Question

1. For the following two systems of linear equations answer the questions 4 + x + 2xy + 2x - 6 3x + 2x + 3x3 + 3x = 11 2x + 2x + 3.5+ 2x- 9 2x + 2x+4x3+5x - 13 3x, +2, +4x3+4x-13 3x+3x+3x2+2x, -11 (1) Solve the above systems of linear equations using naive Gauss elimination (b) solve the above systems of linear equations using Gauss elimination with partial pivoting . Axb 2. For the following matrix and vectors, solve the systems of linear equations, Axb and Axb, using Lu factorization 5 12] 3 2 3 [ 4 8 ] 10 [6] M 3. Solve the following equation using Lu factorization with partial pivoting + xy + 2xy + 2x = 6 + x + 4x + 5x = 11 3x + 2x + 3x + 4x = 12 2 +2x+3x+2x =10

Answer 1

1 ija) -> (1) 4 + Mg + ZM3 tiny = 6 1+302 +423+ 40 =12 3m +242 + 33 +39 = 1 224 +222 +34z + 204 =9 In the first step, to eliminate 3rd and 4th equations, multiply equation successively by a in the from and first -1,-3,-2 and third add respectively to the seconda and fourth equations , we get Y -13) nith+ Zn2+ 24 =6 2 ZM2+ 223 + 204=6 -22-303-304 =-7 - m - 229=-3 second step, to eliminate de from 3rd equation of the system (3), we Now, in

Multiply the and equation by and add to the 3rd- equation, we get MtM+ Zn +2n4=6 202 + 2M3 tana = 6 -243-244-4 I (3) -M3 – 2n4 =-3 Now, in the third step, to eliminate Mz from 4th equation of (3), we multiply the 3rd equation by and add to the 4th equation . Then we get hth2 +243 +224 = 6 l 202 + Znz tzna 26 call -20-204 =-4 any=-1 The equations (4) are equivalent to the given system (1). Hence the solution be na=1 =1, Nazi, m=1. mal; m2=1,83 = nazi

1) ü)a) 4+H2 + Zn₂+2y =6 1 2m +242 +40 +50=13 34+ 2M2 +4n3 tang = 13 34 +3024303 + zna = 1 ) > >(1) successively by Multiply the first equation 22 ,-3 ,-3 and add respectively to 4th equations. Then we the 2nd 3rd and get Mthathrat zna = 6 14 = Tot -n-203 - 209 = -5 -3 m; _414 = -7 from (2), 24 =1. Putting these values on (2) owe get M+22+ ZN 3 24 - 22-Zn3 =-3 -393=-3. :, ang – 2.(1)=-3. . Mezi 4+1 +2:1 = 4 > m= 1

Hence the solution be 14=1, nat, m=1, mail a ) 24+2 + 2z +22426 M +372 +42 +4n = 12 34+ 202 +373 +30=11 244 +242 +30₂ ton = 9 Now, we consider A = Ax =b, where 3 A 4 1 3 2 3 3 2 2 321 Nwo- MN N W WIN : 2 4:12 - [A16] = ri 1 2 4 1 1 1 3 3 2 R2-R1 R3-3R RA-2R1 1 2 Til 2 267 lo 113 oo-2 -2 - 4 1 2 a1 OOO-a 2 216 2 2:61 -3 -367 R2 Reth R 10o-1-27-31 II 2 2 6 - 2k3 1 0 1 1 1 Ls RA - 2 R3 loo Ilie Loo 0 -1:-

- Ric Re 0 ! 13. Re-R37 Too o! Rg+RA Tooloin - 0 - RA TR-R3 | RI-Rq i o o o '17 -000 0-00 o-oo Hence the solution be u=1 mai, mai ,naal. 4&h2 + 2m + 20 = 6 24 +262 +40 +5n =13 34. + 2a + 40 + 40 =13 34+34 +373 +2n =il be consider the system (1) as An=b, where Azril 2 27 1 T 2 T 3 L 3 WAAN WNN- duod b z 1 mm =

2:6 1 R2-24 NNM Nram durd Tenne - ) . 3 3 3 2 u RS-3R, RA-BR Fio o 0111 TII 22 :56 OO-3-4-7 7 R1+R3 looo! o-1-2-2 - 5 R RA o- Loo-3-4-5 -2 -2:05 R2 R3 -5 -2 ( 7 -7 -4 ao OToo 0 Too owno Owno Ti 0 0 ,7 oi- 01-36 o 1+ 2013 -R2 0.0 Iolit 22 or lil Rz+ARA R2 +2R4 To o os | R2-2R3 ri 0 o 0:17 LOOO - o-oo 00-0 000- -- Love - Hence the solution be mz1, 1221 , m= 7, n=1.

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