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Ch.1 Solutions Pg9

# Ch.1 Solutions Pg9 - 34 35 3E 31 ISM Linear Algebra Section...

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Unformatted text preview: 34. 35. 3E. 31'. ISM: Linear Algebra Section 1.1 We write everything in terms of a, revealing e = 3n. and E: = 1 — 4n. So, ﬁt] = atz +[1— 4a]t+ 3&- for an arbitraryr a. m} =at2+bt+e, so ﬂl}=a+b+e=1,_f[3}=9o+3b+c=3. Also, :12} =3,so 2(2}a+b=4a+b=3. a+b+c=1 ﬂa+3b+e=3 4a+b=3 Thus, our system is When we reduce this, however, our last equation becomes {1 = 3, meaning that this system is inconsistent. m} = seat + hem, so ﬁll} = a + 5 =1 and f'[t) = sum“ + 256*, so f’{I]} = 3n + 2-5 = 4. n+b=1 Thus we obtain the system 3n. + 3!? = 4 1| which reveals {1:2 ‘ b=—1' So ﬁt] = 2e“ — a”. ﬁt] 2 aensﬂt} + lisinlﬂt] and 3f[t} + 2f’{t] + f"[t}=1'?ms[3t}. f’{t] = 2b eos[2t] — 3a sin{3t} and f"[t) = —4b sinfﬂt} — 4a oosfﬂt}. So, 17 nos[2t) = 3(a. ensfﬂt} +5 sin{2t}]+3{3-ﬁ ensﬂt} —2a sinﬂt] )—l— [—43 sinfﬂt}—t1u eosfﬂtﬂ = (—40. + 4!:- + 3n} eosfﬂt} + (—4!) — via. + 3b] sinﬁﬂt] = [—a. + 4-5] ens(2t} + (—4-: — b] sin{2t}. . _ —:J.+£1E1I=1'.r So,oursystem1s. —4a—b=ﬂ ‘ This reduces to' a = _1 . b: 4 . So our function is ﬁt] = — ensﬂt] + dsin[3t]. 3 — z = _5";+2b The given system reduces to y + 22 = 4.134 I] = a. — 3!? + c This system has solutions [in fact inﬁnitely many} if a. — 2!: + o = D. The points (a, he} with this property form a plane through the origin. 9 ...
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