{"id":305,"date":"2023-07-06T14:08:28","date_gmt":"2023-07-06T14:08:28","guid":{"rendered":"https:\/\/mathority.org\/pt\/discussao-de-sistemas-de-equacoes-com-parametros\/"},"modified":"2023-07-06T14:08:28","modified_gmt":"2023-07-06T14:08:28","slug":"discussao-de-sistemas-de-equacoes-com-parametros","status":"publish","type":"post","link":"https:\/\/mathority.org\/pt\/discussao-de-sistemas-de-equacoes-com-parametros\/","title":{"rendered":"Discuss\u00e3o de sistemas de equa\u00e7\u00f5es com par\u00e2metros"},"content":{"rendered":"<p>Nesta p\u00e1gina veremos como discutir e resolver um <strong>sistema de equa\u00e7\u00f5es com par\u00e2metros<\/strong> . Al\u00e9m disso, voc\u00ea encontrar\u00e1 exemplos e exerc\u00edcios resolvidos de sistemas de equa\u00e7\u00f5es lineares para praticar.<\/p>\n<p> Por outro lado, para analisar sistemas de equa\u00e7\u00f5es lineares \u00e9 importante que voc\u00ea saiba <a href=\"https:\/\/mathority.org\/pt\/exemplos-de-regras-e-exercicios-resolvidos-de-cramer\/\">o que \u00e9 a regra de Cramer<\/a> e <a href=\"https:\/\/mathority.org\/pt\/teorema-de-de-rouche-frobenius-com-exemplos-e-exercicios-resolvidos\/\">o que \u00e9 o teorema de Rouch\u00e9\u2013Frobenius<\/a> , pois os utilizaremos constantemente.<\/p>\n<h2 class=\"wp-block-heading\"> Exemplo de sistema de equa\u00e7\u00f5es lineares com par\u00e2metros<\/h2>\n<ul>\n<li> Discuta e resolva o seguinte sistema de equa\u00e7\u00f5es em termos do par\u00e2metro <em>m<\/em> :<\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6ab2286d15c20029b98a5ea4622033d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+y+2z= 2 \\\\[1.5ex] -x+my+2z=0 \\\\[1.5ex] 3x+mz = 4\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"155\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> 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-bef8e6b26595703c77c65178cbf90ffc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; m \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; m &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Agora resolvemos o determinante de A usando a regra de Sarrus, para ver qual \u00e9 a classifica\u00e7\u00e3o da matriz:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e823f83f25f798bd854612a7352680d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\begin{aligned}  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; m \\end{vmatrix} &amp; =m^2+6+0-6m-0+m \\\\ &amp; = m^2-5m+6 \\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"110\" width=\"366\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Portanto, o resultado do determinante de A depende do valor de <em>m<\/em> . Veremos, portanto, para quais valores de <em>m<\/em> o determinante desaparece. Para fazer isso, <strong>definimos o resultado igual a 0<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a2101d9a1d88f3e3d29bb4758d2fb8a6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   m^2-5m+6 = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"133\" style=\"vertical-align: -2px;\"><\/p>\n<\/p>\n<p> E resolvemos a equa\u00e7\u00e3o quadr\u00e1tica com a f\u00f3rmula:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-51910dd69f9df6fdde6dd7597f889700_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  m = \\cfrac{-b \\pm \\sqrt{b^2-4ac}}{2a}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"170\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e863082ac1f9b43df4de9fe93f5eb305_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  m = \\cfrac{-(-5) \\pm \\sqrt{(-5)^2-4\\cdot 1 \\cdot 6}}{2 \\cdot 1} = \\cfrac{5 \\pm \\sqrt{25-24}}{2} =\\cfrac{5 \\pm 1}{2} = \\begin{cases} \\bm{m = 3} \\\\[2ex] \\bm{m =2} \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"537\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Ent\u00e3o quando <em>m<\/em> for igual a 2 ou 3, o determinante de A ser\u00e1 0. E quando <em>m<\/em> for diferente de 2 e diferente de 3, o determinante de A ser\u00e1 diferente de 0.<\/p>\n<p> Devemos, portanto, analisar cada caso separadamente:<\/p>\n<p style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m\u22603 e m\u22602:<\/strong><\/span><\/p>\n<p> Como acabamos de ver, quando o par\u00e2metro <em>m<\/em> \u00e9 diferente de 2 e 3, o determinante da matriz A \u00e9 diferente de 0. Portanto, o <strong>posto de A \u00e9 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> Al\u00e9m disso, <strong>o posto da matriz A&#8217; tamb\u00e9m \u00e9 3<\/strong> , porque dentro dela existe uma submatriz 3\u00d73 cujo determinante \u00e9 diferente de 0. E n\u00e3o pode ser de posto 4 pois \u2018n\u00e3o podemos fazer um determinante 4\u00d74.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Ent\u00e3o, como o posto da matriz A \u00e9 igual ao posto da matriz A&#8217; e ao n\u00famero de inc\u00f3gnitas do sistema (3), pelo <strong>teorema de Rouch\u00e9-Frobenius<\/strong> sabemos que se trata de um <strong>Sistema Determinado Compat\u00edvel<\/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 que sabemos que o sistema \u00e9 um Sistema Determinado Compat\u00edvel (DCS), 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-bef8e6b26595703c77c65178cbf90ffc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; m \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; m &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-aac47361358555f733a42cffecabdbe9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; m \\end{vmatrix} = m^2-5m+6\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"268\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular