{"id":183,"date":"2025-10-06T08:15:06","date_gmt":"2025-10-06T06:15:06","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=183"},"modified":"2025-12-09T12:18:03","modified_gmt":"2025-12-09T11:18:03","slug":"alg-determinante-y-menor-de-una-matriz","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=183","title":{"rendered":"ALG: Determinante y menor de una matriz"},"content":{"rendered":"<p>Para que sea m\u00e1s f\u00e1cil definimos los determinantes de forma recursiva, utilizando el valor de un determinante de orden dos y la <a href=\"http:\/\/es.wikipedia.org\/wiki\/Teorema_de_Laplace\" target=\"_blank\" rel=\"noopener noreferrer\">Regla de Laplace<\/a>:<\/p>\n<blockquote>\n<ul>\n<li>Sea \\(A=\\begin{bmatrix}a_{11}&#038;a_{12}\\\\ a_{21}&#038;a_{22}\\end{bmatrix}\\in\\mathcal{M}_2(\\mathbb{K})\\), definimos el determinante de \\(A\\), como \\[|A|=a_{11}a_{22}-a_{12}a_{21}.\\]<\/li>\n<li> Para todo \\(n>2\\) y \\(A\\in\\mathcal{M}_n(\\mathbb{K})\\), definimos \\[|A|=\\sum _{j=1}^{n}a_{1j}\\;A_{1j},\\] donde \\(A_{1j}=(-1)^{(1+j)}\\;\\alpha _{1j}\\), siendo \\(\\alpha _{1j}\\) el determinante de orden \\(n-1\\) que queda tras eliminar de la matriz \\(A\\) la fila 1 y la columna \\(j\\).\n<\/p>\n<\/blockquote>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\[\\begin{bmatrix}1 &#038; 2 &#038; 3\\\\<br \/>\n\\mathop{-}1 &#038; 0 &#038; 1\\\\<br \/>\n2 &#038; \\mathop{-}1 &#038; \\mathop{-}1\\end{bmatrix}\\in\\mathcal{M}_3(\\mathbb{R}),\\]  \u00bfcu\u00e1nto es su determinante? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv42() {\n  var htmlShow42 = document.getElementById(\"html-show42\");\n  if (htmlShow42.style.display === \"none\") {\n    htmlShow42.style.display = \"block\";\n  } else {\n    htmlShow42.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv42()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show42\" style=\"display: none;\">\nApliquemos lo anterior:<\/p>\n<p>\\[\\begin{align*}<br \/>\n\\begin{vmatrix}1 &#038; 2 &#038; 3\\\\<br \/>\n\\mathop{-}1 &#038; 0 &#038; 1\\\\<br \/>\n2 &#038; \\mathop{-}1 &#038; \\mathop{-}1\\end{vmatrix}&#038;=1\\cdot\\begin{bmatrix}0 &#038; 1\\\\<br \/>\n\\mathop{-}1 &#038; \\mathop{-}1\\end{bmatrix}-2\\cdot\\begin{bmatrix}\\mathop{-}1 &#038; 1\\\\<br \/>\n2 &#038; \\mathop{-}1\\end{bmatrix}+3\\cdot\\begin{bmatrix}\\mathop{-}1 &#038; 0\\\\<br \/>\n2 &#038; \\mathop{-}1\\end{bmatrix} \\\\<br \/>\n&#038;=(0\\cdot(-1)-1\\cdot(-1))-2((-1)\\cdot(-1)-1\\cdot 2)+\\\\ &#038;\\quad +3((-1)\\cdot(-1)-0\\cdot 2)\\\\<br \/>\n&#038;=6<br \/>\n\\end{align*}\\]\n<\/p><\/div>\n<hr \/>\n<p>La definici\u00f3n cl\u00e1sica y su significado puede verse en <a href=\"http:\/\/en.wikipedia.org\/wiki\/Determinant\" target=\"_blank\" rel=\"noopener noreferrer\">Determinante<\/a>. En este enlace pod\u00e9is encontrar tambi\u00e9n propiedades importantes. Recordad estas propiedades porque ser\u00e1n muy importantes para aprender bien este tema.<\/p>\n<blockquote>\n<p><strong>Regla de Laplace:<\/strong> El determinante de una matriz es independiente de la fila o columna que elijamos en el paso 2 anterior.