{"id":111,"date":"2023-09-17T05:59:45","date_gmt":"2023-09-17T05:59:45","guid":{"rendered":"https:\/\/mathority.org\/nl\/omgekeerde-matrix\/"},"modified":"2023-09-17T05:59:45","modified_gmt":"2023-09-17T05:59:45","slug":"omgekeerde-matrix","status":"publish","type":"post","link":"https:\/\/mathority.org\/nl\/omgekeerde-matrix\/","title":{"rendered":"Hoe de inverse matrix te berekenen"},"content":{"rendered":"<p>Op deze pagina leert u wat het is en hoe u de inverse van een matrix kunt berekenen met de methode van determinanten (of adjunct-matrix) en met de Gauss-methode. Je zult ook alle eigenschappen van de inverse matrix zien, en je zult ook stap voor stap opgeloste voorbeelden en oefeningen voor elke methode vinden, zodat je ze volledig begrijpt. Ten slotte leggen we een formule uit voor het snel inverteren van een 2&#215;2-matrix en zelfs het grootste nut van deze matrixbewerking: het oplossen van een systeem van lineaire vergelijkingen.<\/p>\n<h2 class=\"wp-block-heading\"> Wat is het omgekeerde van een matrix? <\/h2>\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\"> 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-25b206f25506e6d6f46be832f7119ffa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"13\" style=\"vertical-align: 0px;\"><\/p>\n<p> een vierkante matrix. De <strong>inverse matrix<\/strong> van<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-25b206f25506e6d6f46be832f7119ffa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"13\" style=\"vertical-align: 0px;\"><\/p>\n<p> het is geschreven<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e2b32875906f7ed9c10ffd1b09a6ed5e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A^{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"30\" style=\"vertical-align: 0px;\"><\/p>\n<p> , en het is deze matrix die voldoet aan:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fd42c364eee57f5eada44b8ef06f254a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A \\cdot A^{-1} = I\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" 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-a003e1fd3042f8cd7ec7d3fe7f286f5b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A^{-1}\\cdot A  = I\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p align=\"LEFT\"> Goud<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-18b5e45cb4a1ee02e81b9a980f828db8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> is de Identiteitsmatrix.<\/p>\n<\/div>\n<h2 class=\"wp-block-heading\"> Wanneer kun je een matrix omkeren en wanneer niet?<\/h2>\n<p> De eenvoudigste manier om de invertibiliteit van een matrix te bepalen, is door de determinant ervan te gebruiken:<\/p>\n<ul>\n<li> Als de determinant van de betreffende matrix verschillend is van 0, betekent dit dat de matrix inverteerbaar is. In dit geval zeggen we dat het een reguliere matrix is. Bovendien impliceert dit dat de matrix de maximale rang heeft.<\/li>\n<\/ul>\n<ul>\n<li> Aan de andere kant, als de determinant van de matrix gelijk is aan 0, kan de matrix niet worden omgekeerd. En in dit geval zeggen we dat het een enkelvoudige of gedegenereerde matrix is.<\/li>\n<\/ul>\n<p> Er zijn hoofdzakelijk twee methoden om elke matrix om te keren: de methode van determinanten of adjunct-matrix en de Gauss-methode. Hieronder heb je de uitleg van het eerste, maar je kunt hieronder ook raadplegen hoe je een matrix kunt omkeren met de Gauss-methode.<\/p>\n<h2 class=\"wp-block-heading\"> Een matrix omkeren met behulp van de determinantenmethode (of met behulp van de aangrenzende matrix) <\/h2>\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\"> Om de <strong>inverse van een matrix<\/strong> 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-3017946e4911f6188e04dfdca6f050ba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"30\" style=\"vertical-align: 0px;\"><\/p>\n<p> , moet de volgende formule worden toegepast:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p align=\"LEFT\" style=\"margin-bottom:8px\"> Goud:<\/p>\n<ul>\n<li style=\"margin-bottom:12px\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a80d0312d139244060532c8c78fe6140_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix} A \\end{vmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"21\" style=\"vertical-align: -7px;\"><\/p>\n<p> is de determinant van de matrix<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-25b206f25506e6d6f46be832f7119ffa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"13\" style=\"vertical-align: 0px;\"><\/p>\n<\/li>\n<li style=\"margin-bottom:12px\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e87ef954487ce9371eac7dc25f234613_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adj}(A)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"55\" style=\"vertical-align: -5px;\"><\/p>\n<p> is de adjunct-matrix van<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-25b206f25506e6d6f46be832f7119ffa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"13\" style=\"vertical-align: 0px;\"><\/p>\n<\/li>\n<li> De exposant\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-50d6971192a73f12b183dbddd7c75197_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{t}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: 0px;\"><\/p>\n<p> geeft matrixtransponering aan, dwz dat de bijgevoegde matrix moet worden getransponeerd.<\/li>\n<\/ul>\n<\/div>\n<p> <strong>Opmerking:<\/strong> Sommige boeken gebruiken een enigszins andere inverse matrixformule: ze transponeren eerst matrix A en berekenen vervolgens de bijbehorende matrix, in plaats van eerst de aangrenzende matrix te berekenen en deze vervolgens te transponeren. In werkelijkheid doet de volgorde er niet toe, omdat het resultaat precies hetzelfde is. Hier laten we u de formule achter om een gewijzigde matrix om te keren, voor het geval u deze liever 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\/formule-de-la-matrice-inverse-adjointe-de-transposee-3.webp\" alt=\"formule voor de inverse matrix met de adjunct-matrix van de getransponeerde\" class=\"wp-image-4372\" width=\"238\" height=\"239\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p> We zullen dan zien <strong>hoe we de inverse van een matrix kunnen vinden<\/strong> door een oefening als voorbeeld op te lossen:<\/p>\n<h3 class=\"wp-block-heading\"> Voorbeeld van het berekenen van de inverse matrix met behulp van de determinantenmethode (of adjunct-matrix):<\/h3>\n<ul>\n<li> Bereken de inverse van de volgende matrix:<\/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-c37ec4a7afd5b313bcf3c50d6ce26c6d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A = \\begin{pmatrix} 4 &amp; -2  \\\\[1.1ex] 3 &amp; -1  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"109\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Om de inverse van de matrix te bepalen, moeten we de volgende formule toepassen: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/formule-de-la-matrice-inverse-avec-la-methode-par-determinants-ou-par-la-matrice-adjointe.webp\" alt=\"formule van de inverse matrix met de methode door determinanten of door de adjunct-matrix\" width=\"218\" height=\"59\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p> Maar als de determinant van de matrix nul is, betekent dit dat de matrix niet inverteerbaar is. Daarom is het eerste dat u moet doen het berekenen van de determinant van de matrix en controleren of deze 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-710ccd4e4912dd492b496a742eaf7f56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\lvert A \\rvert  = \\begin{vmatrix}  4 &amp; -2  \\\\[1.1ex] 3 &amp; -1 \\end{vmatrix} = -4- (-6) = 2\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"240\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De <strong>determinant is niet 0<\/strong> , dus <strong>de matrix is inverteerbaar<\/strong> .<\/p>\n<p> Als we daarom de waarde van de determinant in de formule vervangen, zal het omgekeerde van de matrix 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-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9be7ff27e83825750fc7b378f743412f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{2} \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"161\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> We moeten nu de plaatsvervanger van A berekenen. Om dit te doen, moeten we elk element van matrix A vervangen door zijn plaatsvervanger. <\/p>\n<div style=\"background-color:#fffde7;padding-top: 20px; padding-bottom: 0.5px; padding-right: 40px; padding-left: 30px\" class=\"has-background\">\n<p align=\"LEFT\"> Onthoud dat om de <strong>bijlage<\/strong> van 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-41d4a89db3722950dc94351832a1bcd9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a_{ij}\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"20\" style=\"vertical-align: -6px;\"><\/p>\n<p> , dat wil zeggen van het rij-element<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-695d9d59bd04859c6c99e7feb11daab6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"i\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: 0px;\"><\/p>\n<p> en de kolom<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-43c82d5bb00a7568d935a12e3bd969dd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"j\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> , moet de volgende formule worden toegepast:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-dcce4b79a3549da03df7c78b678add31_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de } a_{ij} = (-1)^{i+j} \\bm{\\cdot} \\text{Menor complementario de } a_{ij}\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"430\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p align=\"LEFT\"> Waar de complementaire minor van<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-41d4a89db3722950dc94351832a1bcd9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a_{ij}\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"20\" style=\"vertical-align: -6px;\"><\/p>\n<p> is de determinant van de matrix die de rij elimineert<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-695d9d59bd04859c6c99e7feb11daab6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"i\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: 0px;\"><\/p>\n<p> en de kolom<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-43c82d5bb00a7568d935a12e3bd969dd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"j\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> .<\/p>\n<\/div>\n<p> De plaatsvervangers van de elementen van matrix A zijn dus: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c37ec4a7afd5b313bcf3c50d6ce26c6d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A = \\begin{pmatrix} 4 &amp; -2  \\\\[1.1ex] 3 &amp; -1  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"109\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-981d47faf70cc1377c1abb515419a881_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 4} =\\displaystyle (-1)^{1+1} \\bm{\\cdot} \\begin{vmatrix} -1 \\end{vmatrix} = 1 \\cdot (-1) = \\bm{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"357\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0f1e6a5a5c504b3b6d06e5d3d8e0862e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de -2} =\\displaystyle (-1)^{1+2} \\bm{\\cdot} \\begin{vmatrix} 3 \\end{vmatrix} = -1 \\cdot 3 = \\bm{-3}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"336\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-02a2bf190ba8788264d0326f38cb0a21_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 3}  =\\displaystyle (-1)^{2+1} \\bm{\\cdot} \\begin{vmatrix} -2 \\end{vmatrix} = -1 \\cdot (-2) = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"357\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f01b7eb06a25b50bf15fbfd08e68cd13_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de -1} =\\displaystyle (-1)^{2+2} \\bm{\\cdot} \\begin{vmatrix} 4 \\end{vmatrix} = 1 \\cdot 4 = \\bm{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"309\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p> <strong>Opmerking:<\/strong> Verwar de determinant 1\u00d71 niet met de absolute waarde, want in de determinant 1\u00d71 wordt het getal niet omgezet naar positief.<\/p>\n<p> Zodra de plaatsvervangers zijn berekend, vervangt u eenvoudigweg de elementen van A door hun plaatsvervangers om de <strong>plaatsvervangermatrix van A<\/strong> te vinden:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-08fb7666b4518399c2a469ba445762be_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\displaystyle \\text{Adj}(A) = \\begin{pmatrix} -1 &amp; -3  \\\\[1.1ex] 2 &amp; 4  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"165\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong>Opmerking:<\/strong> op bepaalde plaatsen is de adjunct-matrix de transpositie van de adjunct-matrix die we hier defini\u00ebren.