{"id":304,"date":"2023-07-06T14:13:39","date_gmt":"2023-07-06T14:13:39","guid":{"rendered":"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/"},"modified":"2023-07-06T14:13:39","modified_gmt":"2023-07-06T14:13:39","slug":"exemplos-de-regras-e-exercicios-resolvidos-de-cramer","status":"publish","type":"post","link":"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/","title":{"rendered":"Regra de cramer"},"content":{"rendered":"<p>Nesta p\u00e1gina voc\u00ea ver\u00e1 o que \u00e9 a regra de Cramer e, al\u00e9m disso, encontrar\u00e1 exemplos e exerc\u00edcios com resolu\u00e7\u00e3o de sistemas de equa\u00e7\u00f5es pela regra de Cramer.<\/p>\n<h2 class=\"wp-block-heading\"> Qual \u00e9 a regra de Cramer?<\/h2>\n<p> <strong>A regra de Cramer<\/strong> \u00e9 um m\u00e9todo usado para resolver sistemas de equa\u00e7\u00f5es por determinantes. Vamos ver como \u00e9 usado:<\/p>\n<p> Considere um sistema de equa\u00e7\u00f5es:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e0141f3451719f665ef28e4061489551_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} ax+by+cz= \\color{red}\\bm{j} \\\\[1.5ex] dx+ey+fz=\\color{red}\\bm{k} \\\\[1.5ex] gx+hy+iz = \\color{red}\\bm{l} \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"171\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> A matriz A e a matriz estendida A&#8217; do sistema s\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1d628a13ec7de4b3ba7a301c0a5d8ac6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} a &amp; b &amp; c  \\\\[1.1ex] d &amp; e &amp; f  \\\\[1.1ex] g &amp; h &amp; i  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} a &amp; b &amp; c &amp;  \\color{red}\\bm{j}  \\\\[1.1ex] d &amp; e &amp; f &amp; \\color{red}\\bm{k} \\\\[1.1ex] g &amp; h &amp; i &amp; \\color{red}\\bm{l} \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"384\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong><span style=\"text-decoration: underline;\">A regra de Cramer<\/span><\/strong> diz que a solu\u00e7\u00e3o de um sistema de equa\u00e7\u00f5es \u00e9: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/quelle-est-la-regle-de-cramer.webp\" alt=\"o que \u00e9 a regra de Cramer, explica\u00e7\u00e3o da regra de Cramer\" class=\"wp-image-1062\" width=\"677\" height=\"385\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> Observe que os determinantes dos numeradores s\u00e3o como o determinante da matriz A, mas mudando a coluna de cada inc\u00f3gnita para a coluna dos termos independentes.<\/p>\n<p> Portanto, a regra de Cramer \u00e9 usada para resolver sistemas de equa\u00e7\u00f5es lineares. Mas, como voc\u00ea j\u00e1 sabe, existem muitas maneiras de resolver um sistema de equa\u00e7\u00f5es, por exemplo <a href=\"https:\/\/mathority.org\/pt\/metodo-jordan-gauss-com-exemplos-e-exercicios-resolvidos\/\">, o m\u00e9todo de Gauss Jordan<\/a> \u00e9 bem conhecido.<\/p>\n<p> Abaixo est\u00e3o exemplos de resolu\u00e7\u00e3o de sistemas de equa\u00e7\u00f5es lineares com a regra de Cramer, ou \u00e0s vezes tamb\u00e9m escrita como regra de Kramer.<\/p>\n<h2 class=\"wp-block-heading\"> Exemplo 1: sistema compat\u00edvel determinado (SCD)<\/h2>\n<ul>\n<li> Resolva o seguinte sistema de 3 equa\u00e7\u00f5es com 3 inc\u00f3gnitas usando a regra de Cramer:<\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6013b7e73c89c24fe388f1a5d018f32b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 2x+y+3z= 1 \\\\[1.5ex] 3x-2y-z=0 \\\\[1.5ex] x+3y+2z = 5\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"135\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Primeiro fazemos a matriz A e a matriz estendida A&#8217; do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c710ed86223f47f39b5a25720b5ca19d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; 3 &amp; 1 \\\\[1.1ex] 3 &amp; -2 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 3 &amp; 2 &amp; 5 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Calculamos agora a classifica\u00e7\u00e3o das duas matrizes, para ver que tipo de sistema \u00e9. Para calcular a classifica\u00e7\u00e3o de A, calculamos o determinante 3\u00d73 de toda a matriz (usando a regra de Sarrus) e vemos se d\u00e1 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-ae4a3bb88d113494463df8e670c326c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{vmatrix} =-8-1+27+6+6-6 = 24 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"427\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> O determinante de A \u00e9 diferente de 0, ent\u00e3o <strong>a matriz A tem classifica\u00e7\u00e3o 3.<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Portanto <strong>, a matriz A&#8217; tamb\u00e9m \u00e9 de posto 3<\/strong> , uma vez que n\u00e3o pode ser de posto 4 e deve ter pelo menos o mesmo posto que a matriz A.