{"id":306,"date":"2023-07-06T14:08:28","date_gmt":"2023-07-06T14:08:28","guid":{"rendered":"https:\/\/mathority.org\/id\/pembahasan-sistem-persamaan-dengan-parameter\/"},"modified":"2023-07-06T14:08:28","modified_gmt":"2023-07-06T14:08:28","slug":"pembahasan-sistem-persamaan-dengan-parameter","status":"publish","type":"post","link":"https:\/\/mathority.org\/id\/pembahasan-sistem-persamaan-dengan-parameter\/","title":{"rendered":"Pembahasan sistem persamaan dengan parameter"},"content":{"rendered":"<p>Pada halaman ini kita akan melihat bagaimana membahas dan menyelesaikan <strong>sistem persamaan dengan parameter<\/strong> . Selain itu, Anda akan menemukan contoh dan latihan penyelesaian sistem persamaan linear untuk dipraktikkan.<\/p>\n<p> Di sisi lain, untuk menganalisis sistem persamaan linier, penting bagi Anda untuk mengetahui <a href=\"https:\/\/mathority.org\/id\/contoh-aturan-dan-latihan-cramer-yang-diselesaikan\/\">apa itu aturan Cramer<\/a> dan <a href=\"https:\/\/mathority.org\/id\/teorema-de-rouche-frobenius-dengan-contoh-dan-latihan-yang-diselesaikan\/\">apa itu teorema Rouch\u00e9\u2013Frobenius<\/a> , karena kita akan menggunakannya terus-menerus.<\/p>\n<h2 class=\"wp-block-heading\"> Contoh sistem persamaan linear dengan parameter<\/h2>\n<ul>\n<li> Diskusikan dan selesaikan sistem persamaan berikut dengan parameter <em>m<\/em> :<\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6ab2286d15c20029b98a5ea4622033d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+y+2z= 2 \\\\[1.5ex] -x+my+2z=0 \\\\[1.5ex] 3x+mz = 4\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"155\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> 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-bef8e6b26595703c77c65178cbf90ffc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; m \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; m &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Sekarang kita selesaikan determinan A menggunakan aturan Sarrus, untuk melihat berapa rank matriksnya:<\/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-e823f83f25f798bd854612a7352680d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\begin{aligned}  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; m \\end{vmatrix} &amp; =m^2+6+0-6m-0+m \\\\ &amp; = m^2-5m+6 \\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"110\" width=\"366\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Jadi hasil determinan A bergantung pada nilai <em>m<\/em> . Oleh karena itu, kita akan melihat nilai <em>m<\/em> mana yang determinannya hilang. Untuk melakukan ini, <strong>kami menetapkan hasilnya sama dengan 0<\/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-a2101d9a1d88f3e3d29bb4758d2fb8a6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   m^2-5m+6 = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"133\" style=\"vertical-align: -2px;\"><\/p>\n<\/p>\n<p> Dan kita selesaikan persamaan kuadrat tersebut dengan rumus:<\/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-51910dd69f9df6fdde6dd7597f889700_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  m = \\cfrac{-b \\pm \\sqrt{b^2-4ac}}{2a}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"170\" style=\"vertical-align: -12px;\"><\/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-e863082ac1f9b43df4de9fe93f5eb305_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  m = \\cfrac{-(-5) \\pm \\sqrt{(-5)^2-4\\cdot 1 \\cdot 6}}{2 \\cdot 1} = \\cfrac{5 \\pm \\sqrt{25-24}}{2} =\\cfrac{5 \\pm 1}{2} = \\begin{cases} \\bm{m = 3} \\\\[2ex] \\bm{m =2} \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"537\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Jadi jika <em>m<\/em> sama dengan 2 atau 3 maka determinan A akan sama dengan 0. Dan jika <em>m<\/em> berbeda dengan 2 dan berbeda dengan 3 maka determinan A akan berbeda dengan 0.<\/p>\n<p> Oleh karena itu kita harus menganalisis setiap kasus secara terpisah:<\/p>\n<p style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m\u22603 dan m\u22602:<\/strong><\/span><\/p>\n<p> Seperti yang baru saja kita lihat, jika parameter <em>m<\/em> berbeda dengan 2 dan 3, determinan matriks A berbeda dengan 0. Oleh karena itu, <strong>rank A adalah 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> Selain itu, <strong>rank matriks A&#8217; juga 3<\/strong> , karena di dalamnya terdapat submatriks 3\u00d73 yang determinannya berbeda dengan 0. Dan tidak mungkin berpangkat 4 karena &#8216;kita tidak dapat membuat determinan 4\u00d74.<\/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> Kemudian, karena pangkat dari matriks A sama dengan pangkat dari matriks A&#8217; dan dengan banyaknya sistem yang tidak diketahui (3), berdasarkan <strong>teorema Rouch\u00e9-Frobenius<\/strong> kita mengetahui bahwa matriks tersebut adalah <strong>Depended System 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<p> Setelah kita mengetahui bahwa sistem tersebut adalah Sistem yang Ditentukan Kompatibel (DCS), kita menerapkan <strong>aturan Cramer<\/strong> untuk menyelesaikannya. Untuk melakukannya, ingatlah bahwa matriks A, determinannya, dan matriks A&#8217; 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-bef8e6b26595703c77c65178cbf90ffc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; m \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; m &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"404\" 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-aac47361358555f733a42cffecabdbe9_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] -1 &amp; m &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; m \\end{vmatrix} = m^2-5m+6\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"268\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Untuk