From A First Course in Linear Algebra

Version 2.00

© 2004.

Licensed under the GNU Free Documentation License.

http://linear.ups.edu/

We will motivate our study of linear algebra by considering the problem of
solving several linear equations simultaneously. The word “solve” tends to
get abused somewhat, as in “solve this problem.” When talking about
equations we understand a more precise meaning: find all of the values of
some variable quantities that make an equation, or several equations,
true.

Example STNE

Solving two (nonlinear) equations

Suppose we desire the simultaneous solutions of the two equations,

You can easily check by substitution that $x=\frac{\sqrt{3}}{2},\phantom{\rule{3.26288pt}{0ex}}y=\frac{1}{2}$ and $x=-\frac{\sqrt{3}}{2},\phantom{\rule{3.26288pt}{0ex}}y=-\frac{1}{2}$ are both solutions. We need to also convince ourselves that these are the only solutions. To see this, plot each equation on the $xy$-plane, which means to plot $\left(x,\phantom{\rule{0em}{0ex}}y\right)$ pairs that make an individual equation true. In this case we get a circle centered at the origin with radius 1 and a straight line through the origin with slope $\frac{1}{\sqrt{3}}$. The intersections of these two curves are our desired simultaneous solutions, and so we believe from our plot that the two solutions we know already are indeed the only ones. We like to write solutions as sets, so in this case we write the set of solutions as

$$\begin{array}{llll}\hfill S& =\left\{\left(\frac{\sqrt{3}}{2},\phantom{\rule{0em}{0ex}}\frac{1}{2}\right),\phantom{\rule{0em}{0ex}}\left(-\frac{\sqrt{3}}{2},\phantom{\rule{0em}{0ex}}-\frac{1}{2}\right)\right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ $\u22a0$In order to discuss systems of linear equations carefully, we need a precise definition. And before we do that, we will introduce our periodic discussions about “Proof Techniques.” Linear algebra is an excellent setting for learning how to read, understand and formulate proofs. But this is a difficult step in your development as a mathematician, so we have included a series of short essays containing advice and explanations to help you along. These can be found back in Section PT of Appendix P, and we will reference them as they become appropriate. Be sure to head back to the appendix to read this as they are introduced. With a definition next, now is the time for the first of our proof techniques. Head back to Section PT of Appendix P and study Technique D. We’ll be right here when you get back. See you in a bit.

Definition SLE

System of Linear Equations

A system of linear equations is a collection of
$m$ equations in the
variable quantities ${x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}{x}_{3},\dots ,{x}_{n}$
of the form,

where the values of ${a}_{ij}$, ${b}_{i}$ and ${x}_{j}$ are from the set of complex numbers, ${\u2102}^{}$. $\u25b3$

Don’t let the mention of the complex numbers, ${\u2102}^{}$, rattle you. We will stick with real numbers exclusively for many more sections, and it will sometimes seem like we only work with integers! However, we want to leave the possibility of complex numbers open, and there will be occasions in subsequent sections where they are necessary. You can review the basic properties of complex numbers in Section CNO, but these facts will not be critical until we reach Section O. For now, here is an example to illustrate using the notation introduced in Definition SLE.