<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Calcular\\[\\begin{vmatrix}1 &#038; 2 &#038; 3 &#038; 0\\\\<br \/>\n\\mathop{-}1 &#038; 0 &#038; 1 &#038; 0\\\\<br \/>\n2 &#038; \\mathop{-}1 &#038; \\mathop{-}1 &#038; 0\\\\<br \/>\n8 &#038; 1 &#038; 3 &#038; 1\\end{vmatrix}\\] <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv42w() {\n  var htmlShow42w = document.getElementById(\"html-show42w\");\n  if (htmlShow42w.style.display === \"none\") {\n    htmlShow42w.style.display = \"block\";\n  } else {\n    htmlShow42w.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv42w()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show42w\" style=\"display: none;\">\nLa regla de Lapalace nos dice que podemos utilizar cualquier fila o columna para desarrollar el determinante, luego, si elegimos la cuarta columna, tendremos<br \/>\n\\[\\begin{vmatrix}1 &#038; 2 &#038; 3 &#038; 0\\\\<br \/>\n\\mathop{-}1 &#038; 0 &#038; 1 &#038; 0\\\\<br \/>\n2 &#038; \\mathop{-}1 &#038; \\mathop{-}1 &#038; 0\\\\<br \/>\n8 &#038; 1 &#038; 3 &#038; 1\\end{vmatrix}=\\begin{vmatrix}1 &#038; 2 &#038; 3\\\\<br \/>\n\\mathop{-}1 &#038; 0 &#038; 1\\\\<br \/>\n2 &#038; \\mathop{-}1 &#038; \\mathop{-}1\\end{vmatrix}=6\\]\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\[A=\\begin{bmatrix}<br \/>\ni &#038; 0 &#038; 0 &#038; 0\\\\<br \/>\n1 &#038; i &#038; 0 &#038; 0 \\\\<br \/>\n-1 &#038; 1 &#038; i &#038; 0 \\\\<br \/>\ni &#038; 1 &#038; -1 &#038; i<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_4(\\mathbb{C}),\\]  \u00bfcu\u00e1nto es su determinante? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv4() {\n  var htmlShow4 = document.getElementById(\"html-show4\");\n  if (htmlShow4.style.display === \"none\") {\n    htmlShow4.style.display = \"block\";\n  } else {\n    htmlShow4.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv4()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show4\" style=\"display: none;\">\nSoluci\u00f3n: 1.\n<\/div>\n<hr \/>\n<p>El ejercicio anterior nos lleva a una conclusi\u00f3n muy interesante:<\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> El determinante de una matriz triangular es igual al producto de los elementos de su diagonal principal.<\/p><\/blockquote>\n<p>Esto nos permite resolver un determinante mediante operaciones elementales. Esto se justifica por las siguientes propiedades:<\/p>\n<p>Asumamos \\(A\\) y \\(B\\) dos matrices cuadradas del mismo orden,<\/p>\n<ol>\n<li>Si \\(B\\) es el resultado de hacer una transformaci\u00f3n elemental por fila(columna) a la matriz \\(A\\), \\(A\\overset{f_i+\\lambda f_j\\\\ (c_i+\\lambda c_j)}{\\sim}B\\Rightarrow|A|=|B|\\)<\/li>\n<li>Si \\(B\\) es el resultado de intercambiar una fila(columna) de la matriz \\(A\\), \\(A\\overset{f_i \\leftrightarrow f_j\\\\ (c_i\\leftrightarrow c_j)}{\\sim}B\\Rightarrow|A|=-|B|\\)<\/li>\n<li>Si \\(B\\) es el resultado de multiplicar una fila(columna) de la matriz \\(A\\) por un escalar, \\(A\\overset{f_i = \\lambda f_i\\\\ (c_i=\\lambda c_i)}{\\sim}B\\Rightarrow|B|=\\lambda |A|\\)<\/li>\n<\/ol>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Cu\u00e1l es el valor del determinante  \\[\\begin{vmatrix}<br \/>\n1 &#038; 1 &#038; 1 &#038; 1 &#038; \\cdots  &#038; 1 &#038; 1\\\\<br \/>\n1 &#038; 2 &#038; 2 &#038; 2 &#038; \\cdots  &#038; 2 &#038; 2\\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 3 &#038; \\cdots  &#038; 3 &#038; 3\\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4 &#038; \\cdots &#038; 4 &#038; 4\\\\<br \/>\n\\vdots &#038; \\vdots  &#038; \\vdots  &#038;\\vdots   &#038; \\cdots  &#038; \\vdots  &#038; \\vdots \\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4 &#038; \\cdots  &#038; n-1 &#038; n-1\\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4 &#038; \\cdots  &#038; n-1 &#038; n<br \/>\n\\end{vmatrix}\\] <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv7() {\n  var htmlShow7 = document.getElementById(\"html-show7\");\n  if (htmlShow7.style.display === \"none\") {\n    htmlShow7.style.display = \"block\";\n  } else {\n    htmlShow7.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv7()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show7\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Determinantes: EJ.3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/iWFECtAih2U?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>Consecuencia de las propiedades anteriores son estos resultados:<\/p>\n<blockquote>\n<ol>\n<li>El determinante de una matriz con una fila, o columna, todo ceros vale cero.<\/li>\n<li>El determinante de una matriz con dos filas, o columnas, proporcionales vale cero.<\/li>\n<\/ol>\n<\/blockquote>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Cu\u00e1l es el valor del determinante de la matriz \\[\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 0 &#038; 0 &#038; \\cdots  &#038; 0 &#038; \\mathbf{\\alpha}\\\\<br \/>\n1 &#038; 2 &#038; 0 &#038; 0 &#038; \\cdots  &#038; 0 &#038; 0\\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 0 &#038; \\cdots  &#038; 0 &#038; 0\\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4 &#038; \\cdots &#038; 0 &#038; 0\\\\<br \/>\n\\vdots &#038; \\vdots  &#038; \\vdots  &#038;\\vdots   &#038; \\cdots  &#038; \\vdots  &#038; \\vdots \\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4 &#038; \\cdots  &#038; n-1 &#038; 0\\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4 &#038; \\cdots  &#038; n-1 &#038; n<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_n(R).\\] <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv6() {\n  var htmlShow6 = document.getElementById(\"html-show6\");\n  if (htmlShow6.style.display === \"none\") {\n    htmlShow6.style.display = \"block\";\n  } else {\n    htmlShow6.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv6()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show6\" style=\"display: none;\">\nSoluci\u00f3n: \\(n!\\)<br \/>\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Determinantes. Ej.12 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/dmfXX1VyEuc?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<h3>Propiedades de los determinantes<\/h3>\n<p>Sea \\(A\\) una matrices cuadrada,<\/p>\n<ol>\n<li>\\(|A|=|A^t|\\)<\/li>\n<li>\\(\\begin{vmatrix}a_{11}&amp;a_{12}\\\\ a+b &amp; c+d\\end{vmatrix}=\\begin{vmatrix}a_{11}&amp;a_{12}\\\\ a &amp; c\\end{vmatrix}+\\begin{vmatrix}a_{11}&amp;a_{12}\\\\ b &amp; d\\end{vmatrix}\\). De igual modo podemos hacerlo para toda matriz cuadrada de orden \\(n\\).