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{\bigrm \centerline{Math 54 Solutions - Sections 1.7, 2.1 and 2.2}}
\centerline{(this week by William Stein)}

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\noindent
1.7: 1,2,3,4,6

\noindent
2.1: 1abc,2abc,3,4,5,8,9,10

\noindent
2.2: 2,4,5,9,13

\prob{1.7}{1}
\par (a) The matrix $\pmatrix{2&0\cr0&-5}$ is invertible since all of its
diagonal entries are nonzero and its inverse is 
$\pmatrix{{1\over 2}&0\cr 0&-{1\over5}}$.
\par (b) The matrix $\pmatrix{4&0&0\cr0&0&0\cr0&0&5}$ is not invertible
because one of its diagonal entries is zero. 
\par (c) The matrix is indeed invertible with inverse
$$\pmatrix{-1&0&0\cr 0&{1\over2}&0 \cr 0&0&3}$$ 
\prob{1.7}{2}
\par (a) $\pmatrix{6&3\cr4&-1\cr4&10}$
\par (b) $\pmatrix{-24&-10&12\cr3&-10&0\cr60&20&-16}$
\prob{1.7}{3}
\par (a) $A^2=\pmatrix{1&0\cr 0&4\cr}$, $A^{-2}=\pmatrix{1&0\cr 0&{1\over 4}}$,
       $A^{-k}=\pmatrix{1&0\cr 0&4^{-k}}$
\par (b) $A^2=\pmatrix{{1\over 4}&0&0\cr 0&{1\over 9}&0 \cr 0&0&{1\over 16}\cr}$,
       $A^{-k}=\pmatrix{2^k&0&0\cr 0&3^k&0 \cr 0&0&4^k\cr}$
                                                                   
\prob{1.7}{4}
Only the matrices (b) and (c) are symmetric.
\prob{1.7}{6}
 If $A$ is symmetric then
     $$\eqalign{a-2b+2c &= 3\cr 2a+b+c&=0\cr a+c&=-2\cr}$$
   Solving this system we see that $a=11,b=-9,c=-13.$       


\prob{2.1}{1abc}
(a) $3+0+1+1=5$, (b) $4+2+2+1=9$, (c) $2+1+2+1=6$               
\prob{2.1}{2abc}
 (a) odd, (b) odd, (c) even
\prob{2.1}{3}
 $3\cdot 4 - (-2) \cdot 5 = 22$     
\prob{2.1}{4}
 $4 \cdot 2 - 8 \cdot 1 = 0$     
\prob{2.1}{5}
$(-5) \cdot (-2) - 6 \cdot (-7) = 10 + 42 = 52$    
\prob{2.1}{8}
$-8 - 42 + 240 - [18 + 32 + 140] = 190 - 190 = 0$
\prob{2.1}{9}
$-20-7+72-20-84-6=-65$      
\prob{2.1}{10}
$-5+42 -35 +6$   

\prob{2.2}{2}
(a) $(3)(5)(-2) = -30$ 
(b) $ (\sqrt{2})(\sqrt{2})(-1)(1)=-2$
(c) 0, because two rows are equal. 
(d) 0, because the second row is twice the first.
\prob{2.2}{4}
$$\eqalign{
\left | \matrix{3&6&-9\cr0&0&-2\cr-2&1&5}\right |
&= 3 \left | \matrix{1&2&-3\cr0&0&-2\cr-2&1&5}\right |
= 3 \left | \matrix{1&2&-3\cr0&0&-2\cr0&5&-1}\right | \cr
&= -3 \left | \matrix{1&2&-3\cr0&5&-1\cr0&0&-2}\right |
= (-3)(5)(-2)=30 }$$ 
\prob{2.2}{5}
$$\eqalign{
\left | \matrix{0&3&1\cr1&1&2\cr3&2&4}\right|
&= - \left | \matrix{1&1&2\cr0&3&1\cr3&2&4}\right|
= - \left | \matrix{1&1&2\cr0&3&1\cr0&-1&-2}\right| \cr
&= \left | \matrix{1&1&2\cr0&-1&-2\cr0&3&1}\right|
= \left | \matrix{1&1&2\cr0&-1&-2\cr0&0&-5}\right|
=(1)(-1)(-5)=5}$$
\prob{2.2}{9}
$$\eqalign{
\left | \matrix{2&1&3&1\cr1&0&1&1\cr0&2&1&0\cr0&1&2&3}\right|
&= - \left | \matrix{1&0&1&1\cr2&1&3&1\cr0&2&1&0\cr0&1&2&3}\right|
= - \left | \matrix{1&0&1&1\cr0&1&1&-1\cr0&2&1&0\cr0&1&2&3}\right| \cr
&= - \left | \matrix{1&0&1&1\cr0&1&1&-1\cr0&0&-1&2\cr0&0&1&4}\right| 
= - \left | \matrix{1&0&1&1\cr0&1&1&-1\cr0&0&-1&2\cr0&0&0&6}\right| 
= (1)(1)(-1)(6) = 6
}$$
\prob{2.2}{13}
$$\eqalign{
\left | \matrix{1&1&1\cr{}a&b&c\cr{}a^2&b^2&c^2}\right|
&= \left | \matrix{1&1&1\cr{}0&b-a&c-a\cr{}0&b^2-a^2&c^2-a^2}\right|
= \left | \matrix{1&1&1\cr{}0&b-a&c-a\cr{}0&0&c^2-a^2-(c-a)(b+a)}\right| \cr
&=(b-a)((c^2-a^2)-(c-a)(b+a))=(b-a)(c-a)(c+a-b-a)\cr
&=(b-a)(c-a)(c-b)}$$


\bye