Example NSE

Notation for a system of equations

Given the system of linear equations,

we have $n=4$ variables and $m=3$ equations. Also,

$$\begin{array}{llllllllllllllllllll}\hfill {a}_{11}& =1\phantom{\rule{2em}{0ex}}& \hfill {a}_{12}& =2\phantom{\rule{2em}{0ex}}& \hfill {a}_{13}& =0\phantom{\rule{2em}{0ex}}& \hfill {a}_{14}& =1\phantom{\rule{2em}{0ex}}& \hfill {b}_{1}& =7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{21}& =1\phantom{\rule{2em}{0ex}}& \hfill {a}_{22}& =1\phantom{\rule{2em}{0ex}}& \hfill {a}_{23}& =1\phantom{\rule{2em}{0ex}}& \hfill {a}_{24}& =-1\phantom{\rule{2em}{0ex}}& \hfill {b}_{2}& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{31}& =3\phantom{\rule{2em}{0ex}}& \hfill {a}_{32}& =1\phantom{\rule{2em}{0ex}}& \hfill {a}_{33}& =5\phantom{\rule{2em}{0ex}}& \hfill {a}_{34}& =-7\phantom{\rule{2em}{0ex}}& \hfill {b}_{3}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Additionally, convince yourself that ${x}_{1}=-2$, ${x}_{2}=4$, ${x}_{3}=2$, ${x}_{4}=1$ is one solution (but it is not the only one!). $\u22a0$

We will often shorten the term “system of linear equations” to “system of equations” leaving the linear aspect implied. After all, this is a book about linear algebra.

The next example illustrates the possibilities for the solution set of a system of linear equations. We will not be too formal here, and the necessary theorems to back up our claims will come in subsequent sections. So read for feeling and come back later to revisit this example.

Example TTS

Three typical systems

Consider the system of two equations with two variables,

If we plot the solutions to each of these equations separately on the ${x}_{1}{x}_{2}$-plane, we get two lines, one with negative slope, the other with positive slope. They have exactly one point in common, $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2}\right)=\left(3,\phantom{\rule{0em}{0ex}}-1\right)$, which is the solution ${x}_{1}=3$, ${x}_{2}=-1$. From the geometry, we believe that this is the only solution to the system of equations, and so we say it is unique.

Now adjust the system with a different second equation,

$$\begin{array}{llll}\hfill 2{x}_{1}+3{x}_{2}& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 4{x}_{1}+6{x}_{2}& =6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$A plot of the solutions to these equations individually results in two lines, one on top of the other! There are infinitely many pairs of points that make both equations true. We will learn shortly how to describe this infinite solution set precisely (see Example SAA, Theorem VFSLS). Notice now how the second equation is just a multiple of the first.

One more minor adjustment provides a third system of linear equations,

$$\begin{array}{llll}\hfill 2{x}_{1}+3{x}_{2}& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 4{x}_{1}+6{x}_{2}& =10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$A plot now reveals two lines with identical slopes, i.e. parallel lines. They have no points in common, and so the system has a solution set that is empty, $S=\varnothing $. $\u22a0$

This example exhibits all of the typical behaviors of a system of equations. A subsequent theorem will tell us that every system of linear equations has a solution set that is empty, contains a single solution or contains infinitely many solutions (Theorem PSSLS). Example STNE yielded exactly two solutions, but this does not contradict the forthcoming theorem. The equations in Example STNE are not linear because they do not match the form of Definition SLE, and so we cannot apply Theorem PSSLS in this case.

With all this talk about finding solution sets for systems of linear equations, you might be ready to begin learning how to find these solution sets yourself. We begin with our first definition that takes a common word and gives it a very precise meaning in the context of systems of linear equations.

Definition ESYS

Equivalent Systems

Two systems of linear equations are equivalent if their solution sets are equal.
$\u25b3$

Notice here that the two systems of equations could look very different (i.e. not be equal), but still have equal solution sets, and we would then call the systems equivalent. Two linear equations in two variables might be plotted as two lines that intersect in a single point. A different system, with three equations in two variables might have a plot that is three lines, all intersecting at a common point, with this common point identical to the intersection point for the first system. By our definition, we could then say these two very different looking systems of equations are equivalent, since they have identical solution sets. It is really like a weaker form of equality, where we allow the systems to be different in some respects, but we use the term equivalent to highlight the situation when their solution sets are equal.

With this definition, we can begin to describe our strategy for solving linear systems. Given a system of linear equations that looks difficult to solve, we would like to have an equivalent system that is easy to solve. Since the systems will have equal solution sets, we can solve the “easy” system and get the solution set to the “difficult” system. Here come the tools for making this strategy viable.

