1\font\rm=cmr12
2\font\bf=cmb10 scaled \magstep 1
3\rm
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5
6\hsize = 5.5 in
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8
9\def\prob#1#2{\vskip .15in \par\noindent {\bf #1 \##2} }
10
11{\bigrm \centerline{Math 54 Solutions - Sections 1.7, 2.1 and 2.2}}
12\centerline{(this week by William Stein)}
13
14\vskip .5 in
15
16\noindent
171.7: 1,2,3,4,6
18
19\noindent
202.1: 1abc,2abc,3,4,5,8,9,10
21
22\noindent
232.2: 2,4,5,9,13
24
25\prob{1.7}{1}
26\par (a) The matrix $\pmatrix{2&0\cr0&-5}$ is invertible since all of its
27diagonal entries are nonzero and its inverse is
28$\pmatrix{{1\over 2}&0\cr 0&-{1\over5}}$.
29\par (b) The matrix $\pmatrix{4&0&0\cr0&0&0\cr0&0&5}$ is not invertible
30because one of its diagonal entries is zero.
31\par (c) The matrix is indeed invertible with inverse
32$$\pmatrix{-1&0&0\cr 0&{1\over2}&0 \cr 0&0&3}$$
33\prob{1.7}{2}
34\par (a) $\pmatrix{6&3\cr4&-1\cr4&10}$
35\par (b) $\pmatrix{-24&-10&12\cr3&-10&0\cr60&20&-16}$
36\prob{1.7}{3}
37\par (a) $A^2=\pmatrix{1&0\cr 0&4\cr}$, $A^{-2}=\pmatrix{1&0\cr 0&{1\over 4}}$,
38       $A^{-k}=\pmatrix{1&0\cr 0&4^{-k}}$
39\par (b) $A^2=\pmatrix{{1\over 4}&0&0\cr 0&{1\over 9}&0 \cr 0&0&{1\over 16}\cr}$,
40       $A^{-k}=\pmatrix{2^k&0&0\cr 0&3^k&0 \cr 0&0&4^k\cr}$
41
42\prob{1.7}{4}
43Only the matrices (b) and (c) are symmetric.
44\prob{1.7}{6}
45 If $A$ is symmetric then
46     \eqalign{a-2b+2c &= 3\cr 2a+b+c&=0\cr a+c&=-2\cr}
47   Solving this system we see that $a=11,b=-9,c=-13.$
48
49
50\prob{2.1}{1abc}
51(a) $3+0+1+1=5$, (b) $4+2+2+1=9$, (c) $2+1+2+1=6$
52\prob{2.1}{2abc}
53 (a) odd, (b) odd, (c) even
54\prob{2.1}{3}
55 $3\cdot 4 - (-2) \cdot 5 = 22$
56\prob{2.1}{4}
57 $4 \cdot 2 - 8 \cdot 1 = 0$
58\prob{2.1}{5}
59$(-5) \cdot (-2) - 6 \cdot (-7) = 10 + 42 = 52$
60\prob{2.1}{8}
61$-8 - 42 + 240 - [18 + 32 + 140] = 190 - 190 = 0$
62\prob{2.1}{9}
63$-20-7+72-20-84-6=-65$
64\prob{2.1}{10}
65$-5+42 -35 +6$
66
67\prob{2.2}{2}
68(a) $(3)(5)(-2) = -30$
69(b) $(\sqrt{2})(\sqrt{2})(-1)(1)=-2$
70(c) 0, because two rows are equal.
71(d) 0, because the second row is twice the first.
72\prob{2.2}{4}
73\eqalign{ 74\left | \matrix{3&6&-9\cr0&0&-2\cr-2&1&5}\right | 75&= 3 \left | \matrix{1&2&-3\cr0&0&-2\cr-2&1&5}\right | 76= 3 \left | \matrix{1&2&-3\cr0&0&-2\cr0&5&-1}\right | \cr 77&= -3 \left | \matrix{1&2&-3\cr0&5&-1\cr0&0&-2}\right | 78= (-3)(5)(-2)=30 }
79\prob{2.2}{5}
80\eqalign{ 81\left | \matrix{0&3&1\cr1&1&2\cr3&2&4}\right| 82&= - \left | \matrix{1&1&2\cr0&3&1\cr3&2&4}\right| 83= - \left | \matrix{1&1&2\cr0&3&1\cr0&-1&-2}\right| \cr 84&= \left | \matrix{1&1&2\cr0&-1&-2\cr0&3&1}\right| 85= \left | \matrix{1&1&2\cr0&-1&-2\cr0&0&-5}\right| 86=(1)(-1)(-5)=5}
87\prob{2.2}{9}
88\eqalign{ 89\left | \matrix{2&1&3&1\cr1&0&1&1\cr0&2&1&0\cr0&1&2&3}\right| 90&= - \left | \matrix{1&0&1&1\cr2&1&3&1\cr0&2&1&0\cr0&1&2&3}\right| 91= - \left | \matrix{1&0&1&1\cr0&1&1&-1\cr0&2&1&0\cr0&1&2&3}\right| \cr 92&= - \left | \matrix{1&0&1&1\cr0&1&1&-1\cr0&0&-1&2\cr0&0&1&4}\right| 93= - \left | \matrix{1&0&1&1\cr0&1&1&-1\cr0&0&-1&2\cr0&0&0&6}\right| 94= (1)(1)(-1)(6) = 6 95}
96\prob{2.2}{13}
97\eqalign{ 98\left | \matrix{1&1&1\cr{}a&b&c\cr{}a^2&b^2&c^2}\right| 99&= \left | \matrix{1&1&1\cr{}0&b-a&c-a\cr{}0&b^2-a^2&c^2-a^2}\right| 100= \left | \matrix{1&1&1\cr{}0&b-a&c-a\cr{}0&0&c^2-a^2-(c-a)(b+a)}\right| \cr 101&=(b-a)((c^2-a^2)-(c-a)(b+a))=(b-a)(c-a)(c+a-b-a)\cr 102&=(b-a)(c-a)(c-b)}
103
104
105\bye
106