{"id":5807,"date":"2024-11-22T11:07:08","date_gmt":"2024-11-22T17:07:08","guid":{"rendered":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/?page_id=5807"},"modified":"2024-11-22T11:07:08","modified_gmt":"2024-11-22T17:07:08","slug":"mn3d","status":"publish","type":"page","link":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/unidad-3\/mn3d\/","title":{"rendered":"mn3d Gauss &#8211; Zamora"},"content":{"rendered":"\n<p>El m\u00e9todo de Zamora consiste en ordenar las ecuaciones de tal forma que cada uno de los elementos en diagonal sean unos, y por debajo de ellos s\u00f3lo haya ceros.<\/p>\n\n\n\n<p>Para el sistema<\/p>\n\n\n\n<p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"242\" height=\"114\" src=\"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-content\/uploads\/sites\/89\/2023\/08\/imagen-103.png\" alt=\"\" class=\"wp-image-5882\" \/><\/figure><\/div>\n\n\n<p>La matriz aumentada queda como<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"242\" height=\"103\" src=\"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-content\/uploads\/sites\/89\/2023\/08\/imagen-104.png\" alt=\"\" class=\"wp-image-5883\" \/><\/figure><\/div>\n\n\n<p>Primero hay que dividir R1 por 2 para hacer que el coeficiente en Z sea uno. &nbsp;<\/p>\n\n\n\n<p>R1 = 0.5 &nbsp;0.5 &nbsp;1 &nbsp;3.5<\/p>\n\n\n\n<p>Con este nuevo rengl\u00f3n 1 se hacen cero en la variable Z en los otros renglones,<\/p>\n\n\n\n<p>R2 = R2 + R1 &nbsp; y &nbsp;R3 = R3 + R1<\/p>\n\n\n\n<p>R2 = 2.5 &nbsp; -0.5 &nbsp; 0 &nbsp;1.5 &nbsp; &nbsp;y &nbsp;R3 = 1.5 &nbsp;-1.5 &nbsp; 0 &nbsp; -1.5<\/p>\n\n\n\n<p>El nuevo sistema es ahora<\/p>\n\n\n\n<ol>\n<li>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;0.5 &nbsp; &nbsp; &nbsp;0.5 &nbsp; &nbsp; &nbsp; 1 &nbsp; &nbsp; &nbsp;3.5<\/li>\n\n\n\n<li>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;2.5 &nbsp; &nbsp; -0.5 &nbsp; &nbsp; &nbsp; 0 &nbsp; &nbsp; &nbsp;1.5<\/li>\n\n\n\n<li>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;1.5 &nbsp; &nbsp; -1.5 &nbsp; &nbsp; &nbsp; 0 &nbsp; &nbsp; -1.5<\/li>\n<\/ol>\n\n\n\n<p>Ahora se hace uno el coeficiente de Y en el segundo rengl\u00f3n. &nbsp; <\/p>\n\n\n\n<p>R2 = R2*(-2) &nbsp;= &nbsp;-5 &nbsp; &nbsp;1 &nbsp; 0 &nbsp; -3<\/p>\n\n\n\n<p>Despu\u00e9s se elimina Y en el tercer rengl\u00f3n con R2<\/p>\n\n\n\n<p>R3 &nbsp; = &nbsp; R3 + 1.5R2 &nbsp; &nbsp;= &nbsp; [1.5 &nbsp; -1.5 &nbsp; 0 &nbsp; -1.5 ] + [-7.5 &nbsp; &nbsp;1.5 &nbsp; &nbsp;0 &nbsp; -4.5] &nbsp;= &nbsp; [-6.0 &nbsp; 0 &nbsp; &nbsp;0 &nbsp; \u2013 6.0]<\/p>\n\n\n\n<p><\/p>\n\n\n\n<p>El nuevo sistema es ahora<\/p>\n\n\n\n<ol>\n<li>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;0.5 &nbsp; 0.5 &nbsp; &nbsp;1 &nbsp;  &nbsp;3.5<\/li>\n\n\n\n<li>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;-5 &nbsp; &nbsp; 1 &nbsp; &nbsp; &nbsp; 0 &nbsp;  -3.0<\/li>\n\n\n\n<li>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;-6 &nbsp; &nbsp; 0 &nbsp; &nbsp; &nbsp; 0 &nbsp; &nbsp; -6<\/li>\n<\/ol>\n\n\n\n<p>Haciendo uno en la variable X de la tercera ecuaci\u00f3n, se encuentra que &nbsp; X = -6 \/-6 &nbsp;= 1<\/p>\n\n\n\n<p>sustituyendo