136A5S

# 136A5S - ASSIGNMENT 5 for SECTION 001 Solutions Questions...

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ASSIGNMENT 5 for SECTION 001 Solutions Questions from the textbook Section 3-4: B3 (a), (d) and (e) [3 marks] ; B4 [2 marks] ; B6 [2 marks] ; B10 [3 marks] and D3 [3 marks] Section 4-1: B1 (a), (d) and (e) [3 marks] ; B2 (a) and (b) [2 marks] ; and D4 [6 marks] Section 4-2: B5 [3 marks] , D1 [2 mark] , D3 [3 marks] , D4 [3 marks] and D7 [3 marks] In addition to the questions listed here, do the MATLAB question from Assignment 5 for the other sections; this may be found under Content > Assignments > Assignment 5 [2 marks] Other questions 1. [5 marks] Let x 1 , . . . , x m and y 1 , . . . , y m be vectors in R n such that a 11 x 1 + a 12 x 2 + ··· + a 1 m x m = y 1 a 21 x 1 + a 22 x 2 + ··· + a 2 m x m = y 2 . . . a m 1 x 1 + a m 2 x 2 + ··· + a mm x m = y m and Rank a 11 a 12 ··· a 1 m a 21 a 22 ··· a 2 m . . . . . . . . . . . . a m 1 a m 2 ··· a mm = m. Prove that Span { x 1 , . . . , x m } = Span { y 1 , . . . , y m } . Since the y i are written as linear combinations of the x i , we have Span { x 1 , . . . , x m } ⊇ Span { y 1 , . . . , y m } . It remains to show that Span { x 1 , . . . , x m } ⊆ Span { y 1 , . . . , y m } . Let

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136A5S - ASSIGNMENT 5 for SECTION 001 Solutions Questions...

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