<\/li>\n<li>\\(|A\\,B|=|A|\\cdot |B|\\)<\/li>\n<\/ol>\n<blockquote>\n<p><strong>Proposici\u00f3n:<\/strong> Una matriz cuadrada no tiene inversa si su determinante  es cero.<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\(A\\)= [[1,-5,3],[-1,-3,4],[-3,7,x]], \u00bfcu\u00e1l es el valor de \\(x\\) para que la matriz no tenga inversa?<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv4f1() {\n  var htmlShow4f1 = document.getElementById(\"html-show4f1\");\n  if (htmlShow4f1.style.display === \"none\") {\n    htmlShow4f1.style.display = \"block\";\n  } else {\n    htmlShow4f1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv4f1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show4f1\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Inversa de una matriz. Ej.8 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/JwTufPUzRHI?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<h2>Matrices en bloques<\/h2>\n<blockquote><p><strong>Proposici\u00f3n:<\/strong> Sean \\({\\displaystyle A,B,C,D}\\) matrices de tama\u00f1os \\({\\displaystyle n\\times n,n\\times m,m\\times n,m\\times m}\\) respectivamente. Entonces<br \/>\n\\[{\\displaystyle \\left|{\\begin{array}{cc}A&#038;0\\\\C&#038;D\\end{array}}\\right|= \\left|{\\begin{array}{cc}A&#038;B\\\\0&#038;D\\end{array}}\\right|=|A|\\cdot |D|}\\]<\/p><\/blockquote>\n<p>Si adem\u00e1s, resulta que la matriz \\(A\\) anterior es regular(tiene inversa), se cumple:<\/p>\n<blockquote><p><strong>Proposici\u00f3n:<\/strong><br \/>\n\\[\\left|\\begin{array}{cc}A&#038;B\\\\C&#038;D\\end{array}\\right|=|A|\\cdot |D-CA^{-1}B|\\]<\/p><\/blockquote>\n<blockquote><p><strong>Proposici\u00f3n:<\/strong> (Matrices Triangulares por Bloques) Sea $M$ una matriz cuadrada particionada en $k \\times k$ bloques. Si la matriz $M$ es <strong>triangular por bloques<\/strong>, su determinante es igual al producto de los determinantes de los bloques diagonales.\n<\/p><\/blockquote>\n<p>Por ejemplo, una matriz triangular inferior por bloques que tiene la forma:<br \/>\n\\[<br \/>\nM = \\begin{bmatrix}<br \/>\nA_{11} &#038; 0 &#038; \\cdots &#038; 0 \\\\<br \/>\nA_{21} &#038; A_{22} &#038; \\cdots &#038; 0 \\\\<br \/>\n\\vdots &#038; \\vdots &#038; \\ddots &#038; \\vdots \\\\<br \/>\nA_{k1} &#038; A_{k2} &#038; \\cdots &#038; A_{kk}<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Donde:<\/p>\n<ol>\n<li> \\(\\mathbf{A_{ii}}\\) son <strong>matrices cuadradas<\/strong> de tama\u00f1os compatibles (los bloques diagonales).<\/li>\n<li> Todos los bloques por encima de la diagonal principal, \\((\\mathbf{A_{ij}}\\) con \\(i&lt;j\\)) son <strong>matrices de ceros<\/strong> \\((\\mathbf{0})\\).