix} 4 &  6\\ 1 & 0 \end{vmatrix}\\ &\qquad=(-1).(0-6)=6\\&\alpha_{\displaystyle 13}=(-1)^{\displaystyle 1+3}\left| M_{\displaystyle 13} \right|=1.\begin{vmatrix} 4 &  5\\ 1 & 2 \end{vmatrix}\\ &\qquad=1.(8-5)=3\\&\alpha_{\displaystyle 21}=(-1)^{\displaystyle 2+1}\left| M_{\displaystyle 21} \right|=(-1).\begin{vmatrix} 2 &  3\\ 2 & 0 \end{vmatrix}\\ &\qquad=(-1).(0-6)=6\\&\alpha_{\displaystyle 22}=(-1)^{\displaystyle 2+2}\left| M_{\displaystyle 22} \right|=1.\begin{vmatrix} 1 &  3\\ 1 & 0 \end{vmatrix}\\ &\qquad=1.(0-3)=-3\\&\alpha_{\displaystyle 23}=(-1)^{\displaystyle 2+3}\left| M_{\displaystyle 23} \right|=(-1).\begin{vmatrix} 1 &  2\\ 1 & 2 \end{vmatrix}\\ &\qquad=(-1).(2-2)=0\\&\alpha_{\displaystyle 31}=(-1)^{\displaystyle 3+1}\left| M_{\displaystyle 31} \right|=1.\begin{vmatrix} 2 &  3\\ 5 & 6 \end{vmatrix}\\ &\qquad=1.(12-15)=-3\\&\alpha_{\displaystyle 32}=(-1)^{\displaystyle 3+2}\left| M_{\displaystyle 32} \right|=(-1).\begin{vmatrix} 1 &  3\\ 4 & 6 \end{vmatrix}\\ &\qquad=(-1).(6-12)=6\\&\alpha_{\displaystyle 33}=(-1)^{\displaystyle 3+3}\left| M_{\displaystyle 33} \right|=1.\begin{vmatrix} 1 &  2\\ 4 & 5 \end{vmatrix}\\ &\qquad=1.(5-8)=-3\\&\textrm{Dengan menggunakan data kofaktor-kofaktor}\\ &\textrm{di atas dari matriks B, diperoleh}\\ &\textrm{adj B}=\begin{pmatrix} \alpha_{\displaystyle 11} & \alpha_{\displaystyle 21} & \alpha_{\displaystyle 31} \\ \alpha_{\displaystyle 12} & \alpha_{\displaystyle 22} & \alpha_{\displaystyle 32} \\ \alpha_{\displaystyle 13} & \alpha_{\displaystyle 23} & \alpha_{\displaystyle 33} \end{pmatrix}=\begin{pmatrix} -12 & 6 &  -3\\  6 & -3 & 6 \\ 3 & 0 & -3 \end{pmatrix}\\ &\textrm{Sedangkan nilai det B atau}\:\left| B \right|\\&\begin{aligned}\quad &=\begin{vmatrix} 1 & 2&3\\ 4 & 5&6\\ 1&2&0 \end{vmatrix}\\ &=(1)(5)(0)+\\ &\: \: \quad (2)(6)(1)+\\ &\: \: \quad (3)(2)(4)+\\ &\: \: \quad -(1)(5)(3)\\ &\: \: \quad -(2)(6)(1)\\ &\: \: \quad -(0)(4)(2)\\ &=0+12+24\\ &\: \: \: -15-12-0\\ &=9 \\\end{aligned}\\ &\textrm{Jadi},\: \textrm{B}^{\displaystyle -1}=\displaystyle \frac{1}{\left| B \right|}.\textrm{Adj B}=\displaystyle \frac{1}{9}\begin{pmatrix} -12 & 6 & -3 \\ 6 & -3 & 6 \\ 3 & 0 & -3 \end{pmatrix} \end{aligned}$.

$\begin{aligned}4.\quad&\textrm{Tunjukkan bahwa untuk matriks B pada soal no.3}\\&\textrm{di atas dan matriks inversnya yaitu}\quad B^{\displaystyle -1}\\ &\textrm{jika dikalikan akan mengahsilkan matriks identitas}\\\\ &\textrm{Bukti}:\\ &\textrm{Akan dibuktikan secara langsung bahwa}:B.B^{\displaystyle -1}=I\\ &\Leftrightarrow \begin{pmatrix} 1 & 2 &  3\\ 4 & 5 &  6\\ 1 & 2 & 0 \end{pmatrix}.\displaystyle \frac{1}{9}\begin{pmatrix} -12 & 6 &  -3\\ 6 & -3 &  6\\ 3 & 0 & -3 \end{pmatrix}\\ &=\displaystyle \frac{1}{9}\begin{pmatrix} -12+12+9 & 6-6+0 &  -3+12-9\\ -48+30+18 & 24-15+0 &  -12+30-18\\ -12+12+0 & 6-6+0 & -3+12+0 \end{pmatrix}\\ &=\displaystyle \frac{1}{9}\begin{pmatrix} 9 & 0 &  0\\ 0 & 9 &  0\\ 0 & 0 & 9 \end{pmatrix}=\begin{pmatrix} 1 & 0 &  0\\ 0 & 1 &  0\\ 0 & 0 & 1 \end{pmatrix}\quad \textbf{terbukti} \end{aligned}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono, Kurnianingsih, S. 2005. Matematika untuk SMA dan MA Kelas XII Program Ilmu Alam. Jakarta: GELORA AKSARA PRATAMA.

.