<\/p>\n<p> Daarom vervangen we de bijgevoegde matrix in de inverse matrixformule en deze wordt:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9be7ff27e83825750fc7b378f743412f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{2} \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"161\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0abb4127db9c3c1d0a7b669fbc782605_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{2} \\cdot \\begin{pmatrix} -1 &amp; -3  \\\\[1.1ex] 2 &amp; 4  \\end{pmatrix} ^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"57\" width=\"173\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De exposant<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-50d6971192a73f12b183dbddd7c75197_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{t}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: 0px;\"><\/p>\n<p> Dit vertelt ons dat we <strong>de matrix moeten transponeren<\/strong> . En om een matrix te transponeren moet je <strong>de rijen in kolommen veranderen<\/strong> , dat wil zeggen dat de eerste rij van de matrix de eerste kolom van de matrix wordt, en de tweede rij de tweede kolom:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-22965912cf8aee99610c81cf575c0ecd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{2} \\cdot \\begin{pmatrix} -1 &amp; 2  \\\\[1.1ex] -3 &amp; 4  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"151\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> En ten slotte vermenigvuldigen we elke term van de matrix met<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3a7c03ba828b3d8aef58199ac2c95a47_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{1}{2} :\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"18\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-220748840151b429919c7ce6587b1bc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\begin{pmatrix} \\sfrac{-1}{2} &amp; \\sfrac{2}{2}  \\\\[1.1ex] \\sfrac{-3}{2} &amp; \\sfrac{4}{2}  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/matrice-inverse-de-lexercice-resolu-par-les-determinants-22152.webp\" alt=\"oefening opgeloste inverse matrix met 2x2 determinanten\" width=\"188\" height=\"69\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<h3 class=\"wp-block-heading\"> Opgeloste oefeningen over inverse matrices met de methode van determinanten (of de aangrenzende matrix)<\/h3>\n<h4 class=\"wp-block-heading\"> Oefening 1<\/h4>\n<p> Inverteer de volgende matrix van afmeting 2\u00d72 met behulp van de adjunct-matrixmethode: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bfb0807249e78845b375a402eb23a32b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} 1 &amp; 3 \\\\[1.1ex] 2 &amp; 7  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"95\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> De inverse matrixformule 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-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We berekenen eerst de determinant van de matrix:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1c4e3bac90eb0da0361b4be1a2225146_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}A\\end{vmatrix}=\\begin{vmatrix} 1 &amp; 3 \\\\[1.1ex] 2 &amp; 7 \\end{vmatrix} = 7-6 = 1\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"187\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De determinant is anders dan 0, dus de matrix kan worden omgekeerd.<\/p>\n<p class=\"has-text-align-left\"> Laten we nu de adjunct-matrix van 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-34ac8739bfee66d594eee01b7a2b9205_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 1} =\\displaystyle (-1)^{1+1} \\bm{\\cdot} \\begin{vmatrix} 7 \\end{vmatrix} = 1 \\cdot 7 = \\bm{7}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"303\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4021288fe5f1db07d81dbb43ce15e82a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 3} =\\displaystyle (-1)^{1+2} \\bm{\\cdot} \\begin{vmatrix} 2\\end{vmatrix} = -1 \\cdot 2 = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"329\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-12783f7673a347fc5e0df04917332fa0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 2}  =\\displaystyle (-1)^{2+1} \\bm{\\cdot} \\begin{vmatrix} 3 \\end{vmatrix} = -1 \\cdot 3 = \\bm{-3}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"330\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8d48d00b2e8df51348f8f41c96b9197b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 7} =\\displaystyle (-1)^{2+2} \\bm{\\cdot} \\begin{vmatrix} 1 \\end{vmatrix} = 1 \\cdot 1 = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"302\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3dea8fca2c025ff9b7d7673904344996_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\displaystyle \\text{Adj}(A) = \\begin{pmatrix} 7 &amp; -2  \\\\[1.1ex] -3 &amp; 1 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"165\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Zodra de determinant van de matrix en zijn adjunct zijn berekend, vervangen we hun waarden in de formule: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9475e4162eff7e1ed9c08f363a8279ec_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{1} \\cdot \\begin{pmatrix} 7 &amp; -2 \\\\[1.1ex] -3 &amp; 1 \\end{pmatrix}^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"57\" width=\"173\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We transponeren de bijgevoegde matrix:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a5a6aaa8168e55c6eab1e3be1229a3da_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = 1 \\cdot \\begin{pmatrix} 7 &amp; -3 \\\\[1.1ex] -2 &amp; 1 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"162\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De inverse matrix van A 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-1236ad7262705dbbd9b0a094084ceac5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{A^{-1} =} \\begin{pmatrix} \\bm{7} &amp; \\bm{-3} \\\\[1.1ex] \\bm{-2} &amp; \\bm{1} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"139\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\">Oefening 2<\/h4>\n<p> Inverteer de volgende vierkante matrix met behulp van de determinantenmethode: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eb735917d200ed35918cd44be6bd155b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} -3 &amp; -2 \\\\[1.1ex] 5 &amp; 4  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"122\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> De inverse matrixformule 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-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We berekenen eerst de determinant van de matrix:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-49cd3daf7c50c811e78c29efe036bda4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}A\\end{vmatrix}=\\begin{vmatrix} -3 &amp; -2 \\\\[1.1ex] 5 &amp; 4\\end{vmatrix} = -12+10 = -2\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"260\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De determinant is anders dan 0, dus de matrix kan worden omgekeerd.<\/p>\n<p class=\"has-text-align-left\"> Laten we nu de adjunct-matrix van 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-8bff9fecfd83ca1edacba562d8714cbf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de -3} =\\displaystyle (-1)^{1+1} \\bm{\\cdot} \\begin{vmatrix} 4 \\end{vmatrix} = 1 \\cdot 4 = \\bm{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"309\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b558b2d47ccf4b3065ed8b26ab620502_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de -2} =\\displaystyle (-1)^{1+2} \\bm{\\cdot} \\begin{vmatrix} 5\\end{vmatrix} = -1 \\cdot 5 = \\bm{-5}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"335\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b6d0a26085435c08c6d60ab80f4fbb2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 5}  =\\displaystyle (-1)^{2+1} \\bm{\\cdot} \\begin{vmatrix} -2 \\end{vmatrix} = -1 \\cdot (-2) = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"357\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-79a60c5e3003ea311503867a147c1500_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 4} =\\displaystyle (-1)^{2+2} \\bm{\\cdot} \\begin{vmatrix} -3 \\end{vmatrix} = 1 \\cdot (-3) = \\bm{-3}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"358\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-208ab7161076485ca6928bd1208f6714_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\displaystyle \\text{Adj}(A) = \\begin{pmatrix} 4 &amp; -5  \\\\[1.1ex] 2 &amp; -3 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"151\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Zodra de determinant van de matrix en zijn adjunct zijn gevonden, vervangen we hun waarden in de formule: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-babecc87455bdc54006a77ba5369e540_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{-2} \\cdot \\begin{pmatrix} 4 &amp; -5 \\\\[1.1ex] 2 &amp; -3 \\end{pmatrix}^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"57\" width=\"173\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We transponeren de bijgevoegde matrix:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-17529597656a112a27d136ca212834d8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{-2} \\cdot \\begin{pmatrix} 4 &amp; 2 \\\\[1.1ex] -5 &amp; -3 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We vermenigvuldigen elk element met <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6f70ecad488bad8503fe7f8427180e2e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{1}{-2} :\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"32\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-be52d2df839244cbb0b0ee00c9e45265_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\begin{pmatrix} \\cfrac{4}{-2} &amp; \\cfrac{2}{-2} \\\\[3ex] \\cfrac{-5}{-2} &amp; \\cfrac{-3}{-2} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"163\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De inverse matrix van A 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-13e218c7d075daba3f875345f324d001_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{A^{-1} =} \\begin{pmatrix} \\bm{-2} &amp; \\bm{-1} \\\\[2ex] \\cfrac{\\bm{5}}{\\bm{2}} &amp; \\cfrac{\\bm{3}}{\\bm{2}} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"76\" width=\"141\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\">Oefening 3<\/h4>\n<p> Inverteer de volgende matrix van afmeting 3\u00d73 met behulp van de adjunct-matrixmethode: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d1b6a5f638281754d80983b5a50e15be_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix}2&amp;3&amp;-2\\\\[1.1ex] 1&amp;4&amp;1\\\\[1.1ex] 2&amp;1&amp;-3\\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"136\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> De inverse matrixformule 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-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We lossen eerst de determinant van de matrix op 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-fcac1cb3935b1000b6493a2866e8728a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}A\\end{vmatrix}=\\begin{vmatrix} 2&amp;3&amp;-2\\\\[1.1ex] 1&amp;4&amp;1\\\\[1.1ex] 2&amp;1&amp;-3 \\end{vmatrix} = -24+6-2+16-2+9 = 3\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"381\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De determinant is anders dan 0, dus de matrix kan worden omgekeerd.