{"id":454,"date":"2025-11-10T08:16:46","date_gmt":"2025-11-10T07:16:46","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=454"},"modified":"2025-11-06T06:18:59","modified_gmt":"2025-11-06T05:18:59","slug":"alg-teorema-de-rouche-frobenius","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=454","title":{"rendered":"ALG: Teorema de Rouch\u00e9-Fr\u00f6benius"},"content":{"rendered":"<p>Recordad que todo sistemas de ecuaciones los podemos formular mediante una ecuaci\u00f3n matricial \\[AX=B,\\] donde \\(A\\) es la matriz de coeficiente y \\(B\\) la matriz de t\u00e9rminos independientes. Llamamos matriz ampliada del sistema a la matriz que concatena \\(A\\) y \\(B\\), (\\(A|B\\)) .<\/p>\n<p>El Teorema de Rouch\u00e9-Fr\u00f6benius nos afirma que Existen soluciones para el sistema si y solo si el rango de la matriz de coeficientes es igual al rango de la matriz ampliada.<\/p>\n<p>As\u00ed un sistema ser\u00e1:<\/p>\n<p>\\[\\left\\{\\begin{array}{l}<br \/>\n\\begin{array}{c}<br \/>\nCompatible \\\\<br \/>\nrang(A)=rang(A|B)<br \/>\n\\end{array}\\left\\{\\begin{array}{l}<br \/>\n\\begin{array}{c}<br \/>\nDeterminado \\\\<br \/>\nrang(A)=\\mbox{N\u00famero de inc\u00f3gnitas}<br \/>\n\\end{array} \\\\<br \/>\n\\begin{array}{c}<br \/>\nIndeterminado \\\\<br \/>\nrang(A)<\\mbox{N\u00famero de inc\u00f3gnitas}\n\\end{array} \\\\\n\\end{array}\\right.\\\\\n\\begin{array}{c}\nIncompatible \\\\\nrang(A)\\neq rang(A|B)\n\\end{array}\\\\\n\\end{array}\\right.\\]\n\nPara resolver un sistema compatible s\u00f3lo tenemos que encontrar un menor de \\(A\\) distinto de cero y del mismo orden que en rango de \\(A\\). Supongamos que  \\(\\bar{A}\\) es la submatriz de \\(A\\) cuyo menor es el que buscamos. Entonces \\(A|B\\) se puede transformar mediante operaciones elementales por filas en una matriz\n\n\\[(A|B)\\sim\\left(\\begin{array}{c}\n\\bar{A}\\,\\bar{P}\\\\\n0\\end{array}\\left|\\begin{array}{c}\n\\bar{B}\\\\\n0\\end{array}\\right.\\right)\\]\nDonde \\(\\bar{P}\\) son o \\(0\\) o las columnas de la matriz \\(A\\) tales que \\[rang(A)+\\mbox{n\u00bacolumnas}(\\bar{P})=\\mbox{N\u00famero de inc\u00f3gnitas}.\\]\nDe este modo el sistema tendr\u00e1 por soluci\u00f3n\n\\[\\bar{X}=inv(\\bar{A})\\cdot (\\bar{B}-\\bar{P}K),\\]\ndonde \\(K\\) son las variables, en forma de par\u00e1metros, que faltan en el menor de \\(\\bar{A}\\), y tales que \\(X^t=(\\bar{X}^t K^t)\\).\n\n\n<blockquote><strong>Ejemplo:<\/strong> Determina las ecuaciones param\u00e9tricas del hiperplano de \\(\\mathbb{R}^4\\), \\(x+2y-z-t=1\\).<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1d() {\n  var htmlShow1d = document.getElementById(\"html-show1d\");\n  if (htmlShow1d.style.display === \"none\") {\n    htmlShow1d.style.display = \"block\";\n  } else {\n    htmlShow1d.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv1d()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1d\" style=\"display: none;\">\nComo tenemos una ecuaci\u00f3n habr\u00e1 3 par\u00e1metros. Elijamos un menor de orden 1 distinto de cero y consideremos par\u00e1metros el resto de variables:\\[x=1-2y+z+t,\\]<br \/>\nLuego \\[\\begin{array}{l} x=-2\\lambda +\\mu +\\eta  \\\\ y=\\lambda\\\\ z=\\mu\\\\t=\\eta \\end{array}\\]\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Determina las ecuaciones param\u00e9tricas del plano de \\(\\mathbb{R}^4\\) dado por las ecuaciones impl\u00edcitas \\(x+2y-z-t=1\\),  \\(y+2z=t\\).