{"id":349,"date":"2023-07-06T14:13:39","date_gmt":"2023-07-06T14:13:39","guid":{"rendered":"https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/"},"modified":"2023-07-06T14:13:39","modified_gmt":"2023-07-06T14:13:39","slug":"regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer","status":"publish","type":"post","link":"https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/","title":{"rendered":"De regel van cramer"},"content":{"rendered":"<p>Op deze pagina zie je wat de regel van Cramer is en daarnaast vind je voorbeelden en oefeningen met het oplossen van stelsels vergelijkingen volgens de regel van Cramer.<\/p>\n<h2 class=\"wp-block-heading\"> Wat is de regel van Cramer?<\/h2>\n<p> <strong>De regel van Cramer<\/strong> is een methode die wordt gebruikt om stelsels vergelijkingen op te lossen met behulp van determinanten. Laten we eens kijken hoe het wordt gebruikt:<\/p>\n<p> Beschouw een systeem van vergelijkingen:<\/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-e0141f3451719f665ef28e4061489551_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} ax+by+cz= \\color{red}\\bm{j} \\\\[1.5ex] dx+ey+fz=\\color{red}\\bm{k} \\\\[1.5ex] gx+hy+iz = \\color{red}\\bm{l} \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"171\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De matrix A en de uitgebreide matrix A&#8217; van het systeem zijn:<\/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-1d628a13ec7de4b3ba7a301c0a5d8ac6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} a &amp; b &amp; c  \\\\[1.1ex] d &amp; e &amp; f  \\\\[1.1ex] g &amp; h &amp; i  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} a &amp; b &amp; c &amp;  \\color{red}\\bm{j}  \\\\[1.1ex] d &amp; e &amp; f &amp; \\color{red}\\bm{k} \\\\[1.1ex] g &amp; h &amp; i &amp; \\color{red}\\bm{l} \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"384\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong><span style=\"text-decoration: underline;\">De regel van Cramer<\/span><\/strong> zegt dat de oplossing van een stelsel vergelijkingen is: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/quelle-est-la-regle-de-cramer.webp\" alt=\"wat is de regel van Cramer, uitleg van de regel van Cramer\" class=\"wp-image-1062\" width=\"677\" height=\"385\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> Merk op dat de determinanten van de tellers vergelijkbaar zijn met de determinant van matrix A, maar dat de kolom van elke onbekende verandert in de kolom met onafhankelijke termen.<\/p>\n<p> Daarom wordt de regel van Cramer gebruikt om stelsels van lineaire vergelijkingen op te lossen. Maar zoals je al weet, zijn er veel manieren om een stelsel vergelijkingen op te lossen; <a href=\"https:\/\/mathority.org\/nl\/jordan-gauss-methode-met-voorbeelden-en-opgeloste-oefeningen\/\">de methode van Gauss Jordan<\/a> is bijvoorbeeld algemeen bekend.<\/p>\n<p> Hieronder staan voorbeelden van het oplossen van stelsels van lineaire vergelijkingen met de regel van Cramer, of soms ook geschreven als de regel van Kramer.<\/p>\n<h2 class=\"wp-block-heading\"> Voorbeeld 1: bepaald compatibel systeem (SCD)<\/h2>\n<ul>\n<li> Los het volgende stelsel van 3 vergelijkingen met 3 onbekenden op met behulp van de regel van Cramer:<\/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-6013b7e73c89c24fe388f1a5d018f32b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 2x+y+3z= 1 \\\\[1.5ex] 3x-2y-z=0 \\\\[1.5ex] x+3y+2z = 5\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"135\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> We maken eerst de matrix A en de uitgebreide matrix A&#8217; van het systeem:<\/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-c710ed86223f47f39b5a25720b5ca19d_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; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; 3 &amp; 1 \\\\[1.1ex] 3 &amp; -2 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 3 &amp; 2 &amp; 5 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> We berekenen nu de rangorde van de twee matrices, om te zien wat voor soort systeem het is. Om de rangorde van A te berekenen, berekenen we de 3\u00d73 determinant van de gehele matrix (met behulp van de regel van Sarrus) en kijken of deze 0 oplevert:<\/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-ae4a3bb88d113494463df8e670c326c6_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; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{vmatrix} =-8-1+27+6+6-6 = 24 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"427\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De determinant van A is anders dan 0, dus <strong>matrix A heeft rang 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> <strong>De matrix A&#8217; heeft dus ook rang 3<\/strong> , aangezien deze niet rang 4 kan hebben en op zijn minst dezelfde rang moet hebben als matrix 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-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> De omvang van de matrix A is gelijk aan de omvang van de matrix A&#8217; en het aantal onbekenden van het systeem (3). Daarom weten we volgens de <strong>stelling van Rouch\u00e9-Frobenius<\/strong> dat het een <strong>bepaald compatibel systeem<\/strong> (SCD) 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-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> Zodra we weten dat het systeem een SCD is, passen we <strong>de regel van Cramer<\/strong> toe om het op te lossen. Om dit te doen, bedenk dat de matrix A, zijn determinant en de matrix A&#8217; zijn:<\/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-31b2b3e5865c2264c360fb887d37a5f6_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; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; 3 &amp; \\color{red}\\bm{1} \\\\[1.1ex] 3 &amp; -2 &amp; -1 &amp; \\color{red}\\bm{0} \\\\[1.1ex] 1 &amp; 3 &amp; 2 &amp; \\color{red}\\bm{5} \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"431\" 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-0a604d8f5a3927a47a264d28f7a007b2_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; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{vmatrix} =24\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"187\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Om het onbekende te berekenen<\/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-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de eerste kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-a1fa494ffb5e452d59c4d2dad40f925a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} \\color{red}\\bm{1} &amp; 1 &amp; 3 \\\\[1.1ex] \\color{red}\\bm{0} &amp; -2 &amp; -1 \\\\[1.1ex] \\color{red}\\bm{5} &amp; 3 &amp; 2 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{24}{24} = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"238\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Om het onbekende te berekenen<\/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-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de tweede kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-08e3dabe2f33434eb96658491f67c0b4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 2 &amp; \\color{red}\\bm{1} &amp; 3 \\\\[1.1ex] 3 &amp;  \\color{red}\\bm{0} &amp; -1 \\\\[1.1ex] 1 &amp; \\color{red}\\bm{5} &amp; 2\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{48}{24} = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"223\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Berekenen<\/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-23aa090e6102a41de5ad5515112e4d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  z\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de derde kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-96e76cb8867224755e9c19254678abd4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix} 2 &amp; 1 &amp; \\color{red}\\bm{1} \\\\[1.1ex] 3 &amp; -2 &amp;  \\color{red}\\bm{0} \\\\[1.1ex] 1 &amp; 3 &amp;  \\color{red}\\bm{5}\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-24}{24} = \\bm{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"259\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De oplossing van het stelsel vergelijkingen is daarom:<\/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-be5a19fed42dcb59880c2d0eee8e51f4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x = 1 \\qquad y=2 \\qquad z = -1}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"210\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> Voorbeeld 2: Onbepaald compatibel systeem (ICS)<\/h2>\n<ul>\n<li> Los het volgende stelsel vergelijkingen op met behulp van de regel van Cramer:<\/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-781530aac4d8507fd6c7cbd77c3b4651_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0 \\\\[1.5ex] x+5y+3z = 1 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"149\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> We maken eerst de matrix A en de uitgebreide matrix A&#8217; van het systeem:<\/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-a64800a78bf8e2e2f547be907e6863cb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; 2 &amp; 4 \\\\[1.1ex] -2 &amp; 3 &amp; -1 \\\\[1.1ex] 1 &amp; 5 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 3 &amp; 2 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; 3 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 5 &amp; 3 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Nu berekenen we het bereik van de twee matrices en kunnen we zo zien welk type systeem het is. Om de rangorde van A te berekenen, berekenen we de determinant van de gehele matrix (met behulp van de regel van Sarrus) en controleren we of deze 0 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-581c58cbe0fdd9952e7e25b919ecc33b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; 2 &amp; 4 \\\\[1.1ex] -2 &amp; 3 &amp; -1 \\\\[1.1ex] 1 &amp; 5 &amp; 3\\end{vmatrix} = 27-2-40-12+15+12= 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"407\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De determinant geeft 0, dus matrix A heeft niet rang 3. Maar hij heeft een 2\u00d72 determinant die verschilt van 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-a5d1acad8bc31240f80d8cfbf3605997_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; 2 \\\\[1.1ex] -2 &amp; 3 \\end{vmatrix} =9-(-4)=13\\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"222\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong>Matrix A heeft dus rang 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> Zodra we de omvang van matrix A kennen, berekenen we die van matrix A&#8217;. De determinant van de eerste 3 kolommen geeft 0, dus we proberen de andere mogelijke 3\u00d73 determinanten in de matrix 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-686e7ca635ecee685005f6013c2e64ad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 4 &amp; 1 \\\\[1.1ex] 3 &amp; -1 &amp; 0 \\\\[1.1ex] 5 &amp; 3 &amp; 1 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 3 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 3 &amp; 1 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 3 &amp; 2 &amp; 1 \\\\[1.1ex] -2 &amp; 3 &amp; 0 \\\\[1.1ex] 1 &amp; 5 &amp; 1 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"440\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Alle determinanten van orde 3 geven 0. Maar uiteraard heeft matrix A&#8217; dezelfde niet-0 2\u00d72 determinant als matrix 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-a5d1acad8bc31240f80d8cfbf3605997_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; 2 \\\\[1.1ex] -2 &amp; 3 \\end{vmatrix} =9-(-4)=13\\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"222\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Daarom <strong>heeft de matrix A&#8217; ook rang 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> Dus aangezien de rangorde van matrix A gelijk is aan de rangorde van matrix A&#8217;, maar deze twee kleiner zijn dan het aantal onbekenden van systeem (3), weten we door de <strong>stelling van Rouch\u00e9-Frobenius<\/strong> dat dit een <strong>onbepaald compatibel systeem<\/strong> is. (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 class=\"has-background\" style=\"background-color:#dff6ff\"> Als we een compatibel onbepaald systeem (SCI) willen oplossen, moeten we <strong>het systeem transformeren<\/strong> : we elimineren eerst een vergelijking, dan converteren we een variabele naar \u03bb (meestal de variabele z), en tenslotte voegen we de termen met \u03bb samen met de onafhankelijke termen.