\documentclass[11pt]{amsart}
\usepackage{amsmath,amsthm,amsfonts,latexsym,amssymb,amscd,color}
\pagestyle{headings}
\setlength{\textwidth}{468true pt}
\setlength{\headheight}{0true pt} % needed for 12pt documents
\setlength{\oddsidemargin}{0 truept}
\setlength{\evensidemargin}{0 truept}
\setlength{\textheight}{580true pt}
\newtheorem{thm}{Theorem} %[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lm}[thm]{Lemma}
\newtheorem{prp}[thm]{Proposition}
\newtheorem{clm}[thm]{Claim}
%
\theoremstyle{definition}
\newtheorem{df}[thm]{Definition}

\begin{document}


\noindent
{\bf Solutions to HW 1.}

\bigskip
  
\begin{enumerate}
\item[1.] (Exercise 1.22.)
  For which values of $k$ are there no solutions, many solutions, or a unique
solution to this system?
\[
\begin{array}{ccl}
x&-y &= 1\\
3x&-3y &= k
\end{array}
\]
\end{enumerate}

\bigskip

\color{blue}
{\bf Solution.} The augmented matrix for this system is 
\[
\left[
  \begin{array}{rr|l}
  1&-1&1\\
  3&-3&k
\end{array}
\right].  
\]
The Reduced Row Echelon Form is 
\[
\left[
\begin{array}{rr|l}
1 & -1 & 1\\
3 & -3 & k\\
\end{array}
\right]
{\overset{-3R_1+R_2}{\xrightarrow{\hspace*{1.2cm}}}}
\left[
\begin{array}{rr|c}
1 & -1 & 1\\
0 & 0 & k-3\\
\end{array}
\right]
\]
\begin{enumerate}
\item[(i)] If there is a nonzero pivot to the right of the vertical bar,
  then the system is inconsistent (= no solution).
  This happens if $k-3\neq 0$, or equivalently if $k\neq 3$.
  Hence there is no solution if $k\neq 3$.
\item[(ii)] If $k=3$, then the matrix reduces to   
  \[
\left[
\begin{array}{rr|c}
1 & -1 & 1\\
0 & 0 & 0\\
\end{array}
\right]
\]
The second column does not have a pivot, so the second variable $y$
is a free variable. We can use the first equation to
solve for $x$ in terms of the free variable $y$: $x-y=1$,
so $x = 1+y$. The full solution in vector form is
\[
\left[
\begin{array}{c}
x\\
y
\end{array}
\right]  =
\left[
\begin{array}{c}
1+y\\
y
\end{array}
\right]  =
\left[
\begin{array}{c}
1\\
0
\end{array}
\right]  +
y\left[
\begin{array}{c}
1\\
1
\end{array}
\right].
\]
In this case there are infinitely many solutions, one for every choice of $y$.
\item[(iii)]
  There is no value of $k$ for which the system has a unique
  solution. (All values of $k$ are accounted for in the previous
  cases.)
\end{enumerate}
\color{black}

\bigskip
\bigskip

\begin{enumerate}
\item[2.] (Exercise 1.27.)
  Find the coefficients $a$, $b$, and $c$ so that the graph of
  $f(x) = ax^2 + bx + c$ passes
  through the points $(1, 2)$, $(-1, 6)$, and $(2, 3)$.
\end{enumerate}
\bigskip