menghitung x dengan aturan Cramer, kita ubah kolom pertama determinan matriks A menjadi kolom suku bebas dan membaginya dengan determinan 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-b17f49436fdadbb014011b5c461a4a56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\bm{x} = \\cfrac{\\begin{vmatrix} 2 &amp; 1 &amp; 2\\\\[1.1ex]0&amp;m&amp;2 \\\\[1.1ex] 4 &amp; 0 &amp; m \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{2m^2+8-8m}{m^2-5m+6}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"257\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Untuk menghitung y dengan aturan Cramer, kita ubah kolom kedua determinan A menjadi kolom suku bebas dan membaginya dengan determinan 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-6a2bf75bdabfb2c83870f1869ce19e3d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 0 &amp; 2 \\\\[1.1ex] 3 &amp; 4 &amp; m \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}}=\\cfrac{-4+2m}{m^2-5m+6}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"255\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Untuk menghitung z dengan aturan Cramer, kita ubah kolom ketiga determinan A menjadi kolom suku bebas dan membaginya dengan determinan 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-eebb3c4d280afc8a9aed8877ddcd4ac5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix}  1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-2m+4}{m^2-5m+6}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"254\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Oleh karena itu, penyelesaian sistem persamaan untuk kasus m\u22603 dan m\u22602 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-7cba73a14f41d3314575e075f1229e87_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =} \\cfrac{\\bm{2m^2+8-8m}}{\\bm{m^2-5m+6}} \\qquad \\bm{y=} \\cfrac{\\bm{-4+2m}}{\\bm{m^2-5m+6}} \\qquad \\bm{z =} \\cfrac{\\bm{-2m+4}}{\\bm{m^2-5m+6}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"470\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p> Seperti yang Anda lihat, dalam hal ini solusi sistem persamaan adalah fungsi dari m.<\/p>\n<p> Setelah kita menemukan solusi ketika m berbeda dari 2 dan 3, kita akan menyelesaikan sistem ketika m sama dengan 2:<\/p>\n<div class=\"adsb30\" style=\" margin:px; text-align:\"><\/div>\n<p style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=2:<\/strong><\/span><\/p>\n<p> Sekarang kita akan menganalisis sistem jika parameter <em>m<\/em> sama dengan 2. Dalam hal ini matriks A dan A&#8217; 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-f42ec4801f3e84cd44b4e0b2ae6351cf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; 2 &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; 2 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 2 &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; 2 &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Seperti yang kita lihat sebelumnya, ketika m=2 determinan A adalah 0. Oleh karena itu, matriks A tidak berpangkat 3. Namun di dalamnya terdapat 2\u00d72 determinan selain 0, misalnya:<\/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-55ef6cd148fca7a869e14760007e1f2e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}1 &amp; 1 \\\\[1.1ex] -1 &amp; 2  \\end{vmatrix} = 2 - (-1)=3 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Jadi, dalam hal ini <strong>pangkat A adalah 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 rank matriks A, kita menghitung rank A&#8217;. Penentu dari 3 kolom pertama menghasilkan 0, jadi kita coba kemungkinan determinan 3\u00d73 lainnya pada matriks A&#8217;:<\/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-2c68c742cae37c52ad2566b7feec5301_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 &amp; 2 \\\\[1.1ex] 2 &amp; 2 &amp; 0 \\\\[1.1ex] 0 &amp; 2 &amp; 4 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 2 &amp; 4 \\end{vmatrix}=0\\qquad \\begin{vmatrix} 1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; 4\\end{vmatrix}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"412\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Semua determinan yang mungkin untuk dimensi 3\u00d73 menghasilkan 0. Namun, jelas bahwa matriks A&#8217; mempunyai determinan non-0 2\u00d72 yang sama dengan 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-55ef6cd148fca7a869e14760007e1f2e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}1 &amp; 1 \\\\[1.1ex] -1 &amp; 2  \\end{vmatrix} = 2 - (-1)=3 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Oleh karena itu, <strong>matriks A&#8217; juga 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> Jadi, karena pangkat matriks A sama dengan pangkat matriks A&#8217; tetapi keduanya lebih kecil dari jumlah sistem yang tidak diketahui (3), kita mengetahui melalui <strong>teorema Rouch\u00e9-Frobenius<\/strong> bahwa ini adalah <strong>Sistem yang Kompatibel Tak Pasti<\/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<p> Karena ini adalah ICS, kita perlu mengubah sistem untuk mengatasinya. Untuk melakukan ini, pertama-tama kita harus menghilangkan persamaan dari sistem, dalam hal ini <strong>kita akan menghapus persamaan terakhir:<\/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-10c7facda35cb8894e6bbb236e4953f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+y+2z= 2 \\\\[1.5ex] -x+2y+2z=0 \\\\[1.5ex] \\cancel{3x+2z = 4} \\end{cases} \\longrightarrow \\quad \\begin{cases}  x+y+2z= 2 \\\\[1.5ex] -x+2y+2z=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong>Sekarang mari kita ubah variabel z menjadi \u03bb:<\/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-0155083595420da31a486927e953805c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}x+y+2z= 