{"id":347,"date":"2023-07-06T14:52:33","date_gmt":"2023-07-06T14:52:33","guid":{"rendered":"https:\/\/mathority.org\/nl\/stelling-van-de-rouche-frobenius-met-voorbeelden-en-opgeloste-oefeningen\/"},"modified":"2023-07-06T14:52:33","modified_gmt":"2023-07-06T14:52:33","slug":"stelling-van-de-rouche-frobenius-met-voorbeelden-en-opgeloste-oefeningen","status":"publish","type":"post","link":"https:\/\/mathority.org\/nl\/stelling-van-de-rouche-frobenius-met-voorbeelden-en-opgeloste-oefeningen\/","title":{"rendered":"Stelling van rouche-fr\u00e9benius"},"content":{"rendered":"<p>Op deze pagina zullen we ontdekken wat de <strong>stelling van Rouch\u00e9 Frobenius<\/strong> is en hoe je daarmee de rang van een matrix kunt berekenen. Je vindt er ook voorbeelden en oefeningen die stap voor stap worden opgelost met de stelling van Rouch\u00e9-Frobenius.<\/p>\n<h2 class=\"wp-block-heading\"> Wat is de stelling van Rouch\u00e9-Frobenius?<\/h2>\n<p> <strong>De stelling van Rouch\u00e9-Frobenius is een methode voor het classificeren van stelsels van lineaire vergelijkingen.<\/strong> Met andere woorden, de stelling van Rouch\u00e9-Frobenius wordt gebruikt om erachter te komen hoeveel oplossingen een stelsel vergelijkingen heeft zonder het op te lossen.<\/p>\n<p> Er zijn 3 soorten stelsels vergelijkingen:<\/p>\n<ul>\n<li> <strong>Systeemcompatibel bepaald (SCD):<\/strong> Het systeem heeft een unieke oplossing.<\/li>\n<li> <strong>Onbepaald compatibel systeem (ICS):<\/strong> het systeem heeft oneindige oplossingen.<\/li>\n<li> <strong>Systeem incompatibel (SI):<\/strong> Het systeem heeft geen oplossing.<\/li>\n<\/ul>\n<p> Bovendien zal de stelling van Rouch\u00e9-Frobenius ons later ook in staat stellen <a href=\"https:\/\/mathority.org\/nl\/regelsvoorbeelden-en-opgeloste-oefeningen-van-cramer\/\">systemen op te lossen met behulp van de regel van Cramer<\/a> .<\/p>\n<h2 class=\"wp-block-heading\"> Verklaring van de stelling van Rouch\u00e9-Frobenius<\/h2>\n<p> De stelling van Rouch\u00e9-Frobenius zegt dat<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bd767a13412c19de65e75a6826caee08_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{A}\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"13\" style=\"vertical-align: 0px;\"><\/p>\n<p> is de matrix gevormd door de co\u00ebffici\u00ebnten van de onbekenden van een stelsel vergelijkingen. en de buik<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bf22ad0d457d763be692e97f3bcdf221_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{A'}\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"17\" style=\"vertical-align: 0px;\"><\/p>\n<p> , of <strong>uitgebreide matrix<\/strong> , is de matrix gevormd door de co\u00ebffici\u00ebnten van de onbekenden van een systeem van vergelijkingen en de onafhankelijke termen: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><\/figure>\n<\/div>\n<div style=\"background-color:#dff6ff;padding-top: 20px; padding-bottom: 0.5px; padding-right: 40px; padding-left: 30px\" class=\"has-background\">\n<p align=\"LEFT\" style=\"margin-bottom:20px\"> De <strong>stelling van Rouch\u00e9-Frobenius<\/strong> stelt ons in staat te weten met welk type systeem van vergelijkingen we te maken hebben, afhankelijk van de rangorde van de matrices A en A&#8217;:<\/p>\n<ul style=\"color:#E53935; font-weight: bold;\">\n<li style=\"margin-bottom:20px\"> <span style=\"color:#000000;font-weight: normal;\">Als rang(A) = rang(A&#8217;) = aantal onbekenden \u27f6 Bepaald compatibel systeem (SCD)<\/span><\/li>\n<li style=\"margin-bottom:20px;\"> <span style=\"color:#000000;font-weight: normal;\">Als rang(A) = rang(A&#8217;) &lt; aantal onbekenden \u27f6 Onbepaald compatibel systeem (SCI)<\/span><\/li>\n<li> <span style=\"color:#000000;font-weight: normal;\">als bereik (A)\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-be0e48d5500c7e73c450241ea2197789_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{\\neq}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"13\" style=\"vertical-align: -4px;\"><\/p>\n<p><\/span> bereik (A&#8217;) \u27f6 Incompatibel systeem (SI)<\/li>\n<\/ul>\n<\/div>\n<p> Zodra we weten wat de stelling van Rouch\u00e9-Frobenius zegt, zullen we zien hoe we de stellingoefeningen van Rouch\u00e9-Frobenius kunnen oplossen. Hier zijn 3 voorbeelden: een oefening die wordt opgelost met behulp van de stelling van elk type stelsel vergelijkingen.