en R2 &nbsp; Y = -3 + 5X = -3 + 5(1) &nbsp;= 2<\/p>\n\n\n\n<p>sustituyendo en R1 &nbsp; &nbsp;Z = 3.5 \u2013 0.5Y \u2013 0.5X &nbsp;= &nbsp;3.5 \u2013 0.5(2) \u2013 0.5(1) = 3.5 -1 \u2013 0.5 = 2<\/p>\n\n\n\n<p>La soluci\u00f3n del sistema es X = 1, Y = 2, Z = 2<\/p>\n\n\n\n<p>El m\u00e9todo tiene la ventaja de disminuir la presencia de divisiones entre cero que ocurre en el m\u00e9todo de Gauss, y no confundir el c\u00e1lculo de las variables ya que primero se conoce X y al final la \u00faltima variable, Z.<\/p>\n\n\n\n<p>C\u00f3digo de programaci\u00f3n<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">% GAUSS ZAMORA\nclc;\nR=[1 1 2 7;2 -1 -1 -2;1 -2 -1 -5]\nR1=R(1,:)\nR2=R(2,:)\nR3=R(3,:)\n% Hacer pivote 1 R1\nR1=R1\/R1(1,3)\n% Eliminaci\u00f3n de Z en las dos ecuaciones siguientes\nR2=R2-R1*R2(1,3)\nR3=R3-R1*R3(1,3)\n% Hacer pivote 2 R2\nR2=R2\/R2(1,2)\n% Eliminaci\u00f3n de Y en la ecuaci\u00f3n siguiente\nR3=R3-R2*R3(1,2)\n% Soluci\u00f3n\nx=R3(1,4)\/R3(1,1)\ny=R2(1,4)-R2(1,1)*x\nz=R1(1,4)-R1(1,2)*y-R1(1,1)*x<\/pre>\n\n\n\n<p><\/p>\n\n\n\n<div class=\"is-content-justification-center is-layout-flex wp-container-1 wp-block-buttons\">\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link wp-element-button\" href=\"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/unidad-3\/\">regresar<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>El m\u00e9todo de Zamora consiste en ordenar las ecuaciones de tal forma que cada uno de los elementos en diagonal sean unos, y por debajo de ellos s\u00f3lo haya ceros. Para el sistema &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; La matriz aumentada &hellip; <a href=\"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/unidad-3\/mn3d\/\" class=\"more-link\">Contin\u00faa leyendo <span class=\"screen-reader-text\">mn3d Gauss &#8211; Zamora<\/span><\/a><\/p>\n","protected":false},"author":123458,"featured_media":0,"parent":436,"menu_order":4,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_bbp_topic_count":0,"_bbp_reply_count":0,"_bbp_total_topic_count":0,"_bbp_total_reply_count":0,"_bbp_voice_count":0,"_bbp_anonymous_reply_count":0,"_bbp_topic_count_hidden":0,"_bbp_reply_count_hidden":0,"_bbp_forum_subforum_count":0},"_links":{"self":[{"href":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-json\/wp\/v2\/pages\/5807"}],"collection":[{"href":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-json\/wp\/v2\/users\/123458"}],"replies":[{"embeddable":true,"href":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-json\/wp\/v2\/comments?post=5807"}],"version-history":[{"count":3,"href":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-json\/wp\/v2\/pages\/5807\/revisions"}],"predecessor-version":[{"id":6607,"href":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-json\/wp\/v2\/pages\/5807\/revisions\/6607"}],"up":[{"embeddable":true,"href":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-json\/wp\/v2\/pages\/436"}],"wp:attachment":[{"href":"https:\/\/blogceta.zaragoza.unam.mx\/mnumericos\/wp-json\/wp\/v2\/media?parent=5807"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}