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1%%% First homework for Lenstra's 54.
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8\newcommand{\prob}{\vspace{.2in}\par\noindent {\bf #1}\par}
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11\title{MATH 54\\Linear Algebra and Diff. Eqns.\\
12       Section 1 (Lenstra), assignment \# 1}
13\author{William Stein}
14
15\begin{document}
16\maketitle
17
18\begin{center}
19{\bf
20{\em Anton}: I.1, \# 5,7,9,13; I.2, \# 3,4ad,5bc,6ad,7bd,12
21}
22\end{center}
23
24\section{Chapter 1, Section 1}
25
26\prob{I.1.5}
27(a) \begin{align*}
282x_1 & = 0\\
293x_1 - 4x_2 & = 0\\
30x_2 & = 1
31\end{align*}
32
33(b) \begin{align*}
343x_1 - 2x_3 & = 5\\
357x_1 + x_2 +4x_3 & = -3\\
36-2x_2 + x_3 & = 7
37\end{align*}
38
39(c) \begin{align*}
407x_1+2x_2+x_3-3x_4 & = 5\\
41x_1+2x_2+4x_3 & = 1
42\end{align*}
43
44(d) \begin{align*}
45x_1 & = 7\\
46x_2 & = -2\\
47x_3 & = 3\\
48x_4 & = 4
49\end{align*}
50
51\prob{I.1.7}
52We must find a system of linear equations whose solution $(a,b,c)$
53give the coefficients of $y=ax^2+bx+c$. Since $y=ax^2+bx+c$
54passes through the three indicated points, then by substituting each
55of the three points $(x_1,y_1)$, $(x_2,y_2)$,
56and $(x_3,y_3)$ into the equation $y=ax^2+bx+c$ we see that
57\begin{align*}
58y_1& =ax_1^2+bx_1+c\\
59y_2& =ax_2^2+bx_2+c\\
60y_3& =ax_3^2+bx_3+c.
61\end{align*}
62This system of equations corresponds to the augmented matrix
63\begin{equation*}
64\begin{bmatrix}
65x_1^2&x_1&1&y_1\\
66x_2^2&x_2&1&y_2\\
67x_3^2&x_3&1&y_3
68\end{bmatrix}
69\end{equation*}
70
71\prob{I.1.9}
72(a) The three lines must not all intersect in a common point.
73For example they could form a "triangle" -- or two could be parallel.
74
75(b) The three lines all intersect in
76one common point.
77
78(c) All three of the lines coincide, that is,
79lie on top of each other.
80
81\prob{I.1.13}
82We must show that if the equations $x_1+kx_2=c$ and
83$x_1+lx_2=d$ have the same solution set, then
84the equations are in fact identical. So assume the
85two equations have the same solution set. To show
86that the equations are the same means to
87show that $k=l$ and $c=d$. To do this we first
88subtract the second equation from the first
89to see that $(k-l)x_2=c-d$. If $k\neq l$
90then $k-l\neq 0$ so we may divide through to
91see that $x_2=(c-d)/(k-l)$. But for every possible value
92of $x_2$ there is a corresponding value of $x_1$
93which provides a solution to $x_1+kx_2=c$ and
94hence to $x_1+lx_2=d$ as well (since they have
95the same solutions). Thus $x_2$ can't only take
96on the value $(c-d)/(k-l)$ so our assumption
97that $k\neq l$ must be false and hence $k=l$.
98But then $c-d=(k-l)x_2=0x_2=0$ so $c=d$ as well.
99This completes the proof.
100
101\section{Chapter 1, Section 2}
102
103\prob{I.2.3}
104(a) both
105
106(b) neither since the second and third rows
107violate property 3
108
109(c) both
110
111(d) row-echelon form, but not reduced row-echelon
112form because the second column contains a leading one
113but does not have zeros everywhere else which
114violates property 4
115
116(e) neither, because property 3 is violated by the
117second row
118
119(f) both
120
122(a) Once we recover the system of equations from
123its matrix form it is easy to read off the solutions.
