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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.
 Fall '08
 Olson

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