{"id":137207,"date":"2019-09-29T23:50:27","date_gmt":"2019-09-30T02:50:27","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=137207"},"modified":"2023-06-04T22:33:48","modified_gmt":"2023-06-05T01:33:48","slug":"problema-um-sistema-de-matrizes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-um-sistema-de-matrizes\/","title":{"rendered":".Problema: Um sistema de matrizes"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:25px;\">\n<p class=\"tit\" style=\"line-height:20px; margin-bottom: 10px;\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p>(Matem\u00e1tica- Dante, Vol. \u00danico) Determine as matrizes [tex]X[\/tex] e [tex]Y[\/tex] que s\u00e3o as solu\u00e7\u00f5es do sistema<br \/>\n\\begin{equation}<br \/>\n\\left \\{<br \/>\n\\begin{array}{cc}<br \/>\nX+Y=A+3B\\\\<br \/>\nX-Y=3A-2B\\\\<br \/>\n\\end{array}<br \/>\n\\right.,<br \/>\n\\end{equation}<br \/>\nsendo [tex]A=\\left(<br \/>\n\\begin{array}{c}<br \/>\n1\\\\<br \/>\n0\\\\<br \/>\n2<br \/>\n\\end{array}<br \/>\n\\right)[\/tex] e [tex]B=\\left(<br \/>\n\\begin{array}{c}<br \/>\n4\\\\<br \/>\n2\\\\<br \/>\n0<br \/>\n\\end{array}<br \/>\n\\right)[\/tex].<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\" style=\"line-height:28px;\">\n<p class=\"tit\" style=\"line-height:30px; margin-bottom: 10px;\">Solu\u00e7\u00e3o <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p>\t Notemos que sendo [tex]A[\/tex] e [tex]B[\/tex] matrizes de ordem [tex]3\\times 1[\/tex],  ent\u00e3o as matrizes [tex]X[\/tex] e [tex]Y[\/tex] tamb\u00e9m ter\u00e3o essa ordem.<br \/>\nVamos solucionar o sistema, usando o m\u00e9todo de adi\u00e7\u00e3o. <\/p>\n<ul>\n<li>Somando as duas equa\u00e7\u00f5es do sistema temos que:<\/li>\n<p>[tex] \\qquad 2X=4A+B\\\\<br \/>\n\\qquad X=\\dfrac{1}{2}\\cdot 4A+\\dfrac{1}{2}B \\\\<br \/>\n\\qquad \\boxed{X=2A+\\dfrac{1}{2}B}.\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\textcolor{#800000}{(i)}[\/tex]<\/p>\n<li>Substituindo [tex]\\textcolor{#800000}{(i)}[\/tex] na segunda equa\u00e7\u00e3o do sistema, [tex]X-Y=3A-2B[\/tex], obtemos<\/li>\n<p>[tex] \\qquad 2A+\\dfrac{1}{2}B-Y=3A-2B \\\\<br \/>\n\\qquad -Y=3A-2B-2A-\\dfrac{1}{2}B\\\\<br \/>\n\\qquad  -Y=A-\\dfrac{5}{2}B\\\\<br \/>\n\\qquad  \\boxed{Y=-A+\\dfrac{5}{2}B}.\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\textcolor{#800000}{(ii)}[\/tex]\n<\/ul>\n<p>Sendo  [tex]A=\\left(<br \/>\n\\begin{array}{c}<br \/>\n1\\\\<br \/>\n0\\\\<br \/>\n2<br \/>\n\\end{array}<br \/>\n\\right)[\/tex] e [tex]B=\\left(<br \/>\n\\begin{array}{c}<br \/>\n4\\\\<br \/>\n2\\\\<br \/>\n0<br \/>\n\\end{array}<br \/>\n\\right)[\/tex], segue por [tex]\\textcolor{#800000}{(i)}[\/tex] e por [tex]\\textcolor{#800000}{(ii)}[\/tex] que:<\/p>\n<table  style=\"margin-left:auto; margin-right:auto; width:600px\">\n<tr>\n<td class=\"bf\" style=\"line-height:27px;border-right:2px solid #800000;\">\n[tex]<br \/>\nX=2A+\\dfrac{1}{2}B\\\\<br \/>\n X=2\\cdot \\left(<br \/>\n\\begin{array}{c}<br \/>\n1\\\\<br \/>\n0\\\\<br \/>\n2<br \/>\n\\end{array}<br \/>\n\\right)+\\dfrac{1}{2}\\cdot \\left(<br \/>\n\\begin{array}{c}<br \/>\n4\\\\<br \/>\n2\\\\<br \/>\n0<br \/>\n\\end{array}<br \/>\n\\right)\\\\<br \/>\n X= \\left(<br \/>\n\\begin{array}{c}<br \/>\n2\\\\<br \/>\n0\\\\<br \/>\n4<br \/>\n\\end{array}<br \/>\n\\right)+\\left(<br \/>\n\\begin{array}{c}<br \/>\n2\\\\<br \/>\n1\\\\<br \/>\n0<br \/>\n\\end{array}<br \/>\n\\right)\\\\<br \/>\n X=\\left(<br \/>\n\\begin{array}{c}<br \/>\n4\\\\<br \/>\n1\\\\<br \/>\n4<br \/>\n\\end{array}<br \/>\n\\right).