ans1 - Answers to Practice Problem Set 1 1. (a) 1 −2 0...

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Unformatted text preview: Answers to Practice Problem Set 1 1. (a) 1 −2 0 −1 6 0 01 0 −1 (b) x1 = 2x2 + x4 + 6; x3 = −1; x2 and x4 are free. 0 1 1 2. (a) {b ∈ R2 : b1 = − 1 b2 } (c) S = 0 , 1 , 1 , for instance. 3 0 0 0 x1 x2 = x1 , for instance. (b) 10 (c) 0 −1 (d) f (x) = e1 3. (a) T x1 13 −1 0 x3 4. (a) False. The system x1 = 0, x2 = 0, x3 = 0, x4 = 0 has only one solution. (b) True, since 3 > 2. (c) False. A(−p) = −Ap = −b is not equal to b unless b = 0. (d) True. Solving Ax = b amounts to expressing b as a linear combination of the columns of A. 5. (a) x1 = −7x3 − 2; x2 = −3x3 − x4 + 1; x3 and x4 are free. 2 2 2 1 (b) (i) −2 (ii) c · −2 : c ∈ R (iii) 2 + c · −2 : c ∈ R 2 2 −1 2 0 1 1 0 , 1 , 1 6. (b) 0 0 0 √ √ 1/√2 −1/ 2 √ 7. (b) 1/ 2 1/ 2 (c) (i) No (ii) No (iii) Yes 8. (a) False. For instance, two of the equations might be x1 + x7 = 1 and x1 + x7 = 0. 14 (b) True. In fact T = TA with A = 2 5. 36 1 2 = 2T (c) False. We must have T 1 2 (d) True. If c1 x1 + c2 x2 + c3 x3 = 0 with ci = 0 for some 1 ≤ i ≤ 3 then c1 x1 + c2 x2 + c3 x3 + 0x4 = 0. (e) False. For instance, x4 might be 0. 9. (a) True (b) False (c) True (d) False 5 −6 −2 00 10 1 (b) A = 10. (a) −2 3 ,B= 01 00 0 −1 0 11. (a) −2 3/2 (b) False (c) i. B −1 A−1 ii. (A−1 )5 iii. (A−1 )T 1 −1/2 ...
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This note was uploaded on 12/27/2011 for the course MAS 3114 taught by Professor Olson during the Fall '08 term at University of Florida.

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