<\/p>\n<p> Zodra we het systeem hebben getransformeerd, passen we de regel van Cramer toe en krijgen we de oplossing van het systeem als functie van \u03bb.<\/p>\n<p> In dit geval <strong>zullen we de laatste vergelijking uit het systeem verwijderen<\/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-f0511fecc9c2af695b6b8eccae6b0661_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0 \\\\[1.5ex]\\cancel{x+5y+3z = 1} \\end{cases} \\longrightarrow \\quad \\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong>Laten we nu de variabele z naar \u03bb converteren:<\/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-2d6142d2be611954fd849a032a97245a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} 3x+2y+4\\lambda=1 \\\\[1.5ex] -2x+3y-\\lambda=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"398\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> En we plaatsen <strong>de termen met \u03bb bij de onafhankelijke termen:<\/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-00214205f2334f1c9bc10810c1c1df83_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y=1-4\\lambda \\\\[1.5ex] -2x+3y=\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"adsb30\" style=\" margin:px; text-align:\"><\/div>\n<p> Daarom blijven de matrix A en de matrix A&#8217; van het systeem:<\/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-9c4b47303973b823a1c5628f5448ca79_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; 2  \\\\[1.1ex] -2 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 3 &amp; 2 &amp; 1 -4\\lambda \\\\[1.1ex] -2 &amp; 3 &amp; \\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"363\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Ten slotte passen we, zodra we het systeem hebben getransformeerd, <strong>de regel van Cramer toe<\/strong> . We lossen daarom de determinant van A op:<\/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-d1b79f52dc82f5cfc311867273e78c06_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; 2  \\\\[1.1ex] -2 &amp; 3\\end{vmatrix} = 13\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"148\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Om het onbekende te berekenen<\/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-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de eerste kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-0ff917eaea976c65bd18e0476078d3cb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 1 -4\\lambda &amp; 2  \\\\[1.1ex] \\lambda &amp; 3 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{3(1-4\\lambda) -2\\lambda}{13} = \\cfrac{\\bm{3-14\\lambda} }{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"349\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Om het onbekende te berekenen<\/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-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de tweede kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-155ca520739bbf7e040a6cdc632f7c27_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 3 &amp; 1 -4\\lambda  \\\\[1.1ex]-2&amp;  \\lambda  \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{3\\lambda -\\bigl(-2(1-4\\lambda)\\bigr)}{13}= \\cfrac{3\\lambda -\\bigl(-2+8\\lambda\\bigr)}{13} = \\cfrac{\\bm{2-5\\lambda} }{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"529\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Hoewel de oplossing van het stelsel vergelijkingen een functie is van \u03bb, omdat het een SCI is en daarom oneindig veel oplossingen heeft:<\/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-c9866e045041eb2d8fe103db2309f229_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =} \\cfrac{\\bm{3-14\\lambda} }{\\bm{13}} \\qquad \\bm{y=}\\cfrac{\\bm{2-5\\lambda} }{\\bm{13}} \\qquad \\bm{z = \\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"283\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> De regel van Cramer loste problemen op <\/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=\"wp-block-heading\"> Oefening 1<\/h3>\n<p> Pas de regel van Cramer toe om het volgende stelsel van twee vergelijkingen met twee onbekenden op te lossen: <\/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-pas-a-pas-de-la-regle-de-cramer-22.webp\" alt=\"oefening stap voor stap opgelost met de 2x2 regel van Cramer\" class=\"wp-image-3999\" width=\"137\" height=\"83\" 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>zie oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Het eerste wat je moet doen is de matrix A en de uitgebreide matrix A&#8217; van het systeem:<\/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-2a001db9cf56846150730fee7126dacd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{cc} 2 &amp; 5 \\\\[1.1ex] 1 &amp; 4 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 2 &amp; 5 &amp; 8 \\\\[1.1ex] 1 &amp; 4 &amp; 7 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"294\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We moeten nu de rangorde van matrix A vinden. Om dit te doen, controleren we of de determinant van de gehele matrix anders