<br \/>\n[\/tex]\n<\/td>\n<td class=\"bf\" style=\" padding-left:40px\">\n[tex] Y=-A+\\dfrac{5}{2}B \\\\<br \/>\n Y=-\\left(<br \/>\n\\begin{array}{c}<br \/>\n1\\\\<br \/>\n0\\\\<br \/>\n2<br \/>\n\\end{array}<br \/>\n\\right)+\\dfrac{5}{2}\\cdot \\left(<br \/>\n\\begin{array}{c}<br \/>\n4\\\\<br \/>\n2\\\\<br \/>\n0<br \/>\n\\end{array}<br \/>\n\\right) \\\\<br \/>\n Y=\\left(<br \/>\n\\begin{array}{c}<br \/>\n-1\\\\<br \/>\n0\\\\<br \/>\n-2<br \/>\n\\end{array}<br \/>\n\\right)+\\left(<br \/>\n\\begin{array}{c}<br \/>\n10\\\\<br \/>\n5\\\\<br \/>\n0<br \/>\n\\end{array}<br \/>\n\\right)\\\\<br \/>\n Y=\\left(<br \/>\n\\begin{array}{c}<br \/>\n9\\\\<br \/>\n5\\\\<br \/>\n-2<br \/>\n\\end{array}<br \/>\n\\right).[\/tex]\n<\/td>\n<\/tr>\n<\/table>\n<p>Portanto, as solu\u00e7\u00f5es do sistema s\u00e3o [tex]X=\\left(<br \/>\n\\begin{array}{c}<br \/>\n4\\\\<br \/>\n1\\\\<br \/>\n4<br \/>\n\\end{array}<br \/>\n\\right)[\/tex] e [tex]Y=\\left(<br \/>\n\\begin{array}{c}<br \/>\n9\\\\<br \/>\n5\\\\<br \/>\n-2<br \/>\n\\end{array}<br \/>\n\\right)[\/tex].<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:20px\">\n<table style=\" height:auto; margin-left: auto; margin-right: auto;width:640px;\">\n<tr>\n<td class=\"bf\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"80\" height=\"115\" \/><\/td>\n<td class=\"bf\">\n<p class=\"fala\" style=\"width:520px;background-color: #FFF5EE; padding:10px; font-family: calibri; font-size: 13px;color: #1C1C1C; line-height: 20px; \"> Se for conveniente, voc\u00ea pode obter um arquivo PDF desta p\u00e1gina, com o problema e a solu\u00e7\u00e3o, clicando no bot\u00e3o abaixo.<br \/>\nVoc\u00ea pode abrir o arquivo diretamente no seu navegador (<em>Chrome<\/em>, <em>Edge<\/em>, <em>Firefox<\/em>, <em>Safari<\/em>, entre outros), mas tamb\u00e9m pode utilizar o software gratuito <em><strong>Adobe Acrobat Reader<\/strong><\/em>.<br \/>\nCaso o dispositivo que voc\u00ea est\u00e1 utilizando n\u00e3o tenha o <strong><em>Acrobat Reader<\/em><\/strong> instalado, \u00e9 s\u00f3 clicar <a href=\"http:\/\/get.adobe.com\/br\/reader\/\" rel=\"noopener\" target=\"_blank\"><strong>AQUI<\/strong><\/a> para fazer o download adequado ao seu dispositivo.<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<div style=\" float:right;  width: 100px;\">\n<p style=\"text-align: center; border-style: dotted; border-color :#FFFAFA; border-width :thin; width:90px;  height:40px;  padding: 5px; background-color: #800000;\">\n<a href=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/10\/Problema_-Um-Sistema-de-Matrizes.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Problema_ Um Sistema de Matrizes.pdf\"><span style=\"color: #FFFAFA; font-family: georgia; font-size: 16px;\"><strong>Arquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 2\u00ba ano do E. M.) (Matem\u00e1tica- Dante, Vol. \u00danico) Determine as matrizes [tex]X[\/tex] e [tex]Y[\/tex] que s\u00e3o as solu\u00e7\u00f5es do sistema \\begin{equation} \\left \\{ \\begin{array}{cc} X+Y=A+3B\\\\ X-Y=3A-2B\\\\ \\end{array} \\right., \\end{equation} sendo [tex]A=\\left( \\begin{array}{c} 1\\\\ 0\\\\ 2 \\end{array} \\right)[\/tex] e [tex]B=\\left( \\begin{array}{c} 4\\\\ 2\\\\ 0 \\end{array} \\right)[\/tex]. Solu\u00e7\u00e3o Notemos que sendo &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-um-sistema-de-matrizes\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":721,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-137207","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/137207","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/721"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=137207"}],"version-history":[{"count":14,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/137207\/revisions"}],"predecessor-version":[{"id":193312,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/137207\/revisions\/193312"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=137207"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}