<\/li>\n<\/ol>\n<p>En este caso, el determinante de la matriz \\(M\\) es:<\/p>\n<p>\\[<br \/>\n\\det(M) = \\det(A_{11}) \\cdot \\det(A_{22}) \\cdot \\ldots \\cdot \\det(A_{kk})<br \/>\n\\]<\/p>\n<p>\\[<br \/>\n\\det(M) = \\prod_{i=1}^{k} \\det(A_{ii})<br \/>\n\\]<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Calcular \\[\\begin{vmatrix}1 &#038; 2 &#038; 0 &#038; 0 &#038; 0 &#038; 0\\\\<br \/>\n\\mathop{-}1 &#038; 3 &#038; 0 &#038; 0 &#038; 0 &#038; 0\\\\<br \/>\n1 &#038; 2 &#038; \\mathop{-}1 &#038; 4 &#038; 0 &#038; 0\\\\<br \/>\n0 &#038; 3 &#038; \\mathop{-}2 &#038; 1 &#038; 0 &#038; 0\\\\<br \/>\n\\mathop{-}1 &#038; \\mathop{-}2 &#038; 0 &#038; 1 &#038; 2 &#038; 1\\\\<br \/>\n0 &#038; 1 &#038; 1 &#038; 1 &#038; 1 &#038; 1\\end{vmatrix}\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv4f13() {\n  var htmlShow4f13 = document.getElementById(\"html-show4f13\");\n  if (htmlShow4f13.style.display === \"none\") {\n    htmlShow4f13.style.display = \"block\";\n  } else {\n    htmlShow4f13.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv4f13()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show4f13\" style=\"display: none;\">\nVeamos c\u00f3mo podemos hacer bloque la matriz de partida:<br \/>\n\\[\\left[\\begin{array}{cc|cc|cc}1 &#038; 2 &#038; 0 &#038; 0 &#038; 0 &#038; 0\\\\<br \/>\n\\mathop{-}1 &#038; 3 &#038; 0 &#038; 0 &#038; 0 &#038; 0\\\\ \\hline<br \/>\n1 &#038; 2 &#038; \\mathop{-}1 &#038; 4 &#038; 0 &#038; 0\\\\<br \/>\n0 &#038; 3 &#038; \\mathop{-}2 &#038; 1 &#038; 0 &#038; 0\\\\ \\hline<br \/>\n\\mathop{-}1 &#038; \\mathop{-}2 &#038; 0 &#038; 1 &#038; 2 &#038; 1\\\\<br \/>\n0 &#038; 1 &#038; 1 &#038; 1 &#038; 1 &#038; 1\\end{array}\\right]\\]<\/p>\n<p>Luego<br \/>\n\\[\\begin{vmatrix}1 &#038; 2 &#038; 0 &#038; 0 &#038; 0 &#038; 0\\\\<br \/>\n\\mathop{-}1 &#038; 3 &#038; 0 &#038; 0 &#038; 0 &#038; 0\\\\<br \/>\n1 &#038; 2 &#038; \\mathop{-}1 &#038; 4 &#038; 0 &#038; 0\\\\<br \/>\n0 &#038; 3 &#038; \\mathop{-}2 &#038; 1 &#038; 0 &#038; 0\\\\<br \/>\n\\mathop{-}1 &#038; \\mathop{-}2 &#038; 0 &#038; 1 &#038; 2 &#038; 1\\\\<br \/>\n0 &#038; 1 &#038; 1 &#038; 1 &#038; 1 &#038; 1\\end{vmatrix}=\\begin{vmatrix}1 &#038; 2 \\\\ -1&#038; 3\\end{vmatrix}\\cdot<br \/>\n\\begin{vmatrix}-1 &#038; 4 \\\\ -2&#038; 1\\end{vmatrix}\\cdot \\begin{vmatrix}2 &#038; 1 \\\\ 1&#038; 1\\end{vmatrix}=35\\]\n<\/p><\/div>\n<hr \/>\n<h2>Menor de una matriz<\/h2>\n<p>Un menor de una matriz \\(A\\) es el determinante de una submatriz cuadrada de \\(A\\). <\/p>\n<p>Un menor complementario de una matriz \\(A\\) es el determinante de alguna submatriz, obtenido de \\(A\\) mediante la eliminaci\u00f3n de una o m\u00e1s de sus filas o columnas. De este modo designamos mediante \\(m_{ij}\\) el menor del elemento \\(a_{ij}\\) en la matriz \\(A\\); es decir, el menor resultante de eliminar la fila \\(i\\) y la columna \\(j\\).<\/p>\n<p> La definici\u00f3n de menor nos da pie a otro resultado muy interesante. Podemos extender la definici\u00f3n de menor para una matriz no cuadrada a cualquier determinante de una submatriz cuadrada. En este caso:\n<\/p>\n<blockquote>\n<p><strong> Teorema.<\/strong> Si \\(A\\) es una matriz, el rango de \\(A\\) es el orden del mayor menor de \\(A\\) no nulo.