Definition EO

Equation Operations

Given a system of linear equations, the following three operations will transform
the system into a different one, and each operation is known as an equation
operation.

- Swap the locations of two equations in the list of equations.
- Multiply each term of an equation by a nonzero quantity.
- Multiply each term of one equation by some quantity, and add these terms to a second equation, on both sides of the equality. Leave the first equation the same after this operation, but replace the second equation by the new one.

These descriptions might seem a bit vague, but the proof or the examples that follow should make it clear what is meant by each. We will shortly prove a key theorem about equation operations and solutions to linear systems of equations. We are about to give a rather involved proof, so a discussion about just what a theorem really is would be timely. Head back and read Technique T. In the theorem we are about to prove, the conclusion is that two systems are equivalent. By Definition ESYS this translates to requiring that solution sets be equal for the two systems. So we are being asked to show that two sets are equal. How do we do this? Well, there is a very standard technique, and we will use it repeatedly through the course. If you have not done so already, head to Section SET and familiarize yourself with sets, their operations, and especially the notion of set equality, Definition SE and the nearby discussion about its use.

Theorem EOPSS

Equation Operations Preserve Solution Sets

If we apply one of the three equation operations of Definition EO to a system of linear
equations (Definition SLE), then the original system and the transformed system are
equivalent. $\square $

Proof We take each equation operation in turn and show that the solution sets of the two systems are equal, using the definition of set equality (Definition SE).

- It will not be our habit in proofs to resort to saying statements are “obvious,” but in this case, it should be. There is nothing about the order in which we write linear equations that affects their solutions, so the solution set will be equal if the systems only differ by a rearrangement of the order of the equations.
- Suppose $\alpha \ne 0$
is a number. Let’s choose to multiply the terms of equation
$i$ by
$\alpha $
to build the new system of equations,
$$\begin{array}{llll}\hfill {a}_{11}{x}_{1}+{a}_{12}{x}_{2}+{a}_{13}{x}_{3}+\cdots +{a}_{1n}{x}_{n}& ={b}_{1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{21}{x}_{1}+{a}_{22}{x}_{2}+{a}_{23}{x}_{3}+\cdots +{a}_{2n}{x}_{n}& ={b}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{31}{x}_{1}+{a}_{32}{x}_{2}+{a}_{33}{x}_{3}+\cdots +{a}_{3n}{x}_{n}& ={b}_{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \vdots \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \alpha {a}_{i1}{x}_{1}+\alpha {a}_{i2}{x}_{2}+\alpha {a}_{i3}{x}_{3}+\cdots +\alpha {a}_{in}{x}_{n}& =\alpha {b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \vdots \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{m1}{x}_{1}+{a}_{m2}{x}_{2}+{a}_{m3}{x}_{3}+\cdots +{a}_{mn}{x}_{n}& ={b}_{m}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
Let $S$ denote the solutions to the system in the statement of the theorem, and let $T$ denote the solutions to the transformed system.

- Show $S\subseteq T$.
Suppose $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}\dots ,{x}_{n}\right)=\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in S$
is a solution to the original system. Ignoring the
$i$-th
equation for a moment, we know it makes all the other equations of the
transformed system true. We also know that
$$\begin{array}{llll}\hfill {a}_{i1}{\beta}_{1}+{a}_{i2}{\beta}_{2}+{a}_{i3}{\beta}_{3}+\cdots +{a}_{in}{\beta}_{n}& ={b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \\ \multicolumn{4}{c}{\text{whichwecanmultiplyby}\alpha \text{toget}}\\ \phantom{\rule{2em}{0ex}}\\ \hfill \alpha {a}_{i1}{\beta}_{1}+\alpha {a}_{i2}{\beta}_{2}+\alpha {a}_{i3}{\beta}_{3}+\cdots +\alpha {a}_{in}{\beta}_{n}& =\alpha {b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
This says that the $i$-th equation of the transformed system is also true, so we have established that $\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in T$, and therefore $S\subseteq T$.

