Unformatted text preview: Note Title 13. Here, 55‘ is not in V, as We ﬁnd an inconsistency while attempting to solve the system. 30. Let’s build B “oolumnhyeohuml”: B = [[TWL)]B[TW2)]BETW$}]5]
[H0 2 i W H0 2' 4] W H0 2 4] Hi i
= 2 —1 0 1 2 —1 D 1 2 —1 0 .2
4 4 1 1' 3 4 —4 1 2 B 4 —4 l 4 3
[H H [OH [1”] '
= 1 —l 0 = 0 H]. U .
1 B —2 B s B '. 0 o e 38. We went a basis 5 = (171,232) such that TEE} = as, and TE?) = be; for some scalars a and h. Then the B—matrix of T will he B = [[T{ﬂ'1}]g [Tfﬁzjlg] = [3 g], which 11? for vectors parallel to the
—u'5' = (—lJu'i for vectors perpendicular to L. lei to L and 6'2 is perpendicular, for example, is a diagonal matrix as required. Note that TIE‘U} = r? =
Line L about which we reﬂect, and Tftm 2
Thus, We can pick a. basis where 6', is para] editiﬁi)
HEN; *i:5i=i§i' I—IJ—t a? E 5” Q
5*
+
S1
+
£1.
+
51 I'i 6 The} = Tiﬁo — 61 — 63) = 4%) — T071) — Tfﬁg) =—e—e—e=e
Hence, T is a. rotati 9 ' "
given by on through 120 about the hne spanned by eg. Its matrix, B, is [[TiﬁiisiTiﬁ‘aiis [Thishsi where —I
Tfﬂi)=5’o = at; 452—5330 [21171)]3: [1] —1
0
Tirhj=ﬁgsoITW2JIs= 1
U
1
T(?Ta)_=17*1 50 [T(53}IH= U
U
—1 D 1
andB= —l 1 0 .
l 0 0 33 = Ia since if the tetrahedron rotates through 120° three times, it returns to the
original position. 73. e = 51113, where s = A = [:33 "233] Thus B = A = [2:38; ‘28]  ...
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 Spring '08
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