# 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...

**Read or Download Schaum's Outline of Theory and Problems of Linear Algebra (Schaum's Outlines) PDF**

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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 Â s denote an r Â s matrix. Find the sizes of those matrix products that are de®ned: (a) (b) 2 Â 3 3 Â 4; 4 Â 1 1 Â 2, (c) 1 Â 2 3 Â 1; (d) 5 Â 2 2 Â 3, (e) 4 Â 4 3 Â 3 ( f ) 2 Â 2 2 Â 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 2 (a) (c) not de®ned, (d) 5 Â 3, 3 À1 2 and B 3 (e) not de®ned (f) 2 Â 4 0 À2 À4 . Find: (a) AB, (b) BA. 6 Since A is a 2 Â 2 and B a 2 Â 3 matrix, the product AB is de®ned and is a 2Â 3 matrix.

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

11. 2(ii): A B C AB AC. Let A aij , B bjk , C cjk , and let D B C djk , E AB eik , F AC fik . Then djk bjk cjk ; eik m P j1 aij bjk ; fik m P j1 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 m P j1 aij bjk m P j1 aij cjk m P j1 aij bjk cjk On the other hand, the ij-entry of the matrix AD A B C is ai1 d1k ai2 d2k Á Á Á aim dmk m P j1 aij djk m P j1 aij bjk cjk Thus A B C AB AC, since the corresponding elements are equal.