is dan 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-0c75c1c344c286016bea83237f1f418e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 5 \\\\[1.1ex] 1 &amp; 4 \\end{vmatrix} = 8-5=3 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"216\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Omdat de matrix een 2\u00d72 determinant heeft die verschilt van 0, <strong>heeft matrix A rang 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\"> Zodra we de rang van A kennen, berekenen we de rang van A&#8217;. Dit zal op zijn minst van rang 2 zijn, omdat we zojuist hebben gezien dat er een determinant in zit van orde 2 die verschilt van 0. Bovendien kan het niet van rang 3 zijn, omdat we geen 3\u00d73 determinant kunnen maken. Daarom <strong>heeft de matrix A&#8217; ook rang 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\"> Door de <strong>stelling van Rouch\u00e9-Frobenius toe te passen,<\/strong> weten we daarom dat dit een <strong>compatibel bepaald systeem<\/strong> (SCD) is, omdat het bereik van A gelijk is aan het bereik van A&#8217; en het aantal onbekenden.<\/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-bbd67b16bb6d52a0696e70a77833cd3b_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} = 2 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 2 \\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\"> Zodra we weten dat het systeem een SCD is, passen we <strong>de regel van Cramer<\/strong> toe om het op te lossen.<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de eerste kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-b0adeda8f2ce557661466996038b1148_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 8 &amp; 5 \\\\[1.1ex] 7 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-3}{3} = \\bm{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de tweede kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-59790a66cc31fac07be1d5a7bb556d9e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}2 &amp; 8 \\\\[1.1ex] 1 &amp; 7\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{6}{3} = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"150\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De oplossing van het stelsel vergelijkingen is daarom: <\/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-26fa7c9ed2d05ca07ff62a968ba7ab11_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x = -1 \\qquad y=2}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"133\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Oefening 2<\/h3>\n<p> Vind de oplossing van het volgende stelsel van drie vergelijkingen met 3 onbekenden met behulp van de regel van Cramer: <\/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-systeme-de-regles-de-cramer-des-equations-3-3.webp\" alt=\"Opgeloste oefening van Cramer's regel van een 3x3 systeem van vergelijkingen\" class=\"wp-image-4002\" width=\"181\" height=\"124\" 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>zie oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> We maken eerst de matrix A en de uitgebreide matrix A&#8217; van het systeem:<\/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-eea75fbf6d86ebc3d0b9e236cd2160f5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 3 &amp; 2\\\\[1.1ex] -1 &amp; 5 &amp; -1\\\\[1.1ex] 3 &amp; -1 &amp; 4 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 3 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 5 &amp; -1 &amp; 4 \\\\[1.1ex] 3 &amp; -1 &amp; 4 &amp; 0 \\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\"> We vinden nu de rangorde van matrix A door de determinant van de 3\u00d73 matrix te berekenen met de Sarrusregel:<\/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-73f751f3b5c527c16b5de1b10bf07a4e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 3 &amp; 2 \\\\[1.1ex] -1 &amp; 5 &amp; -1\\\\[1.1ex] 3 &amp; -1 &amp; 4 \\end{vmatrix} = 20-9+2-30-1+12=-6 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"445\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De matrix heeft een determinant van orde 3 die verschilt van 0, <strong>de matrix A heeft rang 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\"> bijgevolg is de matrix A&#8217; ook van rang 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\"> Daarom weten we, met behulp van de <strong>stelling van Rouch\u00e9-Frobenius,<\/strong> dat dit een <strong>compatibel bepaald systeem<\/strong> (SCD) is, omdat het bereik van A gelijk is aan het bereik van A&#8217; en het aantal onbekenden.<\/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\"> Zodra we weten dat het systeem een SCD is, moeten we <strong>de regel van Cramer<\/strong> toepassen om het systeem op te lossen.<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de eerste kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-fc574297f609b68e4fb48466ec6c8077_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 2 &amp; 3 &amp; 2 \\\\[1.1ex] 4 &amp; 5 &amp; -1\\\\[1.1ex]0 &amp; -1 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-18}{-6} = \\bm{3}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"235\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de tweede kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-2544601137d62e217ff1866f278203d6_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; 4 &amp; -1\\\\[1.1ex] 3 &amp; 0 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-6}{-6} = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"224\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Berekenen<\/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-23aa090e6102a41de5ad5515112e4d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  z\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de derde kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-42d7d4adcfc48954185ca14b56b8e128_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix} 1 &amp; 3 &amp; 2 \\\\[1.1ex] -1 &amp; 5 &amp; 4 \\\\[1.1ex] 3 &amp; -1 &amp; 0\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{12}{-6} = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"230\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De oplossing van het stelsel vergelijkingen is daarom: <\/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-685195d3a299f30f6421bb387f7f00e4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =3 \\qquad y=1 \\qquad z=-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"210\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Oefening 3<\/h3>\n<p> Bereken de oplossing van het volgende stelsel van drie vergelijkingen met 3 onbekenden met behulp van de regel van Cramer: <\/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-de-regle-de-cramer.webp\" alt=\"voorbeeld van de regel van Cramer\" class=\"wp-image-4003\" width=\"183\" 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>zie oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> We maken eerst de matrix A en de uitgebreide matrix A&#8217; van het systeem:<\/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-afd359275e5ebaaf3229504c47a5815f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 2 &amp; 5\\\\[1.1ex] 2 &amp; 3 &amp; -1 \\\\[1.1ex] 3 &amp; 4 &amp; -7 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 2 &amp; 5 &amp; 1 \\\\[1.1ex] 2 &amp; 3 &amp; -1 &amp; 5 \\\\[1.1ex] 3 &amp; 4 &amp; -7 &amp; 9 \\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\"> We berekenen de omvang van matrix 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-47ddf17a2b3eed5a680d685900a79b31_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 2 &amp; 5\\\\[1.1ex] 2 &amp; 3 &amp; -1 \\\\[1.1ex] 3 &amp; 4 &amp; -7 \\end{vmatrix} =-21-6+40-45+4+28=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"398\" 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-fdd4380c7c76418bd3ec12c94359f886_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 \\\\[1.1ex] 2 &amp; 3  \\end{vmatrix} = 3-4 = -1 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"186\" 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-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\"> Zodra we de omvang van matrix A kennen, berekenen we die van matrix A&#8217;. De determinant van de eerste 3 kolommen geeft 0, dus we proberen de andere mogelijke 3\u00d73 determinanten in de matrix 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-1addc62130e0462075b3bade26a7e35e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 5 &amp; 1 \\\\[1.1ex]  3 &amp; -1 &amp; 5 \\\\[1.1ex] 4 &amp; -7 &amp; 9 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 1 &amp; 5 &amp; 1 \\\\[1.1ex] 2 &amp; -1 &amp; 5 \\\\[1.1ex] 3 &amp; -7 &amp; 9\\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 3 &amp; 5 \\\\[1.1ex] 3 &amp; 4 &amp; 9 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"412\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Alle determinanten van orde 3 geven 0. De matrix A&#8217; heeft echter dezelfde 2\u00d72 niet-0 determinant als de matrix 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-7de377466bd5afd03f58f9b532324e75_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 \\\\[1.1ex] 2 &amp; 3 \\end{vmatrix} = 3-4 = -1 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Daarom heeft de matrix A&#8217; ook rang 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\"> Omdat de rangorde van matrix A gelijk is aan de rangorde van matrix A&#8217;, maar deze twee kleiner zijn dan het aantal onbekenden van systeem (3), weten we door de <strong>stelling van Rouch\u00e9-Frobenius<\/strong> dat het een <strong>onbepaald compatibel systeem<\/strong> (ICS) 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-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 class=\"has-text-align-left\"> Omdat we een ICS-systeem zijn, moeten we een vergelijking elimineren. In dit geval <strong>zullen we de laatste vergelijking uit het systeem verwijderen<\/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-3a1d067e155540f4345cf56e5c1567d3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+2y+5z=1 \\\\[1.5ex] 2x+3y-z=5 \\\\[1.5ex]\\cancel{3x+4y-7z = 9} \\end{cases} \\longrightarrow \\quad \\begin{cases} x+2y+5z=1 \\\\[1.5ex] 2x+3y-z=5\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"357\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Laten we nu de variabele z naar \u03bb converteren:<\/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-b5fa91777a722d3783b2f887aab44152_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+2y+5z=1 \\\\[1.5ex] 2x+3y-z=5  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} x+2y+5\\lambda=1 \\\\[1.5ex] 2x+3y-\\lambda=5\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"369\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> En we plaatsen de termen met \u03bb bij de onafhankelijke termen:<\/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-76ff21181be050b01c247981298986a7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+2y=1-5\\lambda\\\\[1.5ex] 2x+3y=5+\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"136\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Zodanig dat de matrix A en de matrix A&#8217; van het systeem blijven:<\/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-230e5b28dd467127e63f4f9756cf90da_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 2  \\\\[1.1ex] 2 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 1 &amp; 2 &amp; 1 -5\\lambda \\\\[1.1ex] 2 &amp; 3 &amp;5+\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"335\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ten slotte passen we, zodra we het systeem hebben getransformeerd, <strong>de regel van Cramer toe<\/strong> . We lossen daarom de determinant van A op:<\/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-f127efbd217e2bca8852ec792610732f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 