{"id":303,"date":"2023-07-06T14:52:33","date_gmt":"2023-07-06T14:52:33","guid":{"rendered":"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/"},"modified":"2023-07-06T14:52:33","modified_gmt":"2023-07-06T14:52:33","slug":"teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan","status":"publish","type":"post","link":"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/","title":{"rendered":"Teorema rouche \u2013 fr\u00e9benius"},"content":{"rendered":"<p>Di halaman ini kita akan mempelajari apa itu <strong>teorema Rouch\u00e9 Frobenius<\/strong> dan cara menghitung pangkat suatu matriks dengannya. Anda juga akan menemukan contoh dan latihan yang diselesaikan langkah demi langkah dengan teorema Rouch\u00e9-Frobenius.<\/p>\n<h2 class=\"wp-block-heading\"> Apa yang dimaksud dengan teorema Rouch\u00e9\u2013Frobenius?<\/h2>\n<p> <strong>Teorema Rouch\u00e9-Frobenius adalah metode untuk mengklasifikasikan sistem persamaan linear.<\/strong> Dengan kata lain, teorema Rouch\u00e9-Frobenius digunakan untuk mengetahui berapa banyak solusi yang dimiliki suatu sistem persamaan tanpa harus menyelesaikannya.<\/p>\n<p> Ada 3 jenis sistem persamaan:<\/p>\n<ul>\n<li> <strong>System Kompatibel Ditentukan (SCD):<\/strong> Sistem memiliki solusi unik.<\/li>\n<li> <strong>Sistem kompatibel tak tentu (ICS):<\/strong> sistem yang mempunyai solusi tak terhingga.<\/li>\n<li> <strong>Sistem Tidak Kompatibel (SI):<\/strong> Sistem tidak memiliki solusi.<\/li>\n<\/ul>\n<p> Selain itu, teorema Rouch\u00e9-Frobenius nantinya juga memungkinkan kita <a href=\"https:\/\/mathority.org\/id\/contoh-aturan-dan-latihan-cramer-yang-diselesaikan\/\">menyelesaikan sistem menggunakan aturan Cramer<\/a> .<\/p>\n<h2 class=\"wp-block-heading\"> Pernyataan teorema Rouch\u00e9-Frobenius<\/h2>\n<p> Teorema Rouch\u00e9-Frobenius mengatakan demikian<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bd767a13412c19de65e75a6826caee08_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{A}\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"13\" style=\"vertical-align: 0px;\"><\/p>\n<p> adalah matriks yang dibentuk oleh koefisien-koefisien yang tidak diketahui dari suatu sistem persamaan. dan perut<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bf22ad0d457d763be692e97f3bcdf221_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{A'}\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"17\" style=\"vertical-align: 0px;\"><\/p>\n<p> , atau <strong>matriks diperluas<\/strong> , adalah matriks yang dibentuk oleh koefisien-koefisien yang tidak diketahui dari suatu sistem persamaan dan suku-suku bebasnya: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><\/figure>\n<\/div>\n<div style=\"background-color:#dff6ff;padding-top: 20px; padding-bottom: 0.5px; padding-right: 40px; padding-left: 30px\" class=\"has-background\">\n<p align=\"LEFT\" style=\"margin-bottom:20px\"> <strong>Teorema Rouch\u00e9-Frobenius<\/strong> memungkinkan kita mengetahui jenis sistem persamaan apa yang kita hadapi berdasarkan pangkat matriks A dan A&#8217;:<\/p>\n<ul style=\"color:#E53935; font-weight: bold;\">\n<li style=\"margin-bottom:20px\"> <span style=\"color:#000000;font-weight: normal;\">Jika peringkat(A) = peringkat(A&#8217;) = jumlah yang tidak diketahui \u27f6 Sistem kompatibel yang ditentukan (SCD)<\/span><\/li>\n<li style=\"margin-bottom:20px;\"> <span style=\"color:#000000;font-weight: normal;\">Jika peringkat(A) = peringkat(A&#8217;) &lt; jumlah yang tidak diketahui \u27f6 Sistem kompatibel tak tentu (SCI)<\/span><\/li>\n<li> <span style=\"color:#000000;font-weight: normal;\">jika rentang(A)\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-be0e48d5500c7e73c450241ea2197789_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{\\neq}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"13\" style=\"vertical-align: -4px;\"><\/p>\n<p><\/span> rentang (A&#8217;) \u27f6 Sistem tidak kompatibel (SI)<\/li>\n<\/ul>\n<\/div>\n<p> Setelah kita mengetahui apa yang dikatakan teorema Rouch\u00e9-Frobenius, kita akan melihat cara menyelesaikan latihan teorema Rouch\u00e9-Frobenius. Berikut 3 contohnya: latihan yang diselesaikan menggunakan teorema setiap jenis sistem persamaan.