 

MATRIKS-CONTOH 6

 $\begin{array}{ll}\\ 26.&(\textbf{UM UGM 2004})\\ &\textrm{Nilai-nilai}\: \: x\: \: \textrm{agar matriks}\\ &\qquad\quad\quad\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}\\ &\textrm{tidak memiliki invers adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&4\: \: \textrm{atau}\: \: 5\\ \textrm{b}.&-2\: \: \textrm{atau}\: \: 2\\ \textrm{c}.&-4\: \: \textrm{atau}\: \: 5\\ \textrm{d}.&-6\: \: \textrm{atau}\: \: 4\\ \textrm{e}.&0 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{supaya matriks}\: \: \begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}\\ &\textrm{tidak memiliki invers},\: \textrm{maka}\\ &\textrm{determinan matriks}\: \: \begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}=0\\ &\textrm{Sehingga}\\ &\begin{vmatrix} 5x & 5\\ 4 & x \end{vmatrix}=0\\ &\Leftrightarrow 5x^{2}-20=0\\ &\Leftrightarrow x^{2}=4\\ &\Leftrightarrow x=\pm 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&(\textbf{UM UGM 2005})\\ &\textrm{Matriks}\: \: \begin{pmatrix} x & 1\\ -2 & 1-x \end{pmatrix}\\ &\textrm{tidak memiliki invers untuk}\\ &\textrm{nilai}\: \: x=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{atau}\: \: -2\\ \textrm{b}.&-1\: \: \textrm{atau}\: \: 0\\ \textrm{c}.&-1\: \: \textrm{atau}\: \: 1\\ \textrm{d}.&-1\: \: \textrm{atau}\: \: 2\\ \textrm{e}.&1\: \: \textrm{atau}\: \: 2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Mirip dengan pembahasan no. 26}\\ &\begin{aligned}&\textrm{Nilai}\: \: \begin{vmatrix} x & 1\\ -2 & 1-x \end{vmatrix}=0\\ &\Leftrightarrow x-x^{2}-(-2)=0\\ &\Leftrightarrow 2+x-x^{2}=0\\ &\Leftrightarrow x^{2}-x-2=0\\ &\Leftrightarrow (x-2)(x+1)=0\\ &\Leftrightarrow x=2\: \: \textrm{atau}\: \: x=-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&(\textbf{Mat Das SIMAK UI 2014})\\ &\textrm{Jika matriks}\: \: \textrm{A}\: \: \textrm{adalah invers}\\ &\textrm{dari matriks}\: \: \displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix}\: \: \textrm{dan}\\ &\textrm{A}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 1\\ 3 \end{pmatrix}\: \: \textrm{maka nilai}\: \: 2x+y\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-\displaystyle \frac{10}{3}\\ \textrm{b}.&-\displaystyle \frac{1}{3}\\ \textrm{c}.&1\\ \textrm{d}.&\displaystyle \frac{9}{7}\\ \textrm{e}.&\displaystyle \frac{20}{3} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Misalkan diketahui matriks}\\ &\textrm{B}=\displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix},\\ &\textrm{maka}\: \: \textrm{A}=\left ( \displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix} \right )^{-1}\\ &\textrm{selanjutnya}\: \: \textrm{A}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=A^{-1}\begin{pmatrix} 1\\ 3 \end{pmatrix},\\ & \textrm{ingat bahwa}\: \: \left (\textbf{A}^{-1} \right )^{-1}=\textbf{A}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=\left ( \left ( \displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix} \right )^{-1} \right )^{-1}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3}\begin{pmatrix} -1-9\\ 4+15 \end{pmatrix}=\begin{pmatrix} \displaystyle -\frac{10}{3}\\ \displaystyle \frac{19}{3} \end{pmatrix}\\ &2x+y=2\left ( -\displaystyle \frac{10}{3} \right )+\frac{19}{3}\\ &\qquad\: \: \: \, =\displaystyle \frac{-20+19}{3}=-\displaystyle \frac{1}{3} \end{aligned} \end{array}$

 

