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AME525_Homework_02_Solutions_092107

AME525_Homework_02_Solutions_092107 - H W#0 Section6.5 ’2...

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Unformatted text preview: H W #0). Section6.5 ’2. Suppose a1(2i) + 02(3j) + a3(5i — 12k) + 0146 + j + k) = 0. Then 2411 + 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 21, 3j, 5i—12k, i+j +k are linearly dependent. {8. SuppOSe a1(—1, 1, 0, 0, 0) +a2(0, —1, 1, 0, 0)+a3(0, 1, 1, 1,0) = 0. Then —a1 — a3 .= 0,012 + a3 = 0,03 = 0 from which we see the only solution is a; . the vectors are linearly independent. ' ‘ 10. Suppose 011(3, 0, 0, 4) + (12(2, 0, 0, 8) = 0. Then 3011 + 2012 = 0 and 401 + 80:2 — 0. The only solution of these equations is a] = 012 = 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 3' 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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