<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1c() {\n  var htmlShow1c = document.getElementById(\"html-show1c\");\n  if (htmlShow1c.style.display === \"none\") {\n    htmlShow1c.style.display = \"block\";\n  } else {\n    htmlShow1c.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv1c()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1c\" style=\"display: none;\">\nSi planteamos la matriz ampliada \\[\\begin{bmatrix}  1&#038;2  &#038;-1  &#038;-1  &#038;1 \\\\   0&#038;  1&#038; 2 &#038; -1 &#038; 0 \\end{bmatrix}\\] vemos que el rango es dos y coincide con el rango de la matriz de coeficientes. Luego es un sistema compatible indeterminado. Solo necesitamos establecer un menor de orden dos distinto de cero; por ejemplo, \\[\\begin{vmatrix}  1&#038;2  \\\\   0&#038;  1 \\end{vmatrix}\\] De este modo, consideramos par\u00e1metros \\(z=\\lambda\\) y \\(t=\\mu\\) y tendremos el sistema:\\[\\begin{align*}  x+2y&#038;=1+\\lambda+\\mu  \\\\   y&#038;=-2\\lambda+\\mu \\end{align*}\\]<br \/>\nLas ecuaciones param\u00e9tricas que buscamos son:\\[\\begin{align*}  x&#038;=1+5\\lambda-\\mu  \\\\   y&#038;=-2\\lambda+\\mu\\\\ z&#038;=\\lambda\\\\ t&#038;=\\mu \\end{align*}\\]\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Verificar si [1,1,1,-1,1] pertenece al subespacio director de la variedad \\[\\{(x,y,z,t,u)\\in\\mathbb{R}^5;\\ x-2=y+3=z-1=t+2u\\}\\]<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1b() {\n  var htmlShow1b = document.getElementById(\"html-show1b\");\n  if (htmlShow1b.style.display === \"none\") {\n    htmlShow1b.style.display = \"block\";\n  } else {\n    htmlShow1b.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv1b()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1b\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Ejercicio 1 Ecuaciones param\u00e9tricas de variedades - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/iRHJ6j-7sRs?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Determinar las ecuaciones param\u00e9tricas del \\(\\mathbf{Ker} f\\) de la aplicaci\u00f3n lineal dada por \\[f(x,y,z,t)=(2y+2x+t,2y-3x+z)\\]<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a() {\n  var htmlShow1a = document.getElementById(\"html-show1a\");\n  if (htmlShow1a.style.display === \"none\") {\n    htmlShow1a.style.display = \"block\";\n  } else {\n    htmlShow1a.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv1a()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Ecuaciones param\u00e9tricas del n\u00facleo de una aplicaci\u00f3n lineal. Ej. 1 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/H98B_85GrQ0?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div>\n<hr \/>\n<p>Utilizar las ecuaciones impl\u00edcitas nos sirve para encontrar con m\u00e1s facilidad la intersecci\u00f3n de dos subespacios: \\(S\\cap T\\) estar\u00e1 formado por las ecuaciones impl\u00edcitas de \\(S\\) m\u00e1s la de \\(T\\).