x com a regra de Cramer, trocamos a primeira coluna do determinante da matriz 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-b17f49436fdadbb014011b5c461a4a56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\bm{x} = \\cfrac{\\begin{vmatrix} 2 &amp; 1 &amp; 2\\\\[1.1ex]0&amp;m&amp;2 \\\\[1.1ex] 4 &amp; 0 &amp; m \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{2m^2+8-8m}{m^2-5m+6}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"257\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular y 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-6a2bf75bdabfb2c83870f1869ce19e3d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 0 &amp; 2 \\\\[1.1ex] 3 &amp; 4 &amp; m \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}}=\\cfrac{-4+2m}{m^2-5m+6}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"255\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular z 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-eebb3c4d280afc8a9aed8877ddcd4ac5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix}  1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; m &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-2m+4}{m^2-5m+6}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"254\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Portanto, a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es para o caso m\u22603 e m\u22602 \u00e9:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7cba73a14f41d3314575e075f1229e87_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =} \\cfrac{\\bm{2m^2+8-8m}}{\\bm{m^2-5m+6}} \\qquad \\bm{y=} \\cfrac{\\bm{-4+2m}}{\\bm{m^2-5m+6}} \\qquad \\bm{z =} \\cfrac{\\bm{-2m+4}}{\\bm{m^2-5m+6}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"470\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p> Como voc\u00ea pode ver, neste caso a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es \u00e9 fun\u00e7\u00e3o de m.<\/p>\n<p> Uma vez encontrada a solu\u00e7\u00e3o para quando m \u00e9 diferente de 2 e 3, resolveremos o sistema para quando m \u00e9 igual a 2:<\/p>\n<div class=\"adsb30\" style=\" margin:px; text-align:\"><\/div>\n<p style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=2:<\/strong><\/span><\/p>\n<p> Analisaremos agora o sistema quando o par\u00e2metro <em>m<\/em> for igual a 2. Neste caso as matrizes A e 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-f42ec4801f3e84cd44b4e0b2ae6351cf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; 2 &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; 2 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 2 &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; 2 &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Como vimos anteriormente, quando m=2 o determinante de A \u00e9 0. Portanto, a matriz A n\u00e3o \u00e9 de posto 3. Mas dentro dela possui 2\u00d72 determinantes diferentes de 0, por exemplo:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-55ef6cd148fca7a869e14760007e1f2e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}1 &amp; 1 \\\\[1.1ex] -1 &amp; 2  \\end{vmatrix} = 2 - (-1)=3 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Ent\u00e3o, neste caso <strong>, a classifica\u00e7\u00e3o de A \u00e9 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 classifica\u00e7\u00e3o da matriz A, calculamos a classifica\u00e7\u00e3o de 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-2c68c742cae37c52ad2566b7feec5301_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 &amp; 2 \\\\[1.1ex] 2 &amp; 2 &amp; 0 \\\\[1.1ex] 0 &amp; 2 &amp; 4 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 2 &amp; 4 \\end{vmatrix}=0\\qquad \\begin{vmatrix} 1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; 4\\end{vmatrix}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"412\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Todos os determinantes poss\u00edveis de dimens\u00e3o 3\u00d73 d\u00e3o 0. Mas, obviamente, 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-55ef6cd148fca7a869e14760007e1f2e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}1 &amp; 1 \\\\[1.1ex] -1 &amp; 2  \\end{vmatrix} = 2 - (-1)=3 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> 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> Por se tratar de um ICS, precisamos transformar o sistema para resolv\u00ea-lo. Para isso, devemos primeiro eliminar uma equa\u00e7\u00e3o do sistema, neste caso <strong>iremos deletar a \u00faltima equa\u00e7\u00e3o:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-10c7facda35cb8894e6bbb236e4953f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+y+2z= 2 \\\\[1.5ex] -x+2y+2z=0 \\\\[1.5ex] \\cancel{3x+2z = 4} \\end{cases} \\longrightarrow \\quad \\begin{cases}  x+y+2z= 2 \\\\[1.5ex] -x+2y+2z=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong>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-0155083595420da31a486927e953805c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}x+y+2z= 2 \\\\[1.5ex] -x+2y+2z=0  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} x+y+2\\lambda= 2 \\\\[1.5ex] -x+2y+2\\lambda=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"398\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> 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-b8486baee4be39f417988ee12b5e67c7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}x+y=2-2\\lambda \\\\[1.5ex] -x+2y=-2\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"133\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> 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-8591e8c21bce2f49998311bbb08f7dee_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 