2 \\\\[1.5ex] -x+2y+2z=0  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} x+y+2\\lambda= 2 \\\\[1.5ex] -x+2y+2\\lambda=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"398\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Dan kami menempatkan <strong>suku dengan \u03bb dengan suku independen:<\/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-b8486baee4be39f417988ee12b5e67c7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}x+y=2-2\\lambda \\\\[1.5ex] -x+2y=-2\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"133\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Oleh karena itu, matriks A dan matriks A&#8217; dari sistem tersebut tetap:<\/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-8591e8c21bce2f49998311bbb08f7dee_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 1  \\\\[1.1ex] -1 &amp; 2 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 1 &amp; 1 &amp; 2 -2\\lambda \\\\[1.1ex] -1 &amp; 2 &amp; -2\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"363\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Terakhir, setelah kita mengubah sistem, <strong>kita menerapkan aturan Cramer<\/strong> . Untuk melakukan ini, pertama-tama kita selesaikan determinan 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-c34669d7234c9736c350f793df337bd3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 1  \\\\[1.1ex] -1 &amp; 2\\end{vmatrix} =2-(-1)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"229\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Untuk menghitung <em>x<\/em> dengan aturan Cramer, kita ubah kolom pertama determinan A menjadi kolom suku bebas dan membaginya dengan determinan 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-987ebe052154332042afeb27535996f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 2 -2\\lambda &amp; 1  \\\\[1.1ex] -2\\lambda &amp; 2 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{4-4\\lambda-(-2\\lambda)}{3} = \\cfrac{\\bm{4-2\\lambda}}{\\bm{3}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"345\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Untuk menghitung <em>y<\/em> dengan aturan Cramer, kita ubah kolom kedua determinan A menjadi kolom suku bebas dan membaginya dengan determinan 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-8a3c7b2cd7319f7f9db6df7df79abb50_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 1 &amp; 2 -2\\lambda  \\\\[1.1ex] -1 &amp; -2\\lambda  \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}}=\\cfrac{-2\\lambda -(-2+2\\lambda)}{3} = \\cfrac{\\bm{2-4\\lambda} }{\\bm{3}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"379\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Sehingga ketika m=2 solusi sistem persamaan tersebut adalah fungsi dari \u03bb, karena merupakan SCI dan oleh karena itu memiliki solusi tak terhingga:<\/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-1bc871365fab3194e382053fc6b083b5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =}  \\cfrac{\\bm{4-2\\lambda}}{\\bm{3}}  \\qquad \\bm{y=}\\cfrac{\\bm{2-4\\lambda}}{\\bm{3}} \\qquad \\bm{z=\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"274\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Kita telah menganalisis solusi sistem ketika parameter <em>m<\/em> berbeda dari 2 dan 3, dan ketika sama dengan 2. Oleh karena itu, kita hanya memerlukan kasus terakhir: ketika <em>m<\/em> bernilai 3: <\/p>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-118\"><\/div>\n<\/div>\n<p style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=3:<\/strong><\/span><\/p>\n<p> Sekarang kita akan menganalisis apa yang terjadi jika parameter <em>m<\/em> adalah 3. Dalam hal ini matriks A dan A&#8217; 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-c49bbc0d7d36606aa59be050c2682de5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; 3 &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 3 &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; 3 &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Seperti yang kita lihat sebelumnya, ketika m=3 determinan A adalah 0. Jadi matriks A tidak berpangkat 3. Namun di dalamnya terdapat 2\u00d72 determinan yang berbeda dengan 0, misalnya:<\/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-d88ce42feb4bba9aa74aae98e1062c4a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}1 &amp; 1 \\\\[1.1ex] -1 &amp; 3  \\end{vmatrix} = 3 - (-1)=4 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Jadi, dalam hal ini <strong>pangkat A adalah 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 rank matriks A, kita menghitung rank A&#8217;. Penentu 3 kolom pertama menghasilkan 0, oleh karena itu kita coba determinan 3\u00d73 lain yang ada di dalam matriks A&#8217;, misalnya matriks 3 kolom terakhir:<\/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-5e6f1a5c155ca004c73e51bdcbe5ece9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 &amp; 2 \\\\[1.1ex] 3 &amp; 2 &amp; 0 \\\\[1.1ex] 0 &amp; 3 &amp; 4\\end{vmatrix}=2\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"100\" 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> Jadi, ketika m = 3, pangkat matriks A lebih rendah daripada pangkat matriks A&#8217;. Jadi, dari teorema Rouch\u00e9-Frobenius, kami menyimpulkan bahwa sistem tersebut adalah <strong>Sistem yang Tidak Kompatibel<\/strong> (IS) <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-3454f804b63f3cca9bcf08bc93815f90_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<p> Oleh karena itu, sistem persamaan <strong>tidak memiliki solusi jika m = 3.