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> A extens\u00e3o da matriz A \u00e9 igual \u00e0 extens\u00e3o da matriz A&#8217; e ao n\u00famero de inc\u00f3gnitas do sistema (3), portanto, pelo <strong>teorema de Rouch\u00e9-Frobenius<\/strong> , sabemos que se trata de um <strong>sistema compat\u00edvel determinado<\/strong> (SCD):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-557185e16670c72d23eec5a3ea13b487_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 3 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Uma vez sabendo que o sistema \u00e9 um SCD, aplicamos <strong>a regra de Cramer<\/strong> para resolv\u00ea-lo. Para fazer isso, lembre-se que a matriz A, seu determinante e a matriz A&#8217; s\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-31b2b3e5865c2264c360fb887d37a5f6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; 3 &amp; \\color{red}\\bm{1} \\\\[1.1ex] 3 &amp; -2 &amp; -1 &amp; \\color{red}\\bm{0} \\\\[1.1ex] 1 &amp; 3 &amp; 2 &amp; \\color{red}\\bm{5} \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"431\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0a604d8f5a3927a47a264d28f7a007b2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{vmatrix} =24\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"187\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a primeira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a1fa494ffb5e452d59c4d2dad40f925a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} \\color{red}\\bm{1} &amp; 1 &amp; 3 \\\\[1.1ex] \\color{red}\\bm{0} &amp; -2 &amp; -1 \\\\[1.1ex] \\color{red}\\bm{5} &amp; 3 &amp; 2 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{24}{24} = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"238\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-08e3dabe2f33434eb96658491f67c0b4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 2 &amp; \\color{red}\\bm{1} &amp; 3 \\\\[1.1ex] 3 &amp;  \\color{red}\\bm{0} &amp; -1 \\\\[1.1ex] 1 &amp; \\color{red}\\bm{5} &amp; 2\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{48}{24} = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"223\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Calcular<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-23aa090e6102a41de5ad5515112e4d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  z\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a terceira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96e76cb8867224755e9c19254678abd4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix} 2 &amp; 1 &amp; \\color{red}\\bm{1} \\\\[1.1ex] 3 &amp; -2 &amp;  \\color{red}\\bm{0} \\\\[1.1ex] 1 &amp; 3 &amp;  \\color{red}\\bm{5}\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-24}{24} = \\bm{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"259\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> A solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es \u00e9, portanto:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-be5a19fed42dcb59880c2d0eee8e51f4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x = 1 \\qquad y=2 \\qquad z = -1}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"210\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> Exemplo 2: Sistema Compat\u00edvel Indeterminado (ICS)<\/h2>\n<ul>\n<li> Resolva o seguinte sistema de equa\u00e7\u00f5es usando a regra de Cramer:<\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-781530aac4d8507fd6c7cbd77c3b4651_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0 \\\\[1.5ex] x+5y+3z = 1 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"149\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Primeiro fazemos a matriz A e a matriz estendida A&#8217; do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a64800a78bf8e2e2f547be907e6863cb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; 2 &amp; 4 \\\\[1.1ex] -2 &amp; 3 &amp; -1 \\\\[1.1ex] 1 &amp; 5 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 3 &amp; 2 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; 3 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 5 &amp; 3 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Agora calculamos o contradom\u00ednio das duas matrizes e assim podemos ver que tipo de sistema \u00e9. Para calcular o posto de A, calculamos o determinante de toda a matriz (usando a regra de Sarrus) e verificamos se \u00e9 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-581c58cbe0fdd9952e7e25b919ecc33b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; 2 &amp; 4 \\\\[1.1ex] -2 &amp; 3 &amp; -1 \\\\[1.1ex] 1 &amp; 5 &amp; 3\\end{vmatrix} = 27-2-40-12+15+12= 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"407\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> O determinante d\u00e1 0, ent\u00e3o a matriz A n\u00e3o \u00e9 de posto 3. Mas tem um determinante 2\u00d72 diferente de 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a5d1acad8bc31240f80d8cfbf3605997_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; 2 \\\\[1.1ex] -2 &amp; 3 \\end{vmatrix} =9-(-4)=13\\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"222\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Portanto, <strong>a matriz A tem classifica\u00e7\u00e3o 2<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Uma vez conhecida a extens\u00e3o da matriz A, calculamos a da matriz A&#8217;. O determinante das 3 primeiras colunas d\u00e1 0, ent\u00e3o tentamos os outros determinantes 3\u00d73 poss\u00edveis na matriz A&#8217;:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-686e7ca635ecee685005f6013c2e64ad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 4 &amp; 1 \\\\[1.1ex] 3 &amp; -1 &amp; 0 \\\\[1.1ex] 5 &amp; 3 &amp; 1 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 3 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 3 &amp; 1 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 3 &amp; 2 &amp; 1 \\\\[1.1ex] -2 &amp; 3 &amp; 0 \\\\[1.1ex] 1 &amp; 5 &amp; 1 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"440\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Todos os determinantes de ordem 3 d\u00e3o 0. Mas, obviamente, a matriz A&#8217; tem o mesmo determinante n\u00e3o-0 2\u00d72 que a matriz A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a5d1acad8bc31240f80d8cfbf3605997_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; 