Question

Chapter 4, Section 4.5, Question 17 Find a basis for the subspace of that is spanned by the vectors V1 = (1, 0, 0), v2 = (1, 0, 1), V3 = (4, 0, 1), V4 = (0, 0, -2) O vi and v3 form a basis for span {V1, V2, V3, V4}. O v3 and V4 form a basis for span {V1, V2, V3, V4}. O vi and v2 form a basis for span {V1, V2, V3, V4}. vi and V4 form a basis for span {V1, V2, V3, V4}. O v2 and v3 form a basis for span {V1, V2, V3, V4}. O v2 and v4 form a basis for span {V1, V2, V3, V4}.

sorry for the bad picture, and thanks for your assitance in advance :)

Answer 1

using row echelon form the basis for the span has been found out

and the procedure is shown in detail
a V,= (1,0,0), v = (1,0,1), V3 =(4,0,1) V4=(0,0,-2) - The set s = {1, 12, U3, vu? of vectors in R3 is linearly independent if the only solution of CV + ₂ + ₂ V3 + luvy o is C1,C2, 3, 4y = 0 In this can s forms baris for span s otherwise is' is linearly dependent C. 9- 09 -03 = 10 + 12 + 4z + Oly = 0 oc+ og togtocy=0 ocit 14 +1G-264 = 0 R + R3 ooooL R 0 1 1 -2 Loi 1-2 011-2 as li 140 II-2 RR - R₂ Looool

- 1o 3 27 001-2 - reduced row echelon looool form which corresponds to the system 16+ + 363 +264 = 0 1₂ + 1 - 2 ly = 0 0 - 0 . The leading entities have been in block The columms in the matrix that do not contain leading entities, correspond to unknowry that will be arbitary The system has infinitely many solutions. = -36; -2ly a = 1₂ +2lu (3= Cy= arbitrary. since the valuables , cu are arbitrary , each vector "V3, Vy can be expressed as linear combination of vectors in set T = f V, V } for ex = 1, cu o in civ + 2 Vq + Cg Uz fly V = 2 then W=30+1V2 Since the set T = {v, rue} is lineally independent and it spans span s, then the set a formy basis for spans Therefore V, and a form basis for span {v, U2, U3, U4}