124For this matrix, the corresponding system is
125\begin{equation*}
126\begin{align*}
127x_1 + 0 x_2 + 0 x_3& = -3\\
1280x_1 + x_2 + 0 x_3 & = 0 \\
1290x_1 + 0x_2 +  x_3 & = 7
130\end{align*}
131\end{equation*}
132so we see at one that the solution is
133$x_1=-3$, $x_2=0$, $x_3=7$.
134
135(d) The corresponding system of equations is
136\begin{align*}
137x_1-3x_2& = 0\\
138x_3 & = 0\\
1390 & = 1\\
140\end{align*}
141Since the final equation $0=1$ cannot be satisfied
142for any choice if $x_1, x_2, x_3$ we see that this
143system has no solutions.
144
145\prob{I.2.5bc}
146(b) The corresponding system is
147\begin{align*}
148x_1+8x_3-5x_4&=6\\
149x_2+4x_3-9x_4&=3\\
150x_3+x_4&=2
151\end{align*}
152Solving for the leading variables yields
153\begin{align*}
154x_1 & =  13 x_4 -10  \\
155x_2 & = 13 x_4 - 5 \\
156x_3 & = -x_4 + 2
157\end{align*}
158Since $x_4$ can be assigned an arbitrary value, $t$, there
159are infinitely many solutions. The general solution is
160given by the formulas
161\begin{equation*}
163\end{equation*}
164
165(c) The corresponding system is
166\begin{align*}
167x_1+7x_2-2x_3-8x_5&=-3\\
168x_3+x_4+6x_5&=5\\
169x_4+3x_5&=9\\
1700&=0
171\end{align*}
172Solving for the leading variables yields
173\begin{align*}
174x_1&=-65-7x_2-10x_5\\
175x_3&=-4-3x_4\\
176x_4&=9-3x_5
177\end{align*}
178Letting $x_2=s$ and $x_5=t$, the general solution is
179given by the formulas
180\begin{equation*}
183\end{equation*}
184
185\prob{I.2.6}
186(a) We first form the augmented matrix
187\begin{equation*}
188\begin{bmatrix}
1891&1&2&8\\
190-1&-2&3&1\\
1913&-7&4&10
192\end{bmatrix}
193\end{equation*}
194Reducing to reduced row-echelon form yields
195\begin{equation*}
196\begin{bmatrix}
1971&0&0&3\\
1980&1&0&1\\
1990&0&1&2\\
200\end{bmatrix}
201\end{equation*}
202
203(d) The corresponding matrix is
204\begin{equation*}
205\begin{bmatrix}
2060&-2&3&1\\
2073&6&-3&-2\\
2086&6&3&5\\
209\end{bmatrix}
210\end{equation*}
211Reducing to row-echelon form yields
212\begin{equation*}
213\begin{bmatrix}
2141&2&-1&-2/3\\
2150&-2&3&1\\
2160&0&0&6
217\end{bmatrix}
218\end{equation*}
219
220Since the bottom row corresponds to $0=6$ we
221see the system has no solution.
222
223
224
225\prob{I.2.7}
226(b)
227The corresponding augmented matrix is
228\begin{equation*}
229\begin{bmatrix}
2302&2&2&0\\
231-2&5&2&1\\
2328&1&4&-1
233\end{bmatrix}
234\end{equation*}
235Reducing to row-echelon form yields
236\begin{equation*}
237\begin{bmatrix}
2381&1&1&0\\
2390&7&4&1\\
2400&0&0&0
241\end{bmatrix}
242\end{equation*}
243Back substituting now yields the desired solution
244\begin{equation*}
245x_1=-\frac{3}{7}s-\frac{1}{7}\quad x_2=\frac{1}{7} - \frac{4}{7} s
246\quad x_3 = s
247\end{equation*}
248
249(d) As in I.2.6, the system is inconsistent.
250
251\prob{I.2.12}
252(a) Yes, since there are less equations than unknowns.
253
254(b) No, since there are the same number of equations
255as unknowns and there is no redundancy.
256
257(c) Yes.
258
259(b) Yes, since although there are the same number of equations
260as unknowns there is redundancy.
261
262\end{document}
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