2 \\\\[1.1ex] -2 &amp; 3 \\end{vmatrix} =9-(-4)=13\\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"222\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Portanto, <strong>a matriz A&#8217; tamb\u00e9m \u00e9 de posto 2<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Assim, como o posto da matriz A \u00e9 igual ao posto da matriz A&#8217; mas estes dois s\u00e3o menores que o n\u00famero de inc\u00f3gnitas do sistema (3), sabemos pelo <strong>teorema de Rouch\u00e9-Frobenius<\/strong> que este \u00e9 um <strong>sistema indeterminadamente compat\u00edvel<\/strong> (ICS):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-background\" style=\"background-color:#dff6ff\"> Quando queremos resolver um sistema indeterminado compat\u00edvel (SCI), precisamos <strong>transformar o sistema<\/strong> : primeiro eliminamos uma equa\u00e7\u00e3o, depois convertemos uma vari\u00e1vel em \u03bb (geralmente a vari\u00e1vel z) e, finalmente, colocamos os termos com \u03bb juntos com os termos independentes.<\/p>\n<p> Uma vez transformado o sistema, aplicamos a regra de Cramer e obteremos a solu\u00e7\u00e3o do sistema em fun\u00e7\u00e3o de \u03bb.<\/p>\n<p> Neste caso, <strong>eliminaremos a \u00faltima equa\u00e7\u00e3o<\/strong> do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f0511fecc9c2af695b6b8eccae6b0661_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0 \\\\[1.5ex]\\cancel{x+5y+3z = 1} \\end{cases} \\longrightarrow \\quad \\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong>Agora vamos converter a vari\u00e1vel z em \u03bb:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2d6142d2be611954fd849a032a97245a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} 3x+2y+4\\lambda=1 \\\\[1.5ex] -2x+3y-\\lambda=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"398\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> E colocamos <strong>os termos com \u03bb com os termos independentes:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-00214205f2334f1c9bc10810c1c1df83_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y=1-4\\lambda \\\\[1.5ex] -2x+3y=\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"adsb30\" style=\" margin:px; text-align:\"><\/div>\n<p> Portanto, a matriz A e a matriz A&#8217; do sistema permanecem:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9c4b47303973b823a1c5628f5448ca79_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; 2  \\\\[1.1ex] -2 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 3 &amp; 2 &amp; 1 -4\\lambda \\\\[1.1ex] -2 &amp; 3 &amp; \\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"363\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Finalmente, uma vez transformado o sistema, <strong>aplicamos a regra de Cramer<\/strong> . Portanto, resolvemos o determinante de 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-d1b79f52dc82f5cfc311867273e78c06_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; 2  \\\\[1.1ex] -2 &amp; 3\\end{vmatrix} = 13\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"148\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a primeira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0ff917eaea976c65bd18e0476078d3cb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 1 -4\\lambda &amp; 2  \\\\[1.1ex] \\lambda &amp; 3 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{3(1-4\\lambda) -2\\lambda}{13} = \\cfrac{\\bm{3-14\\lambda} }{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"349\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-155ca520739bbf7e040a6cdc632f7c27_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 3 &amp; 1 -4\\lambda  \\\\[1.1ex]-2&amp;  \\lambda  \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{3\\lambda -\\bigl(-2(1-4\\lambda)\\bigr)}{13}= \\cfrac{3\\lambda -\\bigl(-2+8\\lambda\\bigr)}{13} = \\cfrac{\\bm{2-5\\lambda} }{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"529\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Embora a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es seja fun\u00e7\u00e3o de \u03bb, por ser um SCI e, portanto, possui infinitas solu\u00e7\u00f5es:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c9866e045041eb2d8fe103db2309f229_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =} \\cfrac{\\bm{3-14\\lambda} }{\\bm{13}} \\qquad \\bm{y=}\\cfrac{\\bm{2-5\\lambda} }{\\bm{13}} \\qquad \\bm{z = \\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"283\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> Regra de Cramer resolveu problemas <\/h2>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-118\"><\/div>\n<\/div>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Aplique a regra de Cramer para resolver o seguinte sistema de duas equa\u00e7\u00f5es com 2 inc\u00f3gnitas: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-pas-a-pas-de-la-regle-de-cramer-22.webp\" alt=\"exerc\u00edcio resolvido passo a passo com a regra 2x2 de Cramer\" class=\"wp-image-3999\" width=\"137\" height=\"83\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>veja solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> A primeira coisa a fazer \u00e9 a matriz A e a matriz estendida A&#8217; do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2a001db9cf56846150730fee7126dacd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{cc} 2 &amp; 5 \\\\[1.1ex] 1 &amp; 4 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 2 &amp; 5 &amp; 8 \\\\[1.1ex] 1 &amp; 4 &amp; 7 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"294\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Devemos