Sharedwww / 54ogus1.texOpen in CoCalc
Author: William A. Stein
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{\bigrm \centerline{Math 54 Solutions - Sections 1.7, 2.1 and 2.2}}
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\centerline{(this week by William Stein)}
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\vskip .5 in
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\noindent
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1.7: 1,2,3,4,6
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\noindent
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2.1: 1abc,2abc,3,4,5,8,9,10
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\noindent
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2.2: 2,4,5,9,13
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\prob{1.7}{1}
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\par (a) The matrix $\pmatrix{2&0\cr0&-5}$ is invertible since all of its
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diagonal entries are nonzero and its inverse is
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$\pmatrix{{1\over 2}&0\cr 0&-{1\over5}}$.
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\par (b) The matrix $\pmatrix{4&0&0\cr0&0&0\cr0&0&5}$ is not invertible
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because one of its diagonal entries is zero.
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\par (c) The matrix is indeed invertible with inverse
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$$\pmatrix{-1&0&0\cr 0&{1\over2}&0 \cr 0&0&3}$$
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\prob{1.7}{2}
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\par (a) $\pmatrix{6&3\cr4&-1\cr4&10}$
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\par (b) $\pmatrix{-24&-10&12\cr3&-10&0\cr60&20&-16}$
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\prob{1.7}{3}
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\par (a) $A^2=\pmatrix{1&0\cr 0&4\cr}$, $A^{-2}=\pmatrix{1&0\cr 0&{1\over 4}}$,
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$A^{-k}=\pmatrix{1&0\cr 0&4^{-k}}$
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\par (b) $A^2=\pmatrix{{1\over 4}&0&0\cr 0&{1\over 9}&0 \cr 0&0&{1\over 16}\cr}$,
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$A^{-k}=\pmatrix{2^k&0&0\cr 0&3^k&0 \cr 0&0&4^k\cr}$
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\prob{1.7}{4}
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Only the matrices (b) and (c) are symmetric.
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\prob{1.7}{6}
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If $A$ is symmetric then
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$$\eqalign{a-2b+2c &= 3\cr 2a+b+c&=0\cr a+c&=-2\cr}$$
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Solving this system we see that $a=11,b=-9,c=-13.$
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\prob{2.1}{1abc}
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(a) $3+0+1+1=5$, (b) $4+2+2+1=9$, (c) $2+1+2+1=6$
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\prob{2.1}{2abc}
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(a) odd, (b) odd, (c) even
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\prob{2.1}{3}
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$3\cdot 4 - (-2) \cdot 5 = 22$
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\prob{2.1}{4}
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$4 \cdot 2 - 8 \cdot 1 = 0$
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\prob{2.1}{5}
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$(-5) \cdot (-2) - 6 \cdot (-7) = 10 + 42 = 52$
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\prob{2.1}{8}
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$-8 - 42 + 240 - [18 + 32 + 140] = 190 - 190 = 0$
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\prob{2.1}{9}
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$-20-7+72-20-84-6=-65$
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\prob{2.1}{10}
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$-5+42 -35 +6$
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\prob{2.2}{2}
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(a) $(3)(5)(-2) = -30$
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(b) $ (\sqrt{2})(\sqrt{2})(-1)(1)=-2$
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(c) 0, because two rows are equal.
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(d) 0, because the second row is twice the first.
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\prob{2.2}{4}
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$$\eqalign{
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\left | \matrix{3&6&-9\cr0&0&-2\cr-2&1&5}\right |
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&= 3 \left | \matrix{1&2&-3\cr0&0&-2\cr-2&1&5}\right |
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= 3 \left | \matrix{1&2&-3\cr0&0&-2\cr0&5&-1}\right | \cr
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&= -3 \left | \matrix{1&2&-3\cr0&5&-1\cr0&0&-2}\right |
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= (-3)(5)(-2)=30 }$$
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\prob{2.2}{5}
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$$\eqalign{
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\left | \matrix{0&3&1\cr1&1&2\cr3&2&4}\right|
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&= - \left | \matrix{1&1&2\cr0&3&1\cr3&2&4}\right|
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= - \left | \matrix{1&1&2\cr0&3&1\cr0&-1&-2}\right| \cr
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&= \left | \matrix{1&1&2\cr0&-1&-2\cr0&3&1}\right|
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= \left | \matrix{1&1&2\cr0&-1&-2\cr0&0&-5}\right|
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=(1)(-1)(-5)=5}$$
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\prob{2.2}{9}
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$$\eqalign{
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\left | \matrix{2&1&3&1\cr1&0&1&1\cr0&2&1&0\cr0&1&2&3}\right|
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&= - \left | \matrix{1&0&1&1\cr2&1&3&1\cr0&2&1&0\cr0&1&2&3}\right|
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= - \left | \matrix{1&0&1&1\cr0&1&1&-1\cr0&2&1&0\cr0&1&2&3}\right| \cr
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&= - \left | \matrix{1&0&1&1\cr0&1&1&-1\cr0&0&-1&2\cr0&0&1&4}\right|
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= - \left | \matrix{1&0&1&1\cr0&1&1&-1\cr0&0&-1&2\cr0&0&0&6}\right|
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= (1)(1)(-1)(6) = 6
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}$$
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\prob{2.2}{13}
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$$\eqalign{
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\left | \matrix{1&1&1\cr{}a&b&c\cr{}a^2&b^2&c^2}\right|
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&= \left | \matrix{1&1&1\cr{}0&b-a&c-a\cr{}0&b^2-a^2&c^2-a^2}\right|
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= \left | \matrix{1&1&1\cr{}0&b-a&c-a\cr{}0&0&c^2-a^2-(c-a)(b+a)}\right| \cr
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&=(b-a)((c^2-a^2)-(c-a)(b+a))=(b-a)(c-a)(c+a-b-a)\cr
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&=(b-a)(c-a)(c-b)}$$
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\bye
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