<\/p>\n<h2 class=\"wp-block-heading\"> Voorbeeld van een bepaald compatibel systeem (SCD)<\/h2>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b6b2f93c6308c25e8df2fbb5da2af9a8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 2x+y-3z=0 \\\\[1.5ex] x+2y-z= 1 \\\\[1.5ex] 4x-2y+z = 3\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"135\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De <strong>matrix A<\/strong> en de <strong>uitgebreide matrix A&#8217;<\/strong> 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-4597f5171b586bbcf0915d8512f7b89d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; -3  \\\\[1.1ex] 1 &amp; 2 &amp; -1  \\\\[1.1ex] 4 &amp; -2 &amp; 1  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 1 &amp; 2 &amp; -1 &amp; 1  \\\\[1.1ex] 4 &amp; -2 &amp; 1 &amp; 3\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> We berekenen nu de rangorde van de matrix A. 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-6c95b7158a2e6401cd16aeb708f128ff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; -3  \\\\[1.1ex] 1 &amp; 2 &amp; -1  \\\\[1.1ex] 4 &amp; -2 &amp; 1  \\end{vmatrix} = 25 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"219\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Omdat de matrix een 3\u00d73 determinant heeft die verschilt van 0, <strong>heeft matrix A 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> Zodra we de rang van A kennen, berekenen we de rang van A&#8217;, die op zijn minst rang 3 zal zijn, omdat we zojuist hebben gezien dat deze een determinant van orde 3 heeft die verschilt van 0. Bovendien kan deze niet van rang 4 zijn, aangezien we geen enkele determinant van orde 4 kunnen maken. Daarom <strong>is de matrix A&#8217; ook van 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-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> Omdat de rangorde van matrix A dus gelijk is aan de rangorde van matrix A&#8217; en aan het aantal onbekenden van het systeem (3), weten we door de stelling van Rouch\u00e9 Frobenius dat het een <strong>Compatibel Bepaald 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<h2 class=\"wp-block-heading\"> Voorbeeld van een onbepaald compatibel systeem (ICS)<\/h2>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2360b9a47257f73cf3f5dea63fb24098_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x-y+2z=1 \\\\[1.5ex] 3x+2y+z= 5 \\\\[1.5ex] 2x+3y-z = 4\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"135\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De <strong>matrix A<\/strong> en de <strong>uitgebreide matrix A&#8217;<\/strong> 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-b281235e2702433b447e2586ae3092c9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; -1 &amp; 2  \\\\[1.1ex] 3 &amp; 2 &amp; 1  \\\\[1.1ex] 2 &amp; 3 &amp; -1  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; -1 &amp; 2 &amp; 1 \\\\[1.1ex] 3 &amp; 2 &amp; 1 &amp; 5  \\\\[1.1ex] 2 &amp; 3 &amp; -1 &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> We berekenen nu de rangorde van de matrix A. 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-74cafc27ab41134696c3bf263132b98b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; -1 &amp; 2  \\\\[1.1ex] 3 &amp; 2 &amp; 1  \\\\[1.1ex] 2 &amp; 3 &amp; -1 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De determinant van de gehele matrix A geeft 0 en is dus niet van rang 3. Om te zien of deze van rang 2 is, moeten we een submatrix in A vinden waarvan de determinant verschilt van 0. Bijvoorbeeld die uit de linkerbovenhoek :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-22b2487f7664a70c116593120de2743b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; -1  \\\\[1.1ex] 3 &amp; 2 \\end{vmatrix} = 5 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"123\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> 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> Zodra we de rang van A kennen, berekenen we de rang van A&#8217;. We weten al dat de determinant van de eerste 3 kolommen 0 oplevert, dus proberen we de andere mogelijke 3\u00d73 