<\/p>\n<blockquote><p><strong>Ejemplo:<\/strong> Sean las variedades afines \\[\\begin{align*}r_1&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ x+y=0,\\ y+2z=1,\\ z+3t=1\\}\\\\ r_2&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ \\frac{x-1}{-5}=y+1=z-1=t\\}\\end{align*}\\] \u00bf\\(r_1\\cap r_2\\)? <\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv2a() {\n  var htmlShow2a = document.getElementById(\"html-show2a\");\n  if (htmlShow2a.style.display === \"none\") {\n    htmlShow2a.style.display = \"block\";\n  } else {\n    htmlShow2a.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv2a()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2a\" style=\"display: none;\">\nPara considerar las ecuaciones que definen la intersecci\u00f3n debemos tener las ecuaciones impl\u00edcitas de \\(r_2\\), que est\u00e1 dada en forma continua. Estas ecuaciones las tendremos considerando un lado de la igual fijo e igual\u00e1ndolo al resto; es decir, \\[\\begin{multline*}r_2:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ \\frac{x-1}{-5}=y+1=z-1=t\\}=\\\\ \\{(x,y,z,t)\\in\\mathbb{R}^4;\\ \\frac{x-2}{-5}=t,\\ y+1=t,\\ z-1=t\\}\\end{multline*}\\]<br \/>\nAhora vemos que \\(r_1\\cap r_2\\) vendr\u00e1 dada por el sistema que forman todas las ecuaciones conjuntas; es decir, \\((x,y,z,t)\\in r_1\\cap r_2\\) si \\[\\begin{align*} x+y&#038;=0\\\\ y+2z&#038;=1\\\\ z+3t&#038;=1\\\\ x+5t&#038;=1\\\\ y-t&#038;=-1\\\\ z-t&#038;=1\\end{align*}\\]<br \/>\nEste sistema tendr\u00e1 soluci\u00f3n si el rango de la matriz ampliada coincide con la matriz de coeficientes:<br \/>\n\\[\\mathbf{rank}\\begin{bmatrix}1 &#038; 1 &#038; 0 &#038; 0  \\\\<br \/>\n0 &#038; 1 &#038; 2 &#038; 0  \\\\<br \/>\n0 &#038; 0 &#038; 1 &#038; 3  \\\\<br \/>\n1 &#038; 0 &#038; 0 &#038; 5  \\\\<br \/>\n0 &#038; 1 &#038; 0 &#038; -1  \\\\<br \/>\n0 &#038; 0 &#038; 1 &#038; -1  \\end{bmatrix}=\\mathbf{rank}\\begin{bmatrix}1 &#038; 1 &#038; 0 &#038; 0 &#038; 0\\\\<br \/>\n0 &#038; 1 &#038; 2 &#038; 0 &#038; 1\\\\<br \/>\n0 &#038; 0 &#038; 1 &#038; 3 &#038; 1\\\\<br \/>\n1 &#038; 0 &#038; 0 &#038; 5 &#038; 1\\\\<br \/>\n0 &#038; 1 &#038; 0 &#038; -1 &#038; -1\\\\<br \/>\n0 &#038; 0 &#038; 1 &#038; -1 &#038; 1\\end{bmatrix}=4\\]<br \/>\nLuego, la soluci\u00f3n ser\u00e1 \u00fanica y vendr\u00e1 dada por el sistema formado por cuatro ecuaciones linealmente independientes; por ejemplo, las cuatro primeras, ya que el determinante de esta submatriz de coeficientes es distinto de cero. Elegidas estas cuatro ecuaciones, la soluci\u00f3n ser\u00e1:<br \/>\n\\[\\begin{bmatrix}x\\\\<br \/>\ny\\\\<br \/>\nz\\\\<br \/>\nt\\end{bmatrix}=\\begin{bmatrix}1 &#038; 1 &#038; 0 &#038; 0\\\\<br \/>\n0 &#038; 1 &#038; 2 &#038; 0\\\\<br \/>\n0 &#038; 0 &#038; 1 &#038; 3\\\\<br \/>\n1 &#038; 0 &#038; 0 &#038; 5\\end{bmatrix}^{-1}\\begin{bmatrix}0\\\\<br \/>\n1\\\\<br \/>\n1\\\\<br \/>\n1\\end{bmatrix}=\\begin{bmatrix}-5 &#038; 5 &#038; -10 &#038; 6\\\\<br \/>\n6 &#038; -5 &#038; 10 &#038; -6\\\\<br \/>\n-3 &#038; 3 &#038; -5 &#038; 3\\\\<br \/>\n1 &#038; -1 &#038; 2 &#038; -1\\end{bmatrix}\\begin{bmatrix}0\\\\<br \/>\n1\\\\<br \/>\n1\\\\<br \/>\n1\\end{bmatrix}=\\begin{bmatrix}1\\\\<br \/>\n-1\\\\<br \/>\n1\\\\<br \/>\n0\\end{bmatrix}\\]\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Sean las variedades afines \\[\\begin{align*}r_1&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ x-2y+z-t=4,\\ 2x-3y+2z-3t=-1\\}\\\\ r_2&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ 3x-5y+3z-4t=3,\\ -x+y-z+2t=5\\}\\end{align*}\\] \u00bf\\(r_1\\cap r_2\\)? <\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv2b() {\n  var htmlShow2b = document.getElementById(\"html-show2b\");\n  if (htmlShow2b.style.display === \"none\") {\n    htmlShow2b.style.display = \"block\";\n  } else {\n    htmlShow2b.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv2b()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2b\" style=\"display: none;\">\nLa intersecci\u00f3n, \\(r_1\\cap r_2\\), vendr\u00e1 dada por el sistema que forman todas las ecuaciones conjuntas; es decir, \\((x,y,z,t)\\in r_1\\cap r_2\\) si \\[\\begin{align*} x-2y+z-t&#038;=4\\\\ 2x-3y+2z-3t&#038;=-1 \\\\ 3x-5y+3z-4t&#038;=3\\\\ -x+y-z+2t&#038;=5\\end{align*}\\]<br \/>\nY tendr\u00e1 soluci\u00f3n si<br \/>\n\\[\\mathbf{rank}\\begin{bmatrix}1 &#038; -2 &#038; 1 &#038; -1\\\\<br \/>\n2 &#038; -3 &#038; 2 &#038; -3\\\\<br \/>\n3 &#038; -5 &#038; 3 &#038; -4\\\\<br \/>\n-1 &#038; 1 &#038; -1 &#038; 2\\end{bmatrix}\\overset{?}{=}\\mathbf{rank}\\begin{bmatrix}1 &#038; -2 &#038; 1 &#038; -1 &#038; 4\\\\<br \/>\n2 &#038; -3 &#038; 2 &#038; -3 &#038; -1\\\\<br \/>\n3 &#038; -5 &#038; 3 &#038; -4 &#038; 3\\\\<br \/>\n-1 &#038; 1 &#038; -1 &#038; 2 &#038; 5\\end{bmatrix}\\]<br \/>\nComo ambos rangos coinciden, tiene soluci\u00f3n. Esta se consigue cogiendo un menor de orden igual al rango distinto de cero y considerando el resto de variables como par\u00e1metros. <\/p>\n<p>Vemos que<br \/>\n\\[\\begin{bmatrix}1 &#038; -2 &#038; 1 &#038; -1 &#038; 4\\\\<br \/>\n2 &#038; -3 &#038; 2 &#038; -3 &#038; -1\\\\<br \/>\n3 &#038; -5 &#038; 3 &#038; -4 &#038; 3\\\\<br \/>\n-1 &#038; 1 &#038; -1 &#038; 2 &#038; 5\\end{bmatrix}\\sim\\begin{bmatrix}1 &#038; -2 &#038; 1 &#038; -1 &#038; 4\\\\<br \/>\n0 &#038; 1 &#038; 0 &#038; -1 &#038; -9\\\\<br \/>\n0 &#038; 0 &#038; 0 &#038; 0 &#038; 0\\\\<br \/>\n0 &#038; 0 &#038; 0 &#038; 0 &#038; 0\\end{bmatrix}\\]<\/p>\n<p>Como se observa el menor \\[\\begin{vmatrix}1 &#038; -2 \\\\<br \/>\n0 &#038; 1 \\end{vmatrix}=1\\] es distinto de cero, y nos plantea el sistema con p\u00e1rametro \\(z=\\lambda\\) y \\(t=\\mu\\),<br \/>\n\\[\\begin{bmatrix}1 &#038; -2\\\\<br \/>\n0 &#038; 1<br \/>\n\\end{bmatrix}\\begin{bmatrix}x\\\\<br \/>\ny\\end{bmatrix}=\\begin{bmatrix}4\\\\<br \/>\n-9\\end{bmatrix}-\\begin{bmatrix}1&#038;-1\\\\<br \/>\n0&#038;-1\\end{bmatrix}\\begin{bmatrix}\\lambda\\\\<br \/>\n\\mu\\end{bmatrix}\\]<br \/>\nLo que implica que<br \/>\n\\[\\begin{bmatrix}x\\\\<br \/>\ny\\end{bmatrix}=\\begin{bmatrix}1 &#038; -2\\\\<br \/>\n0 &#038; 1<br \/>\n\\end{bmatrix}^{-1}\\begin{bmatrix}4-\\lambda+\\mu\\\\<br \/>\n-9+\\mu\\end{bmatrix}=\\begin{bmatrix}1 &#038; 2\\\\0 &#038; 1 \\end{bmatrix}\\begin{bmatrix}4-\\lambda+\\mu\\\\<br \/>\n-9+\\mu\\end{bmatrix}=\\begin{bmatrix}-14-\\lambda+3\\mu\\\\<br \/>\n-9+\\mu\\end{bmatrix}\\]<br \/>\nLas ecuaciones param\u00e9tricas son \\[\\begin{align*}x=&#038;-14-\\lambda+3\\mu\\\\<br \/>\ny=&#038;-9+\\mu\\\\ z=&#038;\\lambda \\\\ t=&#038;\\mu\\end{align*}\\]\n<\/p><\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Sean las variedades afines \\[\\begin{align*}r_1&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ x-2y+z-t=4,\\ 2x-3y+2z-3t=-1\\}\\\\ r_2&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ 3x-5y+3z+4t=3,\\ -x+y-z+2t=5\\}\\end{align*}\\] \u00bf\\(r_1\\cap r_2\\)? <\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv2c() {\n  var htmlShow2c = document.getElementById(\"html-show2c\");\n  if (htmlShow2c.style.display === \"none\") {\n    htmlShow2c.style.display = \"block\";\n  } else {\n    htmlShow2c.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv2c()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2c\" style=\"display: none;\">\n<iframe loading=\"lazy\" src=\"https:\/\/uploads.jesussoto.es\/maxima\/EjrALGinterseccion05.html\" width=\"650\" height=\"200\" allow=\"fullscreen\"><\/iframe>\n<\/div>\n<hr \/>\n<p>&nbsp;<\/p>\n<table id=\"yzpi\" border=\"0\" width=\"100%\" cellspacing=\"0\" cellpadding=\"3\" bgcolor=\"#999999\">\n<tbody>\n<tr>\n<td width=\"100%\"><strong>Ejercicio:<\/strong> Dados los vectores normales de las ecuaciones impl\u00edcitas de la imagen de la aplicaci\u00f3n lineal \\(f:\\mathbb{R}^3\\to M_2(\\mathbb{R})\\) dada por \\[f(x,y,z)=\\begin{bmatrix} x-y&#038; y-z\\\\ z-x&#038; x-y\\end{bmatrix},\\] \u00bfcu\u00e1l de los siguientes vectores pertenece al subespacio generado por ellos? <\/td>\n<\/tr>\n<tr>\n<td>\n<div id=\"menu-a\">\n<ul>\n<li>[1,-1,2,2]<\/li>\n<li>[0,1,0,1]<\/li>\n<li>[3,2,2,-1]<\/li>\n<\/ul>\n<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><script>\nfunction showHtmlDiv() {\n  var htmlShow = document.getElementById(\"html-show\");\n  if (htmlShow.style.display === \"none\") {\n    htmlShow.style.display = \"block\";\n  } else {\n    htmlShow.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show\" style=\"display: none;\">\n<p><strong>C.)<\/strong><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Recordad que todo sistemas de ecuaciones los podemos formular mediante una ecuaci\u00f3n matricial \\[AX=B,\\] donde \\(A\\) es la matriz de coeficiente y \\(B\\) la matriz de t\u00e9rminos independientes. Llamamos matriz ampliada del&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"class_list":["post-454","post","type-post","status-publish","format-standard","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/454","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=454"}],"version-history":[{"count":2,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/454\/revisions"}],"predecessor-version":[{"id":456,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/454\/revisions\/456"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=454"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=454"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=454"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}