<\/p>\n<h2 class=\"wp-block-heading\"> Contoh sistem kompatibel yang ditentukan (SCD)<\/h2>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b6b2f93c6308c25e8df2fbb5da2af9a8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 2x+y-3z=0 \\\\[1.5ex] x+2y-z= 1 \\\\[1.5ex] 4x-2y+z = 3\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"135\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong>Matriks A<\/strong> dan <strong>matriks perluasan A&#8217;<\/strong> dari sistem tersebut adalah:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4597f5171b586bbcf0915d8512f7b89d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; -3  \\\\[1.1ex] 1 &amp; 2 &amp; -1  \\\\[1.1ex] 4 &amp; -2 &amp; 1  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 1 &amp; 2 &amp; -1 &amp; 1  \\\\[1.1ex] 4 &amp; -2 &amp; 1 &amp; 3\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Sekarang kita menghitung rank matriks A. Untuk melakukannya, kita periksa apakah determinan seluruh matriks berbeda dari 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6c95b7158a2e6401cd16aeb708f128ff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; -3  \\\\[1.1ex] 1 &amp; 2 &amp; -1  \\\\[1.1ex] 4 &amp; -2 &amp; 1  \\end{vmatrix} = 25 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"219\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Karena matriks mempunyai determinan 3\u00d73 yang berbeda dengan 0, <strong>maka matriks A mempunyai rank 3:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Setelah kita mengetahui pangkat A, kita hitung pangkat A&#8217;, yang mana paling tidak pangkatnya adalah 3 karena kita baru saja melihat bahwa di dalamnya terdapat determinan berorde 3 yang berbeda dengan 0. Selain itu, tidak mungkin pangkatnya 4, karena kita tidak dapat membuat determinan berorde 4. Oleh karena itu, <strong>matriks A&#8217; juga berperingkat 3:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Jadi, karena pangkat dari matriks A sama dengan pangkat dari matriks A&#8217; dan dengan banyaknya sistem yang tidak diketahui (3), kita mengetahui melalui teorema Rouch\u00e9 Frobenius bahwa ini adalah <strong>Sistem Penentuan Kompatibel<\/strong> (SCD) :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-557185e16670c72d23eec5a3ea13b487_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 3 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> Contoh sistem kompatibel tak tentu (ICS)<\/h2>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2360b9a47257f73cf3f5dea63fb24098_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x-y+2z=1 \\\\[1.5ex] 3x+2y+z= 5 \\\\[1.5ex] 2x+3y-z = 4\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"135\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong>Matriks A<\/strong> dan <strong>matriks perluasan A&#8217;<\/strong> dari sistem tersebut adalah:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b281235e2702433b447e2586ae3092c9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; -1 &amp; 2  \\\\[1.1ex] 3 &amp; 2 &amp; 1  \\\\[1.1ex] 2 &amp; 3 &amp; -1  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; -1 &amp; 2 &amp; 1 \\\\[1.1ex] 3 &amp; 2 &amp; 1 &amp; 5  \\\\[1.1ex] 2 &amp; 3 &amp; -1 &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Sekarang kita menghitung rank matriks A. Untuk melakukannya, kita periksa apakah determinan seluruh matriks berbeda dari 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-74cafc27ab41134696c3bf263132b98b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; -1 &amp; 2  \\\\[1.1ex] 3 &amp; 2 &amp; 1  \\\\[1.1ex] 2 &amp; 3 &amp; -1 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Penentu seluruh matriks A menghasilkan 0, sehingga tidak berpangkat 3. Untuk mengetahui apakah matriks tersebut berpangkat 2, kita harus mencari submatriks di A yang determinannya berbeda dengan 0. Misalnya dari pojok kiri atas :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-22b2487f7664a70c116593120de2743b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; -1  \\\\[1.1ex] 3 &amp; 2 \\end{vmatrix} = 5 