1  \\\\[1.1ex] -1 &amp; 2 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 1 &amp; 1 &amp; 2 -2\\lambda \\\\[1.1ex] -1 &amp; 2 &amp; -2\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"363\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Finalmente, uma vez transformado o sistema, <strong>aplicamos a regra de Cramer<\/strong> . Para fazer isso, primeiro 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-c34669d7234c9736c350f793df337bd3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 1  \\\\[1.1ex] -1 &amp; 2\\end{vmatrix} =2-(-1)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"229\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular <em>x<\/em> 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-987ebe052154332042afeb27535996f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 2 -2\\lambda &amp; 1  \\\\[1.1ex] -2\\lambda &amp; 2 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{4-4\\lambda-(-2\\lambda)}{3} = \\cfrac{\\bm{4-2\\lambda}}{\\bm{3}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"345\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular <em>y<\/em> 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-8a3c7b2cd7319f7f9db6df7df79abb50_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 1 &amp; 2 -2\\lambda  \\\\[1.1ex] -1 &amp; -2\\lambda  \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}}=\\cfrac{-2\\lambda -(-2+2\\lambda)}{3} = \\cfrac{\\bm{2-4\\lambda} }{\\bm{3}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"379\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> De modo que quando m=2 a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es \u00e9 fun\u00e7\u00e3o de \u03bb, pois \u00e9 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-1bc871365fab3194e382053fc6b083b5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =}  \\cfrac{\\bm{4-2\\lambda}}{\\bm{3}}  \\qquad \\bm{y=}\\cfrac{\\bm{2-4\\lambda}}{\\bm{3}} \\qquad \\bm{z=\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"274\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> J\u00e1 analisamos a solu\u00e7\u00e3o do sistema quando o par\u00e2metro <em>m<\/em> \u00e9 diferente de 2 e 3, e quando \u00e9 igual a 2. Portanto, precisamos apenas do \u00faltimo caso: quando <em>m<\/em> assume o valor de 3: <\/p>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-118\"><\/div>\n<\/div>\n<p style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=3:<\/strong><\/span><\/p>\n<p> Analisaremos agora o que acontece quando o par\u00e2metro <em>m<\/em> \u00e9 3. Neste caso as matrizes A e 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-c49bbc0d7d36606aa59be050c2682de5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}1 &amp; 1 &amp; 2 \\\\[1.1ex] -1 &amp; 3 &amp; 2 \\\\[1.1ex] 3 &amp; 0 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 3 &amp; 2 &amp; 0 \\\\[1.1ex] 3 &amp; 0 &amp; 3 &amp; 4 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Como vimos anteriormente, quando m=3 o determinante de A \u00e9 0. Portanto a matriz A n\u00e3o \u00e9 de posto 3. Mas dentro dela possui 2\u00d72 determinantes diferentes de 0, por exemplo:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d88ce42feb4bba9aa74aae98e1062c4a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}1 &amp; 1 \\\\[1.1ex] -1 &amp; 3  \\end{vmatrix} = 3 - (-1)=4 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Ent\u00e3o, neste caso <strong>, a classifica\u00e7\u00e3o de A \u00e9 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 classifica\u00e7\u00e3o da matriz A, calculamos a classifica\u00e7\u00e3o de A&#8217;. O determinante das 3 primeiras colunas d\u00e1 0, portanto tentamos outro determinante 3\u00d73 que est\u00e1 dentro da matriz A&#8217;, por exemplo o das 3 \u00faltimas colunas:<\/p>\n<p class=\"has-text-align-center\">\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5e6f1a5c155ca004c73e51bdcbe5ece9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 &amp; 2 \\\\[1.1ex] 3 &amp; 2 &amp; 0 \\\\[1.1ex] 0 &amp; 3 &amp; 4\\end{vmatrix}=2\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"100\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Por outro lado, a matriz A&#8217; cont\u00e9m um determinante cujo resultado \u00e9 diferente de 0, portanto <strong>a matriz A&#8217; \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-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> Assim, quando m = 3, o posto da matriz A \u00e9 inferior ao posto da matriz A&#8217;. Assim, do teorema de Rouch\u00e9-Frobenius, deduzimos que o sistema \u00e9 um <strong>Sistema Incompat\u00edvel<\/strong> (SI) <strong>:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3454f804b63f3cca9bcf08bc93815f90_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas}=3\\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A)=2 \\ \\neq \\ rg(A') = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Portanto, o sistema de equa\u00e7\u00f5es <strong>n\u00e3o tem solu\u00e7\u00e3o quando m = 3.