- Now show $T\subseteq S$.
Suppose $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}\dots ,{x}_{n}\right)=\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in T$
is a solution to the transformed system. Ignoring the
$i$-th
equation for a moment, we know it makes all the other equations of the
original system true. We also know that
$$\begin{array}{llll}\hfill \alpha {a}_{i1}{\beta}_{1}+\alpha {a}_{i2}{\beta}_{2}+\alpha {a}_{i3}{\beta}_{3}+\cdots +\alpha {a}_{in}{\beta}_{n}& =\alpha {b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \\ \multicolumn{4}{c}{\text{whichwecanmultiplyby}\frac{1}{\alpha}\text{,since}\alpha \ne 0\text{,toget}}\\ \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{i1}{\beta}_{1}+{a}_{i2}{\beta}_{2}+{a}_{i3}{\beta}_{3}+\cdots +{a}_{in}{\beta}_{n}& ={b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$
This says that the $i$-th equation of the original system is also true, so we have established that $\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in S$, and therefore $T\subseteq S$. Locate the key point where we required that $\alpha \ne 0$, and consider what would happen if $\alpha =0$.

- Show $S\subseteq T$.
Suppose $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}\dots ,{x}_{n}\right)=\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in S$
is a solution to the original system. Ignoring the
$i$-th
equation for a moment, we know it makes all the other equations of the
transformed system true. We also know that
$$\begin{array}{llll}\hfill {a}_{i1}{\beta}_{1}+{a}_{i2}{\beta}_{2}+{a}_{i3}{\beta}_{3}+\cdots +{a}_{in}{\beta}_{n}& ={b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \\ \multicolumn{4}{c}{\text{whichwecanmultiplyby}\alpha \text{toget}}\\ \phantom{\rule{2em}{0ex}}\\ \hfill \alpha {a}_{i1}{\beta}_{1}+\alpha {a}_{i2}{\beta}_{2}+\alpha {a}_{i3}{\beta}_{3}+\cdots +\alpha {a}_{in}{\beta}_{n}& =\alpha {b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
- Suppose $\alpha $
is a number. Let’s choose to multiply the terms of equation
$i$ by
$\alpha $ and add them
to equation $j$
in order to build the new system of equations,
$$\begin{array}{llll}\hfill {a}_{11}{x}_{1}+{a}_{12}{x}_{2}+\cdots +{a}_{1n}{x}_{n}& ={b}_{1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{21}{x}_{1}+{a}_{22}{x}_{2}+\cdots +{a}_{2n}{x}_{n}& ={b}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{31}{x}_{1}+{a}_{32}{x}_{2}+\cdots +{a}_{3n}{x}_{n}& ={b}_{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \vdots \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \left(\alpha {a}_{i1}+{a}_{j1}\right){x}_{1}+\left(\alpha {a}_{i2}+{a}_{j2}\right){x}_{2}+\cdots +\left(\alpha {a}_{in}+{a}_{jn}\right){x}_{n}& =\alpha {b}_{i}+{b}_{j}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \vdots \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{m1}{x}_{1}+{a}_{m2}{x}_{2}+\cdots +{a}_{mn}{x}_{n}& ={b}_{m}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
Let $S$ denote the solutions to the system in the statement of the theorem, and let $T$ denote the solutions to the transformed system.

