[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-detail-72286-es":3,"doc-seo-72286-110":30,"detail-sidebar-cat-0-es-110":88},{"code":4,"msg":5,"data":6},0,"success",{"doc_id":7,"user_id":8,"nickname":9,"user_avatar":10,"doc_module":4,"category_id":11,"category_name":12,"doc_title":13,"doc_description":14,"doc_content":15,"file_id":16,"file_url":17,"file_type":18,"file_size":19,"view_count":20,"is_deleted":4,"is_public":21,"is_downloadable":21,"audit_status":21,"page_count":20,"language":22,"language_code":23,"site_id":24,"html_lang":23,"table_of_contents":25,"faqs":26,"seo_title":27,"seo_description":14,"update_tm":28,"read_time":29},72286,962084926284,"Aurora","https://ap-avatar.wpscdn.com/davatar_29158cc5080c5b710cf443261637dec0",38,"Examen","Matrices y Determinantes - Andalucía 2008 Matemáticas II","Colección de problemas resueltos de Matemáticas II (Selectividad Andalucía 2008) centrados en matrices y determinantes. Incluye ejercicios con cálculo de una matriz P a partir de una ecuación matricial y una traspuesta, determinación de valores de k para los que (A−kI)^2 resulta la matriz nula, y resolución de casos con inversas: cálculo de inversa de A, identificación de que B no es invertible por determinante cero y obtención de X en una ecuación matricial. También se analiza el rango según k y se halla la inversa para un valor concreto.","PROBLEMAS RESUELTOSSELECTIVIDAD ANDALUCÍA 2008  \nMATEMÁTICAS II  \nTEMA 1: MATRICES Y DETERMINANTES  \n􀁸 Reserva 1, Ejercicio 3, Opción B  \n􀁸 Reserva 3 , Ejercicio 3, Opción A  \n􀁸 Reserva 3, Ejercicio 3, Opción B  \n􀁸 Reserva 4, Ejercicio 3, Opción B  \n[http://emestrada. wordpress.com](http://emestrada. wordpress.com)  \n􀀆1 1 1 􀀇 􀀆 1 0􀀇 Dadas las matrices A = 􀀊􀀈 01 12 02􀀋 , B = 􀀊􀀊 02 − 11􀀋 y C = 􀀊􀀈􀀆􀀆􀀊􀀈 − 21 −− 01 − 11􀀇􀀇􀀇􀀇􀀋􀀋􀀋􀀋􀀉  \nCalcula la matriz P que verifica A ⋅P − B = C t ( C t es la matriz traspuesta de C). MATEMÁTICAS II. 2008. RESERVA 1. EJERCICIO 3.OPCIÓN B.  \nR E S O L U C I Ó N  \nA⋅ P − B = C t ⇒ A⋅ P = C t + B ⇒ P = A −1 ⋅ (C t + B)  \n􀀆1  \n􀀈  \nCalculamos la matriz inversa de A = 􀀈 0  \n􀀊􀀈 1  \n1 1  \n2  \n1 􀀇  \n􀀉  \n0 􀀉  \n2 􀀋􀀉  \n􀀆 2 0 −1 􀀇 t 􀀆 2 0 −1 􀀇  \n􀀈 0 1 −1 􀀉 􀀈 0 1 0 􀀉 􀀆 2 0 −1 􀀇 A−1 = ~~ ~~(AAd~~ ~~)t~~ ~~ = ~~ ~~􀀊􀀈~~ ~~−1~~ ~~01~~ ~~1􀀋􀀉~~ ~~ = ~~ ~~􀀊􀀈~~ ~~−1~~ ~~1~~ ~~1~~ ~~􀀋􀀉 = 􀀊 − 01 − 11 01􀀋  \n􀀆 2 0 −1 􀀇 􀀆 􀀆− 2 1􀀇 􀀆1 0 􀀇􀀇 􀀆 2 0 −1 􀀇 􀀆 −1 1 􀀇 􀀆 − 3 0 􀀇 P = A −1 ⋅ (C t + B) = 􀀈 0 1 0 􀀉 ⋅ 􀀈 􀀈 0 −1 􀀉 + 􀀈 0 −1 􀀉 􀀉 = 􀀈 0 1 0 􀀉 ⋅ 􀀈 0 − 2 􀀉 = 􀀈 0 − 2 􀀉  \n􀀈􀀊 −1 −1 1 􀀋􀀉 􀀊􀀈 􀀈􀀊 −1 1􀀋􀀉 􀀊􀀈 2 1 􀀋􀀉 􀀋􀀉 􀀊􀀈 −1 −1 