MATRIKS-CONTOH 5

 $\begin{array}{ll}\\ 21.&(\textbf{SPMB 2003})\\ &\textrm{Diketahu matriks}\: \: \textrm{A}=\begin{pmatrix}a&b\\ c&d \end{pmatrix}.\\ &\textrm{Jika}\: \: \: \textrm{A}^{t}=\textrm{A}^{-1}\: \: \textrm{dengan}\: \: \textrm{A}^{t}\\ &\textrm{adalah transpose matriks A},\\ &\textrm{maka nilai}\: \: ad-bc=....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1\: \: \textrm{atau}\: \: -\sqrt{2}\\ \textrm{b}.&1\: \: \textrm{atau}\: \: \sqrt{2}\\ \textrm{c}.&-\sqrt{2}\: \: \textrm{atau}\: \: -\sqrt{2}\\ \textrm{d}.&-1\: \: \textrm{atau}\: \: 1\\ \textrm{e}.&1\: \: \textrm{atau}\: \: -\sqrt{2}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{matriks}\: \: \textrm{A}=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{A}^{t}=\textrm{A}^{-1},\: \textrm{maka}\\ &\textrm{A}^{t}=\textrm{A}^{-1}\\ &\begin{pmatrix} a & b\\ c & d \end{pmatrix}^{t}=\displaystyle \frac{1}{ad-bc}\times \textrm{Adjoin Matriks}\: \: \textrm{A}\\ &\begin{pmatrix} a & c\\ b & d \end{pmatrix}=\displaystyle \frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix},\\ & \textrm{didapatkan hubungan}\\ &c=\displaystyle \frac{-b}{ad-bc}\quad ...............(1)\\ &b=\displaystyle \frac{-c}{ad-bc}\quad ...............(2)\\ &\textrm{Persamaan}\: \:  (2)\: \: \textrm{disubstitusikan ke persamaan}\: \: (1)\\ &c=\displaystyle \displaystyle \frac{-\displaystyle \frac{-c}{ad-bc}}{ad-bc}\\ &1=(ad-bc)^{2}\\ &(ad-bc)= -1\: \: \textrm{atau}\: \: 1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Diketahu matriks}\: \: \textrm{H}\\ &\textrm{yang memenuhi persamaan}\\ &\textrm{H}\begin{pmatrix} 3 & 2\\ 1 & 4 \end{pmatrix}=\begin{pmatrix} 7 & 8\\ 4 & 6 \end{pmatrix},\\ &\textrm{maka nilai dari}\: \: \: det\: \textrm{H}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-3\\ \textrm{b}.&-2\\ \textrm{c}.&-1\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|}\hline \textbf{Alternatif 1}\\\hline \begin{aligned}\textrm{H.A}&=\textrm{B}\\ \textrm{H.A.A}^{-1}&=\textrm{B.A}^{-1}\\ \textrm{H}&=\textrm{B.A}^{-1}\\ &=\begin{pmatrix} 7 & 8\\ 4 & 6 \end{pmatrix}.\displaystyle \frac{1}{\begin{vmatrix} 3 & 2\\ 1 & 4 \end{vmatrix}}\begin{pmatrix} 4 & -2\\ -1 & 3 \end{pmatrix}\\ &=\begin{pmatrix} 7 & 8\\ 4 & 6 \end{pmatrix}.\displaystyle \frac{1}{12-2}\begin{pmatrix} 4 & -2\\ -1 & 3 \end{pmatrix}\\ &=\displaystyle \frac{1}{10}\begin{pmatrix} 28+(-8) & (-14)+24\\ 16+(-6) & (-8)+18 \end{pmatrix}\\ \textrm{H}&=\displaystyle \frac{1}{10}\begin{pmatrix} 20 & 10\\ 10 & 10 \end{pmatrix}=\begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix}\\ det\: \textrm{H}&=\begin{vmatrix} 2 & 1\\ 1 & 1 \end{vmatrix}=2.1-1.1=2-1=\color{purple}1 \end{aligned}\\\hline \textbf{Alternatif 2}\\\hline \begin{aligned}\textrm{H.A}&=\textrm{B}\begin{cases} det\: \textrm{H} &=\left | \textrm{H} \right | \\ det\: \textrm{A} &=\left | \textrm{A} \right |=\begin{vmatrix} 3 & 4\\ 2 & 1 \end{vmatrix}\\ &=12-2=10 \\ det\: \textrm{B} &=\left | \textrm{B} \right |=\begin{vmatrix} 7 & 8\\ 4 & 6 \end{vmatrix}\\ &=42-32=10 \end{cases}\\ \left | \textrm{H} \right |.\left | \textrm{A} \right |&=\left | \textrm{B} \right |\\ \left | \textrm{H} \right |&=\displaystyle \frac{\left | \textrm{B} \right |}{\left | \textrm{A} \right |}\\ &=\displaystyle \frac{10}{10}\\ &=1 \end{aligned} \\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 23.&(\textbf{UM UGM 2006})\\ &\textrm{Apabila}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{memenuhi}\\ &\textrm{persamaan matriks}\\ &\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -1\\ 2 \end{pmatrix},\\ &\textrm{maka}\: \: x+y=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ A.X&=B\\ A^{-1}.A.X&=A^{-1}.B\\ A^{0}.X&=A^{-1}.B\\ X&=A^{-1}.B\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}^{-1}\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{\begin{vmatrix} 1 & -2\\ -1 & 3 \end{vmatrix}}&\begin{pmatrix} 3 & 2\\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3-2}&\begin{pmatrix} 3.(-1)+2.2 \\ 1.(-1)+1.2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1\\ 1 \end{pmatrix}\\ x+y&=1+1=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&(\textbf{KSM Matematika Kabupten 2019})\\ &\textrm{Matriks}\: \: A\: \: \textrm{dengan entri bulat dan}\\ &\textrm{berukuran 2x2},\: \textrm{dikalikan dengan matriks}\\ &\begin{pmatrix} 1 & 2\\ 2 & 2 \end{pmatrix}\: \: \textrm{dari kanan menghasilkan matriks}\\ &\textrm{yang semua entrinya bilangan prima}.\\ &\textrm{Jika determinan dari matriks}\: \: A\: \: \textrm{juga}\\ &\textrm{bilangan prima, maka nilai minimum dari}\\ &det\: A\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&5\\ \textrm{d}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\begin{pmatrix} 1 & 2\\ 2 & 2 \end{pmatrix}&\times A_{2\times 2}=\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}\\ \begin{vmatrix} 1 & 2\\ 2 & 2 \end{vmatrix}&\times \left | A_{2\times 2} \right |=\begin{vmatrix} \alpha & \beta \\ \gamma & \delta \end{vmatrix}\\ &\left | A_{2\times 2} \right |=\displaystyle \frac{\begin{vmatrix} \alpha & \beta \\ \gamma & \delta \end{vmatrix}}{\begin{vmatrix} 1 & 2\\ 2 & 2 \end{vmatrix}}\\ &\left | A_{2\times 2} \right |=\displaystyle \frac{(\alpha \delta -\beta \gamma )}{-2}\\ &\left | A_{2\times 2} \right |=\displaystyle \frac{(\beta \gamma -\alpha \delta )}{2}\\ \textrm{Karena}&\: \: \left | A_{2\times 2} \right |\: \: \textrm{bilangan prima}\\ \textrm{akan m}&\textrm{engakibatkan}\: \: ( \beta \gamma -\alpha \delta)\\ \textrm{harus h}&\textrm{abis dibagi}\: \: 2,\: \: \textrm{oleh karenanya}\\ \textrm{menyeb}&\textrm{abkan}\: \: ( \beta \gamma -\alpha \delta)\: \: \textrm{berupa bilangan}\\ \textrm{genap.}\, \, \, &\textrm{Dan karena}\: \: ( \beta \gamma -\alpha \delta)\: \: \textrm{genap},\\ \textrm{maka p}&\textrm{astilah}\: \: \left | A_{2\times 2} \right |\: \: \textrm{juga bernilai genap}\\ \textrm{sehingg}&\textrm{a nilai}\: \: \left | A_{2\times 2} \right |\: \: \textrm{pastilah 2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&(\textbf{UM UGM 2005})\textrm{Jika}\\ &\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{pmatrix}=\begin{pmatrix} \sin A & \cos A \end{pmatrix}\\ &\textrm{dan}\: \: A\: \: \textrm{suatu konstanta, maka}\: \: x+y=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{pmatrix}=\begin{pmatrix} \sin A & \cos A \end{pmatrix}\\ &\begin{pmatrix} x\sin \alpha -y\cos \alpha & x\cos \alpha +y\sin \alpha \end{pmatrix}=\begin{pmatrix} \sin A & \cos A \end{pmatrix}\\ &\begin{cases} \sin A & =x\sin \alpha -y\cos \alpha =\sqrt{x^{2}+y^{2}}\cos \left ( \alpha -\tan ^{-1}\displaystyle \frac{x}{-y} \right ) \\ \cos A & =x\cos \alpha +y\sin \alpha =\sqrt{x^{2}+y^{2}}\cos \left (\alpha -\tan ^{-1}\displaystyle \frac{y}{x} \right ) \end{cases}\\ &\textrm{Supaya}\: \: \cos A=\sqrt{x^{2}+y^{2}}\cos \left (\alpha -\tan ^{-1}\displaystyle \frac{y}{x} \right ),\: \: \textrm{maka}\\ &\begin{cases} \sqrt{x^{2}+y^{2}} & =1 \\ \tan ^{-1}\displaystyle \frac{y}{x} & =0\Rightarrow \begin{cases} y & =0 \\ x & =1 \end{cases} \end{cases}\\ &\textrm{Sehingga}\: \: x+y=1+0=1 \end{aligned} \end{array}$