2 \\\\[1.1ex] 2 &amp; 3\\end{vmatrix} =-1\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"138\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de eerste kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-42652a14362b42e606841b6bb3e77cc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 1-5\\lambda &amp; 2 \\\\[1.1ex] 5+\\lambda &amp; 3 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{3-15\\lambda -(10+2\\lambda)}{-1} = \\cfrac{-7-17\\lambda}{-1} = \\bm{7+17\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"491\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de tweede kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-b95c5870f1762a2d82c9ebcccbca7408_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 1 &amp; 1-5\\lambda \\\\[1.1ex] 2 &amp; 5+\\lambda \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{5+\\lambda -(2-10\\lambda)}{-1}= \\cfrac{3+11\\lambda}{-1} = \\bm{-3-11\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"465\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Hoewel de oplossing van het stelsel vergelijkingen een functie is van \u03bb, omdat het een SCI is en daarom oneindig veel oplossingen heeft: <\/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-5483357f081aca551b07fe7c8f9ebf5d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =7+17\\lambda} \\qquad \\bm{y=-3-11\\lambda} \\qquad \\bm{z = \\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"311\" style=\"vertical-align: -4px;\"><\/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\"> Oefening 4<\/h3>\n<p> Los het volgende probleem op van een stelsel van drie vergelijkingen met drie onbekenden door de regel van Cramer toe te passen: <\/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-61e1c3458f33b863db10750b9e51d09e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} -2x+5y+z=8 \\\\[1.5ex] 6x+2y+4z=4 \\\\[1.5ex] 3x-2y+z = -2 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"149\" 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>zie oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Eerst construeren we de matrix A en de uitgebreide matrix A&#8217; van het systeem:<\/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-743a40010cb4a610e8a3fc6ae5d313b4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}-2 &amp; 5 &amp; 1 \\\\[1.1ex] 6 &amp; 2 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} -2 &amp; 5 &amp; 1 &amp; 8 \\\\[1.1ex] 6 &amp; 2 &amp; 4 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1 &amp; -2 \\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\"> Laten we nu de rangorde van matrix A berekenen door de determinant van de 3&#215;3-matrix te berekenen met behulp van de regel van 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-713c634fbc3e1b1cb228e3891c9bff1c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} -2 &amp; 5 &amp; 1 \\\\[1.1ex] 6 &amp; 2 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1 \\end{vmatrix} = -4+60-12-6-16-30=-8 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"453\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De matrix heeft een determinant van orde 3 die verschilt van 0, <strong>de matrix A heeft rang 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\"> bijgevolg heeft de matrix A&#8217; ook rang 3, aangezien deze op zijn minst dezelfde rang moet hebben als de matrix A en kan deze niet van rang 4 zijn omdat het een matrix is met dimensie 3\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 class=\"has-text-align-left\"> Daarom leiden we met behulp van de <strong>stelling van Rouch\u00e9-Frobenius<\/strong> af dat het een <strong>bepaald compatibel systeem<\/strong> (SCD) is, omdat het bereik van A gelijk is aan het bereik van A&#8217; en het aantal onbekenden.<\/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\"> Zodra we weten dat het systeem een SCD is, moeten we <strong>de regel van Cramer<\/strong> toepassen om het systeem op te lossen.<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de eerste kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-8a290479c69ff806f19dcf29f96e1228_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 8 &amp; 5 &amp; 1 \\\\[1.1ex] 4 &amp; 2 &amp; 4 \\\\[1.1ex] -2 &amp; -2 &amp; 1\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{16}{-8} = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"231\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de tweede kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-8bba0765fbcbcebf0585520af25b4a30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}-2 &amp; 8 &amp; 1 \\\\[1.1ex] 6 &amp; 4 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{0}{-6} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"217\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Berekenen<\/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-23aa090e6102a41de5ad5515112e4d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  z\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de derde kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-5bc157a8c4dfe8ee4651affac68ef878_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix} -2 &amp; 5 &amp; 8 \\\\[1.1ex] 6 &amp; 2 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; -2\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-32}{-8} = \\bm{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"247\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De oplossing voor het stelsel lineaire vergelijkingen is daarom: <\/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-6c004c5c466235d2d1a784707145d952_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =-2 \\qquad y=0 \\qquad z=4}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"211\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Oefening 5<\/h3>\n<p> Los het volgende stelsel lineaire