<\/strong><\/p>\n<h3 class=\"wp-block-heading\"> Ringkasan contoh:<\/h3>\n<p> Seperti yang telah kita lihat, penyelesaian sistem persamaan bergantung pada nilai parameter <em>m<\/em> . Berikut adalah ringkasan dari semua kemungkinan kasus: <\/p>\n<div class=\"wp-block-columns is-layout-flex wp-container-32\">\n<div class=\"wp-block-column is-layout-flow\">\n<ul>\n<li> <span style=\"color:#1976d2;font-size:22px\"><strong>m\u22603 dan m\u22602:<\/strong><\/span> <\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-cf366a55bd307517f94fd8aa00cdf598_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{SCD} \\longrightarrow \\begin{cases} x = \\cfrac{2m^2+8-8m}{m^2-5m+6} \\\\[3.5ex] y =\\cfrac{-4+2m}{m^2-5m+6} \\\\[3.5ex] z = \\cfrac{-2m+4}{m^2-5m+6} \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"171\" width=\"240\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<\/div>\n<div class=\"wp-block-column is-layout-flow\">\n<ul>\n<li> <span style=\"color:#1976d2;font-size:22px\"><strong>m=2:<\/strong><\/span> <\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-94002d4f4d866569ed7d6993dd977b81_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{SCI} \\longrightarrow \\begin{cases} x = \\cfrac{4-2\\lambda}{3} \\\\[3.5ex] y= \\cfrac{2-4\\lambda}{3} \\\\[3.5ex] z = \\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"150\" width=\"175\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<\/div>\n<div class=\"wp-block-column is-layout-flow\">\n<ul>\n<li> <span style=\"color:#1976d2;font-size:22px\"><strong>m=3:<\/strong><\/span><\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9b971789910fd078c90174ed3d662e9a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{SI} \\longrightarrow\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"54\" style=\"vertical-align: -1px;\"><\/p>\n<p> Sistem tidak memiliki solusi.<\/p>\n<\/div>\n<\/div>\n<p> Di sini kita telah melakukan seluruh proses menggunakan teorema Rouche dan aturan Cramer, namun sistem persamaan dengan parameter juga dapat didiskusikan dan diselesaikan dengan<a href=\"https:\/\/mathority.org\/id\/metode-jordan-gauss-dengan-contoh-dan-latihan-yang-diselesaikan\/\">metode Gauss (dengan latihan)<\/a> . Anda dapat mempelajari lebih lanjut tentang metode ini di halaman tertaut, di mana Anda akan menemukan penjelasan rinci tentang prosedur serta contoh dan latihan yang diselesaikan langkah demi langkah.<\/p>\n<h2 class=\"wp-block-heading\"> Pembahasan Soal Pembahasan Sistem Persamaan Linier dengan Parameter<\/h2>\n<h3 class=\"wp-block-heading\"> Latihan 1<\/h3>\n<p> Diskusikan dan selesaikan sistem persamaan linier bergantung parameter berikut: <\/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-de-systemes-dequations-a-parametres.webp\" alt=\"menyelesaikan latihan sistem persamaan dengan parameter\" class=\"wp-image-4014\" width=\"191\" height=\"123\" 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-b641845325965882d4aac899246cffb3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 &amp; 1 \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; -m \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}4 &amp; -1 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp; -m &amp; 0\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"418\" 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 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-4d36c7cffe0248a2f45cd5871abc6ed5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{aligned}\\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 4 &amp; -1 &amp; 1 \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; -m \\end{vmatrix} &amp; =-4m+9-2-3-24-m \\\\ &amp; =-5m-20 \\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"381\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Hasil determinan A bergantung pada nilai m. Oleh karena itu, kita akan melihat nilai m mana yang determinannya hilang. Untuk melakukan ini, kita menyamakan hasil yang dihasilkan dengan 0 dan menyelesaikan persamaan: <\/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-e7d065a700c5f2a7f732719be777027f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-5m-20 = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"109\" 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-bc49744634d23cdf0faf446f33487eac_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-5m = 20\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"79\" 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-ca8fa07f8716b941295db488a99a6425_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m = \\cfrac{20}{-5} = -4\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"110\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, ketika m adalah -4, determinan dari A adalah 0. Dan ketika m berbeda dari -4, determinan dari A akan berbeda dari 0. Oleh karena itu kita harus menganalisis setiap kasus secara terpisah:<\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m\u2260-4:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Seperti yang baru saja kita lihat, jika parameter m berbeda dari -4, maka determinan matriks A berbeda dari 0. Oleh karena itu, rank A adalah 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-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\"> Selain itu, rank matriks A&#8217; juga 3, karena di dalamnya terdapat submatriks 3&#215;3 yang determinannya berbeda dengan 0. Dan tidak mungkin mempunyai rank 4 karena &#8216;kita tidak dapat membuat determinan 4&#215;4.