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"123\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Karena matriks mempunyai determinan 2\u00d72 yang berbeda dengan 0, <strong>maka matriks A mempunyai rank 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Setelah kita mengetahui pangkat A, kita menghitung pangkat A&#8217;. Kita sudah tahu bahwa determinan dari 3 kolom pertama menghasilkan 0, jadi kita coba kemungkinan determinan 3\u00d73 lainnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-17f264ad3859da88ffa6784be24e4143_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}1 &amp; -1 &amp;  1 \\\\[1.1ex] 3 &amp; 2 &amp; 5  \\\\[1.1ex] 2 &amp; 3 &amp; 4\\end{vmatrix} = 0 \\quad \\begin{vmatrix}1 &amp; 2 &amp; 1 \\\\[1.1ex] 3 &amp;  1 &amp; 5  \\\\[1.1ex] 2 &amp; -1 &amp; 4\\end{vmatrix} = 0 \\quad \\begin{vmatrix} -1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 1 &amp; 5  \\\\[1.1ex] 3 &amp; -1 &amp; 4\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Semua determinan matriks A&#8217; berukuran 3\u00d73 adalah 0, sehingga matriks A&#8217; juga tidak menduduki rangking 3. Namun di dalamnya memang terdapat determinan orde 2 yang berbeda dengan 0. Contoh:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-22b2487f7664a70c116593120de2743b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; -1  \\\\[1.1ex] 3 &amp; 2 \\end{vmatrix} = 5 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"123\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Jadi <strong>matriks A&#8217; akan mempunyai rangking 2<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Luas matriks A sama dengan luas matriks A&#8217; tetapi luasnya lebih kecil dari jumlah yang tidak diketahui dalam sistem (3). Oleh karena itu, menurut teorema Rouch\u00e9-Frobenius, ini adalah <strong>sistem kompatibel tak tentu<\/strong> (ICS):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> Contoh sistem yang tidak kompatibel (IS)<\/h2>\n<div class=\"adsb30\" style=\" margin:px; text-align:\"><\/div>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-30e1084dd637eb4371f6b2218af24136_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 2x+y-2z=3 \\\\[1.5ex] 3x-2y+z= 2 \\\\[1.5ex] x+4-5z = 3 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"135\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p><strong>Matriks A<\/strong> dan <strong>matriks perluasan A&#8217;<\/strong> dari sistem tersebut adalah:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b435d86f1466af5748d91e6c9bd813e3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; -2 \\\\[1.1ex] 3 &amp; -2 &amp; 1 \\\\[1.1ex] 1 &amp; 4 &amp; -5 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; -2 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; 1 &amp; 2  \\\\[1.1ex] 1 &amp; 4 &amp; -5 &amp; 3 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Sekarang kita menghitung rank matriks A. Untuk melakukannya, kita periksa apakah determinan seluruh matriks berbeda dari 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-714538c91aa2620a6adb40581245f0e0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; -2 \\\\[1.1ex] 3 &amp; -2 &amp; 1 \\\\[1.1ex] 1 &amp; 4 &amp; -5 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Penentu seluruh matriks A menghasilkan 0, sehingga tidak berpangkat 3. Untuk mengetahui apakah matriks tersebut berpangkat 2, kita harus mencari submatriks di A yang determinannya berbeda dengan 0. Misalnya dari pojok kiri atas :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5a46decda8fd850d9c847922b0c896db_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 1  \\\\[1.1ex] 3 &amp; -2 \\end{vmatrix} = -7 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"136\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Karena matriks mempunyai determinan orde 2 yang berbeda dengan 0, <strong>maka matriks A mempunyai rank 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Setelah kita mengetahui pangkat A, kita menghitung pangkat A&#8217;. Kita sudah tahu bahwa determinan dari 3 kolom pertama menghasilkan 0, jadi sekarang kita coba, misalnya, dengan determinan dari 3 kolom