<\/p>\n<p class=\"has-text-align-left\"> Zodra de determinant is opgelost, vinden we de adjunct-matrix 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-c510482ac77a8c5d511c095de600f1ba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 2} = \\displaystyle (-1)^{1+1} \\bm{\\cdot} \\begin{vmatrix} 4&amp;1\\\\[1.1ex] 1&amp;-3 \\end{vmatrix} = 1 \\cdot (-13) = \\bm{-13}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"403\" 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-fa99e03d34c925098c1ad3ed6f06c745_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 3} = \\displaystyle (-1)^{1+2} \\bm{\\cdot} \\begin{vmatrix}1&amp;1\\\\[1.1ex] 2&amp;-3\\end{vmatrix} = -1 \\cdot (-5) = \\bm{5}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"384\" 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-3bf9f8565b3e4a99ff254c7558699c13_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de -2}  = \\displaystyle (-1)^{1+3} \\bm{\\cdot} \\begin{vmatrix} 1&amp;4\\\\[1.1ex] 2&amp;1 \\end{vmatrix} = 1\\cdot (-7) = \\bm{-7}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"377\" 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-99e2c3f55fbba7b5faa014758b60f4a8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 1} = \\displaystyle (-1)^{2+1} \\bm{\\cdot} \\begin{vmatrix} 3&amp;-2 \\\\[1.1ex] 1&amp;-3 \\end{vmatrix} = -1 \\cdot (-7) = \\bm{7}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"385\" 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-23326bccecf752508e7418cbbc8eacd3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 4} = \\displaystyle (-1)^{2+2} \\bm{\\cdot} \\begin{vmatrix} 2&amp;-2\\\\[1.1ex] 2&amp;-3 \\end{vmatrix} = 1 \\cdot (-2) = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"384\" 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-a9d056af07ce26751783152a67cdedb6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 1} = \\displaystyle (-1)^{2+3} \\bm{\\cdot} \\begin{vmatrix} 2&amp;3\\\\[1.1ex] 2&amp;1\\end{vmatrix} = -1 \\cdot (-4) = \\bm{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"371\" 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-bed501806c35c94e491ad2063b2d0653_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 2}  = \\displaystyle (-1)^{3+1} \\bm{\\cdot} \\begin{vmatrix} 3&amp;-2\\\\[1.1ex] 4&amp;1\\end{vmatrix} = 1 \\cdot 11 = \\bm{11}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"360\" 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-3f108a61eec662b9420708f6920060be_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de 1} = \\displaystyle (-1)^{3+2} \\bm{\\cdot} \\begin{vmatrix} 2&amp;-2\\\\[1.1ex] 1&amp;1\\end{vmatrix} = -1 \\cdot 4 = \\bm{-4}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"371\" 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-77a152a00dbb5f1e0f8702dd9511095a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Adjunto de -3} = \\displaystyle (-1)^{3+3} \\bm{\\cdot} \\begin{vmatrix} 2&amp;3\\\\[1.1ex] 1&amp;4 \\end{vmatrix} = 1 \\cdot 5 = \\bm{5}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"335\" 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-b4642a75697fd30286065cdb4063a7bd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\displaystyle \\text{Adj}(A) = \\begin{pmatrix} -13 &amp; 5 &amp; -7  \\\\[1.1ex] 7 &amp; -2 &amp; 4 \\\\[1.1ex] 11 &amp; -4 &amp; 5 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"215\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Nadat we de determinant van de matrix en zijn adjunct hebben berekend, vervangen we hun waarden in de formule: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fae003a07d40b69690566cde77857c3a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{3} \\cdot \\begin{pmatrix} -13 &amp; 5 &amp; -7 \\\\[1.1ex] 7 &amp; -2 &amp; 4 \\\\[1.1ex] 11 &amp; -4 &amp; 5 \\end{pmatrix}^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"89\" width=\"224\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We transponeren de bijgevoegde matrix:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-55717407766afe98f50ca75f20536edc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{3} \\cdot \\begin{pmatrix} -13 &amp; 7 &amp; 11 \\\\[1.1ex] 5 &amp; -2 &amp; -4 \\\\[1.1ex] -7 &amp; 4 &amp; 5 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"215\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> En de omgekeerde matrix A 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-9835713a5b791ee959d6571d706180f3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{A^{-1} =} \\begin{pmatrix} \\sfrac{\\bm{-13}}{\\bm{3}} &amp; \\sfrac{\\bm{7}}{\\bm{3}} &amp; \\sfrac{\\bm{11}}{\\bm{3}} \\\\[1.1ex] \\sfrac{\\bm{5}}{\\bm{3}} &amp; \\sfrac{\\bm{-2}}{\\bm{3}} &amp; \\sfrac{\\bm{-4}}{\\bm{3}} \\\\[1.1ex] \\sfrac{\\bm{-7}}{\\bm{3}} &amp; \\sfrac{\\bm{4}}{\\bm{3}} &amp; \\sfrac{\\bm{5}}{\\bm{3}}\\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"216\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\">Oefening 4<\/h4>\n<p> Inverteer de volgende matrix van orde 3 met behulp van de adjunct-matrixmethode: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bf71320b51e9514d1c372389aeb3410a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix}4&amp;5&amp;-1\\\\[1.1ex] -1&amp;3&amp;2\\\\[1.1ex] 3&amp;8&amp;1\\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"150\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> De inverse matrixformule 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-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We moeten eerst de determinant van de matrix berekenen, want als de determinant 0 is, betekent dit dat de matrix geen inverse 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-eb7dc647f4121450eeadf2f5b62b4475_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}A\\end{vmatrix}=\\begin{vmatrix} 4&amp;5&amp;-1\\\\[1.1ex] -1&amp;3&amp;2\\\\[1.1ex] 3&amp;8&amp;1 \\end{vmatrix} = 12+30+8+9-64+5 = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"389\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De determinant van A is 0, <strong>dus de matrix kan niet worden omgekeerd.<\/strong><\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\"> Oefening 5<\/h4>\n<p> Inverteer de volgende 3 \u00d7 3 vierkante matrix met de determinantenmatrixmethode: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-92e56e0f8013b6b65c0894a139537cae_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix}1 &amp; 4 &amp; -3 \\\\[1.1ex] -2 &amp; 1 &amp; 0 \\\\[1.1ex] -1 &amp; -2 &amp; 2\\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"164\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> De inverse matrixformule 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-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Allereerst lossen we de determinant van de matrix op 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-07f116ed906c31644ed0513667988e6f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\lvert A \\rvert = \\begin{vmatrix} 1 &amp; 4 &amp; -3 \\\\[1.1ex] -2 &amp; 1 &amp; 0 \\\\[1.1ex] -1 &amp; -2 &amp; 2 \\end{vmatrix} = 2+0-12-3-0+16 = 3\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"392\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De determinant is anders dan 0, dus de matrix kan worden omgekeerd.<\/p>\n<p class=\"has-text-align-left\"> Zodra de determinant is opgelost, vinden we de adjunct-matrix 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-20da2eac0d49b1134b39b1f5c95c5659_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\text{Adjunto de 1} =  (-1)^{1+1} \\bm{\\cdot} \\begin{vmatrix}  1 &amp; 0 \\\\[1.1ex]  -2 &amp; 2 \\end{vmatrix} = 1 \\bm{\\cdot} (2-0) = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"377\" 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-c5b80624f0963dfb1a111d96b4e1ceae_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\text{Adjunto de 4} =  (-1)^{1+2} \\bm{\\cdot} \\begin{vmatrix}  -2 &amp;  0 \\\\[1.1ex] -1 &amp; 2 \\end{vmatrix} = -1 \\bm{\\cdot} (-4-0) = \\bm{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"405\" 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-50dd371e77d1896adb197321b68efd1d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\text{Adjunto de -3} = (-1)^{1+3} \\bm{\\cdot} \\begin{vmatrix} -2 &amp; 1 \\\\[1.1ex] -1 &amp; -2 \\end{vmatrix} = 1 \\bm{\\cdot} \\bigl(4-(-1)\\bigr) = \\bm{5}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"427\" 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-60b779f4366a3ef38ae522fcfca8e7d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\text{Adjunto de -2} =  (-1)^{2+1} \\bm{\\cdot} \\begin{vmatrix}  4 &amp; -3  \\\\[1.1ex]  -2 &amp; 2 \\end{vmatrix} = -1 \\bm{\\cdot} (8-6) = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"424\" 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-51cb00c42e6932810a4220eb85c61acd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\text{Adjunto de 1} = (-1)^{2+2} \\bm{\\cdot} \\begin{vmatrix} 1 &amp;  -3  \\\\[1.1ex] -1 &amp;  2 \\end{vmatrix} = 1 \\bm{\\cdot} (2-3) = \\bm{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"405\" 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-a3b26cbfa55d5567d2dae10c5dfbd158_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\text{Adjunto de 0} =  (-1)^{2+3} \\bm{\\cdot} \\begin{vmatrix} 1 &amp; 4  \\\\[1.1ex] -1 &amp; -2 \\end{vmatrix} = -1 \\bm{\\cdot} \\bigl(-2-(-4)\\bigr) = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"462\" 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-8d9f1bf4f5e01df910cd59bd4b25f816_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\text{Adjunto de -1} = (-1)^{3+1} \\bm{\\cdot} \\begin{vmatrix}  4 &amp; -3 \\\\[1.1ex]  1 &amp; 0  \\end{vmatrix} = 1 \\bm{\\cdot} \\bigl(0-(-3)\\bigr) = \\bm{3}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"414\" 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-8ce129b17734facf076e48fb1928d0e1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\text{Adjunto de -2}   = (-1)^{3+2} \\bm{\\cdot} \\begin{vmatrix} 1 &amp; -3 \\\\[1.1ex] -2 &amp; 0 \\end{vmatrix} = -1 \\cdot (0-6) = \\bm{6}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"419\" 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-3c8b319461dad7880bf2b9f20187b6fb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\text{Adjunto de 2} =  (-1)^{3+3} \\bm{\\cdot} \\begin{vmatrix} 1 &amp; 4 \\\\[1.1ex] -2 &amp; 1 \\end{vmatrix} = 1 \\bm{\\cdot} \\bigl(1-(-8)\\bigr) = \\bm{9}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"408\" 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-748fcb9d9d2a8326379da4d2bd08534a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\displaystyle \\text{Adj}(A) = \\begin{pmatrix} 2 &amp; 4 &amp; 5 \\\\[1.1ex] -2 &amp; -1 &amp; -2 \\\\[1.1ex] 3 &amp; 6 &amp; 9 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"206\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Nadat we de determinant van de matrix en zijn adjunct hebben berekend, vervangen we hun waarden in de formule: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1fe85ec6c4385daba7d2488b0d60ee2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{\\vert A \\vert } \\cdot \\Bigl( \\text{Adj}(A)\\Bigr)^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"175\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3a0fc0e6effb520e22ff82c3034b4d4c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{3} \\cdot \\begin{pmatrix} 2 &amp; 4 &amp; 5 \\\\[1.1ex] -2 &amp; -1 &amp; -2 \\\\[1.1ex] 3 &amp; 6 &amp; 9\\end{pmatrix}^{\\bm{t}}\" title=\"Rendered by QuickLaTeX.com\" height=\"89\" width=\"215\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We transponeren de bijgevoegde matrix:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bba6ddbc8ab9f2c64eb03cdb9fea530a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\cfrac{1}{3} \\cdot \\begin{pmatrix} 2 &amp; -2 &amp; 3 \\\\[1.1ex] 4 &amp; -1 &amp; 6 \\\\[1.1ex] 5 &amp; -2 &amp; 9 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"178\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> En tot slot opereren wij: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-41f999c23e7d5ce129b410b9f486983e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1} = \\begin{pmatrix} \\sfrac{2}{3} &amp; \\sfrac{-2}{3} &amp; \\sfrac{3}{3} \\\\[1.1ex] \\sfrac{4}{3} &amp; \\sfrac{-1}{3} &amp; \\sfrac{6}{3} \\\\[1.1ex] \\sfrac{5}{3} &amp; \\sfrac{-2}{3} &amp; \\sfrac{9}{3} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"181\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-matrice-inverse-par-matrice-adjointe-33.webp\" alt=\"oefening stap voor stap opgelost van de inverse matrix volgens de methode van de adjunct-matrix 3x3\" width=\"232\" height=\"104\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h2 class=\"wp-block-heading\"> Inverteer een matrix met behulp van de Gauss-methode:<\/h2>\n<div class=\"adsb30\" style=\" margin:px; text-align:\"><\/div>\n<p> Om <strong>de inverse van een matrix te berekenen met de Gauss-methode<\/strong> , <strong>moet je bewerkingen uitvoeren op de rijen van een matrix<\/strong> (we zullen dit later zien). Voordat u dus gaat zien hoe u de Gauss-methode kunt gebruiken, is het belangrijk dat u alle bewerkingen kent die kunnen worden uitgevoerd op de rijen van de matrices:<\/p>\n<h3 class=\"wp-block-heading\"> Lijntransformaties toegestaan in de Gaussische methode<\/h3>\n<ul>\n<li> <strong><span style=\"color:#1976d2;\">Verander de volgorde<\/span><\/strong> van de rijen van de matrix.