## Answers

To show i., ii., iii., here e reduced the augmented matrix of the system by Gauss elimination method. The calculation is very taff. But I solve it correctly. Hope you understand. Thank you.

The augmented Page matrix of the system is 2a-3 4 a کیا 2+b 6 4 2 2 2a-2 za o -2 40+2 2a42 4 2 - 1 rad 3 za -2b-2 we now be duce id by hauss elimination methode i 2003 2d 2+ b 4 2a-2 Re -To 1 za 2 2.

4 -2 O 2a+2 2 44+2 29+ 3 -1 2a -26-2 Ph = R2-4 Ri R3 = h - Q R R = R4-2 R 2a-3 2+b 20-3 2a 3 N- 6 ي 2013 -2013 0 243 44+6/3 za 111-6) 212+5) o O - 2043 22+373 22-1 3 - 7648 3 3 R2= R2 za mmmm

Page - 29~3 وادی )- 6 226 6 I b 0 ㅜ o a 0 20N3 (4a+g/3 24 | - 2(2+) (Ab+8) 3 - 219, 204 33 241 8 20 3 2a 3 Rt (2.6~3s 23 2 (2+)/ 0 에 ㅏb CO - 0 0 2 ( 2a - 2 0 O 291 -3642 I R 3을 2(a+1) 2a~3 202 Y2 124비6 6 ㅇ - O I-bla ( 0 a O atl 2 21 136429 Lg - gyg" = Ry - Rz

Page ① 2a-3 2013 Yol 2+5/6 l-bla -1 0 atl a at 1 201 at -(36+2)(4+1) +1 B atl atl (1) bo No0) ald System in has unique solution then 2021 ati cie 2a²1 -(36+2) fa+i) +1 must (36+2)(at)ti ati be defined le -36+27(ert) +1 0 → CatD (36+2) = 1 (11) Infinite numbers of solution if 202120 and ²/3 6+2) fa+1) +120 and I caxi) (36+2) = 1 &a=1 a then so if a= (vź +1) (3b+2) = 1 12 V2+1 35+2=

Page 4 lie zb= v2 V2+1 -2 b= - امرا V2 12+1 -2 -21 simiderady if a = -1 then V2 b= rz V-1 - 2 wt 3 (iii) No bolution if and –(36+2)(a+1) 41=0 c) až (a+)(36+2) = a2_170 +1 V2 This are the nequined condition on an and be

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