determinanten:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-17f264ad3859da88ffa6784be24e4143_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}1 &amp; -1 &amp;  1 \\\\[1.1ex] 3 &amp; 2 &amp; 5  \\\\[1.1ex] 2 &amp; 3 &amp; 4\\end{vmatrix} = 0 \\quad \\begin{vmatrix}1 &amp; 2 &amp; 1 \\\\[1.1ex] 3 &amp;  1 &amp; 5  \\\\[1.1ex] 2 &amp; -1 &amp; 4\\end{vmatrix} = 0 \\quad \\begin{vmatrix} -1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 1 &amp; 5  \\\\[1.1ex] 3 &amp; -1 &amp; 4\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Alle 3\u00d73 determinanten van matrix A&#8217; zijn 0, dus matrix A&#8217; zal ook niet van rang 3 zijn. Binnenin heeft het echter determinanten van orde 2 die verschillen van 0. Bijvoorbeeld:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-22b2487f7664a70c116593120de2743b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; -1  \\\\[1.1ex] 3 &amp; 2 \\end{vmatrix} = 5 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"123\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Dus <strong>de matrix A&#8217; zal van rang 2 zijn<\/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> De omvang van matrix A is gelijk aan de omvang van matrix A&#8217;, maar deze zijn kleiner dan het aantal onbekenden van het systeem (3). Daarom is het volgens de stelling van Rouch\u00e9-Frobenius een <strong>onbepaald compatibel systeem<\/strong> (ICS):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> Voorbeeld van incompatibel systeem (IS)<\/h2>\n<div class=\"adsb30\" style=\" margin:px; text-align:\"><\/div>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-30e1084dd637eb4371f6b2218af24136_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 2x+y-2z=3 \\\\[1.5ex] 3x-2y+z= 2 \\\\[1.5ex] x+4-5z = 3 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"135\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p>De <strong>matrix A<\/strong> en de <strong>uitgebreide matrix A&#8217;<\/strong> 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-b435d86f1466af5748d91e6c9bd813e3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; -2 \\\\[1.1ex] 3 &amp; -2 &amp; 1 \\\\[1.1ex] 1 &amp; 4 &amp; -5 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; -2 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; 1 &amp; 2  \\\\[1.1ex] 1 &amp; 4 &amp; -5 &amp; 3 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> We berekenen nu de rangorde van de matrix A. 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-714538c91aa2620a6adb40581245f0e0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; -2 \\\\[1.1ex] 3 &amp; -2 &amp; 1 \\\\[1.1ex] 1 &amp; 4 &amp; -5 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De determinant van de gehele matrix A geeft 0 en is dus niet van rang 3. Om te zien of deze van rang 2 is, moeten we een submatrix in A vinden waarvan de determinant verschilt van 0. Bijvoorbeeld die uit de linkerbovenhoek :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5a46decda8fd850d9c847922b0c896db_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 1  \\\\[1.1ex] 3 &amp; -2 \\end{vmatrix} = -7 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"136\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Omdat de matrix een determinant van orde 2 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> Zodra we de rang van A kennen, berekenen we de rang van A&#8217;. We weten al dat de determinant van de eerste 3 kolommen 0 oplevert, dus nu proberen we bijvoorbeeld met de determinant van de laatste 3 kolommen:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-47aecdf801b92f21f2287fb96eaaa3f8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; -2 &amp; 3 \\\\[1.1ex]  -2 &amp; 1 &amp; 2  \\\\[1.1ex]  4 &amp; -5 &amp; 3 \\end{vmatrix} = 3 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"161\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Aan de andere kant bevat matrix A&#8217; wel een determinant waarvan het resultaat anders is dan 0, daarom <strong>zal matrix A&#8217; rang 3 hebben<\/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> Omdat de rangorde van matrix A kleiner is dan de rangorde van matrix A&#8217;, leiden we daarom uit de stelling van Rouch\u00e9-Frobenius af dat het een <strong>incompatibel systeem<\/strong> (SI) is <strong>:<\/strong> <\/p>\n<p