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A ns. 1 (a) TRUE The truth that P is not unique exists only when there are all zero rows in R because one can arbitrarily append elementary matrices which multiply the all zero rows and add them on other rows resulting in the same R . That is, P’A = E k+1 PA = E k+1 R = R , where E k+1 stands for the elementary matrix operating on the all zero rows. But rank( A ) = m, which means there is no all zero rows in R . P is unique. Comment 3 (b) TRUE Assume ۯൌ ൥ a ଵଵ ڮa ଵ୬ ڭڰڭ a ୫ଵ ୫୬ ൩ ൌ ሾa …a Ö ۯ T ൌ ൥ a ڭ a ۯ T a a a a a a ൩ൌ۽ a a ൌa ଵଵ ൅ڮ൅a ୫ଵ ൌ0 a ଵଵ ൌڮൌa ୫ଵ The same, a ଵ୬ ୫୬ Therefore, all the elements in A are zeros. 1. 3 2. 只討論 2x2 matrix or 3x3 matrix 而沒有擴展至 general case => 2 3. Element index i, j 標示錯誤 => 1

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(c) Case 1: FALSE If “for every b” is regarded as “for all b in CS( A )”, Ö N( A ) 0 if N( A ) 0 => R( A ) n => det( A ) = 0 and A is not invertible. Case 2: TRUE If “for every b” is regarded as “for all b in R n ”, Ö Column space of A is the whole R n . => Rank of A n => det( A ) 0 and A is invertible. Comment 不管寫的答案是 True or False ，完全看 explanation 是不是夠 支持 True or False ，有符合才給分 (d) FALSE The vectors of S are not necessarily in V. S can be a set of basis of k dimension subspace but does not need to the basis of V. 3 (e) If linear transform T is one to one, Ö A x = b has at most one solution.
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96ä¸‹ç·šä»£æœŸä¸­è€ƒ é›

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