- Show $S\subseteq T$.
Suppose $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}\dots ,{x}_{n}\right)=\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in S$
is a solution to the original system. Ignoring the
$j$-th
equation for a moment, we know this solution makes all the other equations
of the transformed system true. Using the fact that the solution makes the
$i$-th and
$j$-th
equations of the original system true, we find
$$\begin{array}{llll}\hfill & \left(\alpha {a}_{i1}+{a}_{j1}\right){\beta}_{1}+\left(\alpha {a}_{i2}+{a}_{j2}\right){\beta}_{2}+\cdots +\left(\alpha {a}_{in}+{a}_{jn}\right){\beta}_{n}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\left(\alpha {a}_{i1}{\beta}_{1}+\alpha {a}_{i2}{\beta}_{2}+\cdots +\alpha {a}_{in}{\beta}_{n}\right)+\left({a}_{j1}{\beta}_{1}+{a}_{j2}{\beta}_{2}+\cdots +{a}_{jn}{\beta}_{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\alpha \left({a}_{i1}{\beta}_{1}+{a}_{i2}{\beta}_{2}+\cdots +{a}_{in}{\beta}_{n}\right)+\left({a}_{j1}{\beta}_{1}+{a}_{j2}{\beta}_{2}+\cdots +{a}_{jn}{\beta}_{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\alpha {b}_{i}+{b}_{j}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
This says that the $j$-th equation of the transformed system is also true, so we have established that $\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in T$, and therefore $S\subseteq T$.

- Now show $T\subseteq S$.
Suppose $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}\dots ,{x}_{n}\right)=\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in T$
is a solution to the transformed system. Ignoring the
$j$-th
equation for a moment, we know it makes all the other equations of the
original system true. We then find
$$\begin{array}{llll}\hfill & {a}_{j1}{\beta}_{1}+{a}_{j2}{\beta}_{2}+\cdots +{a}_{jn}{\beta}_{n}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}={a}_{j1}{\beta}_{1}+{a}_{j2}{\beta}_{2}+\cdots +{a}_{jn}{\beta}_{n}+\alpha {b}_{i}-\alpha {b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}={a}_{j1}{\beta}_{1}+{a}_{j2}{\beta}_{2}+\cdots +{a}_{jn}{\beta}_{n}+\left(\alpha {a}_{i1}{\beta}_{1}+\alpha {a}_{i2}{\beta}_{2}+\cdots +\alpha {a}_{in}{\beta}_{n}\right)-\alpha {b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}={a}_{j1}{\beta}_{1}+\alpha {a}_{i1}{\beta}_{1}+{a}_{j2}{\beta}_{2}+\alpha {a}_{i2}{\beta}_{2}+\cdots +{a}_{jn}{\beta}_{n}+\alpha {a}_{in}{\beta}_{n}-\alpha {b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\left(\alpha {a}_{i1}+{a}_{j1}\right){\beta}_{1}+\left(\alpha {a}_{i2}+{a}_{j2}\right){\beta}_{2}+\cdots +\left(\alpha {a}_{in}+{a}_{jn}\right){\beta}_{n}-\alpha {b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\alpha {b}_{i}+{b}_{j}-\alpha {b}_{i}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}={b}_{j}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
This says that the $j$-th equation of the original system is also true, so we have established that $\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in S$, and therefore $T\subseteq S$.

Why didn’t we need to require that $\alpha \ne 0$ for this row operation? In other words, how does the third statement of the theorem read when $\alpha =0$? Does our proof require some extra care when $\alpha =0$? Compare your answers with the similar situation for the second row operation. (See Exercise SSLE.T20.)

- Show $S\subseteq T$.
Suppose $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}\dots ,{x}_{n}\right)=\left({\beta}_{1},\phantom{\rule{0em}{0ex}}{\beta}_{2},\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}{\beta}_{3},\phantom{\rule{0em}{0ex}}\dots ,{\beta}_{n}\right)\in S$
is a solution to the original system. Ignoring the
$j$-th
equation for a moment, we know this solution makes all the other equations
of the transformed system true. Using the fact that the solution makes the
$i$-th and
$j$-th
equations of the original system true, we find
$$\begin{array}{llll}\hfill & \left(\alpha {a}_{i1}+{a}_{j1}\right){\beta}_{1}+\left(\alpha {a}_{i2}+{a}_{j2}\right){\beta}_{2}+\cdots +\left(\alpha {a}_{in}+{a}_{jn}\right){\beta}_{n}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\left(\alpha {a}_{i1}{\beta}_{1}+\alpha {a}_{i2}{\beta}_{2}+\cdots +\alpha {a}_{in}{\beta}_{n}\right)+\left({a}_{j1}{\beta}_{1}+{a}_{j2}{\beta}_{2}+\cdots +{a}_{jn}{\beta}_{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\alpha \left({a}_{i1}{\beta}_{1}+{a}_{i2}{\beta}_{2}+\cdots +{a}_{in}{\beta}_{n}\right)+\left({a}_{j1}{\beta}_{1}+{a}_{j2}{\beta}_{2}+\cdots +{a}_{jn}{\beta}_{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\alpha {b}_{i}+{b}_{j}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