terakhir:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-47aecdf801b92f21f2287fb96eaaa3f8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; -2 &amp; 3 \\\\[1.1ex]  -2 &amp; 1 &amp; 2  \\\\[1.1ex]  4 &amp; -5 &amp; 3 \\end{vmatrix} = 3 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"161\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Sebaliknya matriks A&#8217; memang mengandung determinan yang hasilnya berbeda dengan 0, sehingga <strong>matriks A&#8217; mempunyai rank 3<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Oleh karena itu, karena pangkat matriks A lebih kecil daripada pangkat matriks A&#8217;, berdasarkan teorema Rouch\u00e9-Frobenius kita menyimpulkan bahwa matriks tersebut merupakan <strong>Sistem Tidak Kompatibel<\/strong> (SI) <strong>:<\/strong> <\/p>\n<p class=\"has-text-align-center\">\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c3da0513f318d25473e93ba88c51fb42_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = 2 \\ \\neq \\ rg(A') = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\">Memecahkan masalah teorema Rouch\u00e9 \u2013 Frobenius <\/h2>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-118\"><\/div>\n<\/div>\n<h3 class=\"estil_titol_H3 wp-block-heading\"> Latihan 1<\/h3>\n<p> Tentukan jenis sistem persamaan berikut dengan 3 yang tidak diketahui menggunakan teorema Rouch\u00e9-Frobenius: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-theoreme-de-rouche-8211-frebenius-1.webp\" alt=\"Latihan teorema Rouche - frobenius\" class=\"wp-image-3984\" width=\"193\" height=\"122\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Pertama-tama kita buat matriks A dan matriks perluasan A&#8217; dari sistem:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-951ce5c1f0c606d4f060a1de58b60303_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -1 &amp; -1 \\\\[1.1ex] -2 &amp; 4 &amp; 2 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 3 &amp; -1 &amp; -1 &amp; 2 \\\\[1.1ex] -2 &amp; 4 &amp; 2 &amp; 8 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"432\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sekarang kita harus mencari rank matriks A. Untuk melakukannya, kita periksa apakah determinan matriks tersebut berbeda dari 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-15cddb69f7590648d1d6ae61d942471e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -1 &amp; -1 \\\\[1.1ex] -2 &amp; 4 &amp; 2 \\end{vmatrix} = -4+2-36+6+8-6=-30 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"450\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Matriks yang mempunyai determinan orde ketiga selain 0, <strong>matriks A mempunyai rangking 3:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Setelah kita mengetahui pangkat A, kita menghitung pangkat A&#8217;. Ini setidaknya akan berada pada peringkat 3, karena kita baru saja melihat bahwa di dalamnya terdapat determinan berorde 3 yang berbeda dari 0. Selain itu, ia tidak mungkin berada pada peringkat 4, karena kita tidak dapat tidak membuat determinan 4\u00d74. Oleh karena itu, <strong>matriks A&#8217; juga mempunyai rangking 3:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, berkat teorema Rouch\u00e9-Frobenius, kita mengetahui bahwa ini adalah <strong>sistem yang kompatibel dengan determinasi<\/strong> (SCD), karena jangkauan A sama dengan jangkauan A&#8217; dan jumlah yang tidak diketahui. <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-31b495a48a75d7af1f23e38818bf4eca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 3 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Latihan 2<\/h3>\n<p> Klasifikasikan sistem persamaan berikut dengan 3 hal yang tidak diketahui menggunakan teorema Rouch\u00e9-Frobenius: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-theoreme-de-rouche-8211-frebenius-2.webp\" alt=\"Latihan terpecahkan dari teorema Rouche-Frobenius\" class=\"wp-image-3987\" width=\"190\" height=\"121\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Pertama-tama kita buat matriks A dan matriks perluasan A&#8217; dari sistem:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-45e13aabe233ece927df7c9ba0bb3ec1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}3 &amp; -1 &amp; 2  \\\\[1.1ex] 1 &amp; 2 &amp; -2  \\\\[1.1ex] 1 &amp; -5 &amp; 6 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 3 &amp; -1 &amp; 2 &amp; 1 \\\\[1.1ex] 1 &amp; 2 &amp; -2 &amp; 5 \\\\[1.1ex] 1 &amp; -5 &amp; 6 &amp; -9 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sekarang mari kita hitung jangkauan matriks A: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-87bc95df0033834bba0398b8421faac5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; -1 &amp; 2 \\\\[1.1ex] 1 &amp; 2 &amp; -2 \\\\[1.1ex] 1 &amp; -5 &amp; 6 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b9805283b75e2b89f67c7865a1263112_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; -1  \\\\[1.1ex] 1 &amp; 2 \\end{vmatrix} = 7 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"123\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi <strong>matriks A mempunyai rangking 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Setelah kita mengetahui pangkat A, kita menghitung pangkat A&#8217;. Kita sudah tahu bahwa determinan dari 3 kolom pertama menghasilkan 0, jadi kita coba kemungkinan determinan 3\u00d73 lainnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e6457fe3f03722b7f0d955191f318915_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}-1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; -2 &amp; 5 \\\\[1.1ex] -5 &amp; 6 &amp; -9\\end{vmatrix} = 0 \\quad \\begin{vmatrix}3 &amp; 2 &amp; 1 \\\\[1.1ex] 1 &amp; -2 &amp; 5 \\\\[1.1ex] 1 &amp; 6 &amp; -9\\end{vmatrix} = 0 \\quad \\begin{vmatrix} 3 &amp; -1 &amp; 1 \\\\[1.1ex] 1 &amp; 2 &amp; 5 \\\\[1.1ex] 1 &amp; -5 &amp; -9\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"446\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Semua determinan matriks A&#8217; berukuran 3\u00d73 adalah 0, sehingga matriks A&#8217; juga tidak menduduki rangking 3. Namun di dalamnya terdapat banyak determinan orde 2 yang berbeda dengan 0. Contoh:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eafa4747802fae3f0c36350357abbeb2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -1 &amp; 2  \\\\[1.1ex] 2 &amp; -2 \\end{vmatrix} = -2 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"150\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi <strong>matriks A&#8217; akan mempunyai rangking 2<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Pangkat matriks A sama dengan pangkat matriks A&#8217; tetapi keduanya lebih kecil dari jumlah yang tidak diketahui pada sistem (3). Oleh karena itu, berdasarkan teorema Rouch\u00e9-Frobenius kita mengetahui bahwa ini adalah <strong>sistem kompatibel tak tentu<\/strong> (ICS): <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Latihan 3<\/h3>\n<p> Tentukan jenis sistem sistem persamaan berikut menggunakan teorema Rouch\u00e9-Frobenius: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-theoreme-de-rouche-8211-frebenius-3.webp\" alt=\"latihan diselesaikan langkah demi langkah teorema rouche - frobenius\" class=\"wp-image-3990\" width=\"188\" height=\"122\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Pertama-tama kita buat matriks A dan matriks perluasan A&#8217; dari sistem:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1820d31e4fd5c79804c9b6fa15abb469_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 4 &amp; -2 \\\\[1.1ex] 3 &amp; -1 &amp; 3  \\\\[1.1ex] 5 &amp; 7 &amp; -1 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 4 &amp; -2 &amp; 3 \\\\[1.1ex] 3 &amp; -1 &amp; 3 &amp; -2 \\\\[1.1ex] 5 &amp; 7 &amp; -1 &amp; 0 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sekarang mari kita hitung jangkauan matriks A: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4f998260ee4c96673085ea6fd4ca87ba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 4 &amp; -2 \\\\[1.1ex] 3 &amp; -1 &amp; 3 \\\\[1.1ex] 5 &amp; 7 &amp; -1\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-159a1c58fdcd972b4b08e4795950e064_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 4  \\\\[1.1ex] 3 &amp; -1 \\end{vmatrix} = -13 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi <strong>matriks A mempunyai rangking 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Setelah