\n<\/p>\n<\/blockquote>\n<p>Hay un tipo de menores muy interesantes. Consideremos una matriz cuadrada \\(A=[a_{ij}]\\) de orden \\(n\\), los \\(n-1\\) menores formados de la forma:<br \/>\n\\[\\begin{vmatrix}<br \/>\na_{11} &#038; a_{12} \\\\<br \/>\na_{21} &#038; a_{22}<br \/>\n\\end{vmatrix}\\quad\\begin{vmatrix}<br \/>\na_{11} &#038; a_{12} &#038;a_{13}\\\\<br \/>\na_{21} &#038; a_{22} &#038;a_{23}\\\\<br \/>\na_{31} &#038; a_{32} &#038;a_{33}<br \/>\n\\end{vmatrix}<br \/>\n\\quad\\begin{vmatrix}<br \/>\na_{11} &#038; a_{12} &#038;a_{13}&#038;a_{14}\\\\<br \/>\na_{21} &#038; a_{22} &#038;a_{23}&#038;a_{24}\\\\<br \/>\na_{31} &#038; a_{32} &#038;a_{33}&#038;a_{34}\\\\<br \/>\na_{41} &#038; a_{42} &#038;a_{43}&#038;a_{44}<br \/>\n\\end{vmatrix}\\ldots\\;<br \/>\n\\begin{vmatrix}<br \/>\na_{11} &#038; a_{12} &#038; \\ldots &#038; a_{1\\,n-1}\\\\<br \/>\na_{21} &#038; a_{22} &#038; \\ldots &#038; a_{2\\,n-1}\\\\<br \/>\n\\vdots &#038; \\vdots &#038; \\ldots &#038; \\vdots\\\\<br \/>\na_{n-1\\,1} &#038; a_{n-1\\,2} &#038; \\ldots &#038; a_{n-1\\,n-1}<br \/>\n\\end{vmatrix}<br \/>\n\\]<br \/>\nse les denomina menores principales de \\(A\\). Recordad que, en nuestros ejercicios, incluiremos entre los menores principales a \\(|a_{11}|\\) y  \\(|A|\\).<\/p>\n<h2>Matriz adjunta e inversa<\/h2>\n<p>Si consideramos \\(m_{ij}\\) el menor complementario del elemento \\(a_{ij}\\) en la matriz \\(A\\), decimos adjunto(cofactor) del elemento \\(a_{ij}\\) en la matriz \\(A\\), y lo notamos por \\(A_{ij}\\), al resultado \\[A_{ij}=(-1)^{i+j}m_{ij}.\\]<\/p>\n<p>En algunas bibliograf\u00edas tambi\u00e9n lo llaman cofactor. As\u00ed la regla de Laplace quedar\u00eda como:<\/p>\n<p>\\[|A|=\\sum_{k=1}^{n}a_{ik}A_{ik}\\] o \\[|A|=\\sum_{k=1}^{n}a_{kj}A_{kj}\\]<\/p>\n<p>En la bibliograf\u00eda a la matriz \\([A_{ij}]\\) se le suele llamar matriz de cofactores.<\/p>\n<p>De este modo definimos la matriz adjunta como \\[adj(A)=[A_{ji}]=[A_{ij}]^t;\\] es decir, la traspuesta de la matriz de cofactores.<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\[A=\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 0 &#038; 0 \\\\<br \/>\n1 &#038; 2 &#038; 0 &#038; 0 \\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 0 \\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_4(\\mathbb{R}).\\] \u00bfCu\u00e1nto es la traza de \\(adj(A)\\)? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv6() {\n  var htmlShow6 = document.getElementById(\"html-show6\");\n  if (htmlShow6.style.display === \"none\") {\n    htmlShow6.style.display = \"block\";\n  } else {\n    htmlShow6.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv6()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show6\" style=\"display: none;\">\nObservar que la traza de \\[\\mathbf{tr}(adj(A))=\\sum_{i=1}^4A_{ii}=\\begin{vmatrix}<br \/>\n2 &#038; 0 &#038; 0 \\\\<br \/>\n2 &#038; 3 &#038; 0 \\\\<br \/>\n2 &#038; 3 &#038; 4<br \/>\n\\end{vmatrix}+\\begin{vmatrix}<br \/>\n1 &#038; 0 &#038; 0 \\\\<br \/>\n1 &#038; 3 &#038; 0 \\\\<br \/>\n1 &#038; 3 &#038; 4<br \/>\n\\end{vmatrix}+<br \/>\n\\begin{vmatrix}<br \/>\n1 &#038; 0  &#038; 0 \\\\<br \/>\n1 &#038; 2  &#038; 0 \\\\<br \/>\n1 &#038; 2  &#038; 4<br \/>\n\\end{vmatrix}+<br \/>\n\\begin{vmatrix}<br \/>\n1 &#038; 0 &#038; 0  \\\\<br \/>\n1 &#038; 2 &#038; 0  \\\\<br \/>\n1 &#038; 2 &#038; 3<br \/>\n\\end{vmatrix}\\]\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\[A=\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 0 &#038; 0 \\\\<br \/>\n1 &#038; 2 &#038; 0 &#038; 0 \\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 0 \\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_4(\\mathbb{R}).