Question

W20 oubam Part 2/4/lusershutch2&key ASH2NAPVIEM1OtqwdE9OGPRwNdueffectiveUsershutch (5 points) Solve the following system using augmented matrix methods -121, +63+ 121, --18 -21, +lx, +213 = -3 -121, +6x, + 121, --18 (a) The initial matrex is (b) First, perform the Row Operation R, R, The resulting matrix is (c) Next perform the operations +2R₂ + R₂ R₂ +12R, R R The resulting matris

Answer 1

solution: The given system is: - 12x + 6x2 + 12xz 2 - 18 - 224 + 1x2 + 2x2 2 - 3 - 1204 + 6x2 + 12x3 = - 18 (a) The initial marrix is :-- - 12 6 12.1 -187 1 - 2 1 2 1 - 3 -12 6 12 1 -18 ] 1) I RIBA -12 o - 3 rel -12 -1 | 1 - 2 1 2 1 | -12 6 12 | . (c) 2R1 + R2 R2 12 R1 + R3 > Ti -12 -1 Lo 0 0 (d) The reduced matrix is :- -1 | 3/2 I - II 12 00 01 od

from the sow seduced augmented matrix we gat; 1 - -17727 Lo 0 0 J L xz1 Lol 2 24 - Do 122 - kg e 3). So the system and three unknowns. infuntely manny So we get only one equation So, the system will have solutions. I (e) The system has Infinite solutions (f) since % - % 2 xq - X z 23/2. 24 2 1622 + 3 + 3% Taking X a ki and az ake where ki and ko are any scalars we ger 2 1 1 ki + kn + %oki + kn + 3/2 anoo son Answer: The solutions to the system & are: 24 2 % K4 + k2 + 3/2 юмоля ko, k, ) xe a ki s are scalars.) RB 2 ka