1 􀀋􀀉 􀀊􀀈 1 2 􀀋􀀉 􀀊􀀈 2 3 􀀋􀀉  \n\n| 􀀂 0\u003Cbr>Sea I la matriz identidad de orden 3 yA = 􀀄 − 1 􀀄􀀇􀀄􀀇 1\u003Cbr>cual ( A −− kI ) 2 es la matriz nula. | − 1\u003Cbr>0 1 | − 2􀀃\u003Cbr>− 2 􀀅 . Calcula, si existe, el valor de k para el 3􀀈􀀅 |\n| --- | --- | --- |\n| MATEMÁTICAS II. 2008. RESERVA 3. EJERCICIO 3.OPCIÓN A. |  |  |\n\nR E S O L U C I Ó N  \n􀀂 0  \n􀀄 A − kI = 􀀄 −1  \n􀀇􀀄 1  \n−1  \n0 1  \n− 2 􀀃 􀀂k  \n􀀅 􀀄− 2 􀀅 − 􀀄 0 3 􀀈􀀅 􀀄􀀇0  \n0  \nk  \n0  \n0 􀀃 􀀂− k 􀀅 􀀄  \n0 􀀅 = 􀀄 −1 k 􀀈􀀅 􀀄􀀇 1  \n−1  \n− k  \n1  \n− 2 􀀃  \n􀀅  \n− 2 􀀅 3 − k 􀀈􀀅  \n􀀂− k (A − kI ) 2 = 􀀄 −1  \n􀀇􀀄 1  \n−1  \n− k  \n1  \n− 2 􀀃 􀀂− k  \n− 2 􀀅 ⋅ 􀀄 −1 3 − k 􀀈􀀅 􀀇􀀄 1  \n−1 − 2 􀀃 􀀂 k 2 −1  \n􀀅 􀀄  \n− k − 2 􀀅 = 􀀄 2k − 2 1 3 − k 􀀈􀀅 􀀄􀀇 2 − 2k  \n2k − 2 4k − 4 k 2 −1 4k − 4 2 − 2k k2 − 6k +  \n􀀃  \n􀀅  \n􀀅  \n5􀀈􀀅  \n􀀂 k 2 −1 􀀄  \n􀀄 2k − 2 􀀄􀀇 2 − 2k  \n2k − 2  \nk 2 −1  \n2 − 2k  \n4k − 4 􀀃 􀀂0 􀀅 􀀄  \n4k − 4 􀀅 = 􀀄 0 k 2 − 6k + 5􀀈􀀅 􀀄􀀇0  \n0 0  \n0  \n0 􀀃􀀅  \n0 􀀅 ⇒ k = 1 0 􀀈􀀅  \nIgualando cada expresión a cero tenemos tendríamos nueve ecuaciones. La única que verifica todas las expresiones es k = 1 .  \n\n| 􀀁 1\u003Cbr>Dadas las matrices A = 􀀃 1 􀀃􀀆􀀃􀀆 1 | 1 2\u003Cbr>1 | 2􀀂􀀄\u003Cbr>1 􀀄 y 1 􀀇􀀄 | 􀀁 1\u003Cbr>B = 􀀃 2\u003Cbr>􀀃􀀆􀀃􀀆 −− 1 | 0 0\u003Cbr>1 | 2􀀂􀀄\u003Cbr>4 􀀄 1 􀀇􀀄 |\n| --- | --- | --- | --- | --- | --- |\n| a) Calcula, si existen, la matriz inversa de A y la de B.\u003Cbr>b) Resuelve la ecuación matricial A ⋅X + B = A + I , donde I denota la matriz identidad de orden 3.\u003Cbr>MATEMÁTICAS II. 2008. RESERVA 3. EJERCICIO 3.OPCIÓN B. |  |  |  |  |  |\n\nR E S O L U C I Ó N  \n􀀁1  \n􀀃  \na) Calculamos la matriz inversa de A = 􀀃 1  \n􀀆􀀃 1  \n1 2  \n1  \n2 􀀂  \n􀀄  \n1 􀀄  \n1 􀀇􀀄  \n􀀁 1 0 −1 􀀂 t 􀀁 1 1 −3 􀀂  \n􀀃 1 −1 0 􀀄 􀀃 0 −1 1 􀀄 􀀁 −1 A−1 = ~~ ~~(AAd~~ ~~)t~~ ~~ = 􀀆􀀃~~ ~~−3~~ ~~−11~~ ~~1~~ ~~􀀇􀀄~~ ~~ = 􀀆􀀃~~ ~~−1~~ ~~−~~ ~~01~~ ~~1􀀇􀀄 = 􀀆 01  \n−1  \n1 0  \n3 􀀂  \n􀀄  \n−1 􀀄  \n−1 􀀇􀀄  \nLa matriz B no tiene inversa, ya que su determinante vale cero.  \nb)  \nA⋅ X + B = A + I ⇒ A⋅ X = A + I − B ⇒ X = A −1 ⋅(A + I − B)  \n􀀁 −1  \nX = A −1 ⋅(A + I − B) = 􀀃 0  \n􀀆􀀃 1  \n−1  \n1 0  \n3 􀀂 􀀁 􀀁1 −1 􀀄 ⋅ 􀀃 􀀃 1 −1 􀀇􀀄 􀀆􀀃 􀀃􀀆1  \n1 2  \n1  \n2 􀀂 􀀁1  \n􀀄 􀀃 1 􀀄 + 􀀃 0  \n1 􀀇􀀄 􀀆􀀃 0  \n0 1  \n0  \n0 􀀂 􀀁 1  \n􀀄 􀀃 0 􀀄 − 􀀃 2 1 􀀇􀀄 􀀃􀀆 −1  \n0 0  \n1  \n2 􀀂􀀂 􀀁 6  \n􀀄􀀄 􀀃 4 􀀄􀀄 = 􀀃 −3 1 􀀇􀀄 􀀇􀀄 􀀃􀀆 −1  \n−4  \n3 1  \n6 􀀂  \n􀀄  \n−4 􀀄  \n−1 􀀇􀀄  \n\n| 􀀆 1 􀀈\u003Cbr>Dada la matriz A = 􀀈 k 􀀈􀀊􀀈􀀊 1 | 3 1\u003Cbr>7 | k 􀀇\u003Cbr>􀀉\u003Cbr>3 􀀉\u003Cbr>k 􀀋􀀉 |\n| --- | --- | --- |\n| a) Estudia el rango de A en función de los valores del parámetro k.