\\] La afirmaci\u00f3n: &quot;No existen dos indices \\(i,j\\in\\{1,\\ldots,4\\}\\) tales que \\(A_{ij}=16\\)&quot; es verdadera o falsa. <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv62() {\n  var htmlShow62 = document.getElementById(\"html-show62\");\n  if (htmlShow62.style.display === \"none\") {\n    htmlShow62.style.display = \"block\";\n  } else {\n    htmlShow62.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv62()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show62\" style=\"display: none;\">\nVerdadera.\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\[A=\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 0 &#038; 0 &#038;  0&#038;0\\\\<br \/>\n2 &#038; 1 &#038; 0 &#038; 0 &#038;  0&#038;0\\\\<br \/>\n3 &#038; 2 &#038; 1 &#038; 0 &#038;  0&#038;0\\\\<br \/>\n4 &#038; 3 &#038; 2 &#038; 1 &#038;  0&#038;0\\\\<br \/>\n5 &#038; 4 &#038; 3 &#038; 2 &#038;  1&#038;0\\\\<br \/>\n6 &#038; 5 &#038; 4 &#038; 3 &#038;  2&#038;1<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_6(\\mathbb{R}),\\] y \\(Adj(A)=[A_{ij}]\\) su matriz adjunta. \u00bfCu\u00e1ntos \\(A_{ij}=0\\) hay? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv11() {\n  var htmlShow11 = document.getElementById(\"html-show11\");\n  if (htmlShow11.style.display === \"none\") {\n    htmlShow11.style.display = \"block\";\n  } else {\n    htmlShow11.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv11()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show11\" style=\"display: none;\">\nSoluci\u00f3n: 21.\n<\/div>\n<hr \/>\n<blockquote>\n<p> Propiedades de la matriz adjunta de una matriz \\(A\\in\\mathcal{M}_n(\\mathbb{K})\\):<\/p>\n<ol>\n<li>\\(adj(A^t)=adj(A)^t\\) <\/li>\n<li>\\(adj(AB)=adj(B)\\cdot adj(A)\\) <\/li>\n<li>\\(adj(A^k)=adj(A)^k\\) <\/li>\n<li>\\(adj(I)=I\\) <\/li>\n<li>\\(A\\cdot adj(A)=adj(A)\\cdot A=|A|\\cdot I\\) <\/li>\n<li>\\(adj(\\lambda A)=\\lambda^{n-1}adj(A)\\) <\/li>\n<li>\\(adj(adj(A))=|A|^{n-2}A\\) <\/li>\n<li>\\(|A|=tr(A\\cdot adj(A))\/n\\) <\/li>\n<li>\\(|adj(A)|=|A|^{n-1}\\) <\/li>\n<\/ol>\n<\/blockquote>\n<p>Estas definiciones nos permiten usarlas para definir el rango de una matriz cualquiera, como orden del mayor de los menores distinto de cero, y dar una f\u00f3rmula para calcular la inversa de una matriz, en caso de que exista:<br \/>\n\\[A^{-1}=\\frac{1}{|A|}adj(A)\\]<\/p>\n<p>Como consecuencia de lo anterior podemos formular el siguiente resultado:<\/p>\n<blockquote>\n<p><strong>Corolario:<\/strong> Una matriz cuadrada es regular(inversible) si su determinante es distinto de cero.<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\[A=\\begin{bmatrix}<br \/>\n2 &#038; 4 &#038; 1 &#038; 12 \\\\<br \/>\n-1 &#038; 1 &#038; 0 &#038; 3 \\\\<br \/>\n0 &#038; -1 &#038; 9 &#038; -3 \\\\<br \/>\n7 &#038; 3 &#038; 6 &#038; 9<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_4(\\mathbb{R}),\\] \u00bfes regular? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv64() {\n  var htmlShow64 = document.getElementById(\"html-show64\");\n  if (htmlShow64.style.display === \"none\") {\n    htmlShow64.style.display = \"block\";\n  } else {\n    htmlShow64.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv64()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show64\" style=\"display: none;\">\nNo\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\[A=\\begin{bmatrix}<br \/>\n-\\alpha &#038; \\alpha-1 &#038; \\alpha+1 \\\\<br \/>\n 1 &#038; 2 &#038; 3 \\\\<br \/>\n2-\\alpha &#038; \\alpha+3 &#038; \\alpha+7<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_3(\\mathbb{R}),\\] \u00bfpara que valores de \\(\\alpha\\) la matriz no es regular? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv62q() {\n  var htmlShow62q = document.getElementById(\"html-show62q\");\n  if (htmlShow62q.style.display === \"none\") {\n    htmlShow62q.style.display = \"block\";\n  } else {\n    htmlShow62q.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv62q()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show62q\" style=\"display: none;\">\nObervar que<br \/>\n\\[\\begin{bmatrix}-\\alpha  &#038; \\alpha -1 &#038; \\alpha +1\\\\<br \/>\n1 &#038; 2 &#038; 3\\\\<br \/>\n2-\\alpha  &#038; \\alpha +3 &#038; \\alpha +7\\end{bmatrix}\\overset{f_3-f_1}{\\sim }<br \/>\n\\begin{bmatrix}-\\alpha  &#038; \\alpha -1 &#038; \\alpha +1\\\\<br \/>\n1 &#038; 2 &#038; 3\\\\<br \/>\n2 &#038; 4 &#038; 6\\end{bmatrix}\\]\n<\/div>\n<hr \/>\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><br \/>\nSea \\(A\\)=[[-1,1],[1,-2],[1,0]]. \u00bfCu\u00e1nto suman los elementos de su pseudoinversa?<\/tr>\n<tr>\n<td>\n<div id=\"menu-a\">\n<ul>\n<li>1\/3<\/li>\n<li>3\/2<\/li>\n<li>4<\/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>A.)<\/strong><\/p>\n<p>Podemos ver que \\(\\mathbf{rank}(A)=2\\), luego existe \\(L\\) talque \\(LA=I\\). Ahora solo tenemos que calcular \\[L=(A^tA)^{-1}A^t=\\begin{bmatrix}\\mathop{-}\\left( \\frac{1}{3}\\right)  &#038; \\mathop{-}\\left( \\frac{1}{6}\\right)  &#038; \\frac{5}{6}\\\\<br \/>\n0 &#038; \\mathop{-}\\left( \\frac{1}{2}\\right)  &#038; \\frac{1}{2}\\end{bmatrix}\\]\n<\/p><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Para que sea m\u00e1s f\u00e1cil definimos los determinantes de forma recursiva, utilizando el valor de un determinante de orden dos y la Regla de Laplace: Sea \\(A=\\begin{bmatrix}a_{11}&#038;a_{12}\\\\ a_{21}&#038;a_{22}\\end{bmatrix}\\in\\mathcal{M}_2(\\mathbb{K})\\), definimos el determinante de&#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-183","post","type-post","status-publish","format-standard","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/183","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=183"}],"version-history":[{"count":19,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/183\/revisions"}],"predecessor-version":[{"id":230,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/183\/revisions\/230"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=183"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=183"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=183"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}