<\/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\"> Oleh karena itu, dengan menerapkan <strong>teorema Rouch\u00e9-Frobenius,<\/strong> kita mengetahui bahwa ini adalah <strong>sistem determinasi yang 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<p class=\"has-text-align-left\"> Setelah kami mengetahui bahwa sistem tersebut adalah SCD, kami menerapkan aturan Cramer untuk menyelesaikannya. Untuk melakukannya, ingatlah bahwa matriks A, determinannya, dan matriks A&#8217; 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-e9e0bd352ad7713a03824ead1239041c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 &amp; 1  \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; -m \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 4 &amp; -1 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp; -m &amp; 0\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"418\" 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-530cb4576ee1a91d6246ed6cf9dd0fc8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 4 &amp; -1 &amp; 1  \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; -m\\end{vmatrix} =-5m-20\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"253\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Untuk menghitung xatex] dengan aturan Cramer, kita ubah kolom pertama determinan A dengan kolom suku bebas dan membaginya dengan determinan 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-b076bbda8d086abedb459570d74c80a9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 0 &amp; -1 &amp; 1  \\\\[1.1ex] 0 &amp; 1 &amp; -3 \\\\[1.1ex] 0 &amp; -2 &amp; -m\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{0}{-5m-20} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"280\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Untuk menghitung yang tidak diketahui dan dengan aturan Cramer, kita ubah kolom kedua determinan A dengan kolom suku bebas dan kita bagi dengan determinan 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-f642a8cb2fd174e5c383a4df53e11a2e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{y} = \\cfrac{\\begin{vmatrix} 4 &amp; 0 &amp; 1  \\\\[1.1ex] 1 &amp; 0 &amp; -3 \\\\[1.1ex] 3 &amp; 0 &amp; -m \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{0}{-5m-20} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"265\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Untuk menghitung z dengan aturan Cramer, kita ubah kolom ketiga determinan A menjadi kolom suku bebas dan membaginya dengan determinan 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-5020a9ba4995b9715d8d1fb4720952b1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{z} = \\cfrac{\\begin{vmatrix}4 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 1 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp; 0 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{0}{-5m-20} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"258\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Oleh karena itu, penyelesaian sistem persamaan untuk kasus m\u2260-4 adalah:<\/p>\n<p class=\"has-text-align-center\"> <strong>x=0 kamu=0 z=0<\/strong><\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=-4:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Kami sekarang akan menganalisis sistem ketika parameter m adalah -4. Dalam hal ini matriks A dan A&#8217; 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-e585e6465d27ea27ccc2c1a6ec1fe9ae_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 &amp; 1 \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; 4 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}4 &amp; -1 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp; 4 &amp; 0\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Seperti yang kita lihat sebelumnya, ketika m=-4 determinan A adalah 0. Jadi, matriks A tidak berpangkat 3. Namun di dalamnya terdapat 2\u00d72 determinan selain 0, misalnya:<\/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-a62d150aef4ec798814d25c988b0afd7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   \\begin{vmatrix}4 &amp; -1 \\\\[1.1ex] 1 &amp; 1 \\end{vmatrix} =4-(-1)=5 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Karena matriks mempunyai determinan orde 2 yang berbeda dengan 0, maka matriks A mempunyai rank 2:<\/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-39fc49c7a63920c8956703a4851ecfc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -1 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp; -3 &amp; 0 \\\\[1.1ex]  -2 &amp; 4 &amp; 0 \\end{vmatrix} = 0 \\quad \\begin{vmatrix}4 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp;  -3 &amp; 0 \\\\[1.1ex] 3 &amp;  4 &amp; 0  \\end{vmatrix} = 0 \\quad \\begin{vmatrix}4 &amp; -1 &amp;  0 \\\\[1.1ex] 1 &amp; 1 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp;  0\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"404\" 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-a62d150aef4ec798814d25c988b0afd7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   \\begin{vmatrix}4 &amp; -1 \\\\[1.1ex] 1 &amp; 1 \\end{vmatrix} =4-(-1)=5 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi matriks A&#8217; akan mempunyai rangking 2:<\/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\"> Luas matriks A sama dengan luas matriks A&#8217; tetapi keduanya lebih kecil dari jumlah yang tidak diketahui dalam sistem (3), oleh karena itu, menurut teorema Rouch\u00e9-Frobenius, c adalah Sistem Kompatibel Tak tentu (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-f43fdf4978386c61d18f9bb5b5883881_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<p class=\"has-text-align-left\"> Ini adalah sistem ICS, jadi kita perlu mengubah sistem untuk menyelesaikannya. Pertama-tama kita hilangkan satu persamaan, yang dalam hal ini akan menjadi persamaan 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-4d5499fda37d3cbf56fbf6ecbfc6bfba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 4x-y+z= 0 \\\\[1.5ex] x+y-3z=0 \\\\[1.5ex] \\cancel{3x-2y+4z = 0} \\end{cases} \\longrightarrow \\quad \\begin{cases} 4x-y+z= 0 \\\\[1.5ex] x+y-3z=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"349\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sekarang