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Por \ ejemplo, \[DownQuestion]c\[OAcute]mo podemos entresacar la columna 3 de la \ matriz ", StyleBox["b5", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], "?" }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Traspuesta de una matriz", FontColor->RGBColor[0, 1, 0]]], "Subsection", CellDingbat->"\[FilledSmallSquare]"], Cell[BoxData[ \(\(b4 = {{1, 0, 2, \(-1\)}, {2, 1, 3, 0}};\)\)], "Input", CellLabel->"In[23]:=", CellMargins->{{Inherited, 30}, {Inherited, Inherited}}, FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[b4]\)], "Input", CellLabel->"In[24]:=", CellMargins->{{Inherited, 30}, {Inherited, Inherited}}, FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "0", "2", \(-1\)}, {"2", "1", "3", "0"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[24]//MatrixForm="] }, Open ]], Cell[BoxData[ \(\(tb4 = 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MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[6]//MatrixForm="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Producto de matrices", FontColor->RGBColor[0, 1, 0]]], "Subsection", CellDingbat->"\[FilledSmallSquare]"], Cell[TextData[{ StyleBox["Cuidado", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " con el producto de matrices, que se representa con un punto (", StyleBox[".", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ")" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[a . b]\)], "Input", CellLabel->"In[7]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"4", "0", "1"}, {"0", "0", "0"}, {"4", "0", "1"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[7]//MatrixForm="] }, Open ]], Cell[TextData[{ "Si dos matrices ", StyleBox["a1", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " y ", StyleBox["b1", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " tienen las mismas dimensiones e indicamos ", StyleBox["a1*b1", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " o bien ", StyleBox["a1 b1 ", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " nos devolver\[AAcute] otra matriz que no tiene nada que ver con el \ producto de estas dos matrices. ", StyleBox["a1*b1", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " es una operaci\[OAcute]n que puede realizarse entre listas de la misma \ dimensi\[OAcute]n; el resultado es otra lista de la misma dimensi\[OAcute]n \ que obtenemos multiplicando elemento a elemento. Por ejemplo:" }], "Text"], Cell[BoxData[ \(a1 = {{1, 0, 1}, {0, 1, 2}}; b1 = {{1, 1, 1}, {3, \(-1\), 0}};\)], "Input", CellLabel->"In[8]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(a1*b1\)], "Input", CellLabel->"In[9]:=", FontSize->14], Cell[BoxData[ \({{1, 0, 1}, {0, \(-1\), 0}}\)], "Output", CellLabel->"Out[9]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(a1*b1 \[Equal] a1\ b1\)], "Input", CellLabel->"In[10]:=", FontSize->14], Cell[BoxData[ \(True\)], "Output", CellLabel->"Out[10]="] }, Open ]], Cell[TextData[{ "Y sin embargo, no es posible realizar el producto de estas dos matrices, \ como nos lo recuerda ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(a1 . b1\)], "Input", CellLabel->"In[11]:=", FontSize->14], Cell[BoxData[ RowBox[{\(Dot::"dotsh"\), \(\(:\)\(\ \)\), "\\\\"RefGuideLinkText\ \\\", ButtonFrame->None, ButtonData:>\\\"Dot::dotsh\\\"]\\)\"\>"}]], "Message",\ CellLabel->"From In[11]:="], Cell[BoxData[ \({{1, 0, 1}, {0, 1, 2}} . {{1, 1, 1}, {3, \(-1\), 0}}\)], "Output", CellLabel->"Out[11]="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Potencias naturales de matrices cuadradas", FontColor->RGBColor[0, 1, 0]]], "Subsection", CellDingbat->"\[FilledSmallSquare]"], Cell[BoxData[ \(\(MatrixPower[a, 8];\)\)], "Input", CellLabel->"In[12]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[%]\)], "Input", CellLabel->"In[13]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"128", "0", "128"}, {"0", "0", "0"}, {"128", "0", "128"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[13]//MatrixForm="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Potencias enteras de matrices regulares", FontColor->RGBColor[0, 1, 0]]], "Subsection", CellDingbat->"\[FilledSmallSquare]"], Cell[TextData[{ "Si una matriz cuadrada es regular, es decir, admite inversa, \ tambi\[EAcute]n podemos calcular las potencias enteras de esa matriz. Si la \ matriz no es regular,", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " nos lo har\[AAcute] saber:" }], "Text"], Cell[BoxData[ \(\(MatrixPower[b, \(-4\)];\)\)], "Input", CellLabel->"In[14]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[%]\)], "Input", CellLabel->"In[15]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {\(-\(15\/4\)\), \(9\/2\), \(-1\)}, {\(-\(19\/4\)\), \(11\/2\), \(-1\)}, {\(43\/4\), \(-\(51\/4\)\), \(11\/4\)} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[15]//MatrixForm="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(MatrixPower[a, \(-2\)];\)\)], "Input", CellLabel->"In[16]:=", FontSize->14], Cell[BoxData[ RowBox[{\(Inverse::"sing"\), \(\(:\)\(\ \)\), "\\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"General::sing\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[16]:="] }, Open ]], Cell[BoxData[ \(Clear["\", $Line]\)], "Input", CellLabel->"In[17]:=", FontSize->14] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["MATRICES INVERTIBLES. 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es decir, la suma \ de los elementos de su diagonal principal:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Tr[a]\)], "Input", CellLabel->"In[4]:=", FontSize->14], Cell[BoxData[ \(15\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Sum[a[\([i, i]\)], {i, \(Dimensions[a]\)[\([1]\)]}] == Tr[a]\)], "Input",\ CellLabel->"In[5]:=", FontSize->14], Cell[BoxData[ \(True\)], "Output", CellLabel->"Out[5]="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["b.-", FontColor->RGBColor[1, 0, 0]], " ", StyleBox["Hallar la inversa de A.", FontColor->RGBColor[0, 0, 1]] }], "Subsubsection", CellDingbat->None, FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(ia = Inverse[a]; MatrixForm[ia]\)], "Input", CellLabel->"In[6]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {\(-\(21\/22\)\), \(2\/11\), \(-\(4\/11\)\), \(-\(7\/22\)\), \(63\ \/22\)}, {\(27\/22\), \(-\(1\/11\)\), \(2\/11\), \(9\/22\), \ \(-\(59\/22\)\)}, {\(49\/11\), \(-\(13\/11\)\), \(4\/11\), \(20\/11\), \(-\(103\/11\ \)\)}, {\(16\/11\), \(-\(2\/11\)\), \(4\/11\), \(9\/11\), \ \(-\(48\/11\)\)}, {\(-\(34\/11\)\), \(7\/11\), \(-\(3\/11\)\), \(-\(15\/11\)\), \ \(80\/11\)} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[6]//MatrixForm="] }, Open ]], Cell["Podemos comprobar el resultado:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(a . ia \[Equal] IdentityMatrix[5]\)], "Input", CellLabel->"In[7]:=", FontSize->14], Cell[BoxData[ \(True\)], "Output", CellLabel->"Out[7]="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["c.