By Dr. sc. nat. Ernst-Adam Pforr, Dr. paed. Lothar Oehlschlaegel, Oberlehrer Dipl.-Math. Georg Seltmann (auth.)
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Extra resources for Übungsaufgaben zur linearen Algebra und linearen Optimierung
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7. 8. 9. 10. E liegen die Punkte PI(O, 0,1), P2(1, -1,0) und P3(-2, 1, 1). E liegen die Punkte A (1,0, -1), B(2, -1, 1) und C( -1, 1,2). E liegt der Punkt Po(l, -2,1), der Ortsvektor OPo ist senkrecht zu E gerichtet. E liegen die Punkte PI(l, 2, 3) und P2(3, 2,1), E steht senkrecht auf der Ebene 4x- y+ 2z = 7. In E liegt der Punkt Po(2, 1, -1), die Schnittgerade g der Ebenen 2x + y - z = 3 und x + 2y + z = 2 steht senkrecht auf E. In E liegt der Punkt A(I,I, - 3), E verläuft parallel zu den Vektoren a = (-3, -2, 2?
3. Man löse die folgenden Optimierungsaufgaben graphisch: a) z = 3xI Jo max, d) z = 3X2Jo min, f) z = -3xI c) z = 3X2 Jo max, , + 3X2 + 3= min, NB jeweils: - 2 ~ XI - X2 ~ 4, XI - 5x2 ~ 0, XI + X2 s; 1, XI ~ 0, X2 ~ O. 4. 2. Graphische Lösung von Aufgaben 1 700 ~ 1 OOOXI + 2 000X2 + 1 500X3 ~ 2 000, 6XI + 2X2+ 3X3~ 2,8, XI+ X2+ x3=1, 55 Xi:;;; 0, i=1,2,3. 5. a) Man löse die folgende Optimierungsaufgabe graphisch: z = 3XI + 10 ~ 4X2 4, max; Xl + 2X2 ~ 11, 2XI +3X2 ~ 18(*), 5XI + 2X2 ~ 30. b) Wie lautet die Lösung, wenn zusätzlich gefordert wird: Xi ganzzahlig, i = 1, 2?
B»), b) x~ = 1 + xi (vgl. 7. c»), c) Z2 = x2 + x~ + y2 (vgl. g»), 1 1 d) x 2 + y2 + 2z 2 - 2xz - 2yz - 2"x - 2"Y - 1 2"z = 0, e) xi + x~ + 2x~ + 2XIX3 + 2X2X3 + 12x2 + 4 = 0, f) 2xi + x~ + 2x~ - 2xIX3 - 2X2X3 + 3X2 + X3 + 5 = ° stellen Flächen zweiter Ordnung dar. Anhand der in Anhang A 4 angegebenen Kriterien bestimme man deren Gestalt. 11. Eine Fläche zweiter Ordnung habe die Gleichung 14 8 10] xTCx= 1 mit C= [ 8 29 38 . 5. Geometrie im Rn 43 a) Man bestimme, was für eine Fläche damit vorgegeben ist.