III Parcial ord CAL II-2014 sol.pdf - ´ gico de Costa Rica Instituto Tecnolo ´ tica Escuela de Matema Puntaje Total 29 puntos Tiempo 2:20 minutos ´

III Parcial ord CAL II-2014 sol.pdf - ´ gico de Costa Rica...

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Instituto Tecnol´ogico de Costa Rica Puntaje Total: 29 puntos Escuela de Matem´atica Tiempo: 2:20 minutos C ´alculo y ´ Algebra Lineal Tercer examen parcial Soluci´on 17 de noviembre, 2014 Instrucciones: Esta es una prueba de desarrollo, por lo tanto, debe presentar todos los pasos y procedimientos que le permitieron obtener cada una de las respuestas. Trabaje en forma clara, ordenada y NO utilice bol´ ıgrafo de tinta roja. No son procedentes las apelaciones que se realicen sobre ex´amenes resueltos con l´ apiz (parcial o totalmente) o que presenten alg´un tipo de alteraci´on. No se permiten dispositivos con memoria de texto ni conectividad inal´ ambrica durante el examen. Puede utilizar calculadora cient´ ıfica no programable. #1. Sean u = (1 , 3 , 2), v = ( - 1 , 4 , - 2) y w = (2 , - 1 , 5) vectores en R 3 . Exprese el vector z =(4 , 5 , 7) como combinaci´on lineal de u , v y w . (4 pts) ? Soluci´on El vector z es una combinaci´ on lineal de los vectores u , v y w si existen escalares a, b, c en R tales que satisfacen la ecuaci´ on vectorial: z = a u + b v + c w Sustituyendo por los vectores dados se tiene z = a u + b v + c w (4 , 5 , 7) = a (1 , 3 , 2) + b ( - 1 , 4 , - 2) + c (2 , - 1 , 5) (4 , 5 , 7) = ( a, 3 a, 2 a ) + ( - b, 4 b, - 2 b ) + (2 c, - c - 5 c ) (4 , 5 , 7) = ( a - b + 2 c, 3 a + 4 b - c, 2 a - 2 b + 5 c ) De ´ esta ´ultima ecuaci´ on se obtiene el sistema lineal: a - b + 2 c = 4 3 a + 4 b - c = 5 2 a - 2 b + 5 c = 7 cuyas soluciones para las variables a, b y c son: a = 4 , b = - 2 y c = - 1. Finalmente tenemos: (4 , 5 , 7) = 4(1 , 3 , 2) - 2( - 1 , 4 , - 2) - (2 , - 1 , 5)
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#2. Sean u = (0 , 1 , 2 , 1), v = (1 , - 1 , 0 , 1) y w = (3 , 2 , - 1 , - 3) vectores en R 4 . Verifique que los vectores u , v y w son linealmente independientes.
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