agora encontrar o posto da matriz A. Para isso, verificamos se o determinante de toda a matriz \u00e9 diferente de 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0c75c1c344c286016bea83237f1f418e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 5 \\\\[1.1ex] 1 &amp; 4 \\end{vmatrix} = 8-5=3 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"216\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Como a matriz tem um determinante 2\u00d72 diferente de 0, <strong>a matriz A tem posto 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Uma vez conhecida a classifica\u00e7\u00e3o de A, calculamos a classifica\u00e7\u00e3o de A&#8217;. Este ser\u00e1 pelo menos de posto 2, pois acabamos de ver que tem dentro de um determinante de ordem 2 diferente de 0. Al\u00e9m disso, n\u00e3o pode ser de posto 3, pois n\u00e3o podemos deixar de fazer um determinante 3\u00d73. Portanto, <strong>a matriz A&#8217; tamb\u00e9m \u00e9 de posto 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, aplicando o <strong>teorema de Rouch\u00e9-Frobenius,<\/strong> sabemos que este \u00e9 um <strong>sistema determinado compat\u00edvel<\/strong> (SCD), pois o contradom\u00ednio de A \u00e9 igual ao contradom\u00ednio de A&#8217; e ao n\u00famero de inc\u00f3gnitas.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bbd67b16bb6d52a0696e70a77833cd3b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 2 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 2 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Uma vez sabendo que o sistema \u00e9 um SCD, aplicamos <strong>a regra de Cramer<\/strong> para resolv\u00ea-lo.<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a primeira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b0adeda8f2ce557661466996038b1148_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 8 &amp; 5 \\\\[1.1ex] 7 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-3}{3} = \\bm{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-59790a66cc31fac07be1d5a7bb556d9e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}2 &amp; 8 \\\\[1.1ex] 1 &amp; 7\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{6}{3} = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"150\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es \u00e9, portanto: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-26fa7c9ed2d05ca07ff62a968ba7ab11_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x = -1 \\qquad y=2}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"133\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 2<\/h3>\n<p> Encontre a solu\u00e7\u00e3o do seguinte sistema de tr\u00eas equa\u00e7\u00f5es com 3 inc\u00f3gnitas usando a regra de Cramer: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-systeme-de-regles-de-cramer-des-equations-3-3.webp\" alt=\"Exerc\u00edcio resolvido da regra de Cramer de um sistema de equa\u00e7\u00f5es 3x3\" class=\"wp-image-4002\" width=\"181\" height=\"124\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>veja solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro fazemos a matriz A e a matriz estendida A&#8217; do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eea75fbf6d86ebc3d0b9e236cd2160f5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 3 &amp; 2\\\\[1.1ex] -1 &amp; 5 &amp; -1\\\\[1.1ex] 3 &amp; -1 &amp; 4 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 3 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 5 &amp; -1 &amp; 4 \\\\[1.1ex] 3 &amp; -1 &amp; 4 &amp; 0 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"432\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Agora encontramos a classifica\u00e7\u00e3o da matriz A calculando o determinante da matriz 3\u00d73 com a regra de Sarrus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-73f751f3b5c527c16b5de1b10bf07a4e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 3 &amp; 2 \\\\[1.1ex] -1 &amp; 5 &amp; -1\\\\[1.1ex] 3 &amp; -1 &amp; 4 \\end{vmatrix} = 20-9+2-30-1+12=-6 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"445\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A matriz tendo um determinante de ordem 3 diferente de 0, <strong>a matriz A \u00e9 de posto 3:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> conseq\u00fcentemente, a matriz A&#8217; tamb\u00e9m \u00e9 de classifica\u00e7\u00e3o 3:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, utilizando o <strong>teorema de Rouch\u00e9-Frobenius,<\/strong> sabemos que este \u00e9 um <strong>sistema determinado compat\u00edvel<\/strong> (SCD), pois o contradom\u00ednio de A \u00e9 igual ao contradom\u00ednio de A&#8217; e ao n\u00famero de inc\u00f3gnitas.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-31b495a48a75d7af1f23e38818bf4eca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 3 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Uma vez sabendo que o sistema \u00e9 um SCD, precisamos aplicar <strong>a regra de Cramer<\/strong> para resolver o sistema.