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Unformatted text preview: Notes on linear algebra (Monday 17th October, 2016, 23:10) page 179 3. Assume now that none of the values bfl_k+1,bn_k+2,...,bn is nonzero. Thus, all of the values bn_k+1, bn_k+2, . . ., b” are zero. Hence, the last a — k equations in the system (152) have the form 0 = 0, and therefore can be discarded (since they are automatically true). (Notice that if a — k = 0, then we are discarding nothing.) We are thus left with the first k equations: A1,1x1 + 141.2312 + ' ' ' + 141,me = 191; 142.1% + 142.2362 + ' ' ' + 1‘12,me = 192; (153) Akf1x1 + Aux; ‘i— ‘ ‘ ' ‘i— Akfmxm = bk 4. For each i E {1,2,....k}, set 90(1) 2 pivind(rowa). Then, Proposition 3.151 (c) says that <P(1)<¢(2)<m<¢(k)- [...] Proof of Theorem 3.157. We only need to show that in step 2 [...] D I" 1 ...
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