<\/strong><\/p>\n<h3 class=\"wp-block-heading\"> Resumo do exemplo:<\/h3>\n<p> Como vimos, a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es depende do valor do par\u00e2metro <em>m<\/em> . Aqui est\u00e1 o resumo de todos os casos poss\u00edveis: <\/p>\n<div class=\"wp-block-columns is-layout-flex wp-container-32\">\n<div class=\"wp-block-column is-layout-flow\">\n<ul>\n<li> <span style=\"color:#1976d2;font-size:22px\"><strong>m\u22603 e m\u22602:<\/strong><\/span> <\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-cf366a55bd307517f94fd8aa00cdf598_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{SCD} \\longrightarrow \\begin{cases} x = \\cfrac{2m^2+8-8m}{m^2-5m+6} \\\\[3.5ex] y =\\cfrac{-4+2m}{m^2-5m+6} \\\\[3.5ex] z = \\cfrac{-2m+4}{m^2-5m+6} \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"171\" width=\"240\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<\/div>\n<div class=\"wp-block-column is-layout-flow\">\n<ul>\n<li> <span style=\"color:#1976d2;font-size:22px\"><strong>m=2:<\/strong><\/span> <\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-94002d4f4d866569ed7d6993dd977b81_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{SCI} \\longrightarrow \\begin{cases} x = \\cfrac{4-2\\lambda}{3} \\\\[3.5ex] y= \\cfrac{2-4\\lambda}{3} \\\\[3.5ex] z = \\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"150\" width=\"175\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<\/div>\n<div class=\"wp-block-column is-layout-flow\">\n<ul>\n<li> <span style=\"color:#1976d2;font-size:22px\"><strong>m=3:<\/strong><\/span><\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9b971789910fd078c90174ed3d662e9a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{SI} \\longrightarrow\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"54\" style=\"vertical-align: -1px;\"><\/p>\n<p> O sistema n\u00e3o tem solu\u00e7\u00e3o.<\/p>\n<\/div>\n<\/div>\n<p> Aqui fizemos todo o processo usando o teorema de Rouche e a regra de Cramer, mas sistemas de equa\u00e7\u00f5es com par\u00e2metros tamb\u00e9m podem ser discutidos e resolvidos pelo <a href=\"https:\/\/mathority.org\/pt\/metodo-jordan-gauss-com-exemplos-e-exercicios-resolvidos\/\">m\u00e9todo de Gauss (com exerc\u00edcios)<\/a> . Voc\u00ea pode aprender mais sobre este m\u00e9todo na p\u00e1gina do link, onde encontrar\u00e1 uma explica\u00e7\u00e3o detalhada do procedimento, bem como exemplos e exerc\u00edcios resolvidos passo a passo.<\/p>\n<h2 class=\"wp-block-heading\"> Problemas de discuss\u00e3o resolvidos de sistemas de equa\u00e7\u00f5es lineares com par\u00e2metros<\/h2>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Discuta e resolva o seguinte sistema de equa\u00e7\u00f5es lineares dependentes de par\u00e2metros: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-de-systemes-dequations-a-parametres.webp\" alt=\"exerc\u00edcio resolvido de sistemas de equa\u00e7\u00f5es com par\u00e2metros\" class=\"wp-image-4014\" width=\"191\" height=\"123\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>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-b641845325965882d4aac899246cffb3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 &amp; 1 \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; -m \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}4 &amp; -1 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp; -m &amp; 0\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"418\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> 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-4d36c7cffe0248a2f45cd5871abc6ed5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{aligned}\\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 4 &amp; -1 &amp; 1 \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; -m \\end{vmatrix} &amp; =-4m+9-2-3-24-m \\\\ &amp; =-5m-20 \\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"381\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O resultado do determinante de A depende do valor de m. Veremos, portanto, para quais valores de m o determinante desaparece. Para fazer isso, igualamos o resultado resultante a 0 e resolvemos a equa\u00e7\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-e7d065a700c5f2a7f732719be777027f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-5m-20 = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"109\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bc49744634d23cdf0faf446f33487eac_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-5m = 20\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"79\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ca8fa07f8716b941295db488a99a6425_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m = \\cfrac{20}{-5} = -4\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"110\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim, quando m for -4, o determinante de A ser\u00e1 0. E quando m for diferente de -4, o determinante de A ser\u00e1 diferente de 0. Devemos, portanto, analisar cada caso separadamente:<\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m\u2260-4:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Como acabamos de ver, quando o par\u00e2metro m \u00e9 diferente de -4, o determinante da matriz A \u00e9 diferente de 0. Portanto, o posto de A \u00e9 3.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Al\u00e9m disso, o posto da matriz A&#8217; tamb\u00e9m \u00e9 3, porque dentro dela existe uma submatriz 3\u00d73 cujo determinante \u00e9 diferente de 0. E n\u00e3o pode ser de posto 4 pois \u2018n\u00e3o podemos fazer um determinante 4\u00d74.