(* Content-type: application/vnd.wolfram.cdf.text *) (*** Wolfram CDF File ***) (* http://www.wolfram.com/cdf *) (* CreatedBy='Mathematica 9.0' *) (*************************************************************************) (* *) (* The Mathematica License under which this file was created prohibits *) (* restricting third parties in receipt of this file from republishing *) (* or redistributing it by any means, including but not limited to *) (* rights management or terms of use, without the express consent of *) (* Wolfram Research, Inc. For additional information concerning CDF *) (* licensing and redistribution see: *) (* *) (* www.wolfram.com/cdf/adopting-cdf/licensing-options.html *) (* *) (*************************************************************************) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 1063, 20] NotebookDataLength[ 295642, 7568] NotebookOptionsPosition[ 278991, 7019] NotebookOutlinePosition[ 280964, 7085] CellTagsIndexPosition[ 280921, 7082] WindowFrame->Normal*) (* Beginning of Notebook Content *) Notebook[{ Cell[CellGroupData[{ Cell["SISTEMAS DE ECUACIONES LINEALES", "Title"], Cell[CellGroupData[{ Cell[TextData[{ "Guillermo S\[AAcute]nchez (", ButtonBox["http://diarium.usal.es/guillermo", BaseStyle->"Hyperlink", ButtonData->{ URL["http://diarium.usal.es/guillermo"], None}, ButtonNote->"http://diarium.usal.es/guillermo"], ")" }], "ItemParagraph"], Cell["\<\ Departamento de Economia e H\.aa Econ\[OAcute]mica. Universidad de Salamanca. \ \>", "ItemParagraph"], Cell["Actualizado : 2012-10-28", "ItemParagraph"] }, Open ]], Cell[CellGroupData[{ Cell["Sobre el estilo utilizado", "Subsubsection"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " las salidas (", StyleBox["Input", FontSlant->"Italic"], ") por defecto las muestra utilizando el estilo: ", Cell[BoxData[ ButtonBox["StandardForm", BaseStyle->"Link", ButtonData->"paclet:ref/StandardForm"]]], ". En su lugar preferiamos utilizar el estilo ", Cell[BoxData[ ButtonBox["TraditionalForm", BaseStyle->"Link", ButtonData->"paclet:ref/TraditionalForm"]]], " que da una apariencia a las salidas (", StyleBox["Output", FontSlant->"Italic"], ") coincidente con el habitualmente utilizado en la notaci\[OAcute]n cl\ \[AAcute]sica utilizada en las matem\[AAcute]ticas. Esto puede hacerse para \ cada celda a\[NTilde]adiendo // TraditionalForm al final de cada ", StyleBox["input", FontSlant->"Italic"], " . Sin embargo puede hacerse que este estilo (TraditionalForm) se aplique a \ todas las salidas del cuaderno (o notebook) a\[NTilde]adiendo la siguiente \ sentencia (en este caso hemos definido la celda para que se ejecute \ automaticamente al inicio): " }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"SetOptions", "[", RowBox[{ RowBox[{"EvaluationNotebook", "[", "]"}], ",", RowBox[{"CommonDefaultFormatTypes", " ", "->", " ", RowBox[{"{", RowBox[{"\"\\"", " ", "->", " ", "TraditionalForm"}], "}"}]}]}], "]"}], " "}]], "Input", InitializationCell->True] }, Closed]], Cell[CellGroupData[{ Cell["Sistemas de ecuaciones lineales", "Section"], Cell[CellGroupData[{ Cell["\<\ Sistema de ecuaciones lineales en notaci\[OAcute]n matricial\ \>", "Subsection"], Cell["Sea el sistema de ecuaciones donde :", "Text"], Cell[BoxData[ FormBox[ RowBox[{"\[NoBreak]", GridBox[{ { RowBox[{ RowBox[{ RowBox[{ SubscriptBox["a", "11"], " ", SubscriptBox["x", "1"]}], "+", RowBox[{ SubscriptBox["a", "12"], " ", SubscriptBox["x", "2"]}], "+", RowBox[{ SubscriptBox["a", "13"], " ", SubscriptBox["x", "3"]}]}], "=", SubscriptBox["b", "1"]}]}, { RowBox[{ RowBox[{ RowBox[{ SubscriptBox["a", "21"], " ", SubscriptBox["x", "1"]}], "+", RowBox[{ SubscriptBox["a", "22"], " ", SubscriptBox["x", "2"]}], "+", RowBox[{ SubscriptBox["a", "23"], " ", SubscriptBox["x", "3"]}]}], "=", SubscriptBox["b", "2"]}]}, { RowBox[{ RowBox[{ RowBox[{ SubscriptBox["a", "31"], " ", SubscriptBox["x", "1"]}], "+", RowBox[{ SubscriptBox["a", "32"], " ", SubscriptBox["x", "2"]}], "+", RowBox[{ SubscriptBox["a", "33"], " ", SubscriptBox["x", "3"]}]}], "=", SubscriptBox["b", "3"]}]} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]"}], TraditionalForm]], "DisplayFormula", "Input", Evaluatable->False], Cell[TextData[{ "donde ", Cell[BoxData[ FormBox[ SubscriptBox["x", "i"], TraditionalForm]]], " son variables que se desean determinar. 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-> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]", ")"}], "\[LongEqual]", RowBox[{"(", "\[NoBreak]", GridBox[{ {"0"}, {"0"}, {"0"} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]", ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "Al menos tendremos como soluci\[OAcute]n", Cell[BoxData[ FormBox[ RowBox[{" ", RowBox[{ SubscriptBox["x", "i"], "=", "0"}]}], TraditionalForm]]], " {con ", StyleBox["i", FontSlant->"Italic"], " = 1, 2, 3} (frecuentemente llamada soluci\[OAcute]n trivial)" }], "Text"], Cell["\<\ Para resolver el sistema de ecuaciones hay distintos m\[EAcute]todos. Por \ ejemplo.:\ \>", "Text"], Cell["\<\ El m\[EAcute]todo de Gauss que consiste en transformar la matriz en otra cuya \ matriz de coeficientes sea escalonada. \ \>", "Text"], Cell[TextData[{ " El m\[EAcute]todo de la matriz inversa. Consiste en calcular la ecuaci\ \[OAcute]n A.X \[Equal]B es: ", Cell[BoxData[ FormBox[ SuperscriptBox["A", RowBox[{"-", "1"}]], TraditionalForm]]], ".A.X = X =", Cell[BoxData[ FormBox[ SuperscriptBox["A", RowBox[{"-", "1"}]], TraditionalForm]]], ".B " }], "Text"], Cell[CellGroupData[{ Cell["\<\ Expresa el siguiente sistema de ecuaciones en forma de matrices\ \>", "Subsubsection"], Cell[BoxData[{ FormBox[ RowBox[{ RowBox[{"y1", "=", RowBox[{"z1", "+", "z2"}]}], ";"}], TraditionalForm], "\n", FormBox[ RowBox[{ RowBox[{"y2", "=", RowBox[{ RowBox[{"2", " ", "z1"}], "-", "z2"}]}], ";"}], TraditionalForm], "\n", FormBox[ RowBox[{ RowBox[{"y3", "=", RowBox[{"z1", "+", RowBox[{"3", " ", "z2"}]}]}], ";"}], TraditionalForm]}], "DisplayFormula"], Cell[CellGroupData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"(", "\[NoBreak]", GridBox[{ {"y1"}, {"y2"}, {"y3"} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]", ")"}], "\[LongEqual]", RowBox[{ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "2"}, {"2", RowBox[{"-", "1"}]}, {"1", "3"} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]", ")"}], ".", RowBox[{"(", "\[NoBreak]", GridBox[{ {"z1"}, {"z2"} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]", ")"}]}]}], ";"}], TraditionalForm]], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", "\[NoBreak]", GridBox[{ {"y1"}, {"y2"}, {"y3"} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]", ")"}], "\[LongEqual]", RowBox[{"(", "\[NoBreak]", GridBox[{ { RowBox[{"z1", "+", RowBox[{"2", " ", "z2"}]}]}, { RowBox[{ RowBox[{"2", " ", "z1"}], "-", "z2"}]}, { RowBox[{"z1", "+", RowBox[{"3", " ", "z2"}]}]} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]", ")"}]}], TraditionalForm]], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Resuelva el siguiente sistema: x-3y+5 \ z\[Equal]1,2x-7y+2z\[Equal]3,5x-11y+9z\[Equal]7 con ", StyleBox["Mathematica", FontSlant->"Italic"], "." }], "Subsubsection"], Cell[TextData[{ "Las funciones de ", StyleBox["Mathematica:", FontSlant->"Italic"], " ", Cell[BoxData[ ButtonBox["Solve", BaseStyle->"Link", ButtonData->"paclet:ref/Solve"]]], " ", StyleBox["y ", FontSlant->"Italic"], " ", Cell[BoxData[ ButtonBox["Reduce", BaseStyle->"Link", ButtonData->"paclet:ref/Reduce"]]], " permiten resolver directamente el sistema cuando se define \ explicitamente. La funci\[OAcute]n ", Cell[BoxData[ ButtonBox["LinearSolve", BaseStyle->"Link", ButtonData->"paclet:ref/LinearSolve"]]], " lo resuelve cuando este se expresa en notaci\[OAcute]n matricial, esto es \ en la forma: A X = B." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Solve", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ RowBox[{"x", "-", RowBox[{"3", "y"}], "+", RowBox[{"5", " ", "z"}]}], "\[Equal]", "1"}], ",", RowBox[{ RowBox[{ RowBox[{"2", "x"}], "-", RowBox[{"7", "y"}], "+", RowBox[{"2", "z"}]}], "\[Equal]", "3"}], ",", RowBox[{ RowBox[{ RowBox[{"5", "x"}], "-", RowBox[{"11", "y"}], "+", RowBox[{"9", "z"}]}], "\[Equal]", "7"}]}], "}"}], ",", RowBox[{"{", RowBox[{"x", ",", "y", ",", "z"}], "}"}]}], "]"}]], "Input"], Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{"{", RowBox[{ RowBox[{"x", "\[Rule]", FractionBox["13", "8"]}], ",", RowBox[{"y", "\[Rule]", "0"}], ",", RowBox[{"z", "\[Rule]", RowBox[{"-", FractionBox["1", "8"]}]}]}], "}"}], "}"}], TraditionalForm]], "Output"] }, 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Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"A", ".", "X"}], "\[Equal]", "B"}]], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", "\[NoBreak]", GridBox[{ { RowBox[{"x", "+", "y", "-", "z"}]}, { RowBox[{"x", "-", "y", "+", "z"}]}, { RowBox[{ RowBox[{"-", "x"}], "+", "y", "+", "z"}]} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]", ")"}], "\[LongEqual]", RowBox[{"(", "\[NoBreak]", GridBox[{ {"1"}, {"1"}, {"1"} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { 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"Input"], Cell[BoxData[ FormBox["2", TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "Cuando rango[A] = Rango[ ", Cell[BoxData[ FormBox[ OverscriptBox["A", "_"], TraditionalForm]]], " ] el sistema es compatible (Teorema de \ Rouch\[EAcute]-Fr\[ODoubleDot]benius). Como Rango[ ", Cell[BoxData[ FormBox[ OverscriptBox["A", "_"], TraditionalForm]]], " ] = 2, indica que tenemos dos ecuaciones independiente. Por tanto \ tenemos 5 variables con 2 ecuaciones linealmente independientes. Es decir: el \ sitema tiene 2 variables (las que eligamos) que podemos expresarla en funci\ \[OAcute]n de las otras 3, que denominamos par\[AAcute]metros (a veces se \ dice quue el sistema tiene 3 grados de libertad). El sistema es \ indeterminado pues al existir mas inc\[OAcute]gnitas que ecuaciones \ lineamente independientes, y como consecuacias tiene muchas soluciones (de \ hecho infinitas). " }], "Text"], Cell["\<\ El resultado del sistema