Schaum's Outline of Theory and Problems of Linear Algebra by Seymour Lipschutz

By Seymour Lipschutz

This ebook doesn't fairly educate Linear Algebra. it's totally superficial and never attention-grabbing. test Jim Hefferon's publication in its place and you'll have a transparent textual content that teaches Linear Algebra for real.

Anyway, there are numerous workouts within the ebook -- that's one cause to shop for it, I think...

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Extra info for Schaum's Outline of Theory and Problems of Linear Algebra (Schaum's Outlines)

Example text

5. Let …r  s† denote an r  s matrix. Find the sizes of those matrix products that are de®ned: (a) (b) …2  3†…3  4†; …4  1†…1  2†, (c) …1  2†…3  1†; (d) …5  2†…2  3†, (e) …4  4†…3  3† ( f ) …2  2†…2  4† In each case, the product is de®ned if the inner numbers are equal, and then the product will have the size of the outer numbers in the given order. 6. 1 Let A ˆ 2 (a) (c) not de®ned, (d) 5  3, 3 À1   2 and B ˆ 3 (e) not de®ned (f) 2  4 0 À2  À4 . Find: (a) AB, (b) BA. 6 Since A is a 2  2 and B a 2  3 matrix, the product AB is de®ned and is a 2 3 matrix.

2 0 À5 À2 Since A is a 2  3 and B a 3  4 matrix, the product AB is de®ned and is a 2  4 matrix. 8. Find: (a) (a) 1 À3 6 5   2 , À7  (b) 2 À7   6 , 5 1 À3  (c) ‰2; À7Š  1 6 . À3 5 The ®rst factor is 2  2 and the second is 2  1, so the product is de®ned as a 2  1 matrix:        1 6 2 2 À 42 À40 ˆ ˆ À3 5 À7 À6 À 35 À41 (b) The product is not de®ned, since the ®rst factor is 2  1 and the second factor is 2  2. 9. Clearly 0A ˆ 0 and A0 ˆ 0, where the 0's are zero matrices (with possibly different sizes).

11. 2(ii): A…B ‡ C† ˆ AB ‡ AC. Let A ˆ ‰aij Š, B ˆ ‰bjk Š, C ˆ ‰cjk Š, and let D ˆ B ‡ C ˆ ‰djk Š, E ˆ AB ˆ ‰eik Š, F ˆ AC ˆ ‰ fik Š. Then djk ˆ bjk ‡ cjk ; eik ˆ m P jˆ1 aij bjk ; fik ˆ m P jˆ1 aij cjk Lipschutz−Lipson:Schaum’s Outline of Theory and Problems of Linear Algebra, 3/e 2. Algebra of Matrices 46 © The McGraw−Hill Companies, 2004 Text ALGEBRA OF MATRICES [CHAP. 2 Thus the ik-entry of the matrix AB ‡ AC is eik ‡ fik ˆ m P jˆ1 aij bjk ‡ m P jˆ1 aij cjk ˆ m P jˆ1 aij …bjk ‡ cjk † On the other hand, the ij-entry of the matrix AD ˆ A…B ‡ C† is ai1 d1k ‡ ai2 d2k ‡ Á Á Á ‡ aim dmk ˆ m P jˆ1 aij djk ˆ m P jˆ1 aij …bjk ‡ cjk † Thus A…B ‡ C† ˆ AB ‡ AC, since the corresponding elements are equal.

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