vergelijkingen op met behulp van de regel van Cramer: <\/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-comment-resoudre-un-systeme-dequations-avec-la-regle-de-cramer.webp\" alt=\"Voorbeeld van het oplossen van een stelsel vergelijkingen met de regel van Cramer\" class=\"wp-image-4008\" width=\"215\" height=\"127\" 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>zie oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> We maken eerst de matrix A en de uitgebreide matrix A&#8217; van het systeem:<\/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-f5153b5951b768cc3cafa2bb2567ba92_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; -2 &amp; -3 \\\\[1.1ex] -1 &amp; 5 &amp; 4 \\\\[1.1ex] 5 &amp; 1 &amp; -2 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 3 &amp; -2 &amp; -3 &amp; 4 \\\\[1.1ex] -1 &amp; 5 &amp; 4 &amp; -10 \\\\[1.1ex] 5 &amp; 1 &amp; -2 &amp; -2 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"455\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We berekenen de omvang van matrix 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-f3778c9499e2a44ea3834dfed1523163_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; -2 &amp; -3 \\\\[1.1ex] -1 &amp; 5 &amp; 4 \\\\[1.1ex] 5 &amp; 1 &amp; -2 \\end{vmatrix} =-30-40+3+75-12+4=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"426\" 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-03d70742b14ced92f33963df0c86e92f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; -2 \\\\[1.1ex] -1 &amp; 5  \\end{vmatrix} = 15- (2)= 13 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"231\" 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-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\"> Zodra we de omvang van matrix A kennen, berekenen we die van matrix A&#8217;. De determinant van de eerste 3 kolommen geeft 0, dus we proberen de andere mogelijke 3\u00d73 determinanten in de matrix 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-5bed93d532ae4ccd4649a73662f55f0f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -2 &amp; -3 &amp; 4 \\\\[1.1ex] 5 &amp; 4 &amp; -10 \\\\[1.1ex]  1 &amp; -2 &amp; -2 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix}3 &amp; -3 &amp; 4 \\\\[1.1ex] -1 &amp; 4 &amp; -10 \\\\[1.1ex] 5 &amp; -2 &amp; -2\\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 3 &amp; -2 &amp; 4 \\\\[1.1ex] -1 &amp; 5 &amp; -10 \\\\[1.1ex] 5 &amp; 1 &amp;-2\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"535\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Alle determinanten van orde 3 geven 0. Maar uiteraard heeft matrix A&#8217; dezelfde determinant van orde 2 anders dan 0 als matrix 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-858d95d7d252b16706b66c0e6aba09c4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; -2 \\\\[1.1ex] -1 &amp; 5 \\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\"> Daarom heeft de matrix A&#8217; ook rang 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\"> De rangorde van matrix A is gelijk aan de rangorde van matrix A&#8217;, maar deze twee zijn kleiner dan het aantal onbekenden van het systeem (3), dus volgens de <strong>stelling van Rouch\u00e9-Frobenius<\/strong> weten we dat het een <strong>onbepaald systeemcompatibel<\/strong> (SCI) 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-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 class=\"has-text-align-left\"> Omdat we een ICS-systeem zijn, moeten we \u00e9\u00e9n vergelijking elimineren. In dit geval <strong>zullen we de laatste vergelijking uit het systeem verwijderen<\/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-4e10bd826663dff41c4272610cbc07b1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x-2y-3z=4 \\\\[1.5ex] -x+5y+4z=-10 \\\\[1.5ex]\\cancel{5x+y-2z = -2} \\end{cases} \\longrightarrow \\quad \\begin{cases} 3x-2y-3z=4 \\\\[1.5ex] -x+5y+4z=-10\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"423\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Laten we nu de variabele z naar \u03bb converteren:<\/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-2502be450040b38761c08e5d6beaf379_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x-2y-3z=4 \\\\[1.5ex] -x+5y+4z=-10  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} 3x-2y-3\\lambda=4 \\\\[1.5ex] -x+5y+4\\lambda=-10\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"444\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> En we plaatsen de termen met \u03bb bij de onafhankelijke termen:<\/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-80a43d98e6be30965d554e8a89aa5d89_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x-2y=4+3\\lambda \\\\[1.5ex] -x+5y=-10-4\\lambda\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"172\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Zodanig dat de matrix A en de matrix A&#8217; van het systeem blijven:<\/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-3451ce571163983cf41794d4998283d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; -2  \\\\[1.1ex] -1 &amp; 5 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 3 &amp; -2 &amp; 4+3\\lambda \\\\[1.1ex] 1 &amp; 5 &amp;-10-4\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"399\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ten slotte passen we, zodra we het systeem hebben getransformeerd, <strong>de regel van Cramer toe<\/strong> . We lossen daarom de determinant van A op:<\/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-0e7a7d6208ea5e762f5c74a44e6838cf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix}3&amp; -2 \\\\[1.1ex] -1 &amp; 5\\end{vmatrix} =13\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"162\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de eerste kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-8c30fcc0526c2d4112eb4f60a3d8847f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 