# hw7s(1) - MATH 4377 ADVANCED LINEAR ALGEBRA SUMMER 2011 Key...

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1. Problem 3 on Page 151. Solution . 0 0 1 0 1 0 1 0 0 - 1 = 0 0 1 0 1 0 1 0 0 , 1 0 0 0 3 0 0 0 1 - 1 = 1 0 0 0 1 / 3 0 0 0 1 , 1 0 0 0 1 0 - 2 0 1 - 1 = 1 0 0 0 1 0 2 0 1 . 2. (i) Find the rank of the matrices in Problem 2 (b), (f) on Page 165. (ii) For the matrices in Problem 2 (b), (f) on Page 165, ﬁnd a basis for its row space (the vector space spanned by the row vectors of the matrix) and a basis for its cloumn space (the vector space spanned by the column vectors of the matrix), Solution . (i) 2 (b) 1 1 0 2 1 1 1 1 1 -→ 1 1 0 0 - 1 1 0 0 1 -→ 1 0 0 0 1 0 0 0 1 . So its rank is 3 (i.e. the matrix is invertible). 2 (f) 1 2 0 1 1 2 4 1 3 0 3 6 2 5 1 - 4 - 8 1 - 3 1 -→ 1 2 0 1 1 0 0 1 1 - 2 0 0 2 2 - 2 0 0 1 1 5 -→ 1 2 0 1 1 0 0 1 1 - 2 0 0 0 0 2 0 0 0 0 7 -→ 1 2 0 1 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 So its rank is 3. 1

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hw7s(1) - MATH 4377 ADVANCED LINEAR ALGEBRA SUMMER 2011 Key...

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