podemos expresarlo como sigue (utilizamos la matriz \ escalonada que obtuvimos para calcular el rango): \ \>", "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "0", "0", "7", RowBox[{"-", "9"}], "3"}, {"0", "1", RowBox[{"-", "1"}], "20", RowBox[{"-", "25"}], "8"}, {"0", "0", "0", "0", "0", "0"} }, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}}], "\[NoBreak]", ")"}], " ", "=", " ", RowBox[{ RowBox[{"(", "\[NoBreak]", GridBox[{ {"x", RowBox[{"0", " ", "y"}], RowBox[{"0", " ", "z"}], RowBox[{"7", "t"}], RowBox[{ RowBox[{"-", "9"}], " ", "u", " "}], "3"}, { RowBox[{"0", " ", "x"}], RowBox[{"1", " ", "y"}], RowBox[{ RowBox[{"-", 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variables a la \ derecha y el resto de los t\[EAcute]rminos pasan a la izquierda cambiados de \ signo): \ \>", "Text"], Cell[BoxData[{ FormBox[ RowBox[{ RowBox[{"x", "=", " ", RowBox[{ RowBox[{"9", " ", "u"}], " ", "-", RowBox[{"7", "t"}], "+", "3"}]}], ";"}], TraditionalForm], "\n", FormBox[ RowBox[{ RowBox[{"y", "=", " ", RowBox[{ RowBox[{"25", "u"}], "-", RowBox[{"20", "t"}], "+", "z", " ", "+", " ", "8"}]}], ";"}], TraditionalForm]}], "DisplayFormula"], Cell["\<\ Cuando el numero de variables coincide con el de ecuaciones linealmente \ independientes entonces el sistema tiene una \[UAcute]nica soluci\[OAcute]n, \ se dice que el sistema es determinado.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Dado el sistema, \[DownQuestion]Tiene soluci\[OAcute]n?\[DownQuestion]Cual \ es?", "\n\n", Cell[BoxData[{ FormBox[Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"x", "+", "y", "+", RowBox[{"2", " ", "z"}], "+", RowBox[{"3", " ", "t"}], "+"}], 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estudiar la compatibilidad del sistema necesitamos determinar el rango \ de A1 (la matriz formada por las 3 primeras columnas de A1m) y el de A1m.\ \>", "Text"], Cell[TextData[{ "El rango de A1 depende de si valor de ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"-", SuperscriptBox["a", "2"]}], "-", "a", " ", "+", "6"}], "=", "0"}], TraditionalForm]]], " o toma otro valor. 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En otro caso su rango ser\[AAcute] 3 y por tanto el sistema ser\ \[AAcute] determinado (esto es: n\[UAcute]mero de incognitas igual al de \ ecuaciones linealmente independientes). " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ FormBox[ RowBox[{"Reduce", "[", RowBox[{ RowBox[{ RowBox[{ FractionBox[ RowBox[{"5", " ", "a"}], "3"], "+", "9"}], " ", "\[Equal]", "0"}], ",", " ", "a"}], "]"}], TraditionalForm]], "Input"], Cell[BoxData[ FormBox[ RowBox[{"a", "\[LongEqual]", RowBox[{"-", FractionBox["27", "5"]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "En conclusi\[OAcute]n : Si a = ", Cell[BoxData[ FormBox[ RowBox[{"-", FractionBox["27", "5"]}], TraditionalForm]], CellChangeTimes->{3.5599967930564013`*^9, 3.559997673590066*^9}], " tendremos un sistema de dos ecuaciones: {x + y +z = 0, -3 y+5z = 0} cuya \ soluci\[OAcute]n ser\[AAcute] (dejando x como variable independiente):" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Solve", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ RowBox[{"x", " ", "+", " ", "y", " ", "+", "z"}], " ", "==", " ", "0"}], ",", " ", RowBox[{ RowBox[{ RowBox[{ RowBox[{"-", "3"}], " ", "y"}], "+", 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