Question

f reduce 3 19 30 21 1233 622 795 31 3ld-296 31 31 280 40 233e-205b-12ld 31 19 30 21 b 8163cl→ 29 -22 8 -6 d 1 5 2 3 e 233 411 1233 Sa-2c- 0 d +Se-3c 84a+ 54b-12le 56 100 5-1 9 31 8 29 1 a -4 19 -22 5 b 4 30 6 82 c 0 21 3 -6 3 d 56 196a-1466+295c 56 7d-3b-3e f reduce 9-4 31 19) 8 1 29 -22 -1 5 4 0)"(3021尺63八8-6八2 3 1) Is the set of vectors T set? If they are dependent give a specific numerical linear combination that justifies your answer 2) Does T span M? If your answer is no, give a specific vector that lies outside of SpanT but in M 3) Determine a basis of Span T that is a subset of the original vectors. (and state its dimension) 4) Determine a basis of Span T that is not a subset of original vectors but instead has more zeros.

Answer 1

Given Let A = [9 31 8 29 -1 -4 19 1 -22 5 4 30 6 8 2 0 21 3 -6 3 a b c d e] Rightarrow [31 0 0 0 0 19 -1233/31 0 0 0 30 -622/31 230/1233 0 0 21 -795/31 40/411 0 0 b 319-296/31 1233c-203b-121d/1233 5a-2c-d/5 d+ 5e-3c/5] The row reduced echelon form of A is given. Leaving the last column in A, all five rows contains precisely the same entries as are in vector in T \r\n Taking u_1 = [9 4 -4 0], u_2 = [31 30 19 21], u_3 = [8 6 1 3] u_4 = [29 8 -22 -6], u_5 = [-1 2 5 3] (1) T = {u_1, u_2, u_3, u_4, u_5} T is linearly dependent since 5 u_1 + 0 u_2 - 2 u_3 - u_4 + 0 u_5 = 0 -(1) is a non-zero linear combination of vectors in T,