class=\"has-text-align-center\">\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c3da0513f318d25473e93ba88c51fb42_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = 2 \\ \\neq \\ rg(A') = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\">Problemen van de stelling van Rouch\u00e9-Frobenius opgelost <\/h2>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-118\"><\/div>\n<\/div>\n<h3 class=\"estil_titol_H3 wp-block-heading\"> Oefening 1<\/h3>\n<p> Bepaal het type van het volgende stelsel vergelijkingen met 3 onbekenden met behulp van de stelling van Rouch\u00e9-Frobenius: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-theoreme-de-rouche-8211-frebenius-1.webp\" alt=\"Opgeloste oefening van de stelling van Rouche - frobenius\" class=\"wp-image-3984\" width=\"193\" height=\"122\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>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-951ce5c1f0c606d4f060a1de58b60303_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -1 &amp; -1 \\\\[1.1ex] -2 &amp; 4 &amp; 2 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 3 &amp; -1 &amp; -1 &amp; 2 \\\\[1.1ex] -2 &amp; 4 &amp; 2 &amp; 8 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"432\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We moeten nu de rangorde van de matrix A vinden. Om dit te doen, controleren we of de determinant van de 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-15cddb69f7590648d1d6ae61d942471e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -1 &amp; -1 \\\\[1.1ex] -2 &amp; 4 &amp; 2 \\end{vmatrix} = -4+2-36+6+8-6=-30 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"450\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De matrix heeft een derde orde determinant 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\"> Zodra we de rang van A kennen, berekenen we de rang van A&#8217;. Dit zal op zijn minst van rang 3 zijn, omdat we zojuist hebben gezien dat er een determinant in zit van orde 3 die verschilt van 0. Bovendien kan het niet van rang 4 zijn, omdat we geen 4\u00d74-determinant kunnen maken. Daarom <strong>is de matrix A&#8217; ook van 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-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\"> Dankzij de stelling van Rouch\u00e9-Frobenius weten we dus 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<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Oefening 2<\/h3>\n<p> Classificeer het volgende stelsel vergelijkingen met 3 onbekenden met behulp van de stelling van Rouch\u00e9-Frobenius: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-theoreme-de-rouche-8211-frebenius-2.webp\" alt=\"Opgeloste oefening van de stelling van Rouche-Frobenius\" class=\"wp-image-3987\" width=\"190\" height=\"121\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>zie oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> We construeren 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-45e13aabe233ece927df7c9ba0bb3ec1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}3 &amp; -1 &amp; 2  \\\\[1.1ex] 1 &amp; 2 &amp; -2  \\\\[1.1ex] 1 &amp; -5 &amp; 6 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 3 &amp; -1 &amp; 2 &amp; 1 \\\\[1.1ex] 1 &amp; 2 &amp; -2 &amp; 5 \\\\[1.1ex] 1 &amp; -5 &amp; 6 &amp; -9 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Laten we nu het bereik van matrix A 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-87bc95df0033834bba0398b8421faac5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; -1 &amp; 2 \\\\[1.1ex] 1 &amp; 2 &amp; -2 \\\\[1.1ex] 1 &amp; -5 &amp; 6 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b9805283b75e2b89f67c7865a1263112_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; -1  \\\\[1.1ex] 1 &amp; 2 \\end{vmatrix} = 7 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"123\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <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 class=\"has-text-align-left\"> Zodra we de rang van A kennen, berekenen we de rang van A&#8217;. We weten al dat de determinant van de eerste 3 kolommen 0 oplevert, dus proberen we de andere mogelijke 3\u00d73 determinanten:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e6457fe3f03722b7f0d955191f318915_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}-1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; -2 &amp; 5 \\\\[1.1ex] -5 &amp; 6 &amp; -9\\end{vmatrix} = 0 \\quad \\begin{vmatrix}3 &amp; 2 &amp; 1 \\\\[1.1ex] 1 &amp; -2 &amp; 5 \\\\[1.1ex] 1 &amp; 6 &amp; -9\\end{vmatrix} = 0 \\quad \\begin{vmatrix} 3 &amp; -1 &amp; 1 \\\\[1.1ex] 1 &amp; 2 &amp; 5 \\\\[1.1ex] 1 &amp; -5 &amp; -9\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"446\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Alle 3\u00d73 determinanten van matrix A&#8217; zijn 0, dus matrix A&#8217; zal ook niet van rang 3 zijn. Binnenin heeft het echter veel determinanten van orde 2 die verschillen van 0. Bijvoorbeeld:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eafa4747802fae3f0c36350357abbeb2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -1 &amp; 2  \\\\[1.1ex] 2 &amp; -2 \\end{vmatrix} = -2 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"150\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dus <strong>de matrix A&#8217; zal van rang 2 zijn<\/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\"> 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). Daarom weten we door de stelling van Rouch\u00e9-Frobenius 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<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Oefening 3<\/h3>\n<p> Bepaal welk type systeem het volgende stelsel vergelijkingen gebruikt volgens de stelling van Rouch\u00e9-Frobenius: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-theoreme-de-rouche-8211-frebenius-3.webp\" alt=\"oefening stap voor stap opgelost van de rouche-stelling - frobenius\" class=\"wp-image-3990\" width=\"188\" height=\"122\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>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-1820d31e4fd5c79804c9b6fa15abb469_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 4 &amp; -2 \\\\[1.1ex] 3 &amp; -1 &amp; 3  \\\\[1.1ex] 5 &amp; 7 &amp; -1 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 4 &amp; -2 &amp; 3 \\\\[1.1ex] 3 &amp; -1 &amp; 3 &amp; -2 \\\\[1.1ex] 5 &amp; 7 &amp; -1 &amp; 0 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Laten we nu het bereik van matrix A 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-4f998260ee4c96673085ea6fd4ca87ba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 4 &amp; -2 \\\\[1.1ex] 3 &amp; -1 &amp; 3 \\\\[1.1ex] 5 &amp; 7 &amp; -1\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-159a1c58fdcd972b4b08e4795950e064_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 4  \\\\[1.1ex] 3 &amp; -1 \\end{vmatrix} = -13 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <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 class=\"has-text-align-left\"> Zodra we de rang van A kennen, berekenen we de rang van A&#8217;. We weten al dat de determinant van de eerste 3 kolommen 0 oplevert, maar niet de determinant van de laatste 3 kolommen:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c673a5bbbd41933208169fa3e08b7c62_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 4 &amp; -2 &amp; 3 \\\\[1.1ex]-1 &amp; 3 &amp; -2 \\\\[1.1ex] 7 &amp; -1 &amp; 0 \\end{vmatrix} = -40 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"198\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Daarom <strong>heeft de matrix A&#8217; 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-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\"> De rangorde van matrix A is kleiner dan de rangorde van matrix A&#8217;, we kunnen daarom uit de stelling van Rouch\u00e9-Frobenius afleiden dat het een <strong>incompatibel systeem<\/strong> (SI) 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-c3da0513f318d25473e93ba88c51fb42_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = 2 \\ \\neq \\ rg(A') = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-119\"><\/div>\n<\/div>\n<h3 class=\"wp-block-heading\"> Oefening 4<\/h3>\n<p> Bepaal het type van het volgende stelsel vergelijkingen met 3 onbekenden met behulp van de stelling van Rouch\u00e9-Frobenius: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-theoreme-de-rouche-8211-frobenius-3-inconnues-3-equations.webp\" alt=\"Rouche - Stelling van Frobenius opgeloste oefening met 3 onbekenden en 3 