\u003Cbr>b) Para k = 0 , halla la matriz inversa de A.\u003Cbr>MATEMÁTICAS II. 2008. RESERVA 4. EJERCICIO 3.OPCIÓN B. |  |  |\n\nR E S O L U C I Ó N  \na) El rango de A es al menos 2, ya que el determinanteel determinante de A.  \n1 3 1 7  \nes distinto de cero. Vamos a calcular  \n-Si k = ±  3 ⇒ el rango de A es 2.  \n-Si k ≠ ±  3 ⇒ el rango de A es 3.  \n􀀆1  \n􀀈  \nb) Calculamos la matriz inversa de A = 􀀈 0  \n􀀊􀀈 1  \n3 1  \n7  \n0 􀀇  \n􀀉  \n3 􀀉  \n0 􀀋􀀉  \n􀀈􀀆􀀈 − 210 30 −14􀀇􀀉 t 􀀈􀀆􀀈 −213 00 − 93􀀇􀀉 􀀆􀀈  A−1 = ~~ ~~(AAd~~ ~~)t~~ ~~ = 􀀊􀀈~~ ~~9~~ ~~32~~ ~~1~~ ~~􀀋􀀉~~ ~~ = ~~ ~~􀀊􀀈~~ ~~−1~~ ~~42~~ ~~1~~ ","cbCaidVvzYBD9kYo","https://ap.wps.com/l/cbCaidVvzYBD9kYo","pdf",272612,5,1,"Spanish","es",110,"# Matrices y determinantes (Tema 1)\n## Reserva 1: Ejercicio 3 (Opción B)\n## Reserva 3: Ejercicio 3 (Opción A)\n## Reserva 3: Ejercicio 3 (Opción B)\n## Reserva 4: Ejercicio 3 (Opción B)","[{\"question\":\"En la Reserva 3, Ejercicio 3, Opción B, ¿por qué la matriz B no tiene inversa y cómo se calcula X?\",\"answer\":\"La matriz B no tiene inversa porque su determinante vale cero. Para X, se usa A·X + B = A + I, se despeja A·X = A + I − B y se obtiene X = A^−1·(A + I − B).\"},{\"question\":\"En la Reserva 4, Ejercicio 3, Opción B, ¿cómo cambia el rango de A según k y qué ocurre para k=0?\",\"answer\":\"El rango depende del parámetro: si k = ±3, el rango es 2; si k ≠ ±3, el rango es 3. Para k = 0 se halla la inversa de A.\"}]","Matrices y Determinantes - Andalucía 2008 Matemáticas II | PDF",1784036052,8,{"code":4,"msg":31,"data":32},"ok",{"site_id":24,"language":23,"slug":33,"title":13,"keywords":34,"description":14,"schema_data":35,"social_meta":83,"head_meta":85,"extra_data":87,"updated_unix":28},"matrices-and-determinants-andalusia-2008-mathematics-ii","",{"@graph":36,"@context":82},[37,54,69],{"@type":38,"itemListElement":39},"BreadcrumbList",[40,44,48,51],{"item":41,"name":42,"@type":43,"position":21},"https://docshare.wps.com","Home","ListItem",{"item":45,"name":46,"@type":43,"position":47},"https://docshare.wps.com/es/document/","Document",2,{"item":49,"name":12,"@type":43,"position":50},"https://docshare.wps.com/es/document/examen/",3,{"item":52,"name":13,"@type":43,"position":53},"https://docshare.wps.com/es/document/matrices-and-determinants-andalusia-2008-mathematics-ii/72286/",4,{"url":52,"name":13,"@type":55,"author":56,"headline":13,"publisher":58,"fileFormat":61,"inLanguage":23,"description":14,"dateModified":62,"datePublished":63,"encodingFormat":61,"isAccessibleForFree":64,"interactionStatistic":65},"DigitalDocument",{"name":9,"@type":57},"Person",{"url":41,"name":59,"@type":60},"DocShare","Organization","application/pdf","2026-08-06","2026-07-14",true,{"@type":66,"interactionType":67,"userInteractionCount":20},"InteractionCounter",{"@type":68},"ViewAction",{"@type":70,"mainEntity":71},"FAQPage",[72,78],{"name":73,"@type":74,"acceptedAnswer":75},"En