mari kita ubah variabel z menjadi \u03bb:<\/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-96ea68274b072531365282e01d926718_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}4x-y+z= 0 \\\\[1.5ex] x+y-3z=0 \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} 4x-y+\\lambda= 0 \\\\[1.5ex] x+y-3\\lambda=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"353\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan kami menempatkan suku dengan \u03bb dengan suku independen:<\/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-6192715e62cc8e3d3fe4c51da8629c70_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 4x-y=-\\lambda \\\\[1.5ex] x+y=3\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"110\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sehingga matriks A dan matriks A&#8217; sistem tetap: <\/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-99a91208ff1742f81e799aa5ab7f9097_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 \\\\[1.1ex] 1 &amp; 1 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 4 &amp; -1 &amp; -\\lambda \\\\[1.1ex] 1 &amp; 1 &amp; 3\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"337\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Terakhir, setelah kita mengubah sistem, kita menerapkan aturan Cramer. Untuk melakukan ini, pertama-tama kita selesaikan determinan 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-34832b783ddaf4af205302240d0feafb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 4 &amp; -1 \\\\[1.1ex] 1 &amp; 1 \\end{vmatrix} = 4-(-1)=5\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"228\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Untuk menghitung x dengan aturan Cramer, kita ubah kolom pertama determinan A menjadi kolom suku bebas dan membaginya dengan determinan 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-362167d2eaa02d7243dedd5c385d08b1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix}-\\lambda &amp; -1 \\\\[1.1ex] 3\\lambda &amp; 1 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-\\lambda-(-3\\lambda)}{5} =\\cfrac{\\bm{2\\lambda}}{\\bm{5}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"285\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Untuk menghitung yang tidak diketahui dan dengan aturan Cramer, kita ubah kolom kedua determinan A dengan kolom suku bebas dan kita bagi dengan determinan 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-ede5a3a87ac0bb9ceea4232ec7b381fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{y} = \\cfrac{\\begin{vmatrix} 4 &amp; -\\lambda \\\\[1.1ex] 1 &amp; 3\\lambda \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{12\\lambda-(-\\lambda)}{5}=\\cfrac{\\bm{13\\lambda}}{\\bm{5}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"267\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sehingga ketika m=-4 solusi sistem persamaan tersebut adalah fungsi dari \u03bb, karena merupakan SCI dan karenanya memiliki solusi tak terhingga: <\/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-316aed1af52eba51058c3c753717c1af_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =}\\cfrac{\\bm{2\\lambda}}{\\bm{5}}\\qquad \\bm{y=}\\cfrac{\\bm{13\\lambda}}{\\bm{5}} \\qquad \\bm{z=\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"222\" style=\"vertical-align: -12px;\"><\/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 2<\/h3>\n<p> Diskusikan dan temukan solusi sistem persamaan linear bergantung parameter berikut: <\/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-etape-par-etape-d-un-systeme-d-equations-lineaires-avec-parametres.webp\" alt=\"latihan diselesaikan langkah demi langkah sistem persamaan linear dengan parameter\" class=\"wp-image-4020\" width=\"187\" 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\"> Hal pertama yang harus dilakukan adalah 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-e924dd1b3fe5c0da561b92da9bf5da3b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; m\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}m &amp; 2 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; m &amp; 3\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"404\" 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 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-5d0f8dbb7408ac6521e0144ac2f3a8a3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{aligned}\\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix}m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; m\\end{vmatrix} &amp; =4m^2+4-4-4+4m-4m \\\\ &amp; =4m^2-4 \\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"384\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Hasil determinan A bergantung pada nilai m. Oleh karena itu, kita akan melihat nilai m mana yang determinannya hilang. Untuk melakukan ini, kita menyamakan hasil yang dihasilkan dengan 0 dan menyelesaikan persamaan: <\/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-059d7760bb8cee65e13e43a97e156e1a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"4m^2-4 = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"95\" 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-1aae0b3268b7d9e348b8eb80ce957ce8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"4m^2=4\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"65\" 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-1f31504086ef5ec070b337db57ad444f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m^2 = \\cfrac{4}{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"58\" style=\"vertical-align: -12px;\"><\/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-baf4ba1311e2f2ce47aecfb90a411b64_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m^2 = 1\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"55\" 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-c8dda3f9896a1a4dece5059a485cdcfb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m = \\pm 1\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"61\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, jika m adalah +1 atau -1, maka determinan dari A adalah 0. Dan jika m berbeda dari +1 dan -1, maka determinan A akan berbeda dari 0. Oleh karena itu kita harus menganalisis setiap kasus dengan:<\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m\u2260+1 dan m\u2260-1:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Seperti yang baru saja kita lihat, jika parameter m berbeda dari +1 dan -1, determinan matriks A berbeda dari 0. Oleh karena itu, rank A adalah 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-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\"> Selain itu, rank matriks A&#8217; juga 3, karena di dalamnya terdapat submatriks 3&#215;3 yang determinannya berbeda dengan 0. Dan tidak mungkin mempunyai rank 4 karena &#8216;kita tidak dapat membuat determinan 4&#215;4.<\/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\"> Oleh karena itu, dengan menerapkan <strong>teorema Rouch\u00e9-Frobenius,<\/strong> kita mengetahui bahwa ini adalah <strong>sistem determinasi yang 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<p class=\"has-text-align-left\"> Setelah kami mengetahui bahwa sistem tersebut adalah SCD, kami menerapkan aturan Cramer untuk menyelesaikannya. Untuk melakukannya, ingatlah bahwa matriks A, determinannya, dan matriks A&#8217; 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-e924dd1b3fe5c0da561b92da9bf5da3b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; m\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}m &amp; 2 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; m &amp; 3\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"404\" 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-b5114be5e37d2c91f02f22fba22edc42_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix}m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; m\\end{vmatrix}=4m^2-4\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"231\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Untuk menghitung x dengan aturan Cramer, kita ubah kolom pertama determinan A menjadi kolom suku bebas dan membaginya dengan determinan 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-4b7402e02ee62bd78a6f880d3d122119_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 2&amp; 2 &amp; 1 \\\\[1.1ex] 0 &amp; 4 &amp; 2 \\\\[1.1ex] 3 &amp; -2 &amp; m\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{\\bm{8m+8}}{\\bm{4m^2-4}}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"218\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Untuk menghitung yang tidak diketahui dan dengan aturan Cramer, kita ubah kolom kedua determinan A dengan kolom suku bebas dan kita bagi dengan determinan 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-551c66a9530d0195a9a4ff64d42350c4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{y} = \\cfrac{\\begin{vmatrix} m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 0 &amp; 2 \\\\[1.1ex] 1 &amp; 3 &amp; m\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{\\bm{-10m+10}}{\\bm{4m^2-4}}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"235\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Untuk menghitung z dengan aturan Cramer, kita ubah kolom ketiga determinan A menjadi kolom suku bebas dan membaginya dengan determinan 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-28375ce522b7644a745a9adea4c78ae7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{z} = \\cfrac{\\begin{vmatrix}m &amp; 2 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; 3 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{\\bm{12m-28}}{\\bm{4m^2-4}}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"227\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Oleh karena itu, penyelesaian sistem persamaan untuk kasus m\u2260+1 dan m\u2260-1 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-59ecb2f1448989cd1c41af45e7bc4a32_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x = }\\cfrac{\\bm{8m+8}}{\\bm{4m^2-4}} \\qquad \\bm{y=}\\cfrac{\\bm{-10m+10}}{\\bm{4m^2-4}}\\qquad \\bm{z =} \\cfrac{\\bm{12m-28}}{\\bm{4m^2-4}}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"383\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=+1:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Sekarang kita akan menganalisis sistem jika parameter m sama dengan 1. Dalam hal ini matriks A dan A&#8217; 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-f6af272a99ed7c281ee8dd9199698686_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; 1 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}1 &amp; 2 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; 1 &amp; 3\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Seperti yang kita lihat sebelumnya, ketika m=+1 determinan A adalah 0. Jadi matriks A tidak berpangkat 3. Namun di dalamnya terdapat 2\u00d72 determinan selain 0, misalnya:<\/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-ebedf6c9e4316844dc99ceca9472fac5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   \\begin{vmatrix}2 &amp; 4\\\\[1.1ex] 1 &amp; -2 \\end{vmatrix} =-4-4=-8 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Karena matriks mempunyai determinan orde 2 yang berbeda dengan 0, maka matriks A mempunyai rank 2:<\/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 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-4d0109b155be9f87a0cee337ddec5517_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 