vergelijkingen\" class=\"wp-image-3991\" width=\"203\" height=\"122\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>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-f8a0454c53a64f612c689ba1dae1196b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 5 &amp; -3 &amp; -2  \\\\[1.1ex] 1 &amp; 4 &amp; 1  \\\\[1.1ex]-3 &amp; 2 &amp; -2  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 5 &amp; -3 &amp; -2 &amp; -2 \\\\[1.1ex] 1 &amp; 4 &amp; 1 &amp; 7 \\\\[1.1ex]-3 &amp; 2 &amp; -2 &amp; 3 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"446\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We moeten nu de rangorde van de matrix A berekenen. Om dit te doen, lossen we de determinant van de matrix op met de Sarrus-regel:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-420f0d1ee000f39cbfbce88bf122f413_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 5 &amp; -3 &amp; -2 \\\\[1.1ex] 1 &amp; 4 &amp; 1 \\\\[1.1ex]-3 &amp; 2 &amp; -2 \\end{vmatrix} = -40+9-4-24-10-6=-75 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"467\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De matrix heeft een derde orde determinant 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\"> Daarom <strong>heeft de matrix A&#8217; ook rang 3<\/strong> , aangezien deze altijd minimaal rang A heeft en niet rang 4 kan zijn omdat we geen enkele 4\u00d74 determinant kunnen oplossen.<\/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\"> Dankzij de toepassing van de stelling van Rouch\u00e9-Frobenius weten we dus dat het systeem een <strong>Compatible 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<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Oefening 5<\/h3>\n<p> Bepaal welk type systeem het volgende stelsel vergelijkingen de stelling van Rouch\u00e9-Frobenius gebruikt: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exemple-du-theoreme-de-rouche-8211-frebenius.webp\" alt=\"voorbeeld van de stelling van Rouche - frobenius\" class=\"wp-image-3992\" width=\"205\" height=\"122\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>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-3211e276b2b040969c38bc6c69eabd52_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 &amp; 3 \\\\[1.1ex] -1 &amp; 7 &amp; 3 \\\\[1.1ex] -5 &amp; 8 &amp; 0 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 4 &amp; -1 &amp; 3 &amp; 5 \\\\[1.1ex] -1 &amp; 7 &amp; 3 &amp; -3 \\\\[1.1ex] -5 &amp; 8 &amp; 0 &amp; 9 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Laten we nu het bereik van matrix A 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-21004095a3a8ef3edfc15bed5c7853a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 4 &amp; -1 &amp; 3 \\\\[1.1ex] -1 &amp; 7 &amp; 3 \\\\[1.1ex] -5 &amp; 8 &amp; 0\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a58059046b56cf1f8d82c6c8939e44ca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 4 &amp; -1  \\\\[1.1ex]  -1 &amp; 7 \\end{vmatrix} = 27 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Matrix A is daarom van 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;. De determinant van de eerste 3 kolommen die we al kennen geeft 0, maar de determinant van de laatste 3 kolommen geeft niet:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-992718d3b50aedf77c80c262fad5845f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -1 &amp; 3 &amp; 5 \\\\[1.1ex]  7 &amp; 3 &amp; -3 \\\\[1.1ex] 8 &amp; 0 &amp; 9\\end{vmatrix} = -408 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"193\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Daarom <strong>heeft de matrix A&#8217; 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-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\"> En ten slotte passen we het domein toe op de stelling van Rouch\u00e9-Frobenius: het domein van de matrix A is kleiner dan het domein van de matrix A&#8217;, het is daarom een <strong>incompatibel systeem<\/strong> (SI) <strong>:<\/strong> <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c3da0513f318d25473e93ba88c51fb42_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = 2 \\ \\neq \\ rg(A') = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Oefening 6<\/h3>\n<p> Classificeer het volgende systeem van vergelijkingen van orde 3 met de stelling van Rouch\u00e9-Frobenius: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d45e8bc425b08e403a98e01693201681_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 