<\/li>\n<\/ul>\n<p> We kunnen bijvoorbeeld de volgorde van regels 2 en 3 van een matrix wijzigen:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1d3f607625afb96bfb250168bd330818_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc} 3 &amp; 5 &amp; -2 \\\\[2ex] -2 &amp; 4 &amp; -1  \\\\[2ex] 6 &amp; 1 &amp; -3 \\end{array} \\right)  \\begin{array}{c} \\\\[2ex] \\xrightarrow{ f_2 \\rightarrow f_3}} \\\\[2ex] \\xrightarrow{ f_3 \\rightarrow f_2}} \\end{array} \\left( \\begin{array}{ccc} 3 &amp; 5 &amp; -2  \\\\[2ex] 6 &amp; 1 &amp; -3  \\\\[2ex] -2 &amp; 4 &amp; -1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"331\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<ul>\n<li> <strong><span style=\"color:#1976d2;\">Vermenigvuldig of deel<\/span><\/strong> alle termen in een rij met een ander getal dan 0.<\/li>\n<\/ul>\n<p> We kunnen bijvoorbeeld regel 1 vermenigvuldigen met 4 en regel 3 delen met 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-3cca4df71c23b1f005068a0a93b77dfe_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc} 1 &amp; -2 &amp; 3 \\\\[2ex] 3 &amp; -1 &amp; 5  \\\\[2ex] 2 &amp; -4 &amp; -2  \\end{array} \\right) \\begin{array}{c}  \\xrightarrow{4  f_1} \\\\[2ex]  \\\\[2ex] \\xrightarrow{ f_3 \/ 2} \\end{array} \\left( \\begin{array}{ccc} 4 &amp; -8 &amp; 12 \\\\[2ex] 3 &amp; -1 &amp; 5  \\\\[2ex] 1 &amp; -2 &amp; -1  \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"103\" width=\"318\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<ul>\n<li> <strong><span style=\"color:#1976d2;\">Vervang een rij<\/span><\/strong> door de som van dezelfde rij plus een andere rij vermenigvuldigd met een getal.<\/li>\n<\/ul>\n<p> In de volgende matrix voegen we bijvoorbeeld rij 3 vermenigvuldigd met 1 toe aan rij 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-8ca6644f015dd42ddbf4ab159bd10dec_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc} -1 &amp; -3 &amp; 4  \\\\[2ex] 2 &amp; 4 &amp; 1  \\\\[2ex] 1 &amp; -2 &amp; 3  \\end{array} \\right) \\begin{array}{c}   \\\\[2ex]  \\xrightarrow{f_2 + 1\\cdot f_3}  \\\\[2ex] &amp; \\end{array} \\left( \\begin{array}{ccc} -1 &amp; -3 &amp; 4  \\\\[2ex] 3 &amp; 2 &amp; 4  \\\\[2ex] 1 &amp; -2 &amp; 3  \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"96\" width=\"339\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h3 class=\"wp-block-heading\"> Voorbeeld van het berekenen van de inverse matrix met behulp van de Gauss-methode:<\/h3>\n<p> Laten we met een voorbeeld zien hoe we de <strong>Gauss-methode<\/strong> kunnen toepassen om een matrix om te keren:<\/p>\n<ul>\n<li> Bereken de inverse van de volgende matrix:<\/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-71553480cefa679dcb8eb98d97e0c717_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A = \\left( \\begin{array}{ccc} 1 &amp; 0 &amp; 1 \\\\[2ex] 0 &amp; 2 &amp; 1 \\\\[2ex] 1 &amp; 5 &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"96\" width=\"139\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Het eerste dat we moeten doen is <strong>de A-matrix en de Identiteitsmatrix combineren tot \u00e9\u00e9n enkele matrix<\/strong> . De A-matrix aan de linkerkant en de Identiteitsmatrix aan de rechterkant: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d0650812fe7946f6da1e7973709dfde1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   \\bigl( A \\  \\lvert \\ I \\bigr)\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"51\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-matrice-inverse-par-la-methode-de-gauss-32153.webp\" alt=\"oefening stap voor stap opgelost van inverse matrix met de 3x3 Gauss-methode\" width=\"203\" height=\"120\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-background\" style=\"background-color:#dff6ff\"> Om de inverse matrix te berekenen, moeten we <strong>de linkermatrix omzetten in een identiteitsmatrix.<\/strong> En om dat te doen, moeten we transformaties op de rijen toepassen totdat we daar zijn.<\/p>\n<p> We gaan verder met kolommen, dat wil zeggen dat we bewerkingen op de rijen uitvoeren om eerst de getallen in de eerste kolom te transformeren, vervolgens die in de tweede kolom en ten slotte die in de derde kolom. <\/p>\n<div class=\"wp-block-columns is-layout-flex wp-container-35\">\n<div class=\"wp-block-column is-layout-flow\" style=\"flex-basis:66.66%\">\n<p class=\"has-text-align-justify\"> De 1&#8217;s en 0&#8217;s in de eerste kolom zijn al geschikt, aangezien de identiteitsmatrix op deze posities ook een 1 en een 0 bevat. Daarom is het op dit moment niet nodig om een transformatie op deze rijen toe te passen. <\/p>\n<\/div>\n<div class=\"wp-block-column is-layout-flow\" style=\"flex-basis:33.33%\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7f51b3a869dde9c1697be9e57fce1548_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left(  \\begin{array}{ccc|ccc} \\color{blue}\\boxed{\\color{black}1} &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] \\color{blue}\\boxed{\\color{black}0} &amp; 2 &amp; 1 &amp; 0 &amp; 1 &amp; 0  \\\\[2ex] 1 &amp; 5 &amp; 4 &amp;0 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"101\" width=\"255\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<\/div>\n<\/div>\n<p> De identiteitsmatrix heeft echter een 0 in het laatste element van de eerste kolom, waar we nu een 1 hebben. We moeten dus de 1 omzetten naar 0. Om dit te doen, <strong>voegen we rij 1 vermenigvuldigd met \u2013 toe aan rij 3.1:<\/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-30b5442d5c5eac3e62aa7a7cae717e48_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{lrrr|rrr}  &amp; 1 &amp; 5 &amp; 4 &amp;0 &amp; 0 &amp; 1  \\\\ + &amp; -1 &amp; 0 &amp; -1 &amp; -1 &amp; 0 &amp; 0  \\\\ \\hline  &amp; 0 &amp; 5 &amp; 3 &amp; -1 &amp; 0 &amp; 1  \\end{array} \\begin{array}{l} \\color{blue}\\bm{\\leftarrow f_3} \\\\ \\color{blue}\\bm{\\leftarrow -f_1} \\\\ \\phantom{hline} \\\\ \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"68\" width=\"313\" style=\"vertical-align: -29px;\"><\/p>\n<\/p>\n<p> Dus als we deze som berekenen, krijgen we de volgende matrix:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-992a31603c2182a97d31ddf787df4f06_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 2 &amp; 1 &amp; 0 &amp; 1 &amp; 0  \\\\[2ex] 1 &amp; 5 &amp; 4 &amp;0 &amp; 0 &amp; 1 \\end{array} \\right) \\begin{array}{c}   \\\\[2ex]  \\\\[2ex] \\xrightarrow{f_3 - f_1} \\end{array} \\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 2 &amp; 1 &amp; 0 &amp; 1 &amp; 0  \\\\[2ex] \\color{blue}\\boxed{\\color{black}0} &amp; 5 &amp; 3 &amp; -1 &amp; 0 &amp; 1  \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"100\" width=\"520\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> We zijn er dus in geslaagd de 1 om te zetten in 0.<\/p>\n<p> Laten we nu verder gaan naar de tweede kolom van de linkermatrix. Het eerste element is een 0, wat goed is omdat de identiteitsmatrix op dezelfde positie een 0 heeft. In plaats van een 2 zou er echter een 1 moeten zijn, <strong>dus delen we de tweede regel door 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-a86b61ee601f9cd0ff9a70d1a280f887_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 2 &amp; 1 &amp; 0 &amp; 1 &amp; 0  \\\\[2ex] 1 &amp; 5 &amp; 4 &amp;0 &amp; 0 &amp; 1 \\end{array} \\right) \\begin{array}{c}   \\\\[2ex] \\xrightarrow{f_2\/2}\\\\[2ex] &amp; \\end{array}  \\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; \\color{blue}\\boxed{\\color{black}1} &amp; \\sfrac{1}{2} &amp; 0 &amp; \\sfrac{1}{2} &amp; 0  \\\\[2ex] 0 &amp; 5 &amp; 3 &amp; -1 &amp; 0 &amp; 1  \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"100\" width=\"527\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Bovendien moeten we in de tweede kolom ook de 5 in 0 veranderen. Omdat de 5 vijf keer groter is dan de 1 in de tweede rij, <strong>voegen we rij 2 vermenigvuldigd met -5 toe aan rij 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-66dd50ad7ec5e4c45f5011094a0c21b3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{lrrr|rrr}  &amp; 0 &amp; 5 &amp; 3 &amp; -1 &amp; 0 &amp; 1  \\\\ + &amp; 0 &amp; -5 &amp; \\sfrac{-5}{2} &amp; 0 &amp; \\vphantom{\\Bigl(}\\sfrac{-5}{2} &amp; 0  \\\\ \\hline &amp; 0 &amp; 0 &amp;  \\sfrac{1}{2}  &amp; -1 &amp; \\sfrac{-5}{2} \\vphantom{\\Bigl(} &amp; 1  \\end{array} \\begin{array}{l} \\color{blue}\\bm{\\leftarrow f_3} \\\\ \\color{blue}\\bm{\\leftarrow -5f_2}\\vphantom{\\Bigl(} \\\\ \\phantom{hline} \\vphantom{\\Bigl(}  \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"88\" width=\"355\" style=\"vertical-align: -39px;\"><\/p>\n<\/p>\n<p> Door deze bewerking uit te voeren, eindigen we dus met de matrix met een 0 in het laatste element van de tweede kolom:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fcc790f05d73d308cb7d992841ab031a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{1}{2} &amp; 0 &amp; \\sfrac{1}{2} &amp; 0  \\\\[2ex] 0 &amp; 5 &amp; 3 &amp; -1 &amp; 0 &amp; 1  \\end{array} \\right) \\begin{array}{c}   \\\\[2ex] \\\\[2ex] \\xrightarrow{f_3 - 5f_2} \\end{array}  \\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{1}{2} &amp; 0 &amp; \\sfrac{1}{2} &amp; 0  \\\\[2ex]  0 &amp; \\color{blue}\\boxed{\\color{black}0} &amp;  \\sfrac{1}{2}  &amp; -1 &amp; \\sfrac{-5}{2}  &amp; 1  \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"100\" width=\"590\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Ten slotte zullen we de laatste kolom van de matrix naar links transformeren, maar deze keer moeten we onderaan beginnen. Het is daarom noodzakelijk om de<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8b05f3ca9cc1227bdfe634ccc9f60935_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sfrac{1}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> in een 1. Daarom <strong>vermenigvuldigen we de laatste regel met 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-69614cae4dd388b6454ffd9b8d63c9a5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{1}{2} &amp; 0 &amp; \\sfrac{1}{2} &amp; 0  \\\\[2ex]  0 &amp; 0 &amp;  \\sfrac{1}{2}  &amp; -1 &amp; \\sfrac{-5}{2}  &amp; 1  \\end{array} \\right)\\begin{array}{c}   \\\\[2ex] \\\\[2ex] \\xrightarrow{2f_3} \\end{array}  \\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{1}{2} &amp; 0 &amp; \\sfrac{1}{2} &amp; 0  \\\\[2ex]  0 &amp; 0 &amp;  \\color{blue}\\boxed{\\color{black}1}  &amp; -2 &amp; -5  &amp; 2  \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"100\" width=\"562\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> We moeten nu de transformatie transformeren<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8b05f3ca9cc1227bdfe634ccc9f60935_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sfrac{1}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> rest van de laatste kolom als 0. Deze keer kunnen we de rij echter niet met 2 vermenigvuldigen, omdat we de 1 ook naar 2 zouden converteren (wanneer de identiteitsmatrix op die positie een 1 heeft). Daarom <strong>voegen we regel 3 gedeeld door -2 toe aan regel 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-881f9ea3ce2e52ddf332a13aba43bbcf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{lrrr|rcr}  &amp; 0 &amp; 1 &amp;  \\vphantom{\\Bigl(} \\sfrac{1}{2} &amp; 0 &amp; \\sfrac{1}{2} &amp; 0  \\\\ + &amp; 0 &amp; 0 &amp;\\vphantom{\\Bigl(} -\\sfrac{1}{2}  &amp; 1 &amp; \\sfrac{5}{2}  &amp; -1  \\\\ \\hline &amp; 0 &amp; 1 &amp; 0\\phantom{0}  &amp; 1 &amp; 3 \\vphantom{\\Bigl(} &amp; -1  \\end{array} \\begin{array}{l}\\vphantom{\\Bigl(} \\color{blue}\\bm{\\leftarrow f_2} \\\\ \\color{blue}\\bm{\\leftarrow f_3\/(-2)}\\vphantom{\\Bigl(} \\\\ \\phantom{hline} \\vphantom{\\Bigl(}  \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"357\" style=\"vertical-align: -44px;\"><\/p>\n<\/p>\n<p> Dus door deze operatie uit te voeren, slagen we erin om de<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8b05f3ca9cc1227bdfe634ccc9f60935_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sfrac{1}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> in een 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-537958a51f67c7602ef121fa2c997ca8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{1}{2} &amp; 0 &amp; \\sfrac{1}{2} &amp; 0  \\\\[2ex]  0 &amp; 0 &amp;  1  &amp; -2 &amp; -5  &amp; 2  \\end{array} \\right) \\begin{array}{c}   \\\\[2ex] \\xrightarrow{f_2-f_3\/2} \\\\[2ex] &amp; \\end{array}  \\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\color{blue}\\boxed{\\color{black}0} &amp; 1 &amp; 3  &amp; -1  \\\\[2ex]  0 &amp; 0 &amp;  1  &amp; -2 &amp; -5  &amp; 2  \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"100\" width=\"598\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Ten slotte hoeven we alleen maar de 1 in de eerste rij van de derde kolom om te zetten in 0. De derde rij heeft ook een 1 in dezelfde kolom, <strong>dus we voegen rij 3 vermenigvuldigd met -1 toe aan rij 1:<\/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-8854a556147caefb16a2030e0e5e949a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{lrrr|rcr}  &amp; 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\ + &amp; 0 &amp; 0 &amp;  -1  &amp; 2 &amp; 5  &amp; -2  \\\\ \\hline &amp; 1 &amp; 0 &amp; 0  &amp; 3 &amp; 5 &amp; -2  \\end{array} \\begin{array}{l}\\color{blue}\\bm{\\leftarrow f_1} \\\\ \\color{blue}\\bm{\\leftarrow -f_3}\\\\ \\phantom{hline}   \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"68\" width=\"300\" style=\"vertical-align: -29px;\"><\/p>\n<\/p>\n<p> En door deze bewerking uit te voeren, slagen we erin om de 1 om te zetten in een 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-8ddd39df6bc92258ba163c65de4fd59f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\ \\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp;0 &amp; 1 &amp; 3  &amp; -1  \\\\[2ex]  0 &amp; 0 &amp;  1  &amp; -2 &amp; -5  &amp; 2  \\end{array} \\right) \\begin{array}{c} \\xrightarrow{f_1-f_3}  \\\\[2ex]  \\\\[2ex]  &amp; \\end{array}  \\left(  \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; \\color{blue}\\boxed{\\color{black}0}  &amp; 3 &amp; 5 &amp; -2  \\\\[2ex] 0 &amp; 1 &amp; 0 &amp; 1 &amp; 3  &amp; -1  \\\\[2ex]  0 &amp; 0 &amp;  1  &amp; -2 &amp; -5  &amp; 2  \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"99\" width=\"589\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Zodra we de linkermatrix met succes hebben omgezet in een identiteitsmatrix, kennen we ook de inverse matrix. Omdat <strong>de inverse matrix de matrix is die we aan de rechterkant verkrijgen door de linkermatrix om te zetten in identiteitsmatrix<\/strong> . De inverse van de matrix is dus: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exemple-de-matrice-inverse-32153.webp\" alt=\"Voorbeeld van een 3x3 inverse matrix\" width=\"251\" height=\"117\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<h3 class=\"wp-block-heading\"> Opgeloste oefeningen over inverse matrices met de Gauss-methode <\/h3>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-118\"><\/div>\n<\/div>\n<h4 class=\"wp-block-heading\"> Oefening 1<\/h4>\n<p> Inverteer de volgende matrix via de Gauss-methode: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-36886e1ab1007f9a53bdc0dd71a0d15b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} 1 &amp; 2 \\\\[1.1ex] 1 &amp; 3  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"95\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Het eerste dat we moeten doen is de A-matrix en de Identiteitsmatrix combineren tot \u00e9\u00e9n enkele matrix. De A-matrix links en de identiteitsmatrix rechts: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3cbeb2e5edb9eaf9e47efc4cc74b1333_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\left( A \\ | \\ I \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"51\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-matrice-de-gauss-inverse-22152.webp\" alt=\"opgeloste oefening van een inverse matrix met de 2x2 Gauss-methode\" width=\"143\" height=\"66\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Om nu de inverse matrix te berekenen, moeten we de linkerzijdematrix omzetten in identiteitsmatrix. En om dat te doen, moeten we transformaties op de rijen toepassen totdat we daar zijn.<\/p>\n<p class=\"has-text-align-left\"> De eerste term van allemaal, 1, is al hetzelfde als de identiteitsmatrix. Daarom is het op dit moment niet nodig om een transformatie op de eerste rij toe te passen.<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft echter een 0 in het laatste element van de eerste kolom, waar we nu een 1 hebben. We moeten daarom de 1 omzetten naar 0. Hiervoor trekken we rij 1 af van rij 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-247d8605795c43e79b5d7742854cfe6d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{cc|cc}1 &amp; 2 &amp; 1 &amp; 0 \\\\[1.5ex] 1 &amp; 3 &amp; 0 &amp; 1\\end{array} \\right) \\begin{array}{c} \\\\[1.5ex] \\xrightarrow{f_2 - f_1}  \\end{array} \\left( \\begin{array}{cc|cc} 1 &amp; 2 &amp; 1 &amp; 0 \\\\[1.5ex] 0 &amp; 1 &amp; -1 &amp; 1\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"56\" width=\"332\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We gaan verder met de tweede kolom: 1 hieronder is goed. Maar niet de 2 hierboven, aangezien de identiteitsmatrix op die positie een 0 heeft. Om de 2 naar 0 om te zetten, trekken we daarom regel 2 vermenigvuldigd met 2 af van regel 1:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-173a7bdb55ba058e5ae16d1fd8e91564_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{cc|cc} 1 &amp; 2 &amp; 1 &amp; 0 \\\\[1.5ex] 0 &amp; 1 &amp; -1 &amp; 1 \\end{array} \\right) \\begin{array}{c}  \\xrightarrow{f_1 - 2f_2} \\\\[1.5ex] &amp; \\end{array} \\left( \\begin{array}{cc|cc} 1 &amp; 0 &amp; 3 &amp; -2 \\\\[1.5ex] 0 &amp; 1 &amp; -1 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"57\" width=\"367\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De inverse matrix is de matrix die we aan de rechterkant verkrijgen na het omzetten van de matrix aan de linkerkant in een identiteitsmatrix. En nu hebben we de identiteitsmatrix aan de linkerkant. De inverse matrix is dus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-98896d28465c9e1402e1c443375d93fe_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{A^{-1}= \\left(} \\begin{array}{cc}  \\bm{3} &amp; \\bm{-2} \\\\[1.5ex]  \\bm{-1} &amp; \\bm{1} \\end{array}\\bm{ \\right)}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"157\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\">Oefening 2<\/h4>\n<p> Inverteer de volgende matrix met de Gauss-procedure: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7ae5ba4a92a5ddc00ddf5b11775edafd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} 1 &amp; 1 &amp; -4 \\\\[1.1ex]  0 &amp; 3 &amp; 2 \\\\[1.1ex] 0 &amp; 1 &amp; 1  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"136\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Eerst plaatsen we de A-matrix en de Identiteitsmatrix in \u00e9\u00e9n enkele matrix: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3cbeb2e5edb9eaf9e47efc4cc74b1333_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\left( A \\ | \\ I \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"51\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-81db2ef94d2db597cebb4c0c77685526_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 1 &amp; -4 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 3 &amp; 2 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Nu moeten we de rijen transformeren totdat we de linkermatrix in een identiteitsmatrix hebben omgezet.<\/p>\n<p class=\"has-text-align-left\"> De eerste kolom van de linkermatrix is al gelijk aan de eerste kolom van de identiteitsmatrix. Het is daarom niet nodig om de cijfers ervan te wijzigen.<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft echter een 1 in het tweede element van de tweede kolom, waar nu een 3 staat. We moeten de 3 dus omzetten in een 1. Om dit te doen trekken we van regel 2 regel 3 af, vermenigvuldigd met 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-bfd7cb4d4b81a75038807eb28393a83e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 1 &amp; -4 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 3 &amp; 2 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right) \\begin{array}{c} \\\\[2ex] \\xrightarrow{f_2 - 2f_3} \\\\[2ex] &amp;  \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 1 &amp; 4 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; -2 \\\\[2ex] 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"458\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft in het laatste element van de tweede kolom een 0, waar nu een 1 staat. We moeten de 1 dus omzetten in 0. Hiervoor trekken we regel 2 af van regel 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-932479e2f574c19ad7906d3d20e52ad0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 1 &amp; -4 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; -2 \\\\[2ex] 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right) \\begin{array}{c} \\\\[2ex]  \\\\[2ex] \\xrightarrow{f_3 - f_2} \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 1 &amp; -4 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; -2 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; 0 &amp; -1 &amp; 3 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"492\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft in het eerste element van de tweede kolom een 0, waar nu een 1 staat. We moeten de 1 dus omzetten naar 0. Hiervoor trekken we regel 2 af van regel 1:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-566e1453aab03f9792cb281e4c88a68c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 1 &amp; -4 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; -2 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; 0 &amp; -1 &amp; 3 \\end{array} \\right) \\begin{array}{c} \\xrightarrow{f_1 - f_2} \\\\[2ex]  \\\\[2ex] &amp;  \\end{array} \\left( \\begin{array}{ccc|ccc}1 &amp; 0 &amp; -4 &amp; 1 &amp; -1 &amp; 2 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; -2 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; 0 &amp; -1 &amp; 3 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"506\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Het enige wat we nu moeten doen is de -4 omzetten in 0. Om dit te doen, voegen we regel 3 vermenigvuldigd met 4 toe aan regel 1:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6f98a9cabeb101602dd11aa73516b998_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; -4 &amp; 1 &amp; -1 &amp; 2 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; -2 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; 0 &amp; -1 &amp; 3\\end{array} \\right) \\begin{array}{c} \\xrightarrow{f_1 + 4f_3} \\\\[2ex]  \\\\[2ex] &amp;  \\end{array} \\left( \\begin{array}{ccc|ccc}1 &amp; 0 &amp; 0 &amp; 1 &amp; -5 &amp; 14 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; -2 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; 0 &amp; -1 &amp; 3 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"499\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We hebben de identiteitsmatrix al vanaf de linkerkant verkregen. De inverse matrix is dus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e43ce6a7061f0339bd5d44b83afec07f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{A^{-1}= \\left( } \\begin{array}{ccc}  \\bm{1} &amp; \\bm{-5}  &amp; \\bm{14} \\\\[2ex]  \\bm{0} &amp; \\bm{1} &amp; \\bm{-2} \\\\[2ex] \\bm{0} &amp; \\bm{-1 }&amp; \\bm{3} \\end{array} \\bm{ \\right)}\" title=\"Rendered by QuickLaTeX.com\" height=\"96\" width=\"185\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\">Oefening 3<\/h4>\n<p> Inverteer de volgende matrix met behulp van de Gauss-methode: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f02b0186690e68baaa9a630db2c870db_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} 1 &amp; 2 &amp; 1 \\\\[1.1ex]  0 &amp; 1 &amp; 0 \\\\[1.1ex] 2 &amp; 0 &amp; 3 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"122\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Voordat we aan de slag gaan, moeten we de A-matrix en de Identiteitsmatrix in \u00e9\u00e9n enkele matrix plaatsen: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3cbeb2e5edb9eaf9e47efc4cc74b1333_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\left( A \\ | \\ I \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"51\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-aa6dc5af82076e22b1d0cf7ea16d748b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 2 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 2 &amp; 0 &amp; 3 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"172\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We moeten nu de linkermatrix omzetten in een identiteitsmatrix door op de rijen te werken.<\/p>\n<p class=\"has-text-align-left\"> De eerste twee elementen van de eerste kolom zijn al dezelfde als die van de identiteitsmatrix. Het is dus niet nodig om deze cijfers aan te passen.