Question

Solving Systems of Linear Equations Using Linear Transformations In problems 1-5 find a basis for the solution set of the homogeneous linear systems. 2. X1 + x2 + x3 = 0 X1 – X2 – X3 = 0 3. x1 + 3x2 + x3 + x4 = 0 2xı – 2x2 + x3 + 2x4 = 0 x1 – 5x2 + x4 = 0 X1 + 2x2 – 2x3 + x4 = 0 X1 – 2x2 + 2x3 + x4 = 0.

Answer 1

34+92 +93 = 0 =o 24 - n₂-13 The w-efficient matrin A 1 aits apply row-reduction on A, '- - TD - 2 - 2 R₂-RI . I Ria R2 0 - The reduced sustem is sy =o 22 +13=0 5) H2=-3 - [81631-13) Hence x3 =23 212 3 [ng is free 1 sro a basis for the solution is,

the co-efficient madrin is 3. 4 + 302 +43 +244 = 0 284 -27% +13 +224 = 0 2 - 502 try to 2 - 2 1 2 1-5 01 F Rq - 2R, 314 7 -8-|-6 ri 314.7 0-8-1-6 R₃-R2 looos O . A . Rz-RI 0 -3 -1 -3 I & R3 1 1/2 R2 1347 69m frog 7 7 R-BR 101 ☆ § I Ri - 7 / RB J R₂ - 3 R3 rioſo or å ol Looo ù i the reduced system 74 + 5 ng=0 ^2 + Ing=0 m = -52 & M2 = - 12 5- 7 Iso [ns is free] ** [3] [51- Hence basis for the solution is

*4 +242 +243 +0 =0 94-224+2713 + 24 = 0 The nt matrix is,A 2 1 1 2-2 11 A - - R₂-R, w [.-** IOI -10 Ri-ZR2 The reduced system is, my = 0 =34 =-* >n2 =23 22 -13=0 T-7 ro7 To 731 + 3 L na . L ny (1,4, are free] 1 LI Hence solution to the basis ist