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p 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-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, aplicamos a regra de Cramer 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-e9e0bd352ad7713a03824ead1239041c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 &amp; 1  \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; -m \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 4 &amp; -1 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp; -m &amp; 0\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"418\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-530cb4576ee1a91d6246ed6cf9dd0fc8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 4 &amp; -1 &amp; 1  \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; -m\\end{vmatrix} =-5m-20\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"253\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular xatex] com a regra de Cramer, trocamos a primeira coluna do determinante de A pela coluna de 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-b076bbda8d086abedb459570d74c80a9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 0 &amp; -1 &amp; 1  \\\\[1.1ex] 0 &amp; 1 &amp; -3 \\\\[1.1ex] 0 &amp; -2 &amp; -m\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{0}{-5m-20} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"280\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular a inc\u00f3gnita e com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna de 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-f642a8cb2fd174e5c383a4df53e11a2e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{y} = \\cfrac{\\begin{vmatrix} 4 &amp; 0 &amp; 1  \\\\[1.1ex] 1 &amp; 0 &amp; -3 \\\\[1.1ex] 3 &amp; 0 &amp; -m \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{0}{-5m-20} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"265\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular z 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-5020a9ba4995b9715d8d1fb4720952b1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{z} = \\cfrac{\\begin{vmatrix}4 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 1 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp; 0 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{0}{-5m-20} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"258\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es para o caso m\u2260-4 \u00e9:<\/p>\n<p class=\"has-text-align-center\"> <strong>x=0 y=0 z=0<\/strong><\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=-4:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Analisaremos agora o sistema quando o par\u00e2metro m for -4. Neste caso as matrizes A e 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-e585e6465d27ea27ccc2c1a6ec1fe9ae_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 &amp; 1 \\\\[1.1ex] 1 &amp; 1 &amp; -3 \\\\[1.1ex] 3 &amp; -2 &amp; 4 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}4 &amp; -1 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp; 1 &amp; -3 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp; 4 &amp; 0\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Como vimos anteriormente, quando m=-4 o determinante de A \u00e9 0. Portanto, a matriz A n\u00e3o \u00e9 de posto 3. Mas dentro dela possui 2\u00d72 determinantes diferentes de 0, por exemplo:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a62d150aef4ec798814d25c988b0afd7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   \\begin{vmatrix}4 &amp; -1 \\\\[1.1ex] 1 &amp; 1 \\end{vmatrix} =4-(-1)=5 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Como a matriz possui um determinante de ordem 2 diferente de 0, a matriz A \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-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;. J\u00e1 sabemos que o determinante das 3 primeiras colunas d\u00e1 0, ent\u00e3o tentamos os outros determinantes 3\u00d73 poss\u00edveis:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-39fc49c7a63920c8956703a4851ecfc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -1 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp; -3 &amp; 0 \\\\[1.1ex]  -2 &amp; 4 &amp; 0 \\end{vmatrix} = 0 \\quad \\begin{vmatrix}4 &amp; 1 &amp; 0 \\\\[1.1ex] 1 &amp;  -3 &amp; 0 \\\\[1.1ex] 3 &amp;  4 &amp; 0  \\end{vmatrix} = 0 \\quad \\begin{vmatrix}4 &amp; -1 &amp;  0 \\\\[1.1ex] 1 &amp; 1 &amp; 0 \\\\[1.1ex] 3 &amp; -2 &amp;  0\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Todos os determinantes 3\u00d73 da matriz A&#8217; s\u00e3o 0, ent\u00e3o a matriz A&#8217; tamb\u00e9m n\u00e3o ter\u00e1 classifica\u00e7\u00e3o 3. Por\u00e9m, dentro dele possui determinantes de ordem 2 diferentes de 0. Por exemplo:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a62d150aef4ec798814d25c988b0afd7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   \\begin{vmatrix}4 &amp; -1 \\\\[1.1ex] 1 &amp; 1 \\end{vmatrix} =4-(-1)=5 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, a matriz A&#8217; ser\u00e1 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\"> A extens\u00e3o da matriz A \u00e9 igual \u00e0 extens\u00e3o da matriz A&#8217; mas estas duas s\u00e3o menores que o n\u00famero de inc\u00f3gnitas no sistema (3), portanto, de acordo com o teorema de Rouch\u00e9-Frobenius, c \u00e9 um Sistema Compat\u00edvel Indeterminado (ICS):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f43fdf4978386c61d18f9bb5b5883881_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u00c9 um sistema ICS, ent\u00e3o precisamos transformar o sistema para resolv\u00ea-lo. Primeiro eliminamos uma equa\u00e7\u00e3o, que neste caso ser\u00e1 a \u00faltima:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4d5499fda37d3cbf56fbf6ecbfc6bfba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 4x-y+z= 0 \\\\[1.5ex] x+y-3z=0 \\\\[1.5ex] \\cancel{3x-2y+4z = 0} \\end{cases} \\longrightarrow \\quad \\begin{cases} 4x-y+z= 0 \\\\[1.5ex] x+y-3z=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"349\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Agora