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Ch.1 Solutions Pg8

# Ch.1 Solutions Pg8 - Chapter 1 ISM Linear Algebra ﬁt = 3...

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Unformatted text preview: Chapter 1 ISM: Linear Algebra. ﬁt]. = 3;} _ 3g + r + [—2.5p+ 4g —1.5r}t +{I15p — q + {151-}:2. 31. m} is ofthe form ot2+ﬁt+o. So m} = a[12J+I:-{1}+c = 3, and 112} = a{22}+b{2}+c = 6. Also, f'ft] = 2at+ :5, meaning that f’[l] = 2u+h =1. n+ﬁ+c=3 Sowehaveasystemofequntinns: 45+2b+c=ﬁ 2a+b=1 (i=2 whichredunesto b=—3 . c=4 Thus, ﬂtj = ﬂtz — 3t + 4 is the only.r solution. 32. m} is ofthe farm ut2+bt+cn So,f[1]= 04:12] +b[1}+c= 1 and m} = 4n+ﬂb+o= 0. Also, ff f{t}dt = £2th + In: + old: 1: = §t3+§t2+ctﬁ =§a+2b+2c—[§+%+c} _'i' 3 _ n+b+c=1 du+2b+o=ﬂ §a+gb+c=—1 So we have a. system of equations: [1:9 which redunesto b=—28 . c=2ﬂ Thus, ﬁt) = 9:2 — 23¢ + 2D is the only solution. 33. m} is ofthe forrn otz +bt+c.f[1}= a+b+o = 1, m} = 9a+3b+c = 3,:1nd f’[t) =2nt+b, sof’{2j =4u+b= 1. u+b+c=1 Nowweset upour systerntnbe gu+3h+c=3. 4n+b=1 a—§=ﬂ Thisrsduces to ﬁ+§c=1. l]=l] ...
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