6x-2y+4z=1 \\\\[1.5ex] -2x+4y+3z= 7 \\\\[1.5ex] 8x-6y+z = -6\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"158\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>zie oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> We construeren 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-8e779eca9135adc44e4a3a55f368560f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 6 &amp; -2 &amp; 4 \\\\[1.1ex] -2 &amp; 4 &amp; 3 \\\\[1.1ex] 8 &amp; -6 &amp; 1  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 6 &amp; -2 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; 4 &amp; 3 &amp; 7 \\\\[1.1ex] 8 &amp; -6 &amp; 1 &amp; -6 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Laten we nu het bereik van matrix A 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-c2f63f79858eae462547cf2f270fc780_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 6 &amp; -2 &amp; 4 \\\\[1.1ex] -2 &amp; 4 &amp; 3 \\\\[1.1ex] 8 &amp; -6 &amp; 1 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e5fa293b94b8c6acfd998f1e154abf7a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 6 &amp; -2  \\\\[1.1ex] -2 &amp; 4 \\end{vmatrix} = 20  \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <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 class=\"has-text-align-left\"> Zodra we de rang van A kennen, berekenen we de rang van A&#8217;. We weten al dat de determinant van de eerste 3 kolommen 0 oplevert, dus proberen we de andere mogelijke 3\u00d73 determinanten:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-98958f866454a1bf9f1ac078562065cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -2 &amp; 4 &amp; 1 \\\\[1.1ex]4 &amp; 3 &amp; 7 \\\\[1.1ex] -6 &amp; 1 &amp; -6\\end{vmatrix} = 0 \\quad \\begin{vmatrix}6 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; 3 &amp; 7 \\\\[1.1ex] 8 &amp;  1 &amp; -6\\end{vmatrix} = 0 \\quad \\begin{vmatrix} 6 &amp; -2 &amp; 1 \\\\[1.1ex] -2 &amp; 4 &amp; 7 \\\\[1.1ex] 8 &amp; -6 &amp; -6\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"446\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Alle 3\u00d73 determinanten van matrix A&#8217; zijn 0, dus matrix A&#8217; zal ook niet van rang 3 zijn. Binnenin heeft het echter determinanten van orde 2 die verschillen van 0. Bijvoorbeeld:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-58091f1a37a4ef81fdf56f01dd9531a3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 6 &amp; -2 \\\\[1.1ex] -2 &amp; 4 \\end{vmatrix} = 20 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dus <strong>de matrix A&#8217; zal van rang 2 zijn<\/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\"> Ten slotte weten we door toepassing van de stelling van Rouch\u00e9-Frobenius dat het een <strong>onbepaald compatibel systeem<\/strong> (ICS) is, omdat het bereik van matrix A gelijk is aan het bereik van matrix A&#8217;, maar deze twee zijn kleiner dan het aantal onbekenden in de systeem(3): <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Op deze pagina zullen we ontdekken wat de stelling van Rouch\u00e9 Frobenius is en hoe je daarmee de rang van een matrix kunt berekenen. Je vindt er ook voorbeelden en oefeningen die stap voor stap worden opgelost met de stelling van Rouch\u00e9-Frobenius. Wat is de stelling van Rouch\u00e9-Frobenius? De stelling van Rouch\u00e9-Frobenius is een methode &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/nl\/stelling-van-de-rouche-frobenius-met-voorbeelden-en-opgeloste-oefeningen\/\"> <span class=\"screen-reader-text\">Stelling van rouche-fr\u00e9benius<\/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-347","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>Stelling van Rouche-Fr\u00e9benius - Mathority<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathority.org\/nl\/stelling-van-de-rouche-frobenius-met-voorbeelden-en-opgeloste-oefeningen\/\" \/>\n<meta property=\"og:locale\" content=\"nl_NL\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Stelling van Rouche-Fr\u00e9benius - Mathority\" \/>\n<meta property=\"og:description\" content=\"Op deze pagina zullen we ontdekken wat de stelling van Rouch\u00e9 Frobenius is en hoe je daarmee de rang van een matrix kunt berekenen. 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