vamos converter a vari\u00e1vel z em \u03bb:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96ea68274b072531365282e01d926718_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}4x-y+z= 0 \\\\[1.5ex] x+y-3z=0 \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} 4x-y+\\lambda= 0 \\\\[1.5ex] x+y-3\\lambda=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"353\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> 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-6192715e62cc8e3d3fe4c51da8629c70_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 4x-y=-\\lambda \\\\[1.5ex] x+y=3\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"110\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tal que a matriz A e a matriz A&#8217; do sistema permanecem: <\/p>\n<p class=\"has-text-align-center\">\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-99a91208ff1742f81e799aa5ab7f9097_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 4 &amp; -1 \\\\[1.1ex] 1 &amp; 1 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 4 &amp; -1 &amp; -\\lambda \\\\[1.1ex] 1 &amp; 1 &amp; 3\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"337\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Finalmente, uma vez transformado o sistema, aplicamos a regra de Cramer. Para fazer isso, primeiro 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-34832b783ddaf4af205302240d0feafb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 4 &amp; -1 \\\\[1.1ex] 1 &amp; 1 \\end{vmatrix} = 4-(-1)=5\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"228\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular x 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-362167d2eaa02d7243dedd5c385d08b1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix}-\\lambda &amp; -1 \\\\[1.1ex] 3\\lambda &amp; 1 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-\\lambda-(-3\\lambda)}{5} =\\cfrac{\\bm{2\\lambda}}{\\bm{5}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"285\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular a inc\u00f3gnita e com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna de 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-ede5a3a87ac0bb9ceea4232ec7b381fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{y} = \\cfrac{\\begin{vmatrix} 4 &amp; -\\lambda \\\\[1.1ex] 1 &amp; 3\\lambda \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{12\\lambda-(-\\lambda)}{5}=\\cfrac{\\bm{13\\lambda}}{\\bm{5}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"267\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> De modo que quando m=-4 a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es \u00e9 fun\u00e7\u00e3o de \u03bb, pois \u00e9 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-316aed1af52eba51058c3c753717c1af_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =}\\cfrac{\\bm{2\\lambda}}{\\bm{5}}\\qquad \\bm{y=}\\cfrac{\\bm{13\\lambda}}{\\bm{5}} \\qquad \\bm{z=\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"222\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-119\"><\/div>\n<\/div>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 2<\/h3>\n<p> Discuta e encontre a solu\u00e7\u00e3o para o seguinte sistema de equa\u00e7\u00f5es lineares dependentes de par\u00e2metros: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-etape-par-etape-d-un-systeme-d-equations-lineaires-avec-parametres.webp\" alt=\"exerc\u00edcio resolvido sistema passo a passo de equa\u00e7\u00f5es lineares com par\u00e2metros\" class=\"wp-image-4020\" width=\"187\" height=\"122\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>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-e924dd1b3fe5c0da561b92da9bf5da3b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; m\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}m &amp; 2 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; m &amp; 3\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> 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-5d0f8dbb7408ac6521e0144ac2f3a8a3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{aligned}\\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix}m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; m\\end{vmatrix} &amp; =4m^2+4-4-4+4m-4m \\\\ &amp; =4m^2-4 \\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"384\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O resultado do determinante de A depende do valor de m. Veremos, portanto, para quais valores de m o determinante desaparece. Para fazer isso, igualamos o resultado resultante a 0 e resolvemos a equa\u00e7\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-059d7760bb8cee65e13e43a97e156e1a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"4m^2-4 = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"95\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1aae0b3268b7d9e348b8eb80ce957ce8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"4m^2=4\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"65\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1f31504086ef5ec070b337db57ad444f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m^2 = \\cfrac{4}{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"58\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-baf4ba1311e2f2ce47aecfb90a411b64_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m^2 = 1\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"55\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c8dda3f9896a1a4dece5059a485cdcfb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m = \\pm 1\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"61\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim, quando m for +1 ou -1, o determinante de A ser\u00e1 0. E quando m for diferente de +1 e -1, o determinante de A ser\u00e1 diferente de 0. Devemos, portanto, analisar cada caso por:<\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m\u2260+1 e m\u2260-1:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Como acabamos de ver, quando o par\u00e2metro m \u00e9 diferente de +1 e -1, o determinante da matriz A \u00e9 diferente de 0. Portanto, o posto de A \u00e9 3.