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AME525_Homework_02_Solutions_091907

AME525_Homework_02_Solutions_091907 - mu#2#1 u1=(3.0.0...

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Unformatted text preview: mu #2 #1 u1=(3.0.0); llu1||=3 ; e1=(1,0,0) v2=u2~ce1 ; c=(u2.e1) c=2 ; _. _ l l v2=(0,2,2) ; ez—(0,:/—§-,:/-§") _ 5 V3=u3-Cle1-Czez; c1= (u3.e1) c1=l; Cz=(u3.ez) Cz—TZ‘ - — 0 1 in) . e =(0 “lg. 352:) V3—( ,—;7§,Vr2 9 3 a 2 ’ 2 "2. Suppose a1 (2i) + a2(3j) + a3(5i — 12k) + a4(i +j + k) = 0. Then 2011 + 5013 + a4 = 0, 3012 + a4 = 0, —12a3 + a4 = 0. By inspection we see (11 = —17/2,a2 = —4,a3 = 1,014 = 12 is a non-trivial solution of these equations. It follows that 2i, 3j, 5i— 12k, i+j +k are linearly dependent. 0. Then 3011 + 2012 = 0 and 4011 + 8012 = 0. The = a2 = 0. Thus the vectors are linearly independent. 20. A basis for S is the set of vectors (1, 0, 0,1,0),(0, 1, —1,—1, 0), (0,0,0, 0, 1), so S has dimension three. ‘22. A basis for S is the set of vectors (—1, 1,0,0) and (0,0, 1, 2), so S has dimension two. ...
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