MATRIKS-CONTOH 4

 $\begin{array}{ll}\\ 16.&\textrm{Determinan untuk matriks}\: \: \begin{pmatrix} 2 & -5\\ 3 & -1 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-17\\ \textrm{b}.&-13\\ \textrm{c}.&11\\ \textrm{d}.&13\\ \textrm{e}.&17 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Determinan}&\: \textrm{dari matriks}\: \: \begin{pmatrix} 2 & -5\\ 3 & -1 \end{pmatrix}\\ &=\begin{vmatrix} 2 & -5\\ 3 & -1 \end{vmatrix}=2(-1)-3(-5)\\ &=-2+15\\ &=13 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Determinan untuk matriks}\\ &\begin{pmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&10\\ \textrm{b}.&18\\ \textrm{c}.&22\\ \textrm{d}.&30\\ \textrm{e}.&36 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\textrm{Determinan}\: \: \textrm{dari matriks}\\ &\begin{pmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{pmatrix}\\ &=\begin{vmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{vmatrix}\\ &=+(2.4.3)+(-1.-1.1)+(-1.1.-2)\\ &\quad -(1.4.-1)-(-2.-1.2)-(3.1.-1)\\ &=24+1+2+4-24+3\\ &=10 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika diketahu matriks}\\ &\textrm{A}=\begin{pmatrix} x+3&-2\\ -16&2x-6 \end{pmatrix},\\ &\textrm{maka nilai dari}\: \: \: x\: \: \textrm{supaya matriks}\\ &\textrm{A tidak memiliki invers adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&5 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Invers}&\: \textrm{dari matriks A adalah}\: \: \: \textrm{A}^{-1}.\\ \textrm{A}^{-1}&=\displaystyle \frac{1}{det\: \textrm{A}}\times Adjoin\: \textrm{A}.\\ \textrm{Karen}&\textrm{a}\: \textrm{tidak memiliki invers},\\ \textrm{maka}\: \, & det\: \textrm{A}=0,\: \textrm{sehingga}\\ det\: \textrm{A}&=\begin{vmatrix} x+3 & -2\\ -16 & 2x-6 \end{vmatrix}=0\\ &\Leftrightarrow (x+3)(2x-6)-(-16.-2)=0\\ &(\textrm{masing-masing ruas dibagi 2})\\ &\Leftrightarrow (x+3)(x-3)-16=0\\ &\Leftrightarrow x^{2}-9-16=0\\ &\Leftrightarrow x^{2}-25=0\\ &\Leftrightarrow (x+5)(x-5)=0\\ &\Leftrightarrow x+5=0\quad \textrm{atau}\quad x-5=0\\ &\Leftrightarrow \: \: \: \, \, x=-5\quad \textrm{atau}\quad x=5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: \begin{vmatrix} 5^{2x} & -5\\ 1 & 1 \end{vmatrix}=6.5^{x}\\ &\textrm{maka}\: \: 5^{2x}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&625\: \: \textrm{atau}\: \: 1\\ \textrm{b}.&25\: \: \textrm{atau}\: \: 1\\ \textrm{c}.&25\: \: \textrm{atau}\: \: 0\\ \textrm{d}.&5\: \: \textrm{atau}\: \: 1\\ \textrm{e}.&5\: \: \textrm{atau}\: \: 0 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&5^{2x}+5=6.5^{x}\\ &5^{2x}-6.5^{x}+5=0\\ &\left ( 5^{x}-1 \right )\left ( 5^{x}-5 \right )=0\\ &5^{x}-1=0\: \: \textrm{atau}\: \: 5^{x}-5=0\\ &5^{x}=1\: \: \textrm{atau}\: \: 5^{x}=5\\ &5^{x}=5^{0}\: \: \textrm{atau}\: \: 5^{x}=5^{1}\\ &x=0\: \: \textrm{atau}\: \: x=1\\ &\textrm{maka}\\ &5^{2x}=\begin{cases} 5^{2.1} &=5^{2}=25 \\ 5^{2.0} &=5^{0}=1 \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Diketahu determinan suatu}\\ &\textrm{matriks adalah}\: \: \begin{vmatrix} x & 1 & 2\\ x & 1 & x\\ 5 & -3 & 7 \end{vmatrix}=0.\\ &\textrm{Jika}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{adalah akar-akar}\\ &\textrm{yang memenuhi persamaan tersebut}\\ &\textrm{maka nilai dari}\: \: \: p+q\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-3\\ \textrm{b}.&-\displaystyle \frac{1}{3}\\ \textrm{c}.&-1\\ \textrm{d}.&\displaystyle \frac{1}{3}\\ \textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Diketahui ba}&\textrm{hwa}:\\ \begin{vmatrix} x & 1 & 2\\ x & 1 & x\\ 5 & -3 & 7 \end{vmatrix}&=0\\ +(x.1.7)+&(1.x.5)+(2.x.-3)\\ -(5.1.2)&-(-3.x.x)-(7.x.1)=0\\ 7x+5x-6x&-10+3x^{2}-7x=0\\ 3x^{2}-x-10&=0\begin{cases} p & \textrm{salah satu akar} \\ q & \textrm{salah satu akar yang lain}, \end{cases}\\ \textrm{dengan}\: \: \: &\begin{cases} a &=3 \\ b &=-1 \\ c &=-10 \end{cases}.\\ \textrm{maka}\: \: \: p+q\: \: &=-\displaystyle \frac{b}{a}=-\displaystyle \frac{-1}{3}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$