la Reserva 3, Ejercicio 3, Opción B, ¿por qué la matriz B no tiene inversa y cómo se calcula X?","Question",{"text":76,"@type":77},"La matriz B no tiene inversa porque su determinante vale cero. Para X, se usa A·X + B = A + I, se despeja A·X = A + I − B y se obtiene X = A^−1·(A + I − B).","Answer",{"name":79,"@type":74,"acceptedAnswer":80},"En la Reserva 4, Ejercicio 3, Opción B, ¿cómo cambia el rango de A según k y qué ocurre para k=0?",{"text":81,"@type":77},"El rango depende del parámetro: si k = ±3, el rango es 2; si k ≠ ±3, el rango es 3. Para k = 0 se halla la inversa de A.","https://schema.org",{"og:url":52,"og:type":84,"og:title":13,"og:site_name":59,"og:description":14},"article",{"robots":86,"canonical":52},"index,follow",{"doc_id":7,"site_id":24},{"code":4,"msg":5,"data":89},[90,95,99,101,105,109,113,117,121,125],{"id":91,"doc_module":4,"doc_module_name":46,"category_name":92,"show_sort_weight":93,"slug":94},39,"Cómic",60,"comic",{"id":96,"doc_module":4,"doc_module_name":46,"category_name":97,"show_sort_weight":93,"slug":98},43,"Estilo de Vida","lifestyle",{"id":11,"doc_module":4,"doc_module_name":46,"category_name":12,"show_sort_weight":93,"slug":100},"exam",{"id":102,"doc_module":4,"doc_module_name":46,"category_name":103,"show_sort_weight":93,"slug":104},44,"General","general",{"id":106,"doc_module":4,"doc_module_name":46,"category_name":107,"show_sort_weight":93,"slug":108},41,"Investigación e Informes","research-report",{"id":110,"doc_module":4,"doc_module_name":46,"category_name":111,"show_sort_weight":93,"slug":112},37,"Literatura","literature",{"id":114,"doc_module":4,"doc_module_name":46,"category_name":115,"show_sort_weight":93,"slug":116},22,"Relatos y Novelas","story-novel",{"id":118,"doc_module":4,"doc_module_name":46,"category_name":119,"show_sort_weight":93,"slug":120},42,"Religión y Espiritualidad","religion-spirituality",{"id":122,"doc_module":4,"doc_module_name":46,"category_name":123,"show_sort_weight":93,"slug":124},40,"Salud y Atención Médica","healthcare",{"id":126,"doc_module":4,"doc_module_name":46,"category_name":127,"show_sort_weight":93,"slug":128},24,"Tecnología","technology"]