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\title{Summary of Day 1}
\author{William Gunther}
\date{May 19, 2014}
\begin{document}
\maketitle
\section{Objectives}
\begin{itemize}
\item Recognize linear equations.
\item Define a system of linear equations.
\item Build geometric intuition for a solution for a system of linear equations (in $\R^2$ and $\R^3$ anyway).
\item Solve a system back back substitution.
\item Express a system as an augmented matrix.
\end{itemize}
\section{Summary}
\begin{itemize}
\item An equation is \df{linear} if it is of the form \[ a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b \]
we call the $a_i$ the \df{coefficents} and $b$ the \df{constant}.
\example
\begin{itemize}
\item The following is a linear equation: \[ 2x + 3y = 1 \] The coefficients are $2$ and $3$ and the constant is $1$.
\item The following is also a linear equation: \[ \sqrt{2} x + \pi/4 y - \sin(\pi/5) z = z\] The coefficents are $\sqrt{2}, \pi/4$ and $\sin(\pi/5)$. Note: the coefficients can be any real numbers.
\item The following is not a linear equation: \[ 2x^2 + \sqrt{y} + \sin(z) = 1 \] The variables $x$, $y$ and $z$ all non-linear since it cannot be put in the above form.
\item The following appears not to be a linear equation: \[ 3x + \sin^2(x) - 2y = -\cos^2(y)\] But, if you do so algebra and use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ it is: \[ 3x - 2y = -1 \]
\end{itemize}
\remark We slightly contrast the notion of a linear expression with a linear function. We will see linear functions later in the course; they are function which have the property $f(x + y) = f(x) + f(y)$ and $f(c\cdot x) = c\cdot f(x)$. If you take a linear equation and solve it for one variable you do not necessarily get a linear function (can you see why?).
\item A \df{vector} (in the vector space $\R^n$) is an $n$-tuple of real numbers (i.e. a list of $n$ real numbers). We will use bold face for vectors, so $\vec{v}\in\R^n$. In blackboard notation, we will write a vector $\bar{v}$.
To give the \df{coordinates} (or \df{components}--the ordered elements from the $n$-tuple) of vector we either write $\vec{v}$ as a \df{column vector} or a \df{row vector} which we write as:
\[ \vec{v} = \begin{pmatrix} s_1\\s_2\\\vdots\\s_n\end{pmatrix} \qquad\qquad \vec{v} = [s_1, s_2, \ldots, s_n]\]
respectively.
\item A \df{solution} to a linear equation of $n$ variables (as expressed above) is a vector $\vec{v}$ of $\R^n$, $\vec{v} = [s_1, \ldots, s_n]$ where
\[
a_1 s_1 + \cdots a_n s_n = b
\]
i.e., when you replace the variables with the corresponding components of vector then the two sides of the equation are actually equal.
\example The vectors $\vec{v} = [0,3]$ and $\vec{w} = [2,2]$ are solutions to $x+2y = 6$
\item A \df{system of linear equations} is a finite set of linear equations, possibly with overlapping variables. A \df{solution} to a system is a solution to each of the equations in the system. The \df{solution set} for a system of equations is the set of all vectors which are solutions to the system.
\example The following is a system of equations:
\begin{alignat*}{3}
2x&+3y&+z &= 1\\
x&+2y&&=0
\end{alignat*}
The vector $\vec{v} = [0,0,1]$ is a solution, but it is not the entire set of solutions; do you see any other solutions to the system?
\item Geometrically, in if you have two linear equations these can be visualized as lines in the plane $\R^2$. The solution set of this system is the points of intersection. Similarly for $\R^3$, except the equations may also be planes.
\item Two systems are called \df{equivalent} if they have the same solution set.
\item A system is \df{consistent} if it has a solution (i.e. the solution set is nonempty). Otherwise, the system is \df{inconsistent}.
\theorem Every system that is consistent either has one solution or infinitely many solutions.
\example The above system is consistent because it has the solution $[0,0,1]$. The following system is inconsistent:
\begin{alignat*}{3}
2x&+3y&+z &= 1\\
2x&+3y&+z&=0
\end{alignat*}
\item \df{Back substitution} is an algorithm for which you can find the solution to particular systems of equations (see the Algorithms section).
\example The procedure for back substitution requires on a particular form of a system; we can do it when the system is `triangular.' It's best to illustrate it just with an example.
\begin{alignat*}{4}
x &&-y &&+z&=&0\\
&&2y &&-z&=&1\\
&& &&3z&=&-1
\end{alignat*}
We observe that $z = -1/3$, and then substitute that into the next equation from the bottom to get that $2y-(-1/3) = 1$, or that $y = 1/3$. We can then substitute both of those quantities into the top equation to get $x - (1/3) + (-1/3) = 0$, so $x = 0$. This gives us a solution $\vec{v} = [0,1/3,-1/3]$.
\item A \df{$m \times n$ matrix} is a grid of $m$ rows and $n$ columns. Each position in the matrix is filled by a real number, which we call an \df{entry} of the matrix.
\example The following is a $2\times 3$ matrix:
\[
\begin{pmatrix}
1&4&3\\
1&1&1
\end{pmatrix}
\]
\smallskip
\item A \df{coefficient matrix} corresponding to a system of $m$ equations with $n$ variables is a $m\times n$ matrix where with entry in the $i$th row and $j$th column is the coefficient of the $j$th variable in the $i$th equation.
\example the coefficient matrix corresponding to this system:
\begin{alignat*}{4}
x &&-y &&+z&=&0\\
&&2y &&-z&=&1\\
&& &&3z&=&-1
\end{alignat*}
is:
\[
\begin{pmatrix}
1&-1&1\\
0&2&-1\\
0&0&3
\end{pmatrix}
\]
\item An \df{augmented matrix} corresponding to a system of $m$ equations with $n$ variables is a $m\times n+1$ matrix where the left $n$ columns consists of the coefficient matrix and the rightmost column is the constants of each of the equations.
\example The augmented matrix corresponding to the above system is:
\[
\begin{amatrix}{3}
1&-1&1&0\\
0&2&-1&1\\
0&0&3&-1
\end{amatrix}
\]
\end{itemize}
\end{document}