<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a primeira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fc574297f609b68e4fb48466ec6c8077_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 2 &amp; 3 &amp; 2 \\\\[1.1ex] 4 &amp; 5 &amp; -1\\\\[1.1ex]0 &amp; -1 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-18}{-6} = \\bm{3}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"235\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2544601137d62e217ff1866f278203d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 4 &amp; -1\\\\[1.1ex] 3 &amp; 0 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-6}{-6} = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"224\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calcular<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-23aa090e6102a41de5ad5515112e4d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  z\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a terceira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-42d7d4adcfc48954185ca14b56b8e128_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix} 1 &amp; 3 &amp; 2 \\\\[1.1ex] -1 &amp; 5 &amp; 4 \\\\[1.1ex] 3 &amp; -1 &amp; 0\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{12}{-6} = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"230\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es \u00e9, portanto: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-685195d3a299f30f6421bb387f7f00e4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =3 \\qquad y=1 \\qquad z=-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"210\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 3<\/h3>\n<p> Calcule a solu\u00e7\u00e3o do seguinte sistema de tr\u00eas equa\u00e7\u00f5es com 3 inc\u00f3gnitas usando a regra de Cramer: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exemple-de-regle-de-cramer.webp\" alt=\"exemplo da regra de Cramer\" class=\"wp-image-4003\" width=\"183\" height=\"123\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>veja solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro fazemos a matriz A e a matriz estendida A&#8217; do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-afd359275e5ebaaf3229504c47a5815f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 2 &amp; 5\\\\[1.1ex] 2 &amp; 3 &amp; -1 \\\\[1.1ex] 3 &amp; 4 &amp; -7 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 2 &amp; 5 &amp; 1 \\\\[1.1ex] 2 &amp; 3 &amp; -1 &amp; 5 \\\\[1.1ex] 3 &amp; 4 &amp; -7 &amp; 9 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calculamos a extens\u00e3o da matriz A: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-47ddf17a2b3eed5a680d685900a79b31_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 2 &amp; 5\\\\[1.1ex] 2 &amp; 3 &amp; -1 \\\\[1.1ex] 3 &amp; 4 &amp; -7 \\end{vmatrix} =-21-6+40-45+4+28=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"398\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fdd4380c7c76418bd3ec12c94359f886_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 \\\\[1.1ex] 2 &amp; 3  \\end{vmatrix} = 3-4 = -1 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Uma vez conhecida a extens\u00e3o da matriz A, calculamos a da matriz A&#8217;. O determinante das 3 primeiras colunas d\u00e1 0, ent\u00e3o tentamos os outros determinantes 3\u00d73 poss\u00edveis na matriz A&#8217;:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1addc62130e0462075b3bade26a7e35e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 5 &amp; 1 \\\\[1.1ex]  3 &amp; -1 &amp; 5 \\\\[1.1ex] 4 &amp; -7 &amp; 9 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 1 &amp; 5 &amp; 1 \\\\[1.1ex] 2 &amp; -1 &amp; 5 \\\\[1.1ex] 3 &amp; -7 &amp; 9\\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 3 &amp; 5 \\\\[1.1ex] 3 &amp; 4 &amp; 9 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"412\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Todos os determinantes de ordem 3 d\u00e3o 0. No entanto, a matriz A&#8217; tem o mesmo determinante 2\u00d72 diferente de 0 que a matriz A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7de377466bd5afd03f58f9b532324e75_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 \\\\[1.1ex] 2 &amp; 3 \\end{vmatrix} = 3-4 = -1 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, a matriz A&#8217; tamb\u00e9m \u00e9 de posto 2:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Como o posto da matriz A \u00e9 igual ao posto da matriz A&#8217; mas estes dois s\u00e3o menores que o n\u00famero de inc\u00f3gnitas do sistema (3), sabemos pelo <strong>teorema de Rouch\u00e9-Frobenius<\/strong> que se trata de um <strong>Sistema Compat\u00edvel Indeterminado<\/strong> (ICS):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sendo um sistema ICS, temos que eliminar uma equa\u00e7\u00e3o. Neste caso, <strong>eliminaremos a \u00faltima equa\u00e7\u00e3o<\/strong> do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3a1d067e155540f4345cf56e5c1567d3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+2y+5z=1 \\\\[1.5ex] 2x+3y-z=5 \\\\[1.5ex]\\cancel{3x+4y-7z = 9} \\end{cases} \\longrightarrow \\quad \\begin{cases} x+2y+5z=1 \\\\[1.5ex] 2x+3y-z=5\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"357\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Agora vamos converter a vari\u00e1vel z em \u03bb:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b5fa91777a722d3783b2f887aab44152_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+2y+5z=1 \\\\[1.5ex] 2x+3y-z=5  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} x+2y+5\\lambda=1 \\\\[1.5ex] 2x+3y-\\lambda=5\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"369\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E colocamos os termos com \u03bb com os termos independentes:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-76ff21181be050b01c247981298986a7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+2y=1-5\\lambda\\\\[1.5ex] 2x+3y=5+\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"136\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tal que a matriz A e a matriz