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Al\u00e9m disso, o posto da matriz A&#8217; tamb\u00e9m \u00e9 3, porque dentro dela existe uma submatriz 3\u00d73 cujo determinante \u00e9 diferente de 0. E n\u00e3o pode ser de posto 4 pois \u2018n\u00e3o podemos fazer um determinante 4\u00d74.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p 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-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, aplicamos a regra de Cramer 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-e924dd1b3fe5c0da561b92da9bf5da3b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; m\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}m &amp; 2 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; m &amp; 3\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b5114be5e37d2c91f02f22fba22edc42_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix}m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; m\\end{vmatrix}=4m^2-4\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"231\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular x 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-4b7402e02ee62bd78a6f880d3d122119_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 2&amp; 2 &amp; 1 \\\\[1.1ex] 0 &amp; 4 &amp; 2 \\\\[1.1ex] 3 &amp; -2 &amp; m\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{\\bm{8m+8}}{\\bm{4m^2-4}}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"218\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular a inc\u00f3gnita e com a regra de Cramer, trocamos a segunda coluna do determinante de A pela coluna de 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-551c66a9530d0195a9a4ff64d42350c4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{y} = \\cfrac{\\begin{vmatrix} m &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 0 &amp; 2 \\\\[1.1ex] 1 &amp; 3 &amp; m\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{\\bm{-10m+10}}{\\bm{4m^2-4}}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"235\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular z 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-28375ce522b7644a745a9adea4c78ae7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{z} = \\cfrac{\\begin{vmatrix}m &amp; 2 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; 3 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{\\bm{12m-28}}{\\bm{4m^2-4}}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"227\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, a solu\u00e7\u00e3o do sistema de equa\u00e7\u00f5es para o caso m\u2260+1 e m\u2260-1 \u00e9:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-59ecb2f1448989cd1c41af45e7bc4a32_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x = }\\cfrac{\\bm{8m+8}}{\\bm{4m^2-4}} \\qquad \\bm{y=}\\cfrac{\\bm{-10m+10}}{\\bm{4m^2-4}}\\qquad \\bm{z =} \\cfrac{\\bm{12m-28}}{\\bm{4m^2-4}}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"383\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=+1:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Analisaremos agora o sistema quando o par\u00e2metro m for igual a 1. Neste caso as matrizes A e 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-f6af272a99ed7c281ee8dd9199698686_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; 1 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}1 &amp; 2 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; 1 &amp; 3\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Como vimos anteriormente, quando m=+1 o determinante de A \u00e9 0. Portanto a matriz A n\u00e3o \u00e9 de posto 3. Mas dentro dela possui 2\u00d72 determinantes diferentes de 0, por exemplo:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ebedf6c9e4316844dc99ceca9472fac5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   \\begin{vmatrix}2 &amp; 4\\\\[1.1ex] 1 &amp; -2 \\end{vmatrix} =-4-4=-8 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Como a matriz possui um determinante de ordem 2 diferente de 0, a matriz A \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-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;. J\u00e1 sabemos que o determinante das 3 primeiras colunas d\u00e1 0, ent\u00e3o agora tentamos, por exemplo, com o determinante das 3 \u00faltimas colunas:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4d0109b155be9f87a0cee337ddec5517_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 1 &amp; 2 \\\\[1.1ex] 4 &amp; 2 &amp; 0 \\\\[1.1ex]  -2 &amp; 1 &amp; 3 \\end{vmatrix} = 16\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"124\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Por outro lado, a matriz A&#8217; cont\u00e9m um determinante 3\u00d73 cujo resultado \u00e9 diferente de 0, de modo que a matriz A&#8217; \u00e9 de posto 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\"> Assim, quando m=+1 o posto da matriz A \u00e9 menor que o posto da matriz A&#8217;. Assim, do teorema de Rouch\u00e9-Frobenius, deduzimos que o sistema \u00e9 um Sistema Incompat\u00edvel (SI):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b2bb3fec88cf5c6d788afb4480ab1f58_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = 2 \\ \\neq \\ rg(A') = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, o sistema de equa\u00e7\u00f5es <strong>n\u00e3o tem solu\u00e7\u00e3o quando m=+1<\/strong> , pois \u00e9 um sistema incompat\u00edvel.