kita mengetahui pangkat A, kita menghitung pangkat A&#8217;. Kita telah mengetahui bahwa determinan dari 3 kolom pertama menghasilkan 0, tetapi bukan determinan dari 3 kolom terakhir:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c673a5bbbd41933208169fa3e08b7c62_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 4 &amp; -2 &amp; 3 \\\\[1.1ex]-1 &amp; 3 &amp; -2 \\\\[1.1ex] 7 &amp; -1 &amp; 0 \\end{vmatrix} = -40 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"198\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Oleh karena itu, <strong>matriks A&#8217; mempunyai rangking 3<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Pangkat matriks A lebih kecil dari pangkat matriks A&#8217;, oleh karena itu kita dapat menyimpulkan dari teorema Rouch\u00e9-Frobenius bahwa ini merupakan <strong>Sistem Tidak Kompatibel<\/strong> (SI) <strong>:<\/strong> <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c3da0513f318d25473e93ba88c51fb42_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = 2 \\ \\neq \\ rg(A') = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-119\"><\/div>\n<\/div>\n<h3 class=\"wp-block-heading\"> Latihan 4<\/h3>\n<p> Tentukan jenis sistem persamaan berikut dengan 3 yang tidak diketahui menggunakan teorema Rouch\u00e9-Frobenius: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-theoreme-de-rouche-8211-frobenius-3-inconnues-3-equations.webp\" alt=\"Rouche - Teorema Frobenius menyelesaikan latihan dengan 3 hal yang tidak diketahui dan 3 persamaan\" class=\"wp-image-3991\" width=\"203\" height=\"122\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Pertama-tama kita buat matriks A dan matriks perluasan A&#8217; dari sistem:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f8a0454c53a64f612c689ba1dae1196b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 5 &amp; -3 &amp; -2  \\\\[1.1ex] 1 &amp; 4 &amp; 1  \\\\[1.1ex]-3 &amp; 2 &amp; -2  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 5 &amp; -3 &amp; -2 &amp; -2 \\\\[1.1ex] 1 &amp; 4 &amp; 1 &amp; 7 \\\\[1.1ex]-3 &amp; 2 &amp; -2 &amp; 3 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"446\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sekarang kita harus menghitung rank matriks A. Untuk melakukannya, kita menyelesaikan determinan matriks dengan aturan Sarrus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-420f0d1ee000f39cbfbce88bf122f413_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 5 &amp; -3 &amp; -2 \\\\[1.1ex] 1 &amp; 4 &amp; 1 \\\\[1.1ex]-3 &amp; 2 &amp; -2 \\end{vmatrix} = -40+9-4-24-10-6=-75 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"467\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Matriks yang mempunyai determinan orde ketiga selain 0, <strong>matriks A mempunyai rangking 3:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Oleh karena itu, <strong>matriks A&#8217; juga mempunyai rangking 3<\/strong> , karena matriks tersebut paling sedikit selalu mempunyai rangking A dan matriks tersebut tidak mungkin mempunyai rangking 4 karena kita tidak dapat menyelesaikan determinan 4\u00d74 apa pun.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, berkat penerapan teorema Rouch\u00e9-Frobenius, kita mengetahui bahwa sistem tersebut merupakan <strong>Sistem Penentuan Kompatibel<\/strong> (SCD), karena jangkauan A sama dengan jangkauan A&#8217; dan jumlah yang tidak diketahui. <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-31b495a48a75d7af1f23e38818bf4eca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 3 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Latihan 5<\/h3>\n<p> Identifikasi jenis sistem sistem persamaan berikut menggunakan teorema Rouch\u00e9-Frobenius: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exemple-du-theoreme-de-rouche-8211-frebenius.webp\" alt=\"contoh teorema rouche - frobenius\" class=\"wp-image-3992\" width=\"205\" height=\"122\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Pertama-tama kita buat matriks A dan matriks perluasan A&#8217; dari sistem:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3211e276b2b040969c38bc6c69eabd52_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 &amp; 3 \\\\[1.1ex] -1 &amp; 7 &amp; 3 \\\\[1.1ex] -5 &amp; 8 &amp; 0 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 4 &amp; -1 &amp; 3 &amp; 5 \\\\[1.1ex] -1 &amp; 7 &amp; 3 &amp; -3 \\\\[1.1ex] -5 &amp; 8 &amp; 0 &amp; 9 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sekarang mari kita hitung jangkauan matriks A: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-21004095a3a8ef3edfc15bed5c7853a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 4 &amp; -1 &amp; 3 \\\\[1.1ex] -1 &amp; 7 &amp; 3 \\\\[1.1ex] -5 &amp; 8 &amp; 0\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a58059046b56cf1f8d82c6c8939e44ca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 4 &amp; -1  \\\\[1.1ex]  -1 &amp; 7 \\end{vmatrix} = 27 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Oleh karena itu, matriks A berada pada peringkat 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Setelah kita mengetahui pangkat A, kita menghitung pangkat A&#8217;. Penentu 3 kolom pertama yang kita ketahui menghasilkan 0, tetapi determinan 3 kolom terakhir tidak memberikan:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-992718d3b50aedf77c80c262fad5845f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -1 &amp; 3 &amp; 5 \\\\[1.1ex]  7 &amp; 3 &amp; -3 \\\\[1.1ex] 8 &amp; 0 &amp; 9\\end{vmatrix} = -408 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"193\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Oleh karena itu, <strong>matriks A&#8217; mempunyai rangking 3<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan terakhir, kita terapkan domain tersebut pada teorema Rouch\u00e9-Frobenius: domain matriks A lebih kecil dari domain matriks A&#8217;, oleh karena itu merupakan <strong>Sistem Tidak Kompatibel<\/strong> (SI) <strong>:<\/strong> <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c3da0513f318d25473e93ba88c51fb42_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = 2 \\ \\neq \\ rg(A') = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Latihan 6<\/h3>\n<p> Klasifikasikan sistem persamaan orde 3 berikut dengan teorema Rouch\u00e9-Frobenius: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d45e8bc425b08e403a98e01693201681_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 6x-2y+4z=1 \\\\[1.5ex] -2x+4y+3z= 7 \\\\[1.5ex] 8x-6y+z = -6\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"158\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Pertama-tama kita buat matriks A dan matriks perluasan A&#8217; dari sistem:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8e779eca9135adc44e4a3a55f368560f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 6 &amp; -2 &amp; 4 \\\\[1.1ex] -2 &amp; 4 &amp; 3 \\\\[1.1ex] 8 &amp; -6 &amp; 1  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 6 &amp; -2 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; 4 &amp; 3 &amp; 7 \\\\[1.1ex] 8 &amp; -6 &amp; 1 &amp; -6 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sekarang mari kita hitung jangkauan matriks A: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c2f63f79858eae462547cf2f270fc780_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 6 &amp; -2 &amp; 4 \\\\[1.1ex] -2 &amp; 4 &amp; 3 \\\\[1.1ex] 8 &amp; -6 &amp; 1 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e5fa293b94b8c6acfd998f1e154abf7a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 6 &amp; -2  \\\\[1.1ex] -2 &amp; 4 \\end{vmatrix} = 20  \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi <strong>matriks A mempunyai rangking 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Setelah kita mengetahui pangkat A, kita menghitung pangkat A&#8217;. Kita sudah tahu bahwa determinan dari 3 kolom pertama menghasilkan 0, jadi kita coba kemungkinan determinan 3\u00d73 lainnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-98958f866454a1bf9f1ac078562065cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -2 &amp; 4 &amp; 1 \\\\[1.1ex]4 &amp; 3 &amp; 7 \\\\[1.1ex] -6 &amp; 1 &amp; -6\\end{vmatrix} = 0 \\quad \\begin{vmatrix}6 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; 3 &amp; 7 \\\\[1.1ex] 8 &amp;  1 &amp; -6\\end{vmatrix} = 0 \\quad \\begin{vmatrix} 6 &amp; -2 &amp; 1 \\\\[1.1ex] -2 &amp; 4 &amp; 7 \\\\[1.1ex] 8 &amp; -6 &amp; -6\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"446\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Semua determinan matriks A&#8217; berukuran 3\u00d73 adalah 0, sehingga matriks A&#8217; juga tidak menduduki rangking 3. Namun di dalamnya memang terdapat determinan orde 2 yang berbeda dengan 0. Contoh:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-58091f1a37a4ef81fdf56f01dd9531a3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 6 &amp; -2 \\\\[1.1ex] -2 &amp; 4 \\end{vmatrix} = 20 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi <strong>matriks A&#8217; akan mempunyai rangking 2<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Terakhir, dengan menerapkan teorema Rouch\u00e9-Frobenius, kita mengetahui bahwa ini adalah <strong>Sistem Kompatibel Tak tentu<\/strong> (ICS), karena jangkauan matriks A sama dengan jangkauan matriks A&#8217; tetapi keduanya lebih kecil dari jumlah yang tidak diketahui di dalam sistem. sistem(3): <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Di halaman ini kita akan mempelajari apa itu teorema Rouch\u00e9 Frobenius dan cara menghitung pangkat suatu matriks dengannya. Anda juga akan menemukan contoh dan latihan yang diselesaikan langkah demi langkah dengan teorema Rouch\u00e9-Frobenius. Apa yang dimaksud dengan teorema Rouch\u00e9\u2013Frobenius? Teorema Rouch\u00e9-Frobenius adalah metode untuk mengklasifikasikan sistem persamaan linear. Dengan kata lain, teorema Rouch\u00e9-Frobenius digunakan untuk &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/\"> <span class=\"screen-reader-text\">Teorema rouche \u2013 fr\u00e9benius<\/span> Selengkapnya &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[51],"tags":[],"class_list":["post-303","post","type-post","status-publish","format-standard","hentry","category-sistem-pendidikan"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Teorema Rouche \u2013 Fr\u00e9benius - Mathority<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/\" \/>\n<meta property=\"og:locale\" content=\"id_ID\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Teorema Rouche \u2013 Fr\u00e9benius - Mathority\" \/>\n<meta property=\"og:description\" content=\"Di halaman ini kita akan mempelajari apa itu teorema Rouch\u00e9 Frobenius dan cara menghitung pangkat suatu matriks dengannya. 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Anda juga akan menemukan contoh dan latihan yang diselesaikan langkah demi langkah dengan teorema Rouch\u00e9-Frobenius. Apa yang dimaksud dengan teorema Rouch\u00e9\u2013Frobenius? Teorema Rouch\u00e9-Frobenius adalah metode untuk mengklasifikasikan sistem persamaan linear. Dengan kata lain, teorema Rouch\u00e9-Frobenius digunakan untuk &hellip; Teorema rouche \u2013 fr\u00e9benius Selengkapnya &raquo;","og_url":"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/","article_published_time":"2023-07-06T14:52:33+00:00","og_image":[{"url":"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bd767a13412c19de65e75a6826caee08_l3.png"}],"author":"Tim Mathority","twitter_card":"summary_large_image","twitter_misc":{"Ditulis oleh":"Tim Mathority","Estimasi waktu membaca":"7 menit"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/#article","isPartOf":{"@id":"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/"},"author":{"name":"Tim Mathority","@id":"https:\/\/mathority.org\/id\/#\/schema\/person\/ea4523caf53a07e2ebf32e306a925b38"},"headline":"Teorema rouche \u2013 fr\u00e9benius","datePublished":"2023-07-06T14:52:33+00:00","dateModified":"2023-07-06T14:52:33+00:00","mainEntityOfPage":{"@id":"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/"},"wordCount":1375,"commentCount":0,"publisher":{"@id":"https:\/\/mathority.org\/id\/#organization"},"articleSection":["Sistem pendidikan"],"inLanguage":"id","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/","url":"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/","name":"Teorema Rouche \u2013 Fr\u00e9benius - 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