A&#8217; do sistema permanecem:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-230e5b28dd467127e63f4f9756cf90da_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 2  \\\\[1.1ex] 2 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 1 &amp; 2 &amp; 1 -5\\lambda \\\\[1.1ex] 2 &amp; 3 &amp;5+\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"335\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Finalmente, uma vez transformado o sistema, <strong>aplicamos a regra de Cramer<\/strong> . Portanto, resolvemos o determinante de 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-f127efbd217e2bca8852ec792610732f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 2 \\\\[1.1ex] 2 &amp; 3\\end{vmatrix} =-1\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"138\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a primeira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-42652a14362b42e606841b6bb3e77cc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 1-5\\lambda &amp; 2 \\\\[1.1ex] 5+\\lambda &amp; 3 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{3-15\\lambda -(10+2\\lambda)}{-1} = \\cfrac{-7-17\\lambda}{-1} = \\bm{7+17\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"491\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b95c5870f1762a2d82c9ebcccbca7408_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 1 &amp; 1-5\\lambda \\\\[1.1ex] 2 &amp; 5+\\lambda \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{5+\\lambda -(2-10\\lambda)}{-1}= \\cfrac{3+11\\lambda}{-1} = \\bm{-3-11\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"465\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Embora a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es seja fun\u00e7\u00e3o de \u03bb, por ser um SCI e, portanto, possui infinitas solu\u00e7\u00f5es: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5483357f081aca551b07fe7c8f9ebf5d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =7+17\\lambda} \\qquad \\bm{y=-3-11\\lambda} \\qquad \\bm{z = \\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"311\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-119\"><\/div>\n<\/div>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 4<\/h3>\n<p> Resolva o seguinte problema de um sistema de tr\u00eas equa\u00e7\u00f5es com 3 inc\u00f3gnitas aplicando a regra de Cramer: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-61e1c3458f33b863db10750b9e51d09e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} -2x+5y+z=8 \\\\[1.5ex] 6x+2y+4z=4 \\\\[1.5ex] 3x-2y+z = -2 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"149\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>veja solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro, constru\u00edmos a matriz A e a matriz estendida A&#8217; do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-743a40010cb4a610e8a3fc6ae5d313b4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}-2 &amp; 5 &amp; 1 \\\\[1.1ex] 6 &amp; 2 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} -2 &amp; 5 &amp; 1 &amp; 8 \\\\[1.1ex] 6 &amp; 2 &amp; 4 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1 &amp; -2 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Agora vamos calcular a classifica\u00e7\u00e3o da matriz A calculando o determinante da matriz 3&#215;3 usando a regra de Sarrus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-713c634fbc3e1b1cb228e3891c9bff1c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} -2 &amp; 5 &amp; 1 \\\\[1.1ex] 6 &amp; 2 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1 \\end{vmatrix} = -4+60-12-6-16-30=-8 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"453\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A matriz tendo um determinante de ordem 3 diferente de 0, <strong>a matriz A \u00e9 de posto 3:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> conseq\u00fcentemente, a matriz A&#8217; tamb\u00e9m \u00e9 de posto 3, pois deve ser pelo menos do mesmo posto que a matriz A e n\u00e3o pode ser de posto 4 porque \u00e9 uma matriz de dimens\u00e3o 3\u00d74.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, utilizando o <strong>teorema de Rouch\u00e9-Frobenius,<\/strong> deduzimos que se trata de um <strong>sistema compat\u00edvel determinado<\/strong> (SCD), pois o contradom\u00ednio de A \u00e9 igual ao contradom\u00ednio de A&#8217; e ao n\u00famero de inc\u00f3gnitas.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-31b495a48a75d7af1f23e38818bf4eca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 3 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Uma vez sabendo que o sistema \u00e9 um SCD, precisamos aplicar <strong>a regra de Cramer<\/strong> para resolver o sistema.<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a primeira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8a290479c69ff806f19dcf29f96e1228_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 8 &amp; 5 &amp; 1 \\\\[1.1ex] 4 &amp; 2 &amp; 4 \\\\[1.1ex] -2 &amp; -2 &amp; 1\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{16}{-8} = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"231\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8bba0765fbcbcebf0585520af25b4a30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}-2 &amp; 8 &amp; 1 \\\\[1.1ex] 6 &amp; 4 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{0}{-6} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"217\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calcular<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-23aa090e6102a41de5ad5515112e4d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  