-", FontColor->RGBColor[1, 0, 0]], " ", StyleBox["Hallar el menor complementario", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ RowBox[{" ", StyleBox[ SubscriptBox[ StyleBox["M", FontSlant->"Plain", FontColor->RGBColor[0, 0, 1]], "43"], FontColor->RGBColor[0, 0, 1]]}], TraditionalForm]]], " ", StyleBox["del elemento ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ StyleBox[ SubscriptBox[ StyleBox["a", FontFamily->"Helvetica", FontSlant->"Plain"], "43"], FontColor->RGBColor[0, 0, 1]], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], StyleBox["= 2 ", FontColor->RGBColor[0, 0, 1]], StyleBox["de A.", FontColor->RGBColor[0, 0, 1]] }], "Subsubsection", CellDingbat->None, FontSize->14], Cell[TextData[{ "Primero entresacamos la matriz que resulta de suprimir en ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " la fila 4 y la columna 3:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(a43 = a[\([{1, 2, 3, 5}, {1, 2, 4, 5}]\)]; MatrixForm[a43]\)], "Input", CellLabel->"In[8]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "3", "0", "2"}, {"2", "4", "1", "0"}, {"0", "2", "4", "7"}, {"1", "1", "1", "2"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[8]//MatrixForm="] }, Open ]], Cell["Y ahora calculamos su determinante:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(m43 = Det[a43]\)], "Input", CellLabel->"In[9]:=", FontSize->14], Cell[BoxData[ \(\(-40\)\)], "Output", CellLabel->"Out[9]="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["d.-", FontColor->RGBColor[1, 0, 0]], " ", StyleBox["Calcular la inversa de la matriz B siguiente, si es que existe:\n\ \n\t\t\t\t\t\tB=", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ StyleBox[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {\(-1\), "2", "0"}, {"2", "3", "1"}, {"3", "1", "1"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]], FontFamily->"Helvetica", FontColor->RGBColor[0, 0, 1]]], FontFamily->"Helvetica"] }], "Subsubsection", CellDingbat->None, FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(b = {{\(-1\), 2, 0}, {2, 3, 1}, {3, 1, 1}}; MatrixForm[b]\)], "Input", CellLabel->"In[10]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {\(-1\), "2", "0"}, {"2", "3", "1"}, {"3", "1", "1"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[10]//MatrixForm="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Inverse[b]\)], "Input", CellLabel->"In[11]:=", FontSize->14], Cell[BoxData[ RowBox[{\(Inverse::"sing"\), \(\(:\)\(\ \)\), "\\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"General::sing\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[11]:="], Cell[BoxData[ \(Inverse[{{\(-1\), 2, 0}, {2, 3, 1}, {3, 1, 1}}]\)], "Output", CellLabel->"Out[11]="] }, Open ]], Cell[TextData[{ "Luego ", StyleBox["B", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " es una matriz singular y, por tanto, no admite inversa, como podemos \ comprobar al calcular su determinante:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Det[b]\)], "Input", CellLabel->"In[12]:=", FontSize->14], Cell[BoxData[ \(0\)], "Output", CellLabel->"Out[12]="] }, Open ]], Cell[BoxData[ \(Clear["\", $Line]\)], "Input", CellLabel->"In[13]:=", FontSize->14] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["MATRICES EQUIVALENTES POR FILAS. EJERCICIO 2.2", FontColor->RGBColor[0, 0, 1]]], "SectionFirst", FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[TextData[{ StyleBox["Dada la matriz \n\t\t\t\t\t", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ StyleBox[ RowBox[{"A", "=", TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "3", "2", "5", "1", "1"}, {"3", "1", \(-2\), \(-1\), "0", \(-5\)}, {"4", "3", \(-1\), "2", "1", \(-5\)}, {"2", "5", "3", "8", \(-1\), "1"}, {\(-1\), "0", "1", "1", "1", "0"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]}], FontFamily->"Helvetica", FontColor->RGBColor[0, 0, 1]]], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14] }], "Subsubsection", CellDingbat->"\[FilledSmallSquare]", FontSize->14, FontWeight->"Bold"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["a.-", FontColor->RGBColor[1, 0, 0]], " ", StyleBox["Calcular ", FontColor->RGBColor[0, 0, 1]], StyleBox["el rango de A.", FontColor->RGBColor[0, 0, 1]], "\t" }], "Subsubsection", CellDingbat->None, FontSize->14], Cell[TextData[{ "Para calcular el rango de una matriz cuyas elementos no dependen de ning\ \[UAcute]n par\[AAcute]metro, ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], ", dispone de la funci\[OAcute]n ", StyleBox["RowReduce", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", que nos devolver\[AAcute] una matriz escalonada (", StyleBox["E", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ") equivalente a la matriz ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " con las entradas principales todas iguales a 1. Adem\[AAcute]s las \ columnas en las que se encuentran las entradas principales tienen ceros \ encima y debajo de las entradas principales.\n\nEsta funci\[OAcute]n \ solamente se puede utilizar con matrices cuyos elementos no dependan de par\ \[AAcute]metros." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(a = {{1, 3, 2, 5, 1, 1}, {3, 1, \(-2\), \(-1\), 0, \(-5\)}, {4, 3, \(-1\), 2, 1, \(-5\)}, {2, 5, 3, 8, \(-1\), 1}, {\(-1\), 0, 1, 1, 1, 0}}\)], "Input", CellLabel->"In[1]:=", FontSize->14], Cell[BoxData[ \({{1, 3, 2, 5, 1, 1}, {3, 1, \(-2\), \(-1\), 0, \(-5\)}, {4, 3, \(-1\), 2, 1, \(-5\)}, {2, 5, 3, 8, \(-1\), 1}, {\(-1\), 0, 1, 1, 1, 0}}\)], "Output", CellLabel->"Out[1]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(e = RowReduce[a]\)], "Input", CellLabel->"In[2]:=", FontSize->14], Cell[BoxData[ \({{1, 0, \(-1\), \(-1\), 0, 0}, {0, 1, 1, 2, 0, 0}, {0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, 0}}\)], "Output", CellLabel->"Out[2]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[e]\)], "Input", CellLabel->"In[3]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "0", \(-1\), \(-1\), "0", "0"}, {"0", "1", "1", "2", "0", "0"}, {"0", "0", "0", "0", "1", "0"}, {"0", "0", "0", "0", "0", "1"}, {"0", "0", "0", "0", "0", "0"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[3]//MatrixForm="] }, Open ]], Cell[TextData[{ "Podemos concluir que como ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " y ", StyleBox["E", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " son equivalentes tienen el mismo rango. El rango de ", StyleBox["E", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", al ser una matriz escalonada, es muy f\[AAcute]cil de hallar; basta con \ contar el n\[UAcute]mero de filas no nulas de la matriz ", StyleBox["E", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", por lo tanto:" }], "Text"], Cell[TextData[StyleBox["r(A)=r(E)=n\[UAcute]mero de filas no nulas de E=4", FontColor->RGBColor[0, 0, 1]]], "Text", TextAlignment->Center, FontSize->14, FontWeight->"Bold"], Cell[TextData[{ "La funci\[OAcute]n ", StyleBox["RowReduce", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " nos permite calcular el rango de una matriz pero adem\[AAcute]s nos \ proporciona m\[AAcute]s informaci\[OAcute]n que puede ser muy \[UAcute]til en \ el tema de espacios vectoriales. Si solamente queremos calcular el rango de \ una matriz sin par\[AAcute]metros podemos tambi\[EAcute]n utilizar la funci\ \[OAcute]n ", StyleBox["MatrixRank", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], StyleBox[".", FontFamily->"Courier"] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixRank[a]\)], "Input", CellLabel->"In[4]:=", FontSize->14], Cell[BoxData[ \(4\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixRank[a] == MatrixRank[RowReduce[a]]\)], "Input", CellLabel->"In[5]:=", FontSize->14], Cell[BoxData[ \(True\)], "Output", CellLabel->"Out[5]="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Las funciones ", FontColor->RGBColor[0, 1, 0]], StyleBox["RowReduce", FontFamily->"Courier", FontColor->RGBColor[0, 1, 0]], StyleBox[" y ", FontColor->RGBColor[0, 1, 0]], StyleBox["MatrixRank", FontFamily->"Courier", FontColor->RGBColor[0, 1, 0]], StyleBox[" y matrices con par\[AAcute]meros.", FontColor->RGBColor[0, 1, 0]] }], "Subsection", CellDingbat->"\[FilledSmallSquare]"], Cell[TextData[{ "Vamos a trabajar con la matriz:\n\t\t\t\t\t\t\t\tM=", Cell[BoxData[ FormBox[ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ { StyleBox["1", FontFamily->"Times New Roman"], StyleBox["1", FontFamily->"Times New Roman"]}, { StyleBox["0", FontFamily->"Times New Roman"], StyleBox["b", FontFamily->"Times New Roman", FontSlant->"Plain"]} }], "\[NegativeThinSpace]", ")"}], TraditionalForm]]], "\nEl rango de esta matriz M depende obviamente del par\[AAcute]metro ", StyleBox["b", FontFamily->"Times New Roman"], ":\n\t\t\t\t\t\t\t\t", Cell[BoxData[ FormBox[ RowBox[{\(rg(M)\), "=", TagBox[ RowBox[{ TagBox[ StyleBox[ RowBox[{"{", StyleBox[GridBox[{ {\(2\ \ \ \ \ \ \ \ si\ \ \ b \[NotEqual] 0\)}, {\(1\ \ \ \ \ \ \ \ \ si\ \ \ \ b = 0\)} }], ShowAutoStyles->True]}], ShowAutoStyles->False], (#&)], StyleBox[" ", ShowAutoStyles->True]}], (#&)]}], TraditionalForm]]], "\nSin embargo, si aplicamos las funciones ", StyleBox["MatrixRank", FontFamily->"Courier", FontWeight->"Bold"], " o ", StyleBox["Reduce", FontFamily->"Courier", FontWeight->"Bold"], ", ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " nos contestar\[AAcute] que el rango es 2 en el primer caso y en el \ segundo caso nos devolver\[AAcute] la matriz unidad de orden 2: " }], "Text"], Cell[BoxData[ \(\(m = {{1, 1}, {0, b}};\)\)], "Input", CellLabel->"In[6]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixRank[m]\)], "Input", CellLabel->"In[7]:=", FontSize->14], Cell[BoxData[ \(2\)], "Output", CellLabel->"Out[7]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(RowReduce[m]\)], "Input", CellLabel->"In[8]:=", FontSize->14], Cell[BoxData[ \({{1, 0}, {0, 1}}\)], "Output", CellLabel->"Out[8]="] }, Open ]], Cell[TextData[{ "Por lo tanto, no debemos aplicar las funciones ", StyleBox["MatrixRank", FontFamily->"Courier", FontWeight->"Bold"], " o ", StyleBox["Reduce", FontFamily->"Courier", FontWeight->"Bold"], " para calcular el rango de una matriz cuyos elementos dependen de par\ \[AAcute]metros. " }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Espacios vectoriales y matrices", FontColor->RGBColor[0, 1, 0]]], "Subsection", CellDingbat->"\[FilledSmallSquare]"], Cell["\", "Text"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["b.-", FontColor->RGBColor[1, 0, 0]], " ", StyleBox["Hallar una base del subespacio vectorial S engendrado por los \ vectores:\n", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["u", FontSlant->"Plain"], "TraditionalForm"], "_"], "1"], TraditionalForm]], FontSize->12, FontColor->RGBColor[0, 0, 1]], StyleBox[" = (1,3,2,5,1,1) ; ", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["u", FontSlant->"Plain"], "TraditionalForm"], "_"], "2"], TraditionalForm]], FontSize->12, FontColor->RGBColor[0, 0, 1]], StyleBox[" = (2,5,3,8,-1,1) ; ", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["u", FontSlant->"Plain"], "TraditionalForm"], "_"], "3"], TraditionalForm]], FontSize->12, FontColor->RGBColor[0, 0, 1]], StyleBox[" = (3,1,-2,-1,0,-5) ; ", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["u", FontSlant->"Plain"], "TraditionalForm"], "_"], "4"], TraditionalForm]], FontSize->12, FontColor->RGBColor[0, 0, 1]], StyleBox["= (4,3,-1,2,1,-5) ; ", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["u", FontSlant->"Plain"], "TraditionalForm"], "_"], "5"], TraditionalForm]], FontSize->12, FontColor->RGBColor[0, 0, 1]], StyleBox["= (-1,0,1,1,1,0)", FontSize->12, FontColor->RGBColor[0, 0, 1]] }], "Subsubsection", CellDingbat->None, FontSize->14], Cell[TextData[{ "Como estos vectores son las filas de la matriz ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", vamos a utilizar un argumento que ya hemos empleado en las clases de \ problemas. No vamos a repetir el razonamiento, aunque se debe hacer, pero \ sabemos por lo expuesto en las clases pr\[AAcute]cticas que una base de ", StyleBox["S", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " estar\[AAcute] formada por los vectores fila no nulos de la matriz \ equivalente ", StyleBox["E", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ". \n\nCon lo cual podemos concluir que:" }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ StyleBox[ SubscriptBox[ StyleBox["B", FontSlant->"Plain", FontColor->RGBColor[0, 0, 1]], StyleBox["S", FontSlant->"Plain"]], FontColor->RGBColor[0, 0, 1]], TraditionalForm]]], StyleBox["= {", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["v", FontSlant->"Plain"], "TraditionalForm"], "_"], "1"], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], StyleBox[" = (", FontColor->RGBColor[0, 0, 1]], "1,0,-1,-1,0,0", StyleBox[") ; ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["v", FontSlant->"Plain"], "TraditionalForm"], "_"], "2"], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], StyleBox[" = (", FontColor->RGBColor[0, 0, 1]], "0,1,1,2,0,0", StyleBox[") ; ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["v", FontSlant->"Plain"], "TraditionalForm"], "_"], "3"], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], StyleBox[" = (", FontColor->RGBColor[0, 0, 1]], "0,0,0,0,1,0", StyleBox[") ; ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["v", FontSlant->"Plain"], "TraditionalForm"], "_"], "4"], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], StyleBox["= (", FontColor->RGBColor[0, 0, 1]], "0,0,0,0,0,1", StyleBox[")}", FontColor->RGBColor[0, 0, 1]] }], "Text", TextAlignment->Center, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[TextData[{ "Cuidado con dar como base de ", StyleBox["S", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " la familia de los 4 primeros vectores fila de la matriz ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ". No son necesariamente linearmente independientes." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(a1 = Delete[a, 5]; MatrixForm[a1]\)], "Input", CellLabel->"In[9]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "3", "2", "5", "1", "1"}, {"3", "1", \(-2\), \(-1\), "0", \(-5\)}, {"4", "3", \(-1\), "2", "1", \(-5\)}, {"2", "5", "3", "8", \(-1\), "1"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[9]//MatrixForm="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(RowReduce[a1]\)], "Input", CellLabel->"In[10]:=", FontSize->14], Cell[BoxData[ \({{1, 0, \(-1\), \(-1\), 0, \(-2\)}, {0, 1, 1, 2, 0, 1}, {0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0}}\)], "Output", CellLabel->"Out[10]="] }, Open ]], Cell[TextData[{ "Luego efectivamente los cuatro primeros vectores fila de la matriz ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " son linealmente dependientes." }], "Text"], Cell[TextData[{ "Otra forma de demostrar con ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " que los cuatro primeros vectores fila de la matriz ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " no son linealmente independientes:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[ a[\([1]\)] x1 + a[\([2]\)] x2 + a[\([3]\)] x3 + a[\([4]\)] x4 \[Equal] 0, {x1, x2, x3, x4}]\)], "Input", CellLabel->"In[11]:=", FontSize->14], Cell[BoxData[ RowBox[{\(Solve::"svars"\), \(\(:\)\(\ \)\), "\\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Solve::svars\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[11]:="], Cell[BoxData[ \({{x1 \[Rule] \(-6\)\ x4, x2 \[Rule] \(-8\)\ x4, x3 \[Rule] 7\ x4}}\)], "Output", CellLabel->"Out[11]="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["c.