Question

3. (6 marks) For b1,b2, 63, 64 € R, consider the linear system of equations + 21 + 22 -11 + 22 23 624 = bi 3.13 + 2014 b2 23 + 2:04 b3 603 + 20:24 = 54 *) 21 + 2:01 + 4.22 where 11, 12, 13, 14 ER. (a) Find a system of equations that bı, b2, 63, 64 must satisfy in order for (+) to be consistent. bi [ must satisfy so (+) is (b) Using your answer in part (a), find a vector equation that 6 = consistent. (c) If b3 = b4 = 1, find bi, b2 so that (+) is consistent. 64

Answer 1

som a consider the augmented mateix: 1 1 6 bi - 1 - -3 2 b2 0 1 2 bz by 2 4 -6 20 1 6 ܝ 0 2 -4 8 bi b + b2 -b + b₃ -2 b +by R21 R2+R, R37 R₃-Ri Rut Ry-2R 0 -) 2 -4 8 2 -4 6 1 . Ry Ry-R2 R21 R2 0 -2 b) (b +62/2 -b, tb3 -b-b₂ t by 4 w 0 -1 2 -4 0 0 D 2 - 1 0 -2 4 - 0 - b2)/2 RITR, R₂ Tb, tb2)/2 R₂ + R₂ + R2 - by + b2 tb3k -3b-b2 + by 0 0 a 0

from the last two lowdo we can say system ů consistent if that -b1 b2 tb₂ =0 a b-b₂-2hzo 2 *3 b byt by=0 * 3b. tb-by=0 Thes, me fare a system of equations: b. - by - 2 bz = 3b + b₂ - by zo b) we have Ib, by - 263=0 3b. tbr - by o 4b - 2by-by=0 4b = 25z + by b = + b3 + t by b) ½ b3 + 1/4 by b2 = b2 by b3 Ibu. by 1/29 74 o by b2 0 1 be + b3 + by 0 6 -- 0 0 bz O 0 by.

af bz= by = 4 then, from the get b. - b -2.(1)=0 b-b2 = 2 36, +by -1=0 * 3b + b₂ = 1 solving by = b.by/= 2 3b + b = 1 46,= 3 b. = 3/5 b₂=2 ther 3 bz = 3/44 -2 = -- b = =-5 4