<\/p>\n<p class=\"has-text-align-left\"> Maar de identiteitsmatrix heeft een 0 in het derde element van de eerste kolom, waar nu een 2 staat. We moeten de 2 daarom omzetten in een 0. Om dit te doen trekken we van regel 3 regel 1 vermenigvuldigd met 2 af:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-680a314b8cc900e01886291af12145e4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc}1 &amp; 2 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 2 &amp; 0 &amp; 3 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right) \\begin{array}{c} \\\\[2ex] \\\\[2ex] \\xrightarrow{f_3 - 2f_1}   \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 2 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; -4 &amp; 1 &amp; -2 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"458\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft een 0 in het eerste element van de tweede kolom, waar nu een 2 staat. We moeten de 2 daarom omzetten in een 0. Hiervoor trekken we van regel 1 regel 2 vermenigvuldigd met 2 af:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f87cbc594287f7ea4938091878562b4c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 2 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; -4 &amp; 1 &amp; -2 &amp; 0 &amp; 1 \\end{array} \\right) \\begin{array}{c} \\xrightarrow{f_1 -2f_2} \\\\[2ex]  \\\\[2ex] &amp; \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; -2 &amp; 0\\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; -4 &amp; 1 &amp; -2 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"499\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft een 0 in het laatste element van de tweede kolom, waar nu een -4 staat. We moeten dus de -4 omzetten in 0. Om dit te doen, voegen we regel 2 vermenigvuldigd met 4 toe aan regel 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-b8cf2c3878d2d35656953a55bb3baf94_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; -2 &amp; 0\\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; -4 &amp; 1 &amp; -2 &amp; 0 &amp; 1 \\end{array} \\right) \\begin{array}{c} \\\\[2ex]  \\\\[2ex] \\xrightarrow{f_3 +4f_2} \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; 1 &amp; -2 &amp; 0\\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; -2 &amp; 4 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"499\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Het enige wat we nu moeten doen is het eerste element van de derde kolom omzetten naar 0. Om dit te doen, voegen we regel 3 vermenigvuldigd met -1 toe aan regel 1:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-aac851b05c2dc25af3d7b9ecc622c9f6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc}1 &amp; 0 &amp; 1 &amp; 1 &amp; -2 &amp; 0\\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; -2 &amp; 4 &amp; 1 \\end{array} \\right) \\begin{array}{c} \\xrightarrow{f_1 - f_3} \\\\[2ex]  \\\\[2ex] &amp;  \\end{array} \\left( \\begin{array}{ccc|ccc}1 &amp; 0 &amp; 0 &amp; 3 &amp; -6  &amp; -1\\\\[2ex]  0 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; -2 &amp; 4 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"492\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We hebben ons al gerealiseerd dat de matrix aan de linkerkant de identiteitsmatrix is. Dus het omgekeerde van de matrix<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-25b206f25506e6d6f46be832f7119ffa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"13\" style=\"vertical-align: 0px;\"><\/p>\n<p> Oosten:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-161fbe4a4d4dcc4fc503b6e3a9e0bfeb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{A^{-1}= \\left( } \\begin{array}{ccc}  \\bm{3} &amp; \\bm{-6}  &amp; \\bm{-1} \\\\[2ex]  \\bm{0} &amp; \\bm{1} &amp; \\bm{0} \\\\[2ex] \\bm{-2} &amp; \\bm{4}&amp; \\bm{1} \\end{array} \\bm{ \\right)}\" title=\"Rendered by QuickLaTeX.com\" height=\"96\" width=\"198\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\">Oefening 4<\/h4>\n<p> Inverteer de volgende matrix met behulp van de Gauss-methode: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-47ad7ccd6aafab72255c96f2bc9148a2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} 1 &amp; -2 &amp; 0 \\\\[1.1ex]  1 &amp; 2 &amp; 2 \\\\[1.1ex] 0 &amp; 3 &amp; 2 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"136\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Het eerste dat we moeten doen is de A-matrix en de Identiteitsmatrix samenvoegen tot \u00e9\u00e9n enkele matrix: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3cbeb2e5edb9eaf9e47efc4cc74b1333_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\left( A \\ | \\ I \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"51\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a832ceb9f09dfa88238c570b46b74d92_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc}1 &amp; -2 &amp; 0 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 1 &amp; 2 &amp; 2 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; 3 &amp; 2 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We moeten nu de matrix aan de linkerkant omzetten in een identiteitsmatrix door rijbewerkingen toe te passen.<\/p>\n<p class=\"has-text-align-left\"> Het eerste element van de eerste kolom is al hetzelfde als dat van de identiteitsmatrix. Het is daarom niet nodig om deze te wijzigen.<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft echter een 0 in het tweede element van de eerste kolom, waar nu een 1 staat. We moeten de 1 dus omzetten in 0. Hiervoor trekken we regel 1 af van regel 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-83933b5a2315a4dcbc770bf92bf3831b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc}1 &amp; -2 &amp; 0 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 1 &amp; 2 &amp; 2 &amp; 0 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; 3 &amp; 2 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right) \\begin{array}{c} \\\\[2ex] \\xrightarrow{f_2 - f_1} \\\\[2ex] &amp;  \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; -2 &amp; 0 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 4 &amp; 2 &amp; -1 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; 3 &amp; 2 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"465\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We gaan verder met de tweede kolom: we transformeren eerst de 4 in een 1 door de tweede rij door 4 te delen:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-298984c72a249e2b5c98740cc0c1a11e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; -2 &amp; 0 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 4 &amp; 2 &amp; -1 &amp; 1 &amp; 0 \\\\[2ex] 0 &amp; 3 &amp; 2 &amp; 0 &amp; 0 &amp; 1\\end{array} \\right) \\begin{array}{c} \\\\[2ex] \\xrightarrow{f_2\/4} \\\\[2ex] &amp;  \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; -2 &amp; 0 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{-1}{4} &amp; \\sfrac{1}{4} &amp; 0 \\\\[2ex] 0 &amp; 3 &amp; 2 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"495\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft een 0 in het eerste element van de tweede kolom, waar nu een -2 staat. We moeten dus -2 omzetten naar 0. Om dit te doen, tellen we regel 2 vermenigvuldigd met 2 op bij regel 1: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ce876446d5d01a152e39480d69affd8c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{lrrr|rcr} &amp; 1 &amp; -2 &amp; 0 &amp; 1 &amp; 0 &amp; 0 \\\\ + &amp; 0 &amp; 2 &amp; 1 &amp; \\vphantom{\\Bigl(}\\sfrac{-2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\\\ \\hline &amp; 1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} \\vphantom{\\Bigl(}&amp; 0 \\end{array} \\begin{array}{l} \\color{blue}\\bm{\\leftarrow f_1} \\\\ \\color{blue}\\bm{\\leftarrow 2f_2}\\vphantom{\\Bigl(} \\\\ \\phantom{hline} \\vphantom{\\Bigl(} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"88\" width=\"313\" style=\"vertical-align: -39px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3dfcdcb586eed87861b3ac0ea46bea2f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; -2 &amp; 0 &amp; 1 &amp; 0 &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{-1}{4} &amp; \\sfrac{1}{4} &amp; 0 \\\\[2ex] 0 &amp; 3 &amp; 2 &amp; 0 &amp; 0 &amp; 1\\end{array} \\right) \\begin{array}{c} \\xrightarrow{f_1 +2f_2} \\\\[2ex]  \\\\[2ex] &amp; \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{-1}{4} &amp; \\sfrac{1}{4} &amp; 0 \\\\[2ex] 0 &amp; 3 &amp; 2 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"525\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft een 0 in het laatste element van de tweede kolom, waar nu een 3 staat. We moeten de 3 daarom omzetten in een 0. Om dit te doen trekken we van regel 3 regel 2 vermenigvuldigd met 3 af: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-210ca8df473a00d9f205470ed2aa19a7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{lrrr|crr} &amp; 0 &amp; 3 &amp; 2 &amp; 0 &amp; 0\\phantom{0} &amp; 1 \\\\ + &amp; 0 &amp; -3 &amp; \\vphantom{\\Bigl(}\\sfrac{-6}{4} &amp; \\sfrac{3}{4} &amp; \\sfrac{-3}{4} &amp; 0 \\\\ \\hline &amp; 0 &amp; 0 &amp; \\vphantom{\\Bigl(}\\sfrac{2}{4} &amp; \\sfrac{3}{4} &amp; \\sfrac{-3}{4} &amp; 1 \\end{array} \\begin{array}{l} \\color{blue}\\bm{\\leftarrow f_3} \\\\ \\color{blue}\\bm{\\leftarrow -3f_2}\\vphantom{\\Bigl(} \\\\ \\phantom{hline} \\vphantom{\\Bigl(} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"88\" width=\"350\" style=\"vertical-align: -39px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-94ed5a1b9cf1db0bfb99ce79d0a6d36b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{-1}{4} &amp; \\sfrac{1}{4} &amp; 0 \\\\[2ex] 0 &amp; 3 &amp; 2 &amp; 0 &amp; 0 &amp; 1 \\end{array} \\right) \\begin{array}{c} \\\\[2ex]  \\\\[2ex] \\xrightarrow{f_3 -3f_2} \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{-1}{4} &amp; \\sfrac{1}{4} &amp; 0 \\\\[2ex] 0 &amp; 0 &amp;\\sfrac{2}{4} &amp; \\sfrac{3}{4} &amp; \\sfrac{-3}{4} &amp; 1  \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"525\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We gaan verder met de derde kolom: we moeten de laatste transformeren<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-cd7d08f65ca5dd13d94128372d3b6c95_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sfrac{2}{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"18\" style=\"vertical-align: 0px;\"><\/p>\n<p> omzetten in een 1. Om dit te doen, vermenigvuldigen we de derde regel met 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-a8134938726d3b48fe3d7d789260b128_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{-1}{4} &amp; \\sfrac{1}{4} &amp; 0 \\\\[2ex] 0 &amp; 0 &amp;\\sfrac{2}{4} &amp; \\sfrac{3}{4} &amp; \\sfrac{-3}{4} &amp; 1   \\end{array} \\right) \\begin{array}{c} \\\\[2ex]  \\\\[2ex] \\xrightarrow{2f_3 } \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{-1}{4} &amp; \\sfrac{1}{4} &amp; 0 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; \\sfrac{6}{4} &amp; \\sfrac{-6}{4} &amp; 2   \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"515\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De identiteitsmatrix heeft een 0 in het tweede element van de laatste kolom. Het is daarom noodzakelijk om de<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-cd7d08f65ca5dd13d94128372d3b6c95_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sfrac{2}{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"18\" style=\"vertical-align: 0px;\"><\/p>\n<p> omzetten in een 0. Om dit te doen, trekken we van regel 2 regel 3 gedeeld door 2 af: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-dc74ebe003751fd9ae3a5a77b2f589c8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{lrrr|ccr} &amp; 0 &amp; 1 &amp; \\vphantom{\\Bigl(} \\sfrac{2}{4} &amp; \\sfrac{-1}{4} &amp; \\sfrac{1}{4} &amp; 0 \\\\ + &amp; 0 &amp; 0 &amp; \\vphantom{\\Bigl(} \\sfrac{-1}{2} &amp; \\sfrac{-6}{8} &amp; \\sfrac{6}{8} &amp; -1  \\\\ \\hline &amp; 0 &amp; 1 &amp; 0\\phantom{0} &amp; -1 &amp; 1 &amp; -1\\vphantom{\\Bigl(} \\end{array} \\begin{array}{l} \\color{blue}\\bm{\\leftarrow f_2}\\vphantom{\\Bigl(}  \\\\ \\color{blue}\\bm{\\leftarrow -f_3\/2}\\vphantom{\\Bigl(} \\\\ \\phantom{hline} \\vphantom{\\Bigl(} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"358\" style=\"vertical-align: -44px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8b91b71183a50e41e9be5c7305f8cf3e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{-1}{4} &amp; \\sfrac{1}{4} &amp; 0 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; \\sfrac{6}{4} &amp; \\sfrac{-6}{4} &amp; 2 \\end{array} \\right) \\begin{array}{c} \\\\[2ex] \\xrightarrow{f_2-f_3\/2 } \\\\[2ex] &amp; \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; 0 &amp; -1 &amp; 1 &amp; -1 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; \\sfrac{6}{4} &amp; \\sfrac{-6}{4} &amp; 2   \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"542\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Het enige wat we nu nog moeten doen is het eerste element van de derde kolom omzetten naar 0. Om dit te doen trekken we rij 3 af van rij 1: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-38796ed093a0fef52426fb5559931586_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{lrrr|rcr} &amp; 1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\vphantom{\\Bigl(} \\\\ + &amp; 0 &amp; 0 &amp; -1 &amp; \\sfrac{-6}{4} &amp; \\sfrac{6}{4} &amp; -2 \\vphantom{\\Bigl(}  \\\\ \\hline &amp; 1 &amp; 0 &amp; 0 &amp; -1 &amp; 2 &amp; -2 \\vphantom{\\Bigl(} \\end{array} \\begin{array}{l} \\color{blue}\\bm{\\leftarrow f_1}\\vphantom{\\Bigl(}  \\\\ \\color{blue}\\bm{\\leftarrow -f_3}\\vphantom{\\Bigl(} \\\\ \\phantom{hline} \\vphantom{\\Bigl(} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"332\" style=\"vertical-align: -44px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2023374b9885dd33fe4d3c12e5a4de59_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left( \\begin{array}{ccc|ccc}1 &amp; 0 &amp; 1 &amp; \\sfrac{2}{4} &amp; \\sfrac{2}{4} &amp; 0 \\\\[2ex] 0 &amp; 1 &amp; 0 &amp; -1 &amp; 1 &amp; -1 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; \\sfrac{6}{4} &amp; \\sfrac{-6}{4} &amp; 2 \\end{array} \\right) \\begin{array}{c} \\xrightarrow{f_1-f_3 }  \\\\[2ex] \\\\[2ex] &amp; \\end{array} \\left( \\begin{array}{ccc|ccc} 1 &amp; 0 &amp; 0 &amp; -1 &amp; 2 &amp; -2 \\\\[2ex] 0 &amp; 1 &amp; 0 &amp; -1 &amp; 1 &amp; -1 \\\\[2ex] 0 &amp; 0 &amp; 1 &amp; \\sfrac{6}{4} &amp; \\sfrac{-6}{4} &amp; 2   \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"98\" width=\"524\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De inverse matrix is dus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0854e7cb80ba561b6e0c724a9a9b5fff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A^{-1}= \\left(  \\begin{array}{ccc}  -1  &amp; 2 &amp; -2 \\\\[2ex]  -1 &amp; 1 &amp; -1 \\\\[2ex] \\sfrac{6}{4} &amp;\\sfrac{-6}{4} &amp; 2 \\end{array} \\bm{ \\right)}\" title=\"Rendered by QuickLaTeX.com\" height=\"96\" width=\"207\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ten slotte kunnen de fracties van de inverse matrix worden vereenvoudigd:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6c7ef6b6cdca2f4a808ed9457bde3b3f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{A^{-1}= \\left( } \\begin{array}{ccc}  \\bm{-1} &amp; \\bm{2}  &amp; \\bm{-2} \\\\[2ex]  \\bm{-1} &amp; \\bm{1} &amp; \\bm{-1} \\\\[2ex] \\sfrac{\\bm{3}}{\\bm{2}} &amp;\\sfrac{\\bm{-3}}{\\bm{2}} &amp; \\bm{2} \\end{array} \\bm{ \\right)}\" title=\"Rendered by QuickLaTeX.com\" height=\"96\" width=\"207\" 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<h2 class=\"wp-block-heading\"> Inverse matrixeigenschappen<\/h2>\n<p> De inverse matrix heeft de volgende kenmerken:<\/p>\n<ul>\n<li> De inverse van een matrix is <span style=\"color:#1976d2;\"><strong>uniek<\/strong><\/span> .<\/li>\n<\/ul>\n<ul>\n<li> De <span style=\"color:#1976d2;\"><strong>inverse van de inverse matrix<\/strong><\/span> is de oorspronkelijke matrix:<\/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-caac2cfeece17b627e46c7ec04020319_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left(A^{-1}\\right)^{-1} = A\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"101\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<ul>\n<li> De <span style=\"color:#1976d2;\"><strong>inverse van de vermenigvuldiging<\/strong><\/span> van twee matrices is gelijk aan het product van de inverse van de matrices, maar verandert hun volgorde.<\/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-8f0dd4094bdc2faa4449008d1d8ee8c9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left(A \\cdot B)^{-1} = B^{-1} \\cdot A^{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"171\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<ul>\n<li> <span style=\"color:#1976d2;\"><strong>Het transponeren van een matrix<\/strong><\/span> en vervolgens de inverse van de matrix uitvoeren, is hetzelfde als eerst de inversie van de matrix uitvoeren en deze vervolgens transponeren.<\/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-f2dc6d83dd3d9b9dacec6e7806c9c0e5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left(A^t\\right)^{-1} = \\left(A^{-1}\\right)^{t}\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"128\" style=\"vertical-align: -7px;\"><\/p>\n<\/p>\n<ul>\n<li> Om de <span style=\"color:#1976d2;\"><strong>determinant van de inverse van een matrix<\/strong><\/span> op te lossen, kunnen we de determinant van de matrix berekenen en vervolgens de inverse ervan uitvoeren, aangezien de twee bewerkingen hetzelfde resultaat opleveren.<\/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-e32fcd8a6c25d8c863947e6cc31efdc6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle det\\left(A^{-1}\\right) =\\bigl( det(A) \\bigr) ^{-1} = \\cfrac{1}{det(A)}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"261\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> Formule om snel de inverse van een 2&#215;2 matrix te berekenen<\/h2>\n<p> Zoals we hebben gezien kan elke matrix worden omgekeerd door de methode van determinanten of door de Gauss-methode. Maar afzonderlijk bestaat er ook een <strong>formule om heel snel de inverse van een 2\u00d72 matrix te vinden<\/strong> : <\/p>\n<div class=\"wp-block-image estil_requadre_foto\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/formule-de-matrice-inverse-22152.webp\" alt=\"formule om de inverse van een 2x2-matrix te vinden, inverse matrixformule 2x2\" class=\"wp-image-673\" width=\"475\" height=\"75\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p> Zoals je kunt zien, is het inverteren van een 2&#215;2-matrix eenvoudig: los gewoon de determinant van de matrix 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-3c046b53b17b87e9ca0f447d664754ba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"(|A|)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"35\" style=\"vertical-align: -5px;\"><\/p>\n<p> , wissel de positie van de elementen van de hoofddiagonaal af en verander het teken van de elementen van de secundaire diagonaal.<\/p>\n<h3 class=\"wp-block-heading\"> Voorbeeld van hoe u een 2 \u00d7 2 inverse matrix kunt krijgen met de formule<\/h3>\n<div class=\"adsb30\" style=\" margin:px; text-align:\"><\/div>\n<p> Bereken de inverse van de volgende 2 \u00d7 2 vierkante matrix:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-599baee27c05b5610a8714363e1260eb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A = \\begin{pmatrix} 3 &amp; 5 \\\\[1.1ex] -2 &amp; -4 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"122\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De determinant van matrix A 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-ab99f7b87d01c670a8598df6364ab58f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{aligned}\\begin{vmatrix}A\\end{vmatrix} = \\begin{vmatrix} 3 &amp; 5 \\\\[1.1ex] -2 &amp; -4 \\end{vmatrix} &amp; = 3 \\cdot (-4)- (-2) \\cdot 5 \\\\ &amp; = -12-(-10) \\\\[2ex] &amp; =-12+10\\\\[2ex] &amp;=-2\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"160\" width=\"281\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Nu passen we <strong>de inverse matrixformule<\/strong> toe:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7d5308484309da4485a3d9b92af86e7d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A = \\begin{pmatrix} a &amp; b \\\\[1.1ex] c &amp; d \\end{pmatrix}\\longrightarrow A^{-1} = \\cfrac{1}{|A|} \\begin{pmatrix} d &amp; -b \\\\[1.1ex] -c &amp; a \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"305\" 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-68fd6e830b576af8abf55be1e11fbafb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A = \\begin{pmatrix} 3 &amp; 5 \\\\[1.1ex] -2 &amp; -4 \\end{pmatrix}\\longrightarrow A^{-1} = \\cfrac{1}{-2} \\begin{pmatrix} -4 &amp; -5 \\\\[1.1ex] 2 &amp; 3 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"333\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> En we vermenigvuldigen de matrix met de breuk:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-41da8ef6bef1d339337717ed4ad86ae5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A^{-1} =\\begin{pmatrix} \\cfrac{-4}{-2} &amp; \\cfrac{-5}{-2} \\\\[3ex] \\cfrac{2}{-2} &amp; \\cfrac{3}{-2} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"163\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De omgekeerde matrix A 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-29da2a64f6da927857de112ca8363ba5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{A^{-1} =}\\begin{pmatrix} \\bm{2} &amp; \\cfrac{\\bm{5}}{\\bm{2}} \\\\[3ex] \\bm{-1} &amp; \\bm{-}\\cfrac{\\bm{3}}{\\bm{2}} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"143\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Zoals u kunt zien, gaat het omkeren van een matrix met deze formule veel sneller, maar deze formule kan alleen worden gebruikt op matrices met afmeting 2&#215;2.<\/p>\n<h3 class=\"wp-block-heading\"> Opgeloste oefeningen van 2\u00d72 inverse matrices met de formule<\/h3>\n<h4 class=\"wp-block-heading\"> Oefening 1<\/h4>\n<p> Inverteer de volgende matrix met afmeting 2\u00d72: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-dc06e21fc1c3c54f9b3fc0dcd4912a8f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} 2 &amp; 5 \\\\[1.1ex] 1 &amp; 3 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"95\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> De determinant van matrix A 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-1b0ae510ea7a336cbe5ea56a554da719_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{aligned}\\begin{vmatrix}A\\end{vmatrix} = \\begin{vmatrix} 2 &amp; 5 \\\\[1.1ex] 1 &amp; 3 \\end{vmatrix} &amp; = 2 \\cdot 3- 1 \\cdot 5 \\\\ &amp; = 6-5 \\\\[2ex] &amp; =1\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"118\" width=\"198\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Nu passen we de formule toe om de inverse matrix te vinden: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7d5308484309da4485a3d9b92af86e7d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A = \\begin{pmatrix} a &amp; b \\\\[1.1ex] c &amp; d \\end{pmatrix}\\longrightarrow A^{-1} = \\cfrac{1}{|A|} \\begin{pmatrix} d &amp; -b \\\\[1.1ex] -c &amp; a \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"305\" 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-b8f18178c829fd38360a04a947d52017_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A=\\begin{pmatrix} 2 &amp; 5 \\\\[1.1ex] 1 &amp; 3 \\end{pmatrix} \\longrightarrow A^{-1} = \\cfrac{1}{1} \\begin{pmatrix} 3 &amp; -5 \\\\[1.1ex] -1 &amp; 2 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"292\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De inverse van matrix A 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-237fe82cd91972f667f6751fa4735534_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{A^{-1} =}\\begin{pmatrix} \\bm{3} &amp; \\bm{-5} \\\\[1.1ex] \\bm{-1} &amp; \\bm{2} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"139\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\">Oefening 2<\/h4>\n<p> Bereken de inverse van de volgende matrix van orde 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-f2289d1c5c9aeb87016f719305d900a7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} 2 &amp; 6 \\\\[1.1ex] -1 &amp; -2 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"122\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> De determinant van matrix A 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-a3fef2cc00702131123994cc588bf7ea_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{aligned}\\begin{vmatrix}A\\end{vmatrix} = \\begin{vmatrix} 2 &amp; 6 \\\\[1.1ex] -1 &amp; -2 \\end{vmatrix} &amp; = 2 \\cdot (-2)- (-1) \\cdot 6 \\\\ &amp; = -4-(-6) \\\\[2ex] &amp; =-4+6 \\\\[2ex] &amp; =2\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"160\" width=\"282\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We passen nu de formule toe om de inverse matrix van dimensie 2\u00d72 op te lossen: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7d5308484309da4485a3d9b92af86e7d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A = \\begin{pmatrix} a &amp; b \\\\[1.1ex] c &amp; d \\end{pmatrix}\\longrightarrow A^{-1} = \\cfrac{1}{|A|} \\begin{pmatrix} d &amp; -b \\\\[1.1ex] -c &amp; a \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"305\" 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-2de7166a0cf59e0f8c5b7750e1947f04_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A=\\begin{pmatrix} 2 &amp; 6 \\\\[1.1ex] -1 &amp; -2 \\end{pmatrix} \\longrightarrow A^{-1} = \\cfrac{1}{2} \\begin{pmatrix} -2 &amp; -6 \\\\[1.1ex] 1 &amp; 2 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"319\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> En tenslotte doen we de vermenigvuldiging: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f6a5973078468914beb4bd4d85a40331_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A^{-1} = \\begin{pmatrix} \\cfrac{-2}{2} &amp; \\cfrac{-6}{2} \\\\[3ex] \\cfrac{1}{2} &amp; \\cfrac{2}{2} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"163\" 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-a540a077ee9a24da96fa988410aef429_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{A^{-1} =}\\begin{pmatrix} \\bm{-1} &amp; \\bm{-3} \\\\[2ex] \\cfrac{\\bm{1}}{\\bm{2}} &amp; \\bm{1} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"76\" width=\"141\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\">Oefening 