<\/p>\n<p class=\"has-text-align-left\" style=\"font-size:26px\"> <span style=\"color:#1976d2;\"><strong>m=-1:<\/strong><\/span><\/p>\n<p class=\"has-text-align-left\"> Analisaremos agora o sistema quando o par\u00e2metro m for -1. Neste caso as matrizes A e 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-46b0a00ef38d0e5a433b418de7eb1ec3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} -1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 4 &amp; 2 \\\\[1.1ex] 1 &amp; -2 &amp; -1 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c}-1 &amp; 2 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 4 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp; -2 &amp; -1 &amp; 3\\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"432\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Como vimos anteriormente, quando m=-1 o determinante de A \u00e9 0. Portanto, a matriz A n\u00e3o \u00e9 de posto 3. Mas dentro dela possui 2\u00d72 determinantes diferentes de 0, por exemplo:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ff5373c7e7901f253421efbbd52d192e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle   \\begin{vmatrix}-1 &amp; 2\\\\[1.1ex] 2 &amp; 4 \\end{vmatrix} =-4-4=-8 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"213\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Como a matriz possui um determinante de ordem 2 diferente de 0, a matriz A \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-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;. J\u00e1 sabemos que o determinante das 3 primeiras colunas d\u00e1 0, ent\u00e3o agora tentamos, por exemplo, com o determinante das colunas 1, 3 e 4:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a95e30910bd64db920f3c2bcb5f2ff62_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -1 &amp; 1 &amp; 2 \\\\[1.1ex] 2 &amp; 2 &amp; 0 \\\\[1.1ex] 1 &amp;  -1 &amp; 3\\end{vmatrix} = -20\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"152\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Por outro lado, a matriz A&#8217; cont\u00e9m um determinante 3\u00d73 cujo resultado \u00e9 diferente de 0, de modo que a matriz A&#8217; \u00e9 de posto 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\"> Assim, quando m = -1, o posto da matriz A \u00e9 inferior ao posto da matriz A&#8217;. Assim, do teorema de Rouch\u00e9-Frobenius, deduzimos que o sistema \u00e9 um Sistema Incompat\u00edvel (SI):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b2bb3fec88cf5c6d788afb4480ab1f58_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = 2 \\ \\neq \\ rg(A') = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, o sistema de equa\u00e7\u00f5es <strong>n\u00e3o tem solu\u00e7\u00e3o quando m=-1<\/strong> , pois \u00e9 um sistema incompat\u00edvel.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Nesta p\u00e1gina veremos como discutir e resolver um sistema de equa\u00e7\u00f5es com par\u00e2metros . Al\u00e9m disso, voc\u00ea encontrar\u00e1 exemplos e exerc\u00edcios resolvidos de sistemas de equa\u00e7\u00f5es lineares para praticar. Por outro lado, para analisar sistemas de equa\u00e7\u00f5es lineares \u00e9 importante que voc\u00ea saiba o que \u00e9 a regra de Cramer e o que \u00e9 o &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/pt\/discussao-de-sistemas-de-equacoes-com-parametros\/\"> <span class=\"screen-reader-text\">Discuss\u00e3o de sistemas de equa\u00e7\u00f5es com par\u00e2metros<\/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":[14],"tags":[],"class_list":["post-305","post","type-post","status-publish","format-standard","hentry","category-explicacoes-matematicas"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Discuss\u00e3o de sistemas de equa\u00e7\u00f5es com par\u00e2metros -<\/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\/discussao-de-sistemas-de-equacoes-com-parametros\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Discuss\u00e3o de sistemas de equa\u00e7\u00f5es com par\u00e2metros -\" \/>\n<meta property=\"og:description\" content=\"Nesta p\u00e1gina veremos como discutir e resolver um sistema de equa\u00e7\u00f5es com par\u00e2metros . Al\u00e9m disso, voc\u00ea encontrar\u00e1 exemplos e exerc\u00edcios resolvidos de sistemas de equa\u00e7\u00f5es lineares para praticar. Por outro lado, para analisar sistemas de equa\u00e7\u00f5es lineares \u00e9 importante que voc\u00ea saiba o que \u00e9 a regra de Cramer e o que \u00e9 o &hellip; Discuss\u00e3o de sistemas de equa\u00e7\u00f5es com par\u00e2metros Leia mais &raquo;\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mathority.org\/pt\/discussao-de-sistemas-de-equacoes-com-parametros\/\" \/>\n<meta property=\"article:published_time\" content=\"2023-07-06T14:08:28+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6ab2286d15c20029b98a5ea4622033d4_l3.png\" \/>\n<meta name=\"author\" content=\"Equipe Mathoridade\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"Equipe Mathoridade\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. tempo de leitura\" \/>\n\t<meta name=\"twitter:data2\" content=\"12 minutos\" \/>\n<script 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