20f_95s_2s

20f_95s_2s - A is the set of possible constant terms and is...

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20F In Class Exam 2/22/95 Solutions and References 1. See pages 212 and 107, respectively. 2. Since columns 1, 2 and 5 are pivot columns, we can compute the dimensions: dim(Col A ) = dim(Row A ) = 3 and dim(Nul A )=2 . The first, second and fifth columns of A are a basis for Col A . The nonzero rows of the reduced echelon form are a basis for Row A . A basis for Nul A is 0 - 1 1 0 0 and - 3 2 0 1 0 . 3. (a) There is not enough information because an elementary row operation may multiply a row by a constant. For example A + I 6 and the matrix that equals I 6 except that a 1 , 1 = - 4 both have I 6 for reduced row echelon form. The determinants are 1 and - 4. (b) Yes. Reduced echelon form being the identity is equivalent to the existence of the inverse. (Theorem 3.7 or 3.8b) (c) Yes. We know that the existence of the inverse guarantees unique solutions to A~x = ~ b . (Theorem 3.5) There are other arguments, too. (d) Yes. We are given dim(Nul A ) = 3. Hence dim(Col A )=8 - 3 = 5. Since Col
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Unformatted text preview: A is the set of possible constant terms and is not all of R 6 , inconsistent ~ b exist. Instead of the last sentence one could observe that since there are only 5 pivots, the echelon form of A will have 6-5 = 1 rows that are all zero. 4. (a) Let the i th component of ~ b j be b i,j . The j th column of AB is A ~ b j = b 1 ,j ~a 1 + . . . + b n,j ~a n , a linear combination of the columns of A . Hence A ~ b j is in Col A . (b) Since every column of AB is in Col A , it follows that Col ( AB ) is a subspace of Col A . Hence dim(Col ( AB )) dim(Col A ). Since dim Col is rank, we are done. (c) We have rank ( AB ) = rank (( AB ) T ) = rank ( B T A T ). Now use (b) with A and B replaced by B T and A T respectively to obtain rank ( B T A T ) rank ( B T ). We also have rank ( B T ) = rank ( B ). Putting all this together, rank ( AB ) rank ( B )....
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