3<\/h4>\n<p> Inverteer de volgende 2&#215;2-matrix: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-36e230a808c42411a9cfd2d9eb44543d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} 4 &amp; 1 \\\\[1.1ex] 5 &amp; 2 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"95\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> De determinant van matrix A 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-e7a6c5ef316ae51b43c90863c6245780_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{aligned}\\begin{vmatrix}A\\end{vmatrix} = \\begin{vmatrix} 4 &amp; 1 \\\\[1.1ex] 5 &amp; 2\\end{vmatrix} &amp; = 4 \\cdot 2 - 5 \\cdot 1 \\\\ &amp; = 8-5 \\\\[2ex] &amp;  =3\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"118\" width=\"198\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> We passen nu de formule toe om de inverse matrix van dimensie 2\u00d72 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-7d5308484309da4485a3d9b92af86e7d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A = \\begin{pmatrix} a &amp; b \\\\[1.1ex] c &amp; d \\end{pmatrix}\\longrightarrow A^{-1} = \\cfrac{1}{|A|} \\begin{pmatrix} d &amp; -b \\\\[1.1ex] -c &amp; a \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"305\" 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-e2f359bd166c295b869a8cf04d927097_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A=\\begin{pmatrix} 4 &amp; 1 \\\\[1.1ex] 5 &amp; 2 \\end{pmatrix} \\longrightarrow A^{-1} = \\cfrac{1}{3} \\begin{pmatrix} 2 &amp; -1 \\\\[1.1ex] -5 &amp; 4 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"292\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> En tenslotte doen we het product tussen de breuk en de matrix:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6a02ea2e547dcc21081ae80df407a4e0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A^{-1} = \\begin{pmatrix} \\cfrac{\\bm{2}}{\\bm{3}} &amp; \\bm{-}\\cfrac{\\bm{1}}{\\bm{3}} \\\\[3ex] \\bm{-}\\cfrac{\\bm{5}}{\\bm{3}} &amp; \\cfrac{\\bm{4}}{\\bm{3}} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"147\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h4 class=\"wp-block-heading\">Oefening 4<\/h4>\n<p> Zoek de inverse van de volgende tweede-orde matrix: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-422fcd6f391a2682e4b546c9e9c05b55_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix} -2 &amp; 5 \\\\[1.1ex] -3 &amp; 10 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"117\" 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 de oplossing<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> De determinant van matrix A 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-9e9997751e16d3b976454be828cb914d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{aligned}\\begin{vmatrix}A\\end{vmatrix} = \\begin{vmatrix} -2 &amp; 5 \\\\[1.1ex] -3 &amp; 10\\end{vmatrix} &amp; = (-2) \\cdot 10- (-3) \\cdot 5 \\\\ &amp; = -20-(-15) \\\\[2ex] &amp; =-20+15 \\\\[2ex] &amp; =-5\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"160\" width=\"285\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Nu passen we de formule toe om de inverse matrix van dimensie 2\u00d72 te cre\u00ebren: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7d5308484309da4485a3d9b92af86e7d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A = \\begin{pmatrix} a &amp; b \\\\[1.1ex] c &amp; d \\end{pmatrix}\\longrightarrow A^{-1} = \\cfrac{1}{|A|} \\begin{pmatrix} d &amp; -b \\\\[1.1ex] -c &amp; a \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"305\" 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-7c0c614039614bd9125b2920da8698eb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A=\\begin{pmatrix} -2 &amp; 5 \\\\[1.1ex] -3 &amp; 10\\end{pmatrix} \\longrightarrow A^{-1} = \\cfrac{1}{-5} \\begin{pmatrix} 10 &amp; -5 \\\\[1.1ex] 3 &amp; -2 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"323\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> En tenslotte doen we de vermenigvuldiging: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-edb1dfc870b3045eaefc1716a80e2ca2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A^{-1} = \\begin{pmatrix} \\cfrac{10}{-5} &amp; \\cfrac{-5}{-5} \\\\[3ex] \\cfrac{3}{-5} &amp; \\cfrac{-2}{-5} \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"155\" 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-5c49e161c701254cfbe20353c11980eb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{A^{-1} =}\\begin{pmatrix} \\bm{-2} &amp; \\bm{1} \\\\[2ex] \\bm{-}\\cfrac{\\bm{3}}{\\bm{5}} &amp; \\cfrac{\\bm{2}}{\\bm{5}} \\ \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"76\" width=\"137\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h2 class=\"wp-block-heading\">Los een stelsel vergelijkingen op met de inverse matrix<\/h2>\n<p> Het is moeilijk om de echte toepassingen van de inverse van een matrix te begrijpen. Sterker nog, je vraagt je waarschijnlijk af&#8230; waar wordt de inverse matrix voor gebruikt? Wordt het echt ergens voor gebruikt?<\/p>\n<p> Welnu, een van de toepassingen van de inverse matrix is <strong>het oplossen van stelsels van lineaire vergelijkingen<\/strong> . En ja, ook al lijken het misschien twee heel verschillende concepten, het is mogelijk om de oplossing van een stelsel vergelijkingen te vinden door een matrix om te keren.<\/p>\n<p> Laten we met een voorbeeld zien hoe dit wordt gedaan:<\/p>\n<ul>\n<li> Bereken de oplossing van het volgende stelsel vergelijkingen met de inverse matrix:<\/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-200c0f994f86752e7d650621a0d4100f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left. \\begin{array}{r} x+3y=5 \\\\[2ex] 2x+4y=6 \\end{array} \\right\\}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"112\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Allereerst moet worden opgemerkt dat een systeem van vergelijkingen kan worden uitgedrukt in de vorm van matrices:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4b9c9f181fc16a501799145c516a9747_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{pmatrix} 1 &amp; 3 \\\\[1.1ex] 2 &amp; 4 \\end{pmatrix}\\begin{pmatrix} x \\\\[1.1ex]y \\end{pmatrix} = \\begin{pmatrix} 5 \\\\[1.1ex] 6 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"156\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> We kunnen verifi\u00ebren dat deze matrixvorm van het systeem equivalent is aan de uitdrukking met vergelijkingen: als we de matrices vermenigvuldigen, zullen we zien dat we de twee vergelijkingen van het systeem verkrijgen.<\/p>\n<p> Om de volgende stappen te vereenvoudigen, zullen we bellen<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-25b206f25506e6d6f46be832f7119ffa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"13\" style=\"vertical-align: 0px;\"><\/p>\n<p> naar de matrix die de co\u00ebffici\u00ebnten van de onbekenden heeft,<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d4ee28752517d6062a3ca0314890342d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"X\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> naar de matrixkolommen met de onbekenden, en<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-770fd1447ccf2fc229801b486b0d8f8a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"B\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"14\" style=\"vertical-align: 0px;\"><\/p>\n<p> naar de kolommatrix met 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-ec1e9c04147230526534e694fb54f316_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle AX=B\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"67\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De matrix dus<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d4ee28752517d6062a3ca0314890342d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"X\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> is de onbekende van de matrixvergelijking.<\/p>\n<p> Om deze matrixvergelijking op te lossen, moet u een procedure volgen die we hier niet zo gedetailleerd zullen uitleggen. Als je het helemaal wilt begrijpen, kun je kijken hoe je <a href=\"https:\/\/mathority.org\/nl\/voorbeelden-van-matrixvergelijkingen-en-opgeloste-oefeningen-van-2x2-en-3x3-matrices\/\">vergelijkingen met matrices<\/a> oplost, waar we het hele proces stap voor stap uitleggen.<\/p>\n<p> Deze procedure is gebaseerd op een eigenschap van inverse matrices: elke matrix vermenigvuldigd met zijn inverse is gelijk aan de identiteitsmatrix (of eenheidsmatrix). Daarom kan de onbekende matrix eenvoudig worden opgelost<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d4ee28752517d6062a3ca0314890342d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"X\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> door beide zijden van de vergelijking te vermenigvuldigen met de inverse 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-ec1e9c04147230526534e694fb54f316_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle AX=B\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"67\" 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-e20a8dfa638cb0fa47765a784dc47a61_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A^{-1}\\cdot AX=A^{-1}\\cdot B\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"156\" 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-218f48c32d9bfd298c1e9559e8059a82_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle IX=A^{-1}\\cdot B\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"107\" 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-acfded1a5d11f4b183ac34c85df906fc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle X=A^{-1}\\cdot B\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"98\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> En zodra we de matrix hebben ge\u00efsoleerd<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d4ee28752517d6062a3ca0314890342d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"X\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> , berekenen we het omgekeerde van<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-25b206f25506e6d6f46be832f7119ffa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"13\" style=\"vertical-align: 0px;\"><\/p>\n<p> en we lossen het product van matrices 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-9a1290e37a9e3f56fc6b288bc7686d66_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle X=\\left.\\begin{pmatrix} 1 &amp; 3 \\\\[1.1ex] 2 &amp; 4 \\end{pmatrix}\\right.^{-1}\\cdot \\begin{pmatrix} 5 \\\\[1.1ex] 6 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"58\" width=\"170\" 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-21471fc8a4c04aac3121519e8ef874e5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle X=\\cfrac{1}{-2} \\begin{pmatrix} 4 &amp; -3 \\\\[1.1ex] -2 &amp; 1 \\end{pmatrix}\\cdot \\begin{pmatrix} 5 \\\\[1.1ex] 6 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"202\" 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-b9457fedf68c4bdfea898922e465eeb8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle X= \\begin{pmatrix} -1 \\\\[1.1ex] 2 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"86\" 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-c2748b49f967580a0871d8739ee0d4f4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{x=-1} \\qquad \\bm{y=2}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"133\" style=\"vertical-align: -4px;\"><\/p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Op deze pagina leert u wat het is en hoe u de inverse van een matrix kunt berekenen met de methode van determinanten (of adjunct-matrix) en met de Gauss-methode. Je zult ook alle eigenschappen van de inverse matrix zien, en je zult ook stap voor stap opgeloste voorbeelden en oefeningen voor elke methode vinden, zodat &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/nl\/omgekeerde-matrix\/\"> <span class=\"screen-reader-text\">Hoe de inverse matrix te berekenen<\/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":[39],"tags":[],"class_list":["post-111","post","type-post","status-publish","format-standard","hentry","category-determinant-van-een-matrix"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Hoe de inverse matrix te berekenen -<\/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\/omgekeerde-matrix\/\" \/>\n<meta property=\"og:locale\" content=\"nl_NL\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Hoe de inverse matrix te berekenen -\" \/>\n<meta property=\"og:description\" content=\"Op deze pagina leert u wat het is en hoe u de inverse van een matrix kunt berekenen met de methode van determinanten (of adjunct-matrix) en met de Gauss-methode. 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