4+3\\lambda &amp; -2 \\\\[1.1ex]-10-4\\lambda &amp; 5\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{20+15\\lambda -(20+8\\lambda)}{13} = \\cfrac{\\bm{7\\lambda}}{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"394\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Om het onbekende te berekenen<\/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-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Met de regel van Cramer veranderen we de tweede kolom van de determinant van A door de kolom met onafhankelijke termen en delen we deze door de determinant van 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-fdb22a54274e019c811c9051502c474a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 3 &amp; 4+3\\lambda \\\\[1.1ex] -1 &amp; -10-4\\lambda\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-30-12\\lambda -(-4-3\\lambda)}{13}= \\cfrac{\\bm{-26-9\\lambda}}{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"473\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De oplossing van het stelsel vergelijkingen is dus een functie van \u03bb, aangezien het een SCI is en het systeem daarom oneindig veel oplossingen heeft:<\/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-d0e525b9aca6bd683491ab7950f039e3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x=} \\cfrac{\\bm{7\\lambda}}{\\bm{13}} \\qquad \\bm{y=} \\cfrac{\\bm{-26-9\\lambda}}{\\bm{13}} \\qquad \\bm{z = \\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"266\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Op deze pagina zie je wat de regel van Cramer is en daarnaast vind je voorbeelden en oefeningen met het oplossen van stelsels vergelijkingen volgens de regel van Cramer. Wat is de regel van Cramer? De regel van Cramer is een methode die wordt gebruikt om stelsels vergelijkingen op te lossen met behulp van determinanten. &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/\"> <span class=\"screen-reader-text\">De regel van cramer<\/span> Lees meer &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-349","post","type-post","status-publish","format-standard","hentry","category-onderwijssystemen"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Regel van Cramer - 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\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/\" \/>\n<meta property=\"og:locale\" content=\"nl_NL\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Regel van Cramer - Mathority\" \/>\n<meta property=\"og:description\" content=\"Op deze pagina zie je wat de regel van Cramer is en daarnaast vind je voorbeelden en oefeningen met het oplossen van stelsels vergelijkingen volgens de regel van Cramer. 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Wat is de regel van Cramer? De regel van Cramer is een methode die wordt gebruikt om stelsels vergelijkingen op te lossen met behulp van determinanten. &hellip; De regel van cramer Lees meer &raquo;","og_url":"https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/","article_published_time":"2023-07-06T14:13:39+00:00","og_image":[{"url":"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e0141f3451719f665ef28e4061489551_l3.png"}],"author":"Redactioneel Team","twitter_card":"summary_large_image","twitter_misc":{"Geschreven door":"Redactioneel Team","Geschatte leestijd":"12 minuten"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/","url":"https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/","name":"Regel van Cramer - Mathority","isPartOf":{"@id":"https:\/\/mathority.org\/nl\/#website"},"datePublished":"2023-07-06T14:13:39+00:00","dateModified":"2023-07-06T14:13:39+00:00","author":{"@id":"https:\/\/mathority.org\/nl\/#\/schema\/person\/19b550cef1a9fbd238be112b7b7bbf64"},"breadcrumb":{"@id":"https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/#breadcrumb"},"inLanguage":"nl-NL","potentialAction":[{"@type":"ReadAction","target":["https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/mathority.org\/nl\/"},{"@type":"ListItem","position":2,"name":"De regel van cramer"}]},{"@type":"WebSite","@id":"https:\/\/mathority.org\/nl\/#website","url":"https:\/\/mathority.org\/nl\/","name":"","description":"Waar nieuwsgierigheid en berekening elkaar ontmoeten!","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/mathority.org\/nl\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"nl-NL"},{"@type":"Person","@id":"https:\/\/mathority.org\/nl\/#\/schema\/person\/19b550cef1a9fbd238be112b7b7bbf64","name":"Redactioneel Team","image":{"@type":"ImageObject","inLanguage":"nl-NL","@id":"https:\/\/mathority.org\/nl\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/8a35e4c8616d1c34c03ca02862b580f4372c5650665668489db53a09579bbc4f?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/8a35e4c8616d1c34c03ca02862b580f4372c5650665668489db53a09579bbc4f?s=96&d=mm&r=g","caption":"Redactioneel Team"},"sameAs":["http:\/\/mathority.org\/nl"]}]}},"yoast_meta":{"yoast_wpseo_title":"","yoast_wpseo_metadesc":"","yoast_wpseo_canonical":""},"_links":{"self":[{"href":"https:\/\/mathority.org\/nl\/wp-json\/wp\/v2\/posts\/349","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathority.org\/nl\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathority.org\/nl\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathority.org\/nl\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathority.org\/nl\/wp-json\/wp\/v2\/comments?post=349"}],"version-history":[{"count":0,"href":"https:\/\/mathority.org\/nl\/wp-json\/wp\/v2\/posts\/349\/revisions"}],"wp:attachment":[{"href":"https:\/\/mathority.org\/nl\/wp-json\/wp\/v2\/media?parent=349"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathority.org\/nl\/wp-json\/wp\/v2\/categories?post=349"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathority.org\/nl\/wp-json\/wp\/v2\/tags?post=349"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}