-", FontColor->RGBColor[1, 0, 0]], " ", StyleBox["Ampliar la base ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ StyleBox[ SubscriptBox[ StyleBox["B", FontSlant->"Plain", FontColor->RGBColor[0, 0, 1]], StyleBox["S", FontSlant->"Plain"]], FontColor->RGBColor[0, 0, 1]], TraditionalForm]]], StyleBox["del subespacio vectorial S hallada en el apartado anterior hasta \ conseguir una base de ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ StyleBox[\(\[DoubleStruckCapitalR]\^6\), FontColor->RGBColor[0, 0, 1]], TraditionalForm]]], "." }], "Subsubsection", CellDingbat->None, FontSize->14], Cell["\", "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(b = Delete[e, 5]; 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", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["v", FontSlant->"Plain"], "TraditionalForm"], "_"], "2"], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], StyleBox["; ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["v", FontSlant->"Plain"], "TraditionalForm"], "_"], "3"], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], StyleBox["; ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["v", FontSlant->"Plain"], "TraditionalForm"], "_"], "4"], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], "; ", Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["e", FontSlant->"Plain"], "TraditionalForm"], "_"], "3"], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], "; ", Cell[BoxData[ FormBox[ SubscriptBox[ OverscriptBox[ FormBox[ StyleBox["e", FontSlant->"Plain"], "TraditionalForm"], "_"], "4"], TraditionalForm]], FontColor->RGBColor[0, 0, 1]], StyleBox["}", FontColor->RGBColor[0, 0, 1]] }], "Text", TextAlignment->Center, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(Clear["\", $Line]\)], "Input", CellLabel->"In[15]:=", FontSize->14] }, Closed]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["EJERCICIO 2.3 (Rango de una matriz cuyos elementos \ dependen de par\[AAcute]metros)", FontColor->RGBColor[0, 0, 1]]], "SectionFirst", FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Sea la matriz real \n\t\t\t\t\t\t", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ StyleBox[ RowBox[{"A", "=", TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"q", "0", "1", "q"}, {"1", "1", \(-p\), "q"}, {\(p + q\), "p", "0", \(2\ q\)} }], "\[NoBreak]", ")"}], (MatrixForm[ #]&)]}], FontFamily->"Helvetica", FontColor->RGBColor[0, 0, 1]]], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14], "\n\t\t\t\t\t\t\n", StyleBox["Hallar el rango de la matriz A seg\[UAcute]n los distintos \ valores de los par\[AAcute]metros reales p y q.", FontColor->RGBColor[0, 0, 1]] }], "Subsubsection", CellDingbat->None, FontSize->14, FontWeight->"Bold"], Cell[TextData[{ "Hemos dicho anteriormente que la funci\[OAcute]n ", StyleBox["RowReduce", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " se puede utilizar siempre que los elementos de la matriz no dependan de \ par\[AAcute]metros. Para ver que sucede cuando hay par\[AAcute]metros vamos a \ utilizar la funci\[OAcute]n ", StyleBox["RowReduce", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(a = {{q, 0, 1, q}, {1, 1, \(-p\), q}, {p + q, p, 0, 2 q}}; MatrixForm[a]\)], "Input", CellLabel->"In[1]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"q", "0", "1", "q"}, {"1", "1", \(-p\), "q"}, {\(p + q\), "p", "0", \(2\ q\)} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[1]//MatrixForm="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(RowReduce[a]\)], "Input", CellLabel->"In[2]:=", FontSize->14], Cell[BoxData[ \({{1, 0, 0, \(2 + p\)\/\(1 + p\)}, {0, 1, 0, \(\(-2\) - p + q\)\/\(1 + p\)}, {0, 0, 1, \(-\(q\/\(1 + p\)\)\)}}\)], "Output", CellLabel->"Out[2]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixRank[a]\)], "Input", CellLabel->"In[3]:=", FontSize->14], Cell[BoxData[ \(3\)], "Output", CellLabel->"Out[3]="] }, Open ]], Cell[TextData[{ "Razonando como en el ejercico anterior llegar\[IAcute]amos a la conclusi\ \[OAcute]n err\[OAcute]nea de que el rango de ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " es siempre 3, independiente de los valores de los par\[AAcute]metros \ reales ", StyleBox["p", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " y ", StyleBox["q", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", y sin embargo:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(a1 = a /. {p \[Rule] 1, q \[Rule] 1}; MatrixForm[a1]\)], "Input", CellLabel->"In[4]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "0", "1", "1"}, {"1", "1", \(-1\), "1"}, {"2", "1", "0", "2"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[4]//MatrixForm="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(RowReduce[a1]\)], "Input", CellLabel->"In[5]:=", FontSize->14], Cell[BoxData[ \({{1, 0, 1, 1}, {0, 1, \(-2\), 0}, {0, 0, 0, 0}}\)], "Output", CellLabel->"Out[5]="] }, Open ]], Cell[TextData[{ "Luego no podemos utilizar con esta matriz la funci\[OAcute]n que ", StyleBox["RowReduce ", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], "ni ", StyleBox["MatrixRank.", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " ", StyleBox["\[DownQuestion]Qu\[EAcute] hacemos?\n", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox["Podemos utilizar la misma t\[EAcute]cnica que la utilizada para \ resolver este tipo de ejercicios con l\[AAcute]piz y papel, buscando un menor \ de orden 2 no nulo y orlando el menor no nulo con la fila que no intervenga \ en ese menor.\nAntes de realizar este ejercicio de ese modo vamos a explicar \ otra t\[EAcute]cnica que resulta m\[AAcute]s sencilla de utilizar con ", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox["Mathematica", FontSize->14, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], StyleBox[". ", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]] }], "Text"], Cell[TextData[{ "Hemos resuelto este ejercicio siguiendo los pasos que \ realizar\[IAcute]amos al intentar resolver el ejercicio con l\[AAcute]piz y \ papel. Con ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], ", es m\[AAcute]s c\[OAcute]modo calcular todos los menores del mayor orden \ posible, es decir, de orden 3 en nuestro ejemplo. Para ello utilizaremos la \ funci\[OAcute]n ", StyleBox["Minors", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], "." }], "Text"], Cell[TextData[{ "De esta forma calculamos todos los menores de orden 3 de la matriz ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ". Obviamente al ser ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " una matriz de 3 filas y cuatro columnas, su m\[AAcute]ximo rango posible \ es el m\[IAcute]nimo entre su n\[UAcute]mero de filas y columnas, es decir: \ 3." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Minors[a, Min[Dimensions[a]]]\)], "Input", CellLabel->"In[6]:=", FontSize->14], Cell[BoxData[ \({{\(-q\) + p\^2\ q, q\^2 - p\ q\^2, \(-2\)\ q + p\ q + p\^2\ q + q\^2 - p\ q\^2, \(-2\)\ q + p\ q + p\^2\ q}}\)], "Output", CellLabel->"Out[6]="] }, Open ]], Cell[TextData[{ "Escribimos el conjunto de todos los menores de orden 3 de la matriz ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " como una lista:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(l = Flatten[%]\)], "Input", CellLabel->"In[7]:=", FontSize->14], Cell[BoxData[ \({\(-q\) + p\^2\ q, q\^2 - p\ q\^2, \(-2\)\ q + p\ q + p\^2\ q + q\^2 - p\ q\^2, \(-2\)\ q + p\ q + p\^2\ q}\)], "Output", CellLabel->"Out[7]="] }, Open ]], Cell["\", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(s = Solve[Table[l[\([i]\)] \[Equal] 0, {i, Length[l]}], {p, q}]\)], "Input",\ CellLabel->"In[8]:=", FontSize->14], Cell[BoxData[ RowBox[{\(Solve::"svars"\), \(\(:\)\(\ \)\), "\\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Solve::svars\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[8]:="], Cell[BoxData[ \({{p \[Rule] 1}, {q \[Rule] 0}, {q \[Rule] 0}}\)], "Output", CellLabel->"Out[8]="] }, Open ]], Cell[TextData[{ "Ahora estamos en condiciones de saber cuando el rango de la matriz ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " es 3 y cuando es menor que 3:" }], "Text"], Cell[TextData[Cell[BoxData[{ RowBox[{ RowBox[{ RowBox[{ StyleBox["\[SixPointedStar]", FontColor->RGBColor[1, 0, 0]], StyleBox[ RowBox[{" ", StyleBox[" ", FontFamily->"Times", FontWeight->"Bold"]}]], StyleBox["p", FontFamily->"Times", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]]}], 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base de ", StyleBox["S", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " encontrada en el ejercicio hasta conseguir una base de ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^5\)]], "." }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["CLASIFICACI\[CapitalOAcute]N Y RESOLUCI\ \[CapitalOAcute]N DE SISTEMAS DE ECUACIONES LINEALES", FontColor->RGBColor[0, 0, 1]]], "SectionFirst", FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[TextData[{ "Si queremos resolver una ecuaci\[OAcute]n o un sistema de ecuaciones no \ necesariamente lineales con ", StyleBox["Mathematica", FontSlant->"Italic"], ", disponemos de las funciones:\t\t", StyleBox["Solve\tNSolve\tLinearSolve\t\tReduce\t", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], "\n\nLas funciones\n\t\t", StyleBox["Solve\tNSolve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], "\nse utlizan para ecuaciones y sistemas de ecuaciones sin \ 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FontColor->RGBColor[0, 0, 1]], " para aprender a manejar correctamente estas funciones. " }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Funciones ", FontColor->RGBColor[0, 0, 1]], StyleBox["Solve", FontFamily->"Courier", FontColor->RGBColor[0, 0, 1]], StyleBox[" y ", FontColor->RGBColor[0, 0, 1]], StyleBox["LinearSolve", FontFamily->"Courier", FontColor->RGBColor[0, 0, 1]] }], "SectionFirst", FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[TextData[{ "Mediante el siguiente ejemplo vamos a intentar explicar las funciones ", StyleBox["Solve ", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], "y ", StyleBox["LinearSolve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " de ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], ". Aplicaremos ambas funciones a un mismo sistema de ecuaciones lineales.\n\ La funci\[OAcute]n ", StyleBox["Solve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " requiere como m\[IAcute]nimo dos argumentos. Para aplicar la funci\ \[OAcute]n ", StyleBox["Solve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " a la resoluci\[OAcute]n de un cierto sistema de ecuaciones lineales, \ necesitamos poner como primer argumento de la funci\[OAcute]n las distintas \ ecuaciones del sistema en la forma ", StyleBox["lhs", "TI"], " ", StyleBox["==", "MR"], " ", StyleBox["rhs", "TI"], ". Las distintas ecuaciones o bien se listan o bien aparecen separadas por \ &&. El segundo argumento es la inc\[OAcute]gnita, si solamente hay una inc\ \[OAcute]gnita o bien una lista con todas las inc\[OAcute]gnitas.\nVeamos un \ ejemplo:" }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Resolver, si es posible, el sistema: \n\t\t\t", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ RowBox[{ StyleBox["x", FontSlant->"Plain"], "+", "2"}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ StyleBox["y", FontSlant->"Plain"], "+", RowBox[{"3", StyleBox["z", FontSlant->"Plain"]}]}], TraditionalForm]]], " = 6\n\t\t\t", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["x", FontSlant->"Plain"], "+", "4"}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ StyleBox["y", FontSlant->"Plain"], "+", RowBox[{"9", StyleBox["z", FontSlant->"Plain"]}]}], TraditionalForm]]], " = 14\n\t\t\t ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["x", FontSlant->"Plain"], "+", "8"}], TraditionalForm]]], Cell[BoxData[ FormBox[ StyleBox["y", FontSlant->"Plain"], TraditionalForm]]], " = 10/7" }], "Subsubsection", CellDingbat->None, TextAlignment->AlignmentMarker, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[{x + 2 y + 3 z \[Equal] 6, x + 4 y + 9 z \[Equal] 14, x + 8 y \[Equal] 10/7}, {x, y, z}]\)], "Input", CellLabel->"In[1]:=", FontSize->14], Cell[BoxData[ \({{x \[Rule] 102\/49, y \[Rule] \(-\(4\/49\)\), z \[Rule] 200\/147}}\)], "Output", CellLabel->"Out[1]="] }, Open ]], Cell[TextData[{ "Como vemos nos presenta la soluci\[OAcute]n como reglas de asignaci\ \[OAcute]n en la forma ", Cell[BoxData[ \(x \[Rule] sol\)]], ". Si hay m\[AAcute]s de una inc\[OAcute]gnita, la soluci\[OAcute]n aparece \ como una lista de asignaciones para cada una de las variables. Y si el \ sistema tiene m\[AAcute]s de una soluci\[OAcute]n, nos presenta la lista de \ las soluciones" }], "Text"], Cell["\", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({x + 2 y + 3 z \[Equal] 6, x + 4 y + 9 z \[Equal] 14, x + 8 y \[Equal] 10/7} /. %[\([1]\)]\)], "Input", CellLabel->"In[2]:=", FontSize->14], Cell[BoxData[ \({True, True, True}\)], "Output", CellLabel->"Out[2]="] }, Open ]], Cell["Otra forma de escribir las euaciones:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[ x + 2 y + 3 z \[Equal] 6 && x + 4 y + 9 z \[Equal] 14 && x + 8 y \[Equal] 10/7, {x, y, z}]\)], "Input", CellLabel->"In[3]:=", FontSize->14], Cell[BoxData[ \({{x \[Rule] 102\/49, y \[Rule] \(-\(4\/49\)\), z \[Rule] 200\/147}}\)], "Output", CellLabel->"Out[3]="] }, Open ]], Cell["Comprobaci\[OAcute]n del resultado:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(x + 2 y + 3 z \[Equal] 6 && x + 4 y + 9 z \[Equal] 14 && x + 8 y \[Equal] 10/7 /. %[\([1]\)]\)], "Input", CellLabel->"In[4]:=", FontSize->14], Cell[BoxData[ \(True\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell[TextData[{ "Para entender un poco m\[AAcute]s sobre las operaciones l\[OAcute]gicas en \ ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], ", y teniendo en cuenta que la \[UAcute]nica soluci\[OAcute]n del sistema \ es la soluci\[OAcute]n que nos devuelve en", StyleBox[" Out[1]", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", preguntamos a ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " si al sustituir (", StyleBox["/.", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ") en las tres ecuaciones del sistema ", StyleBox["x", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " por ", StyleBox["-2", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", ", StyleBox["y", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " por ", StyleBox["4", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " y ", StyleBox["z", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " por ", StyleBox["0", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", la expresi\[OAcute]n resultante es verdadera (", StyleBox["True", FontFamily->"Courier", FontSize->14], ") o falsa (", StyleBox["False", FontFamily->"Courier", FontSize->14], "). Observemos que dependiendo de la forma en que escribimos el sistema la \ respuesta tiene un aspecto diferente:" }], "Text"], Cell[TextData[{ StyleBox["&&", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " es la funci\[OAcute]n l\[OAcute]gica ", StyleBox["Y", FontWeight->"Bold"], " (", StyleBox["\[And]", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], "). ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " eval\[UAcute]a en orden y en el momento que una de las expresiones es \ falsa nos delvolver\[AAcute] ", StyleBox["False", FontFamily->"Courier", FontSize->14], ". Si tosas las expresiones son ciertas nos devolver\[IAcute]a ", StyleBox["True", FontFamily->"Courier", FontSize->14], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(x + 2 y + 3 z \[Equal] 6 && x + 4 y + 9 z \[Equal] 14 && x + 8 y \[Equal] 10/7 /. {x \[Rule] \(-2\), y \[Rule] 4, z \[Rule] 0}\)], "Input", CellLabel->"In[5]:=", FontSize->14], Cell[BoxData[ \(False\)], "Output", CellLabel->"Out[5]="] }, Open ]], Cell[TextData[{ "Si queremos qu\[EAcute] ecuaciones no se satisafcen para los valores de \ las inc\[OAcute]gnitas que estamos probando, podemos escribir las ecuaciones \ del sistema como una lista. ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " nos devolver\[AAcute] una lista de ", StyleBox["True", FontFamily->"Courier", FontSize->14], " o ", StyleBox["False", FontFamily->"Courier", FontSize->14], ", dependiendo de si la ecuaci\[OAcute]n correspondiente se satisface o no: \ " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({x + 2 y + 3 z \[Equal] 6, x + 4 y + 9 z \[Equal] 14, x + 8 y \[Equal] 10/7} /. {x \[Rule] \(-2\), y \[Rule] 4, z \[Rule] 0}\)], "Input", CellLabel->"In[6]:=", FontSize->14], Cell[BoxData[ \({True, True, False}\)], "Output", CellLabel->"Out[6]="] }, Open ]], Cell[TextData[{ "Si ", StyleBox["m", FontFamily->"Courier", FontSize->14], " es la matriz de coeficientes del sistema, podemos escribir \ tambi\[EAcute]n: " }], "Text"], Cell[BoxData[ \(\(m = {{1, 2, 3}, {1, 4, 9}, {1, 8, 0}};\)\)], "Input", CellLabel->"In[7]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[m . {x, y, z} \[Equal] {6, 14, 10/7}, {x, y, z}]\)], "Input", CellLabel->"In[8]:=", FontSize->14], Cell[BoxData[ \({{x \[Rule] 102\/49, y \[Rule] \(-\(4\/49\)\), z \[Rule] 200\/147}}\)], "Output", CellLabel->"Out[8]="] }, Open ]], Cell["\", "Text"], Cell[TextData[{ "Los argumentos de la funci\[OAcute]n ", StyleBox["LinearSolve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " son distintos. Esta funci\[OAcute]n nos permite resolver una ecuaci\ \[OAcute]n del tipo ", StyleBox["A\[CenterDot]X=B", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", donde ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " es una matrix y ", StyleBox["B", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " es un vector o una matriz. Los argumentos de la funci\[OAcute]n ", StyleBox["LinearSolve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " son ", StyleBox["A", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " ", StyleBox["y", FontWeight->"Bold"], " ", StyleBox["B", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " en este orden, si estamos resolviendo la ecuaci\[OAcute]n ", StyleBox["A\[CenterDot]X=B", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ". Para ello escribiremos:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(LinearSolve[m, {6, 14, 10/7}]\)], "Input", CellLabel->"In[9]:=", FontSize->14], Cell[BoxData[ \({102\/49, \(-\(4\/49\)\), 200\/147}\)], "Output", CellLabel->"Out[9]="] }, Open ]], Cell[TextData[{ "Si ", StyleBox["B", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " es una matrix tambi\[EAcute]n podemos utilizar la funci\[OAcute]n ", StyleBox["LinearSolve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " para resolver la ecuaci\[OAcute]n ", StyleBox["A\[CenterDot]X=B", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ". La soluci\[OAcute]n ser\[AAcute] una matriz:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(LinearSolve[m, {{1, 1}, {0, 2}, {0, 1}}]\)], "Input", CellLabel->"In[10]:=", FontSize->14], Cell[BoxData[ \({{12\/7, 3\/7}, {\(-\(3\/14\)\), 1\/14}, {\(-\(2\/21\)\), 1\/7}}\)], "Output", CellLabel->"Out[10]="] }, Open ]], Cell[TextData[{ "Tambi\[EAcute]n podemos utilizar como \[UAcute]nico argumento de la funci\ \[OAcute]n ", StyleBox["LinearSolve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " la matriz ", StyleBox["m", FontFamily->"Courier", FontSize->14], " y luego aplicar el resultado a diferenes vectores o matrices de \ t\[EAcute]rminos independientes. por ejemplo:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(ls = LinearSolve[m]\)], "Input", CellLabel->"In[11]:=", FontSize->14], Cell[BoxData[ TagBox[\(LinearSolveFunction[{3, 3}, ""]\), False, Editable->False]], "Output", CellLabel->"Out[11]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(ls[{6, 14, 10/7}]\)], "Input", CellLabel->"In[12]:=", FontSize->14], Cell[BoxData[ \({102\/49, \(-\(4\/49\)\), 200\/147}\)], "Output", CellLabel->"Out[12]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(ls[{{1, 1}, {0, 2}, {0, 1}}]\)], "Input", CellLabel->"In[13]:=", FontSize->14], Cell[BoxData[ \({{12\/7, 3\/7}, {\(-\(3\/14\)\), 1\/14}, {\(-\(2\/21\)\), 1\/7}}\)], "Output", CellLabel->"Out[13]="] }, Open ]], Cell[TextData[{ "Si el sistema es indeterminando la funci\[OAcute]n ", StyleBox["LinearSolve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " solamente proporciona una de las soluciones, sin embargo la \ funci\[OAcute]n ", StyleBox["Solve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " proporciona todas las soluciones y adem\[AAcute]s nos permite identificar \ las inc\[OAcute]gnitas principales si estamos resolviendo un sistema de \ ecuaiones lineales." }], "Text"], Cell[TextData[{ "Veamos como act\[UAcute]a la funci\[OAcute]n ", StyleBox["LinearSolve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " ante un sistema de ecuaciones lineales compatible indeterminado:" }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Resolver, si es posible, el sistema: \n\t\t\t ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ RowBox[{ StyleBox["x", FontSlant->"Plain"], "+"}], TraditionalForm]]], Cell[BoxData[ FormBox[ StyleBox["y", FontSlant->"Plain"], TraditionalForm]]], " = 1\n\t\t\t", Cell[BoxData[ \(TraditionalForm\`-\)]], Cell[BoxData[ FormBox[ RowBox[{ StyleBox["y", FontSlant->"Plain"], "+", RowBox[{"2", StyleBox["z", FontSlant->"Plain"]}]}], TraditionalForm]]], " = -1\n\t\t ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["x", FontSlant->"Plain"], "+", "2"}], TraditionalForm]]], Cell[BoxData[ FormBox[ StyleBox["z", FontSlant->"Plain"], TraditionalForm]]], " = 0" }], "Subsubsection", CellDingbat->None, TextAlignment->AlignmentMarker, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell[BoxData[ \(LinearSolve[{{1, 1, 0}, {0, \(-1\), 2}, {1, 0, 2}}, {1, \(-1\), 0}]\)], "Input", CellLabel->"In[14]:=", FontSize->14], Cell[BoxData[ \({0, 1, 0}\)], "Output", CellLabel->"Out[14]="] }, Open ]], Cell[TextData[{ "Si utilizamos la funci\[OAcute]n ", StyleBox["Solve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " ante un sistema de ecuaciones lineales compatible indeterminado:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[{x + y \[Equal] 1, \(-y\) + 2 z \[Equal] \(-1\), x + 2 z \[Equal] 0}, {x, y, z}]\)], "Input", CellLabel->"In[15]:=", FontSize->14], Cell[BoxData[ RowBox[{\(Solve::"svars"\), \(\(:\)\(\ \)\), "\\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Solve::svars\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[15]:="], Cell[BoxData[ \({{x \[Rule] \(-2\)\ z, y \[Rule] 1 + 2\ z}}\)], "Output", CellLabel->"Out[15]="] }, Open ]], Cell[TextData[{ StyleBox["Importante:", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], StyleBox[" ", FontSize->14, FontColor->RGBColor[1, 0, 1]], "Las inc\[OAcute]gnitas que aparecen a la izquierda de las flechas son las \ inc\[OAcute]gnitas principales, el resto de las inc\[OAcute]gnitas son inc\ \[OAcute]gnitas libres. Esto nos permite en este caso conocer el rango de la \ matriz de coeficientes del sistema y tambi\[EAcute]n el rango de la matriz \ ampliada (pues el sistema es compatible)." }], "Text"], Cell[TextData[{ "Si empleamos la funci\[OAcute]n ", StyleBox["LinearSolve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " ante un sistema de ecuaciones lineales para saber si es compatible \ determinado o indeterminado, podemos emplear la funci\[OAcute]n ", StyleBox["NullSpace", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " que nos permita saber si el sistema homog\[EAcute]neo asociado es \ compatible determinado (en este caso la respuesta ser\[AAcute] ", StyleBox["{ }", FontFamily->"Courier", FontWeight->"Bold"], ") o si es compatible indeterminado:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(NullSpace[{{1, 1, 0}, {0, \(-1\), 2}, {1, 0, 2}}]\)], "Input", CellLabel->"In[16]:=", FontSize->14], Cell[BoxData[ \({{\(-2\), 2, 1}}\)], "Output", CellLabel->"Out[16]="] }, Open ]], Cell["Sin embargo:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(NullSpace[m]\)], "Input", CellLabel->"In[17]:=", FontSize->14], Cell[BoxData[ \({}\)], "Output", CellLabel->"Out[17]="] }, Open ]], Cell[BoxData[ \(Clear["\", $Line]\)], "Input", CellLabel->"In[18]:=", FontSize->14] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Funciones ", FontColor->RGBColor[0, 0, 1]], StyleBox["Solve", FontFamily->"Courier", FontColor->RGBColor[0, 0, 1]], StyleBox[" y ", FontColor->RGBColor[0, 0, 1]], StyleBox["Reduce", FontFamily->"Courier", FontColor->RGBColor[0, 0, 1]] }], "SectionFirst", FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[TextData[{ "Las funciones ", StyleBox["Solve ", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], "y ", StyleBox["Reduce", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " de ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " se utilizan para resolver ecuaciones o sistemas de ecuaciones no \ necesariamente lineales.\nLa funci\[OAcute]n ", StyleBox["Solve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " se utiliza con ecuaciones o sistemas de ecuaciones sin \ par\[AAcute]metros. No es aconsejable emplear la funci\[OAcute]n ", StyleBox["Solve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " para resolver sistemas de ecuaciones o ecuaciones con par\[AAcute]metros; \ en este caso utilizaremos la funci\[OAcute]n ", StyleBox["Reduce", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ".\nEn los ejemplos siguientes se explican con m\[AAcute]s detalle ambas \ funciones." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["EJERCICIO 2.5 (Un S.E.L. sin par\[AAcute]metros)", FontColor->RGBColor[0, 0, 1]]], "SectionFirst", FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Clasificar y resolver, si es posible, el sistema: \n\t\t\t", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "1"], "+", "2"}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "2"], "-", SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "3"], "+", SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "4"]}], TraditionalForm]]], "=1\n\t\t\t2", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "1"], "+"}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "2"], "+", RowBox[{"2", SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "3"]}], "-", SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "4"]}], TraditionalForm]]], "= 2\n\t\t\t", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "1"], "+", "5"}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "2"], "-", RowBox[{"5", SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "3"]}], "+", RowBox[{"4", SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "4"]}]}], TraditionalForm]]], "=1" }], "Subsubsection", CellDingbat->None, TextAlignment->AlignmentMarker, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[{x1 + 2 x2 - x3 + x4 \[Equal] 1, 2 x1 + x2 + 2 x3 - x4 \[Equal] 2, x1 + 5 x2 - 5 x3 + 4 x4 \[Equal] 1}, {x1, x2, x3, x4}]\)], "Input",\ CellLabel->"In[1]:=", FontSize->14], Cell[BoxData[ RowBox[{\(Solve::"svars"\), \(\(:\)\(\ \)\), "\\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Solve::svars\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[1]:="], Cell[BoxData[ \({{x1 \[Rule] 1 - \(5\ x3\)\/3 + x4, x2 \[Rule] \(4\ x3\)\/3 - x4}}\)], "Output", CellLabel->"Out[1]="] }, Open ]], Cell[TextData[{ "Si ", StyleBox["m", FontFamily->"Courier", FontSize->14], " es la matriz de coeficientes del sistema, podemos escribir \ tambi\[EAcute]n: " }], "Text"], Cell[BoxData[ \(\(m = {{1, 2, \(-1\), 1}, {2, 1, 2, \(-1\)}, {1, 5, \(-5\), 4}};\)\)], "Input", CellLabel->"In[2]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[ m . {x1, x2, x3, x4} \[Equal] {1, 2, 1}, {x1, x2, x3, x4}]\)], "Input", CellLabel->"In[3]:=", FontSize->14], Cell[BoxData[ RowBox[{\(Solve::"svars"\), \(\(:\)\(\ \)\), "\\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Solve::svars\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[3]:="], Cell[BoxData[ \({{x1 \[Rule] 1 - \(5\ x3\)\/3 + x4, x2 \[Rule] \(4\ x3\)\/3 - x4}}\)], "Output", CellLabel->"Out[3]="] }, Open ]], Cell[TextData[{ StyleBox["Importante:", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], StyleBox[" ", FontSize->14, FontColor->RGBColor[1, 0, 1]], "Las inc\[OAcute]gnitas que aparecen a la izquierda de las flechas son las \ inc\[OAcute]gnitas principales, el resto de las inc\[OAcute]gnitas son inc\ \[OAcute]gnitas libres. Esto nos permite en este caso conocer el rango de la \ matriz de coeficientes del sistema y tambi\[EAcute]n el rango de la matriz \ ampliada (pues el sistema es compatible).