MATRIKS-INVERS MATRIKS ORDO 2x2

 $\textrm{F. Invers Matriks ordo 2x2}$

Perhatikanlah kembali materi sebelumnya berkaitan determinan matriks 2x2, yaitu

$\begin{array}{|c|}\hline \begin{aligned}&\textrm{Jika matriks}\: \: A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\ &\textrm{maka determinan matriks}\: \: A\\ &\textrm{ditentukan dengan}\\ &det\: \: A=\begin{vmatrix} a & b\\ c & d \end{vmatrix}=ad-bc \end{aligned}\\\hline \end{array}$

Jika  $det\: \: A$  bernilai tidak sama dengan nol, maka invers matriks ordo 2x2 yang selanjutnya dilambangkan dengan  $A^{\displaystyle -1}$  dapat ditentukan dengan formula:

$\LARGE\boxed{A^{\displaystyle -1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix}}$

$\begin{aligned}&\textbf{Sebagai}\: \: \textrm{CONTOH}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 2x2,\: \: \textrm{yaitu}:\\ &A=\begin{pmatrix} -3 & 2\\ -1 & 4 \end{pmatrix},\: \: \textrm{maka}\: \: A^{^{\displaystyle -1}}\: \: \textrm{adalah}:\\ &A^{^{\displaystyle -1}}=\displaystyle \frac{1}{\begin{vmatrix} -3 & 2\\ -1 & 4 \end{vmatrix}}\begin{pmatrix} 4 & -2\\ 1 & -3 \end{pmatrix}\\ &\: \qquad =\displaystyle \frac{1}{12-(-2)}\begin{pmatrix} 4 & -2\\ 1 & -3 \end{pmatrix}\\ &\: \qquad=\displaystyle \frac{1}{14}\begin{pmatrix} 4 & -2\\ 1 & -3 \end{pmatrix}\\ &\: \qquad =\begin{pmatrix} \displaystyle \frac{4}{14} & \displaystyle \frac{-2}{14}\\ \displaystyle \frac{1}{14} & \displaystyle \frac{-3}{14} \end{pmatrix}\\ &\: \qquad=\begin{pmatrix} \displaystyle \frac{2}{7} & -\displaystyle \frac{1}{7}\\ \displaystyle \frac{1}{14} & -\displaystyle \frac{3}{14} \end{pmatrix} \end{aligned}$