z\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a terceira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5bc157a8c4dfe8ee4651affac68ef878_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix} -2 &amp; 5 &amp; 8 \\\\[1.1ex] 6 &amp; 2 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; -2\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-32}{-8} = \\bm{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"247\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es lineares \u00e9, portanto: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6c004c5c466235d2d1a784707145d952_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =-2 \\qquad y=0 \\qquad z=4}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"211\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 5<\/h3>\n<p> Resolva o seguinte sistema de equa\u00e7\u00f5es lineares usando a regra de Cramer: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exemple-comment-resoudre-un-systeme-dequations-avec-la-regle-de-cramer.webp\" alt=\"Exemplo de resolu\u00e7\u00e3o de um sistema de equa\u00e7\u00f5es com a regra de Cramer\" class=\"wp-image-4008\" width=\"215\" height=\"127\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>veja solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro fazemos a matriz A e a matriz estendida A&#8217; do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f5153b5951b768cc3cafa2bb2567ba92_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; -2 &amp; -3 \\\\[1.1ex] -1 &amp; 5 &amp; 4 \\\\[1.1ex] 5 &amp; 1 &amp; -2 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 3 &amp; -2 &amp; -3 &amp; 4 \\\\[1.1ex] -1 &amp; 5 &amp; 4 &amp; -10 \\\\[1.1ex] 5 &amp; 1 &amp; -2 &amp; -2 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"455\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calculamos a extens\u00e3o da matriz A: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f3778c9499e2a44ea3834dfed1523163_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; -2 &amp; -3 \\\\[1.1ex] -1 &amp; 5 &amp; 4 \\\\[1.1ex] 5 &amp; 1 &amp; -2 \\end{vmatrix} =-30-40+3+75-12+4=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-03d70742b14ced92f33963df0c86e92f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; -2 \\\\[1.1ex] -1 &amp; 5  \\end{vmatrix} = 15- (2)= 13 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"231\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Uma vez conhecida a extens\u00e3o da matriz A, calculamos a da matriz A&#8217;. O determinante das 3 primeiras colunas d\u00e1 0, ent\u00e3o tentamos os outros determinantes 3\u00d73 poss\u00edveis na matriz A&#8217;:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5bed93d532ae4ccd4649a73662f55f0f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -2 &amp; -3 &amp; 4 \\\\[1.1ex] 5 &amp; 4 &amp; -10 \\\\[1.1ex]  1 &amp; -2 &amp; -2 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix}3 &amp; -3 &amp; 4 \\\\[1.1ex] -1 &amp; 4 &amp; -10 \\\\[1.1ex] 5 &amp; -2 &amp; -2\\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 3 &amp; -2 &amp; 4 \\\\[1.1ex] -1 &amp; 5 &amp; -10 \\\\[1.1ex] 5 &amp; 1 &amp;-2\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"535\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Todos os determinantes de ordem 3 d\u00e3o 0. Mas, obviamente, a matriz A&#8217; tem o mesmo determinante de ordem 2 diferente de 0 que a matriz A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-858d95d7d252b16706b66c0e6aba09c4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; -2 \\\\[1.1ex] -1 &amp; 5 \\end{vmatrix} = 13 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, a matriz A&#8217; tamb\u00e9m \u00e9 de posto 2:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O posto da matriz A \u00e9 igual ao posto da matriz A&#8217; mas estes dois s\u00e3o menores que o n\u00famero de inc\u00f3gnitas do sistema (3), ent\u00e3o pelo <strong>teorema de Rouch\u00e9-Frobenius<\/strong> sabemos que se trata de um <strong>Sistema Indeterminado Compat\u00edvel<\/strong> (SCI) :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sendo um sistema ICS, temos que eliminar uma equa\u00e7\u00e3o. Neste caso, <strong>eliminaremos a \u00faltima equa\u00e7\u00e3o<\/strong> do sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4e10bd826663dff41c4272610cbc07b1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x-2y-3z=4 \\\\[1.5ex] -x+5y+4z=-10 \\\\[1.5ex]\\cancel{5x+y-2z = -2} \\end{cases} \\longrightarrow \\quad \\begin{cases} 3x-2y-3z=4 \\\\[1.5ex] -x+5y+4z=-10\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"423\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Agora vamos converter a vari\u00e1vel z em \u03bb:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2502be450040b38761c08e5d6beaf379_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x-2y-3z=4 \\\\[1.5ex] -x+5y+4z=-10  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} 3x-2y-3\\lambda=4 \\\\[1.5ex] -x+5y+4\\lambda=-10\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"444\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E colocamos os termos com \u03bb com os termos independentes:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80a43d98e6be30965d554e8a89aa5d89_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x-2y=4+3\\lambda \\\\[1.5ex] -x+5y=-10-4\\lambda\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"172\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tal que a matriz A e a matriz A&#8217; do sistema