\n", StyleBox["Pregunta extra: ", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], "Utilizando exclusivamente el segmento de ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " anterior razonar cu\[AAcute]l es el rango de la matriz de coeficientes \ del sistema y el rango de la matriz ampliada." }], "Text"], Cell[TextData[{ "Tambi\[EAcute]n podemos utilizar la funci\[OAcute]n ", StyleBox["LinearSolve[", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], StyleBox["m", FontFamily->"Courier", FontSize->14], StyleBox[",", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], StyleBox["b", FontFamily->"Courier", FontSize->14], StyleBox["]", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", siendo ", StyleBox["m", FontFamily->"Courier", FontSize->14], " la matriz de coeficientes del sistema y ", StyleBox["b", FontFamily->"Courier", FontSize->14], " la matriz columna de los t\[EAcute]rminos independientes, que hay que \ cargarla como una lista o tambi\[EAcute]n se puede cargar como una matriz \ columna. Esta funci\[OAcute]n nos devuelve una soluci\[OAcute]n del sistema, \ si es compatible. La soluci\[OAcute]n aparece como una lista." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(LinearSolve[m, {1, 2, 1}]\)], "Input", CellLabel->"In[4]:=", FontSize->14], Cell[BoxData[ \({1, 0, 0, 0}\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(LinearSolve[m, {{1}, {2}, {1}}]\)], "Input", CellLabel->"In[5]:=", FontSize->14], Cell[BoxData[ \({{1}, {0}, {0}, {0}}\)], "Output", CellLabel->"Out[5]="] }, Open ]], Cell[BoxData[ \(Clear["\", $Line]\)], "Input", CellLabel->"In[6]:=", FontSize->14] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["EJERCICIO 2.6 (Un S.E.L. con par\[AAcute]metros)", FontColor->RGBColor[0, 0, 1]]], "SectionFirst", FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Clasificar y resolver, cuando sea posible, el sistema: \n\t\t\t", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "1"], "+"}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "2"], "+", SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "3"]}], TraditionalForm]]], "=1\n\t\t\t-", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "1"], "+"}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox[\(p x\), FontSlant->"Plain"], "2"], "-", SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "3"]}], TraditionalForm]]], "= -1\n\t\t\t", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "1"], "+"}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "2"], "+", SubscriptBox[ StyleBox[\(p x\), FontSlant->"Plain"], "3"]}], TraditionalForm]]], "=1\n\t\t\t-", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "1"], "+"}], TraditionalForm]]], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox[\(p x\), FontSlant->"Plain"], "2"], "-", SubscriptBox[ StyleBox["x", FontSlant->"Plain"], "3"]}], TraditionalForm]]], "= p+1" }], "Subsubsection", CellDingbat->None, TextAlignment->AlignmentMarker, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[TextData[{ StyleBox["Importante:", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], StyleBox[" ", FontSize->14, FontColor->RGBColor[1, 0, 1]], "Como el S.E.L. depende de par\[AAcute]metros debemos utilizar la funci\ \[OAcute]n ", StyleBox["Reduce", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Reduce[ x1 + x2 + x3 \[Equal] 1 && \(-x1\) + p\ x2 - x3 \[Equal] \(-1\) && x1 + x2 + p\ x3 \[Equal] 1 && \(-x1\) + p\ x2 - x3 \[Equal] p + 1, {x1, x2, x3}]\)], "Input", CellLabel->"In[1]:=", FontSize->14], Cell[BoxData[ \(p \[Equal] \(-2\) && x1 \[Equal] 1 && x2 \[Equal] 0 && x3 \[Equal] 0\)], "Output", CellLabel->"Out[1]="] }, Open ]], Cell[BoxData[ \(Clear["\", $Line]\)], "Input", CellLabel->"In[2]:=", FontSize->14], Cell[TextData[{ StyleBox["Importante:", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], StyleBox[" ", FontSize->14, FontColor->RGBColor[1, 0, 1]], "La respuesta que obtenemos al utilizar la funci\[OAcute]n ", StyleBox["Reduce", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], " se presenta como combinaciones de ecuaciones simples. Recordemos que && \ significa que las ecuaciones han de ser ciertas simult\[AAcute]neamente y || \ significa que se cumplen unas u otras.\nAl igual que la funci\[OAcute]n ", StyleBox["Solve", FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], ", las inc\[OAcute]gnitas que aparecen listadas a la izquierda de == son \ las inc\[OAcute]gnitas principales. 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El \ resto, si es que hay alguna, son inc\[OAcute]gnitas libres.\n", StyleBox["Preguntas extra: ", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], "Utilizando exclusivamente el segmento de ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " anterior razonar responder a las siguientes cuestiones:\n", StyleBox["a.-", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontSize->14, FontWeight->"Bold"], StyleBox["\[DownQuestion]Qu\[EAcute] relaci\[OAcute]n existe entre r(A) y \ r(AM), siendo A la matriz de coeficientes del sistema y AM la matriz \ ampliada?\n", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox["b.-", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontSize->14, FontWeight->"Bold"], StyleBox["\[DownQuestion]Para qu\[EAcute] valores de los par\[AAcute]metros \ p y q es regular la matriz A?\n", FontSize->14, FontWeight->"Bold", 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El \ resto, si es que hay alguna, son inc\[OAcute]gnitas libres.\n", StyleBox["Importante:", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], StyleBox[" ", FontSize->14, FontColor->RGBColor[1, 0, 1]], "Presentar la soluci\[OAcute]n de la forma m\[AAcute]s sencilla posible.\n\ ", StyleBox["Preguntas extra: ", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], "Utilizando exclusivamente el segmento de ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " anterior razonar responder a las siguientes cuestiones:\n", StyleBox["a.-", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontSize->14, FontWeight->"Bold"], StyleBox["\[DownQuestion]Qu\[EAcute] relaci\[OAcute]n existe entre r(A) y \ r(AM), siendo A la matriz de coeficientes del sistema y AM la matriz \ ampliada?\n", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox["b.-", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontSize->14, FontWeight->"Bold"], StyleBox["\[DownQuestion]Para qu\[EAcute] valores de los par\[AAcute]metros \ p y q es regular la matriz A?\n", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox["c.-", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontSize->14, FontWeight->"Bold"], StyleBox["Razonar cuando forman base de", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ RowBox[{" ", SuperscriptBox["\[DoubleStruckCapitalR]", StyleBox["3", FontWeight->"Bold"]]}], TraditionalForm]], FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontSize->14, FontColor->RGBColor[0, 0, 1]], StyleBox["los vectores fila de la matriz de coeficientes A del sistema.", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]] }], "Text"] }, Closed]] }, Closed]] }, Open ]] }, FrontEndVersion->"5.2 for Microsoft Windows", ScreenRectangle->{{0, 1280}, {0, 937}}, WindowSize->{738, 670}, WindowMargins->{{102, Automatic}, {Automatic, 0}}, StyleDefinitions -> "ArticleModern.nb" ] (******************************************************************* Cached data follows. 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