$\LARGE\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah invers matriks berikut}\\ &\textrm{a}.\quad B=\begin{pmatrix} 5 & -3\\ 4 & -2 \end{pmatrix}\\ &\textrm{b}.\quad C=\begin{pmatrix} -3 & -5\\ 6 & 9 \end{pmatrix}\\ &\textrm{c}.\quad P=\begin{pmatrix} -1 & 2\\ -3 & 6 \end{pmatrix}\\ &\textrm{d}.\quad Q=\begin{pmatrix} 6 & 9\\ 2 & 3 \end{pmatrix}\\\\ &\textrm{Jawab}:\: \textrm{yang dibahas poin a saja}\\ &B^{-1}=\displaystyle \frac{1}{det\: B}\begin{pmatrix} -2 & 3\\ -4 & 5 \end{pmatrix}\\ &\qquad=\displaystyle \frac{1}{-10-(-12)}\begin{pmatrix} -2 & 3\\ -4 & 5 \end{pmatrix}=\displaystyle \frac{1}{2}\begin{pmatrix} -2 & 3\\ -4 & 5 \end{pmatrix}\\ &\qquad =\begin{pmatrix} -1 & \displaystyle \frac{3}{2}\\ -2 & \displaystyle \frac{5}{2} \end{pmatrix}\\ &\textrm{b. Silahkan dicoba sendiri}\\ &\textrm{c. Silahkan dicoba sendiri}\\ &\textrm{d. Silahkan dicoba sendiri} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui matriks}\\ & E=\begin{pmatrix} 4 & 2\\ 5 & 3 \end{pmatrix}\\ &\textrm{a}.\quad \textrm{Tentukanlah}\\ &\qquad (\textrm{i})\: \: E^{-1}\qquad\qquad\qquad (\textrm{iii})\: \: \left ( E^{-1} \right )^{t}\\ &\qquad (\textrm{i})\: \: E^{t}\qquad\qquad\qquad\: \: \: (\textrm{iv})\: \: \left ( E^{t} \right )^{-1}\\ &\textrm{b}.\quad \textrm{Dengan menggunakan hasil-hasil}\\ &\qquad \textrm{pada a. apakah}\: \: \left ( E^{-1} \right )^{t}=\left ( E^{t} \right )^{-1}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad(\textrm{i})\: \: E^{-1}&=\displaystyle \frac{1}{2}\begin{pmatrix} 3 & -2\\ -5 & 4 \end{pmatrix}=\begin{pmatrix} \displaystyle \frac{3}{2} & -1\\ -\displaystyle \frac{5}{2} & 2 \end{pmatrix}\\ (\textrm{ii})\: \: \: \: E^{t}&=\begin{pmatrix} 4 & 5\\ 2 & 3 \end{pmatrix}\\ (\textrm{iii})\: \: \: \quad&\left (E^{-1} \right )^{t}=\begin{pmatrix} \displaystyle \frac{3}{2} & -\displaystyle \frac{5}{2}\\ -1 & 2 \end{pmatrix}\\ (\textrm{iv})\: \: \: \quad&\left (E^{t} \right )^{-1}=\displaystyle \frac{1}{2}\begin{pmatrix} 3 & -5\\ -2 & 4 \end{pmatrix}=\begin{pmatrix} \displaystyle \frac{3}{2} & -\displaystyle \frac{5}{2}\\ -1 & 2 \end{pmatrix}\\ \textrm{b}.\quad \textrm{Dari ha}&\textrm{sil yang didapat dapat disimpulkan}\\ &\left ( E^{-1} \right )^{t}=\left ( E^{t} \right )^{-1} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 2 Kelas 1 Semester 2. Jakarta: ERLANGGA

MATRIKS-DETERMINAN MATRIKS ORDO 2x2 dan ORDO 3x3

 $\textrm{E. Determinan Matriks}$

$\textrm{1. Ordo 2x2}$

Misalkan A adalah matriks persegi berordo 2x2 dan dituliskan dengan  $A=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}$  dengan  $a_{11}\: \: \textrm{dan}\: \: a_{22}$ sebagai elemen dari diagonal utama dan $a_{12}\: \: \textrm{dan}\: \: a_{21}$ adalah elemen yang menempati diagonal samping, perhatikan lagi matriks A berikut:

$A=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}$

maka determinan dari matriks A yang berordo 2x2 adalah perkalian elemen pada diagonal utama dikurangi dengan hasil kali perkalian diagonal samping dan di tuliskan dengan det A atau tanda |...|. Sehingga dari pengertian tersebut kita dapat menuliskan  bahwa determinan dari matriks A dalah:

$\textrm{det}.\: A=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}$ sama dengan

$\textrm{det}.\: A=\begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix}=a_{11}\times a_{22}-a_{12}\times a_{21}$

$\begin{aligned}&\textbf{Sebagai}\: \: \textrm{CONTOH}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 2x2,\: \: \textrm{yaitu}:\\ &A=\begin{pmatrix} -3 & 2\\ -1 & 4 \end{pmatrix},\: \: \textrm{maka}\: \:  det\: A\: \: \textrm{adalah}:\\ &det\: A=\begin{vmatrix} -3 & 2\\ -1 & 4 \end{vmatrix}=(-3)\times (-4)-(-1)\times (2)\\ &\: \qquad =12-(-2)=12+2=14 \end{aligned}$

$\textrm{2. Ordo 3x3}$

Ada dua buah cara minimal dalam menentukan determinan matriks ordo 3x3, yaitu:

• cara menjabarkan mengikuti baris atau kolom(ekspansi kofaktor)
• aturan Sarrus

Adapun penjelasan lebih lanjut adalah sebagai berikut

$\begin{aligned}&\textrm{Misalkan diberikan matriks ordo}\: 3x3\\ &A=\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{pmatrix}\\ \end{aligned}$

$\textrm{2.1 Menjabarkan mengikuti baris atau kolom}$

$\begin{aligned}\textrm{det A}&=a_{11}\begin{vmatrix} a_{22} & a_{23}\\ a_{32} & a_{33} \end{vmatrix}-a_{12}\begin{vmatrix} a_{21} & a_{23}\\ a_{31} & a_{33} \end{vmatrix}+a_{13}\begin{vmatrix} a_{21} & a_{22}\\ a_{31} & a_{32} \end{vmatrix}\\ &\\ &\textbf{Catatan}:\\ &\textrm{tanda}\: a_{ij}=\textrm{positif jika}\: i+j\: \textrm{genap}\\ &\textrm{tanda}\: a_{ij}=\textrm{negatif jika}\: i+j\: \textrm{ganjil} \end{aligned}$

Anda juga bisa menjabarkan mengikuti baris yang lain termasuk juga menjabarkan mengikuti kolom. Sehingga total cara menjabarkan ini, karena ada 3 baris dan 3 kolom total akan ada sebanyak 6 cara menentukan determinan dari matriks A tersebut.