\color{blue}
{\bf Solution.}
To pass through the points
we must have $f(1)=a(1)^2+b(1)+c=2, f(-1)=a(-1)^2+b(-1)+c=6$, and
$f(2)=a(2)^2+b(2)+c=3$, which means that
\[
\begin{array}{rrrl}
  a&+b&+c&=2\\
  a&-b&+c&=6\\
  4a&+2b&+c&=3.
\end{array}  
\]
The augmented matrix for this system is
\[
\left[
\begin{array}{rrr|c}
1 & 1 & 1 & 2\\
1&-1&1&6\\
4&2&1&3
\end{array}
\right]
\]
Apply Gausian Elimination to put this matrix in Row Echelon Form.
\[
\left[
\begin{array}{rrr|r}
1 & 1 & 1 & 2\\
1&-1&1&6\\
4&2&1&3
\end{array}
\right]
{\underset{-4R_1+R_3}{\overset{-R_1+R_2}{\xrightarrow{\hspace*{1.2cm}}}}}
\left[
\begin{array}{rrr|r}
1 & 1 & 1 & 2\\
0&-2&0&4\\
0&-2&-3&-5
\end{array}
\right]
{\overset{-R_2+R_3}{\xrightarrow{\hspace*{1.2cm}}}}
\left[
\begin{array}{rrr|r}
1 & 1 & 1 & 2\\
0&-2&0&4\\
0&0&-3&-9
\end{array}
\right]
\]
We can go a bit further and put the matrix in Reduced
Row Echelon Form.
\[
\left[
\begin{array}{rrr|r}
1 & 1 & 1 & 2\\
0&-2&0&4\\
0&0&-3&-9
\end{array}
\right]
{\underset{-(1/3)R_3}{\overset{-(1/2)R_2}{\xrightarrow{\hspace*{1.2cm}}}}}
\left[
\begin{array}{rrr|r}
1 & 1 & 1 & 2\\
0&1&0&-2\\
0&0&1&3
\end{array}
\right]
{\underset{-R_3+R_1}{\overset{-R_2+R_1}{\xrightarrow{\hspace*{1.2cm}}}}}
\left[
\begin{array}{rrr|r}
1 & 0 & 0 & 1\\
0&1&0&-2\\
0&0&1&3
\end{array}
\right].
\]
From the final matrix we can read off $c=3, b=-2, a=1$,
so $f(x) = x^2-2x+3$.
\color{black}

\bigskip
\bigskip

\begin{enumerate}
\item[3.] Find all triples $(x,y,z)$
of positive real numbers satisfying the nonlinear system
\[
\begin{array}{rl}
x^2y&=2\\
xy^2z&=4\\
yz^2&=8
\end{array}
\]
using the following method. First, take the
logarithm of each equation. Then replace
$\log(x)$ with a new symbol $X$,
$\log(y)$ with $Y$, $\log(z)$ with $Z$,
and solve the resulting linear system for $(X, Y, Z)$.
Use your answer to solve the original system.
(You can use a log to any base, but it is probably easiest
to use base $2$.)
\end{enumerate}

\bigskip

\color{blue}
Following the hint, we take base $2$ logarithm of each equation:
\[
\begin{array}{rrrl}
2\log_2(x)&+\log_2(y)& &=1\\
\log_2(x)&+2\log_2(y)&+\log_2(z)&=2\\
&\log_2(y)&+2\log_2(z)&=3
\end{array}
\]
Replace $\log_2(x)$ with $X$, $\log_2(y)$ with $Y$,
$\log_2(z)$ with $Z$ 
to obtain
\[
\begin{array}{rrrl}
2X&+Y& &=1\\
X&+2Y&+Z&=2\\
&Y&+2Z&=3.
\end{array}
\]
Solve:
\[
\left[
\begin{array}{rrr|r}
2 & 1 & 0 & 1\\
1 & 2 & 1 & 2\\
0 & 1 & 2 & 3
\end{array}
\right]
{\underset{-(2/3)R_2+R_3}{\overset{-(1/2)R_1+R_2}{\xrightarrow{\hspace*{1.2cm}}}}}
\left[
\begin{array}{rrr|r}
2 & 1 & 0 & 1\\
0&3/2&1&3/2\\
0&0&4/3&2
\end{array}
\right]
{\underset{(3/4)R_3}
{\overset{(1/2)R_1, (2/3)R_2}{\xrightarrow{\hspace*{1.2cm}}}}}
\left[
\begin{array}{rrr|r}
1 & 1/2 & 0 & 1/2\\
0&1&2/3&1\\
0&0&1&3/2
\end{array}
\right]
\]
We can put the last matrix in RRE Form as follows:
\[
\left[
\begin{array}{rrr|r}
1 & 1/2 & 0 & 1/2\\
0&1&2/3&1\\
0&0&1&3/2
\end{array}
\right]
{\underset{-(1/2)R_2+R_1}{\overset{-(2/3)R_3+R_2}{\xrightarrow{\hspace*{1.2cm}}}}}
\left[
\begin{array}{rrr|r}
1 & 0 & 0 & 1/2\\
0&1&0&0\\
0&0&1&3/2
\end{array}
\right].
\]
We read off the solution from the final matrix: $Z=3/2, Y = 0, X = 1/2$.
Therefore $\log_2(z)=3/2, \log_2(y)=0, \log_2(x)=1/2$.
Therefore $z = 2^{3/2}, y = 2^0 = 1, x = 2^{1/2}$.
\color{black}
\end{document}
