Question

Problem 1 Let A= 3 2 13 1 5 7 11 8 -3 9 10 -6 -4 12 8 a) [4 pts) Find a basis for N(A) in rational format. b) (3 pts) Find a particular solution to the matrix equation A*x= 5 -2 14 c) [3 pts] Use your answers in a), b) and the Superposition Principle to express the general solution in vector form to the matrix equation in b).

Answer 1

Let AS () N ) = x + Rº: Ao 3 2. 1 11 را) 5 12 where A 2. 쁨 13 1 3 5 7 -4- 2. 14 8 -3 12, 13 9 10 -6 8 35 Ref =(%,X2, M3Y4,75) 5 1 -4. 8 3 12 2 3 13 9 10 - 6 as Ax=0 -4 ។ 1 14 & - 3 - 8 10 13 9 Ret -413 1 3 기 3 | 3 12 2. 11 -3 3 9 10 8 -6 R2tR2-2 R1 1 113 13 기3 -413 3415 14/3 -233 4419 13 8. 10 g - R3 + R3-13RL 1 113 기3 -403 13 31J3 143 -23/3 4413 1411 0 -353 -lo313 754 R2 K 3 R2 31 119 59 기3 -413) 1 14/31 -23/31 443 143 -353 -10913 Ret Re 1- 1 41131 8131 - 5/34] 1 1434 -23 [31 441311 14) -3513 -109/3 7413 CScanned with CamScanner 0 2 R2 0 0

3-4 R2 0 0 1 o 0 이 0 31 -500 0 108 0 0 이 R3GR3 - 1 0 47751 8131 -57131 1 1431 -23131 4413 1 o 2427의 1일을 Seol34 R35-3L RS 427 이 4791 8131 - St/11 14 131 -233 -4/s1] 1 fol9 -S8o +27 427 Rat Ra-47 R3 - 443 108 3L 14 427 | 427 1 -23 44 31 31 tols R2GR2-1 R3 421 427 31. 1 -443 +27 427 1 -11 124 6 1 1019 -580 so reduced systems is 421 421 1 - 443 loB 427 427 1 124 C1 CL 0 1 (19 4. -580 +21 25 Let u4=t, ag=te we get 427 1 5 443 t - 108 CL 427 CI 427 3 -11g t + 580 u where t and u are arbitary g0 +21 +27 443 1 427 +21 421 +27 114 tu 124 it GA 61 3 l09 +21 +21 것 1 5 u CScanned with CamScanner 이 이 기 2 는 0 0 13 ao 000 0 품은 1월 8 443-168 1 2 CL 380 분 t - 124 u -1019 t + 580u 421 t 01

5-108 (443 427 so baris NCA) ot lil 9 s í -1019 427 427 -124 61 5 80 427 o Anwar 5 14 ط - 3 i 2 11 5 7 - 4:57 8 -3 12;-2 lo -6 8 14 (6) A*X=-3 Consider augmented matrix = [A16] = we have to find a particulas solution 9 13 ! of Ax= b so -405 5 30 +12+503 +7*4 = -2 2x1 + 1142 +883–314 +12855 -64 tous =14 130, + 9x2 + long fet ny=0, as=0 then 3%,+ a2 +503 2M,+11*2 +803 = -2 134, +992 +10M3 = 14 no 3 1 5 5 = 5 2 11 - 2 13 10 1 14 -By converting it into how reduced Echelon form, wegen 1 o 1 538/4-27 -42161 i 163/427 0 0 1. L so particular solution xp= 21 12 538/427 - 42 61 163427 713 Amwen 14 15 0 CScanned with CamScanner

(C) The general solution in given by antap where XHEN(A) 538 » x=CA -108 421 + C2 427 -42111 163 -124 443 427 111 61 -100g 721 1 427 61 580 427 0 1 o o 538 427 Am: a. where C1 and ca are arbitrary contants. (473) 4-4(e) c2 + (hot) (4- (124)62 - 42 (-4023 141 + ( ) C2 + 23 지 12 13 14 427 ns С. C2 CScanned with CamScanner
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