permanecem:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3451ce571163983cf41794d4998283d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; -2  \\\\[1.1ex] -1 &amp; 5 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 3 &amp; -2 &amp; 4+3\\lambda \\\\[1.1ex] 1 &amp; 5 &amp;-10-4\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"399\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Finalmente, uma vez transformado o sistema, <strong>aplicamos a regra de Cramer<\/strong> . Portanto, resolvemos o determinante de 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-0e7a7d6208ea5e762f5c74a44e6838cf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix}3&amp; -2 \\\\[1.1ex] -1 &amp; 5\\end{vmatrix} =13\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"162\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a primeira coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8c30fcc0526c2d4112eb4f60a3d8847f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 4+3\\lambda &amp; -2 \\\\[1.1ex]-10-4\\lambda &amp; 5\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{20+15\\lambda -(20+8\\lambda)}{13} = \\cfrac{\\bm{7\\lambda}}{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"394\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular o desconhecido<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna dos termos independentes e dividimos pelo determinante de A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fdb22a54274e019c811c9051502c474a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 3 &amp; 4+3\\lambda \\\\[1.1ex] -1 &amp; -10-4\\lambda\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-30-12\\lambda -(-4-3\\lambda)}{13}= \\cfrac{\\bm{-26-9\\lambda}}{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"473\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim, a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es \u00e9 fun\u00e7\u00e3o de \u03bb, pois \u00e9 um SCI e, portanto, o sistema possui infinitas solu\u00e7\u00f5es:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d0e525b9aca6bd683491ab7950f039e3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x=} \\cfrac{\\bm{7\\lambda}}{\\bm{13}} \\qquad \\bm{y=} \\cfrac{\\bm{-26-9\\lambda}}{\\bm{13}} \\qquad \\bm{z = \\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"266\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Nesta p\u00e1gina voc\u00ea ver\u00e1 o que \u00e9 a regra de Cramer e, al\u00e9m disso, encontrar\u00e1 exemplos e exerc\u00edcios com resolu\u00e7\u00e3o de sistemas de equa\u00e7\u00f5es pela regra de Cramer. Qual \u00e9 a regra de Cramer? A regra de Cramer \u00e9 um m\u00e9todo usado para resolver sistemas de equa\u00e7\u00f5es por determinantes. Vamos ver como \u00e9 usado: Considere &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/\"> <span class=\"screen-reader-text\">Regra de cramer<\/span> Leia mais &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":[24],"tags":[],"class_list":["post-304","post","type-post","status-publish","format-standard","hentry","category-sistemas-educacionais"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Regra de Cramer - Matoridade<\/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\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Regra de Cramer - Matoridade\" \/>\n<meta property=\"og:description\" content=\"Nesta p\u00e1gina voc\u00ea ver\u00e1 o que \u00e9 a regra de Cramer e, al\u00e9m disso, encontrar\u00e1 exemplos e exerc\u00edcios com resolu\u00e7\u00e3o de sistemas de equa\u00e7\u00f5es pela regra de Cramer. Qual \u00e9 a regra de Cramer? A regra de Cramer \u00e9 um m\u00e9todo usado para resolver sistemas de equa\u00e7\u00f5es por determinantes. 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Matoridade","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/","og_locale":"pt_BR","og_type":"article","og_title":"Regra de Cramer - Matoridade","og_description":"Nesta p\u00e1gina voc\u00ea ver\u00e1 o que \u00e9 a regra de Cramer e, al\u00e9m disso, encontrar\u00e1 exemplos e exerc\u00edcios com resolu\u00e7\u00e3o de sistemas de equa\u00e7\u00f5es pela regra de Cramer. Qual \u00e9 a regra de Cramer? A regra de Cramer \u00e9 um m\u00e9todo usado para resolver sistemas de equa\u00e7\u00f5es por determinantes. Vamos ver como \u00e9 usado: Considere &hellip; Regra de cramer Leia mais &raquo;","og_url":"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/","article_published_time":"2023-07-06T14:13:39+00:00","og_image":[{"url":"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e0141f3451719f665ef28e4061489551_l3.png"}],"author":"Equipe Mathoridade","twitter_card":"summary_large_image","twitter_misc":{"Escrito por":"Equipe Mathoridade","Est. tempo de leitura":"11 minutos"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/#article","isPartOf":{"@id":"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/"},"author":{"name":"Equipe Mathoridade","@id":"https:\/\/mathority.org\/pt\/#\/schema\/person\/26defeb7b79f5baaedafa33a1ac6ac00"},"headline":"Regra de cramer","datePublished":"2023-07-06T14:13:39+00:00","dateModified":"2023-07-06T14:13:39+00:00","mainEntityOfPage":{"@id":"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/"},"wordCount":2154,"commentCount":0,"publisher":{"@id":"https:\/\/mathority.org\/pt\/#organization"},"articleSection":["Sistemas educacionais"],"inLanguage":"pt-BR","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/","url":"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/","name":"Regra de Cramer - 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