$\textrm{2.2 Aturan Sarrus}$

$\begin{aligned}\textrm{det A}&=a_{11}.a_{22}.a_{33}\\ &\quad\: +a_{12}.a_{23}.a_{31}\\ &\quad\: +a_{13}.a_{21}.a_{32}\\ &\quad -a_{31}a_{22}.a_{13}\\ &\quad -a_{32}.a_{23}.a_{11}\\ &\quad -a_{33}.a_{21}.a_{12} \end{aligned}$

$\begin{aligned}&\textbf{Sebagai}\: \: \textrm{CONTOH MENJABARKAN}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 3x3,\: \: \textrm{yaitu}:\\ &A=\begin{pmatrix} 1 & 2&3\\ 1 &3& 4\\ 1&4&3 \end{pmatrix},\: \: \textrm{maka}\: \: det\: A\: \: \textrm{dengan}\\ &\textrm{menjabarkan baris pertama adalah}:\\ &det\: A=1\begin{vmatrix} 3 & 4\\ 4 & 3 \end{vmatrix}-2\begin{vmatrix} 1 & 4\\ 1 & 3 \end{vmatrix}+3\begin{vmatrix} 1 & 3\\ 1 & 4 \end{vmatrix}\\ &\: \qquad=(9-16)-2(3-4)+3(4-3)\\ &\: \qquad=-7+2+3\\ &\: \qquad=-2 \end{aligned}$

$\begin{aligned}&\textbf{Dan berikut}\: \: \textrm{CONTOH aturan SARRUS}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 3x3,\: \: \textrm{yaitu}:\\ &B=\begin{pmatrix} 2 & 1&3\\ 3 &1& 4\\ 4&1&3 \end{pmatrix},\: \: \textrm{maka}\: \: det\: B\: \: \textrm{dengan}\\ &\textrm{metode SARRUS adalah}:\\ &det\: B=(2.1.3)+(1.4.4)+(3.1.3)\\ &\qquad\: \: \: \: \: \: \: -(4.1.3)-(1.4.2)-(3.1.3)\\ &\qquad\: \, =6+16+9-12-8-9=2 \end{aligned}$

$\LARGE\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Diketahui matriks-matriks persegi berikut}\\ &\textrm{a}.\: \: \begin{pmatrix} 2 & 3\\ 6 & 7 \end{pmatrix}\qquad\qquad \textrm{c}.\: \: \begin{pmatrix} -2 & -3\\ 6 & 7 \end{pmatrix}\\ &\textrm{b}.\: \: \begin{pmatrix} 0 & 4\\ -3 & 6 \end{pmatrix}\: \quad\quad\quad \textrm{d}.\: \: \begin{pmatrix} \sqrt{3} & 3\sqrt{3}\\ \sqrt{2} & -2\sqrt{2} \end{pmatrix}\\ &\\ &\textrm{Tentukanlah determinan dari}\\ &\textrm{matriks-matriks persegi di atas}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad &\begin{vmatrix} 2 & 3\\ 6 & 7 \end{vmatrix}\\ &=(2).(7)-(3).(6)\\ &=14-18\\ &=-4\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad &\begin{vmatrix} 0 & 4\\ -3 & 6 \end{vmatrix}\\ &=(0).(6)-(4).(-3)\\ &=0-(-12)\\ &=12\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad &\begin{vmatrix} -2 & -3\\ 6 & 7 \end{vmatrix}\\ &=(-2).(7)-(-3).(6)\\ &=(-14)-(-18)\\ &=-14+18\\ &=4 \end{aligned}&\begin{aligned}\textrm{d}.\quad &\begin{vmatrix} \sqrt{3} & 3\sqrt{3}\\ \sqrt{2} & -2\sqrt{2} \end{vmatrix}\\ &=(\sqrt{3}).(-2\sqrt{2})\\ &\quad-(3\sqrt{3}).(\sqrt{2})\\ &=-2\sqrt{6}-3\sqrt{6}\\ &=-5\sqrt{6}\\ & \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\begin{vmatrix} 1-x & 3\\ 2 & 3-x \end{vmatrix}=2\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\begin{vmatrix} 1-x & 3\\ 2 & 3-x \end{vmatrix}&=2\\ \left ( 1-x \right )\left ( 3-x \right )-(3)(2)&=2\\ 3-x-3x+x^{2}-6&=2\\ x^{2}-4x-3&=2\\ x^{2}-4x-5&=0\\ \left ( x-5 \right )\left ( x+1 \right )&=0\\ x-5=0\: \: \textrm{atau}\: \:x+1&=0\\ x=5\: \: \textrm{atau}\: \: x=-1& \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui matriks-matriks persegi berikut}\\ &(\textrm{i}).\: \: \begin{pmatrix} 1 & 2&3\\ 2 & 4&5\\ 3&5&4 \end{pmatrix}\quad\quad\quad\quad\: \: \, (\textrm{iii}).\: \: \begin{pmatrix} 1 & 2&3\\ 4 & 5&6\\ 7&8&9 \end{pmatrix}\\ &(\textrm{ii}).\: \: \begin{pmatrix} -1 & -2&-3\\ -2 & 6&0\\ -3&0&6 \end{pmatrix}\quad\quad (\textrm{iv}).\: \: \begin{pmatrix} 2 & 1&1\\ 1 & 2&1\\ 1&1&2 \end{pmatrix}\\ &\\ &\textrm{Tentukanlah determinan matriks-matriks}\\ &\textrm{di atas dengan cara}\\ &\textrm{a}.\quad Sarrus\\ &\textrm{b}.\quad \textrm{Menjabarkan baris pertama}\\ &\textrm{c}.\quad \textrm{Menjabarkan baris kedua}\\ &\textrm{d}.\quad \textrm{Menjabarkan baris ketiga}\\ &\textrm{e}.\quad \textrm{Menjabarkan kolom pertama}\\ &\textrm{f}.\quad \textrm{Menjabarkan kolom kedua}\\ &\textrm{g}.\quad \textrm{Menjabarkan kolom ketiga} \end{array}$

