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Unformatted text preview: 4.3 .10 0100 00 m (rum.
2.1,etBbetl1ebasls [0 0].[0 o].[1 o],[0 1]ofR .Thenthecoormavs l l
1 l 1 1 2 _ 2 2 3
the given matrices with respecthnre l 1 B = 1 . 3 4 B— 3 . 5 7 B
1 2 l l l 2 1 l U l} —l
3 1 4 4 . . 1 2 3 4 U l U 4
5] .[6 3L — 6 . Finding Rd 1 3 5 6 — U u l _1 7E I4, we
?' 8 l 4 7 8 U 0 0 0
l l 2 l
1 2 3 4 .
muclude that the four Vumors 1 , 3 , r, . 6 are hnearly dependent. and so are
l 4 T H
. . l 4 1 1 l 2 2 3
the four given matrices. In fad. Li 8]  — [I 1] + 4 [3 4] — [5 7]. 48. Use a diagram: aoos(t)+bsin(t) T bcos{f)—asin(t] l J
[*3] T [—3] 0 l
ThusB— [_l 0]. 5). B = [[(b — l] 0050) — usi11(£)]a [at05(1) + (b — l) sinUﬂB] = [ita] bf: I]. 52. Recall that. was“  6) = cos(6)cos(r) + sin(6) slum and “mu — a) = cwMJSinU)  si:1(§)ms(f). Also, coshr/4) = 51mm) = Jim.
Thus 3 = [[wsu — arr/4)]; [sh(r — «x4113! = [[sgicostt} + !?siu(t)]a [—’?°03m+ jgﬂwﬂsl = a“? 1] U —1 ll
66. a. B: [[T(Jt‘f)]B [T‘IIIQHB [THEHB] = [2 U ~2J. U l {J
l U —1 0 —l
I). Note that Ire[(8) = 0 1 I) . Now im(B) = spam 2 . {l and
0 0 U 0 l
l
ker(B) = span 0 . Thus 23.1.22. —xf + is a basis of im(T) and If + 4:3 is a basis
1 of kerﬂ‘). 0.1. 11]. fi 1? = 2 + 31. + 4 = 6 + 3.1. The two Vectors enclose a right angle if E ii = 6 + :11; I
that is. in: = —2. 16. You may be able to find the solutiom by educated guming. Here is the systematic
approach: we ﬁrst find all vectors .i' that are orthogonal to 51,53, and 63. then we identify
the unit Wetors among them. Finding the vectors 53' with z"  5'. = E  12 = .ir'  (I; = U amounts to solving the system n+r2+xs+n=0
x.+zg—:3—.r4=0
ail—12+I3—qu=0 (we can Inuit all the coeﬂicienls {n f
. .. 272 —f
The solutions are of the form .1: = $3 = _t .
I; I
Since ijll = 2m, we have a unit vector if t = i or t = —%. Thus there are two possible choices for 174: 1 1 2 '2 _$ d %
nu . “i i __ I 2 'i 18. n. lffﬂ2 = 1+§+ﬁ+é+~  . = i—jr = i (use the formula for a geometric series, with n = so that llfﬂ = a: 1.155. I). Ifwe lot II: (13.0....) and t7: (1%. then 9 = arems 317"" = met.ij = = ﬁ(= 30°]. c. .i'.‘ = (I, 7'5.    . 7%,   does the job. since the harmonic series 1+ & +3'§+   diverges (a fact dimImam] in intnniuetory calculus classes].
d. Ifwr let 5: (1.0.0....}.£= (1. g.§.) and a: ﬁll = a? (14.5") then projﬂ: (s as: 5‘, (1.1. }.). 28. Sinoe the three given vectors in the subspace are orthogonal, we have the orthonormal! Imsls
l l l
.. l .. l _. l
ul=% l 1)::3' _l ru3=é _l
l —l I Now we con use Fact 5.1.5, with f = E", :projvf =01] wail + (172  5):}; + (53  53173 =
3 . 1
5 1
1 29. By the Pythagorean theorem (Feet 5.1.9). "in? = “7s. _ 3s, + 2.13, + s. — in“? .
= Il'i'ﬁ'lll2 + "352112 + “2'53"2 + Ill—114“2 + I1552_
= 49 + 9 + 4 + l + l
= 64. so that ;i" = 8. 40 “'52” = V172 "172 = M02 : 3.
42 “51+52H= x/(ﬁ'n +52)  (6: +172 = m. 46. Write the projection as a linear combination of if; and {£2 : (161 + ago). Now we want
1'33 i on?! + C252 to be perpendicular to V. that is, perpendicular to both 51 and 133. Using
dot products, this boils down to two linear equations in two unknowns, 11 = 3c1 + 5c;
and 20 = 561 + 902. with the solution C; = %,02 = Thus, the answer is —%'t71 + ...
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 Fall '05
 HUI
 Linear Algebra, Algebra

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