1 &amp; 2 \\\\[1.1ex] 4 &amp; 2 &amp; 0 \\\\[1.1ex]  -2 &amp; 1 &amp; 3 \\end{vmatrix} = 16\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"124\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sebaliknya matriks A&#8217; memang mengandung determinan 3\u00d73 yang hasilnya berbeda dengan 0, sehingga matriks A&#8217; mempunyai rank 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-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, ketika m=+1 pangkat matriks A lebih kecil dari pangkat matriks A&#8217;. Jadi, dari teorema Rouch\u00e9-Frobenius, kami menyimpulkan bahwa sistem tersebut adalah Sistem yang Tidak Kompatibel (IS):<\/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-b2bb3fec88cf5c6d788afb4480ab1f58_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<p class=\"has-text-align-left\"> Oleh karena itu, sistem persamaan <strong>tidak memiliki solusi ketika m=+1<\/strong> , karena merupakan sistem yang tidak kompatibel.<\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=-1:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Sekarang kita akan menganalisis sistem ketika parameter m adalah -1. Dalam hal ini matriks A dan A&#8217; 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-46b0a00ef38d0e5a433b418de7eb1ec3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} -1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; -1 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}-1 &amp; 2 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; -1 &amp; 3\\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\"> Seperti yang kita lihat sebelumnya, ketika m=-1 determinan A adalah 0. Jadi, matriks A tidak berpangkat 3. Namun di dalamnya terdapat 2\u00d72 determinan selain 0, misalnya:<\/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-ff5373c7e7901f253421efbbd52d192e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   \\begin{vmatrix}-1 &amp; 2\\\\[1.1ex] 2 &amp; 4 \\end{vmatrix} =-4-4=-8 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Karena matriks mempunyai determinan orde 2 yang berbeda dengan 0, maka matriks A mempunyai rank 2:<\/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 sekarang kita coba, misalnya, dengan determinan kolom 1, 3 dan 4:<\/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-a95e30910bd64db920f3c2bcb5f2ff62_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -1 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp;  -1 &amp; 3\\end{vmatrix} = -20\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"152\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sebaliknya matriks A&#8217; memang mengandung determinan 3\u00d73 yang hasilnya berbeda dengan 0, sehingga matriks A&#8217; mempunyai rank 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-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, ketika m = -1, pangkat matriks A lebih rendah daripada pangkat matriks A&#8217;. Jadi, dari teorema Rouch\u00e9-Frobenius, kami menyimpulkan bahwa sistem tersebut adalah Sistem yang Tidak Kompatibel (IS):<\/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-b2bb3fec88cf5c6d788afb4480ab1f58_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<p class=\"has-text-align-left\"> Oleh karena itu, sistem persamaan <strong>tidak memiliki solusi ketika m=-1<\/strong> , karena merupakan sistem yang tidak kompatibel.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Pada halaman ini kita akan melihat bagaimana membahas dan menyelesaikan sistem persamaan dengan parameter . Selain itu, Anda akan menemukan contoh dan latihan penyelesaian sistem persamaan linear untuk dipraktikkan. Di sisi lain, untuk menganalisis sistem persamaan linier, penting bagi Anda untuk mengetahui apa itu aturan Cramer dan apa itu teorema Rouch\u00e9\u2013Frobenius , karena kita akan &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/id\/pembahasan-sistem-persamaan-dengan-parameter\/\"> <span class=\"screen-reader-text\">Pembahasan sistem persamaan dengan parameter<\/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":[41],"tags":[],"class_list":["post-306","post","type-post","status-publish","format-standard","hentry","category-penjelasan-matematis"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Pembahasan sistem persamaan dengan parameter -<\/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\/pembahasan-sistem-persamaan-dengan-parameter\/\" \/>\n<meta property=\"og:locale\" content=\"id_ID\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Pembahasan sistem persamaan dengan parameter -\" \/>\n<meta property=\"og:description\" content=\"Pada halaman ini kita akan melihat bagaimana membahas dan menyelesaikan sistem persamaan dengan parameter . Selain itu, Anda akan menemukan contoh dan latihan penyelesaian sistem persamaan linear untuk dipraktikkan. Di sisi lain, untuk menganalisis sistem persamaan linier, penting bagi Anda untuk mengetahui apa itu aturan Cramer dan apa itu teorema Rouch\u00e9\u2013Frobenius , karena kita akan &hellip; Pembahasan sistem persamaan dengan parameter Selengkapnya &raquo;\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mathority.org\/id\/pembahasan-sistem-persamaan-dengan-parameter\/\" \/>\n<meta property=\"article:published_time\" content=\"2023-07-06T14:08:28+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6ab2286d15c20029b98a5ea4622033d4_l3.png\" \/>\n<meta name=\"author\" content=\"Tim Mathority\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Ditulis oleh\" \/>\n\t<meta name=\"twitter:data1\" content=\"Tim Mathority\" \/>\n\t<meta name=\"twitter:label2\" content=\"Estimasi waktu membaca\" \/>\n\t<meta name=\"twitter:data2\" content=\"10 menit\" \/>\n<script 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