$.\qquad\:  \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline (\textrm{i}).\quad \begin{pmatrix} 1 & 1&3\\ 2 & 4&5\\ 3&5&4 \end{pmatrix}&(\textrm{ii}).\quad \begin{pmatrix} -1 & -2&-3\\ -2 & 6&0\\ -3&0&6 \end{pmatrix}\\\hline (\textrm{iii}).\quad \begin{pmatrix} 1 & 2&3\\ 4 & 5&6\\ 7&8&9 \end{pmatrix}&(\textrm{iv}).\quad \begin{pmatrix} 2 & 1&1\\ 1 & 2&1\\ 1&1&2 \end{pmatrix}\\\hline \begin{aligned}(\textrm{i}).\quad &\begin{vmatrix} 1 & 1&3\\ 2 & 4&5\\ 3&5&4 \end{vmatrix}\\ &=(1)(4)(4)+\\ &\: \: \quad (1)(5)(3)+\\ &\: \: \quad (3)(2)(5)+\\ &\: \: \quad -(3)(4)(3)\\ &\: \: \quad -(5)(5)(1)\\ &\: \: \quad -(4)(2)(1)\\ &=16+15+30\\ &\: \: \: -36-25-8\\ &=-8 \\\end{aligned} &\begin{aligned}(\textrm{ii}).\quad &\begin{vmatrix} -1 & -2&-3\\ -2 & 6&0\\ -3&0&6 \end{vmatrix}\\ &=(-1)(6)(6)+\\ &\: \: \quad (-2)(0)(-3)+\\ &\: \: \quad (-3)(-2)(0)+\\ &\: \: \quad -(-3)(6)(-3)\\ &\: \: \quad -(0)(0)(-1)\\ &\: \: \quad -(6)(-2)(-2)\\ &=-36+0+0\\ &\: \: \: -54-0-24\\ &=-114 \\\end{aligned} \\\hline \begin{aligned}(\textrm{iii}).\quad &\begin{vmatrix} 1 & 2&3\\ 4 & 5&6\\ 7&8&9 \end{vmatrix}\\ &=(1)(5)(9)+\\ &\: \: \quad (2)(6)(7)+\\ &\: \: \quad (3)(4)(8)+\\ &\: \: \quad -(7)(5)(3)\\ &\: \: \quad -(8)(6)(1)\\ &\: \: \quad -(9)(4)(2)\\ &=45+84+96\\ &\: \: \: -105-48-72\\ &=0 \\\end{aligned} &\begin{aligned}(\textrm{iv}).\quad &\begin{vmatrix} 2 & 1&1\\ 1 & 2&1\\ 1&1&2 \end{vmatrix}\\ &=(2)(2)(2)+\\ &\: \: \quad (1)(1)(1)+\\ &\: \: \quad (1)(1)(1)+\\ &\: \: \quad -(1)(2)(1)\\ &\: \: \quad -(1)(1)(2)\\ &\: \: \quad -(2)(1)(1)\\ &=8+1+1\\ &\: \: \: -2-2-2\\ &=4 \\\end{aligned} \\\hline \end{array}\\ &\textrm{yang belum dibahas silahkan dibuat latihan}  \end{aligned}$

DAFTAR PUSTAKA

1. Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 2 Kelas 1 Semester 2. Jakarta: ERLANGGA

MATRIKS-CONTOH 3

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

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

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

$\begin{array}{ll}\\ 14.&\textrm{Diketahui matriks}\: \: \textrm{Z}=\begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}\\ & \textrm{dan}\: \: f(x)=x^{2}-x.\\ &\textrm{Jika}\: \: f(\textrm{Z})=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix},\\ & \textrm{maka nilai}\: \: p^{2}-q^{2}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&5\\ \textrm{b}.&7\\ \textrm{c}.&9\\ \textrm{d}.&12\\ \textrm{e}.&15 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}f(\textrm{Z})&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \color{red}\textrm{Z}^{2}-\textrm{Z}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}\times \begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}-\begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \begin{pmatrix} 4-18 & -12+30\\ 6-15 & -18+25 \end{pmatrix}-\begin{pmatrix} -2 & 6\\ -3 & 5 \end{pmatrix}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \begin{pmatrix} -12 & 12\\ -6 & 2 \end{pmatrix}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \end{aligned} \\ &\begin{aligned} \color{black}\textrm{Sehingga}&\\ -12&=-3p-8q\quad.................(1)\\ -1&=p+q\quad......................(2)\\ \textrm{persamaan}&\: (2)\: \: \textrm{ke persamaan}\: \: (1)\\ -12&=-3p-3q-5q=-3(p+q)-5q\\ -12&=-3(-1)-5q\\ -12&=3-5q\\ 5q&=3+12\\ q&=3\quad........................(3)\\ \textrm{persamaan}&\: \: (3)\: \: \textrm{ke persamaan}\: \: (2)\\ p+q&=-1\\ p&=-1-q=-1-3=-4\\ p^{2}-q^{2}&=(-4)^{2}-3^{2}=16-9\\ &=7 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&(\textbf{SBMPTN 2013})\\ &\textrm{Jika}\: \: A=\begin{pmatrix} 2 & -1 & 1\\ a & b & c \end{pmatrix},\\ &B=\begin{pmatrix} -2 & 1\\ 1 & -1\\ 0 & 2 \end{pmatrix}\: \: \textrm{dan}\\ &AB=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\textrm{maka nilai}\: \: 2c-a=\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&5\\ \textrm{e}.&6 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&AB=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\begin{pmatrix} 2 & -1 & 1\\ a & b & c \end{pmatrix}\begin{pmatrix} -2 & 1\\ 1 & -1\\ 0 & 2 \end{pmatrix}=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\begin{pmatrix} -5 & 5\\ -2a+b&a-b+2c \end{pmatrix}=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\begin{array}{lllll}\\ -2a+b&=3&\\ a-b+2c&=-3&+\\\hline \qquad \color{red}2c-a&=0 \end{array} \end{aligned} \end{array}$

 

DAFTAR PUSTAKA

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.