# F10HW4 - F 2010 Homework 4 Section 2.3 1.b B is obtained...

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F 2010. Homework 4. Section 2.3. 1.b) B is obtained from A by interchanging the ﬁrst and the third rows. 4. A 1 0 0 - 2 3 0 1 0 - 1 0 0 0 1 1 0 0 0 0 0 0 while B 1 0 0 - 2 0 0 1 0 - 1 0 0 0 1 1 0 0 0 0 0 1 A and B don’t have the same reduced echelon form. They are not row equivalent. (If they were augmented matrices of linear systems, the ﬁrst one would be consistent, the second one wouldn’t.) 6. b) m > n , corollary 2.6 does not apply. - 1 4 - 19 5 1 - 11 4 - 5 - 32 2 1 - 2 1 - 2 - 11 1 0 - 3 0 1 4 0 0 0 0 0 0 0 0 0 Rank = 2 < 3, from Th.2.5 there should be non trivial solutions.As a matter of fact, the solution set is given by x 1 = 3 c x 2 = - 4 c x 3 = c S = { c 3 - 4 1 | c R } . 8 b) This problem is equivalent to solving for y the system: y 1 2 1 4 + y 2 1 - 1 3 + y 3 3 2 5 = 5 9 5 which in turn is equivalent to solving the linear system associated with the augmented matrix: 2 1 3 | 5 1 - 1 2 | 9 4 3 5 | 5 . By Gauss elimination, this matrix is row equivalent to 1 - 1 2 | 9 0 3 - 1 | - 13 0 - 7 - 3 | - 31 1 - 1 2 | 9 0 3 - 1 | - 13 0 0 - 2 / 3 | - 2 / 3 1 0 0 | 3 0 1 0 | - 4 0 0 1 | 1 y 1 = 3 , y 2 = - 4 , y 3 = 1, x = 3 a 1 - 4 a 2 + a 3 . 8.d) This problem is equivalent to solving for y the system: y 1 6 - 2 3 + y 2 0 - 5 - 1 + y 3 - 2 1 2 = 2 2 3 1

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F10HW4 - F 2010 Homework 4 Section 2.3 1.b B is obtained...

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