Theorem EOPSS is the necessary tool to complete our strategy for solving systems of equations. We will use equation operations to move from one system to another, all the while keeping the solution set the same. With the right sequence of operations, we will arrive at a simpler equation to solve. The next two examples illustrate this idea, while saving some of the details for later.

Example US

Three equations, one solution

We solve the following system by a sequence of equation operations.

This is now a very easy system of equations to solve. The third equation requires that ${x}_{3}=4$ to be true. Making this substitution into equation 2 we arrive at ${x}_{2}=-3$, and finally, substituting these values of ${x}_{2}$ and ${x}_{3}$ into the first equation, we find that ${x}_{1}=2$. Note too that this is the only solution to this final system of equations, since we were forced to choose these values to make the equations true. Since we performed equation operations on each system to obtain the next one in the list, all of the systems listed here are all equivalent to each other by Theorem EOPSS. Thus $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}{x}_{3}\right)=\left(2,-3,4\right)$ is the unique solution to the original system of equations (and all of the other intermediate systems of equations listed as we transfomed one into another). $\u22a0$

Example IS

Three equations, infinitely many solutions

The following system of equations made an appearance earlier in this section
(Example NSE), where we listed one of its solutions. Now, we will try to find all
of the solutions to this system. Don’t concern yourself too much about why we
choose this particular sequence of equation operations, just believe that the work
we do is all correct.

What does the equation $0=0$ mean? We can choose any values for ${x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}{x}_{4}$ and this equation will be true, so we only need to consider further the first two equations, since the third is true no matter what. We can analyze the second equation without consideration of the variable ${x}_{1}$. It would appear that there is considerable latitude in how we can choose ${x}_{2},\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}{x}_{4}$ and make this equation true. Let’s choose ${x}_{3}$ and ${x}_{4}$ to be anything we please, say ${x}_{3}=a$ and ${x}_{4}=b$.

Now we can take these arbitrary values for ${x}_{3}$ and ${x}_{4}$, substitute them in equation 1, to obtain

$$\begin{array}{llll}\hfill {x}_{1}+2a-3b& =-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{1}& =-1-2a+3b\phantom{\rule{2em}{0ex}}& \hfill & \\ \multicolumn{4}{c}{\text{Similarly,equation2becomes}}\\ \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{2}-a+2b& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{2}& =4+a-2b\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$So our arbitrary choices of values for ${x}_{3}$ and ${x}_{4}$ ($a$ and $b$) translate into specific values of ${x}_{1}$ and ${x}_{2}$. The lone solution given in Example NSE was obtained by choosing $a=2$ and $b=1$. Now we can easily and quickly find many more (infinitely more). Suppose we choose $a=5$ and $b=-2$, then we compute

$$\begin{array}{llll}\hfill {x}_{1}& =-1-2\left(5\right)+3\left(-2\right)=-17\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{2}& =4+5-2\left(-2\right)=13\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$and you can verify that $\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}{x}_{4}\right)=\left(-17,\phantom{\rule{0em}{0ex}}13,\phantom{\rule{0em}{0ex}}5,\phantom{\rule{0em}{0ex}}-2\right)$ makes all three equations true. The entire solution set is written as

$$S=\left\{\left.\left(-1-2a+3b,\phantom{\rule{0em}{0ex}}4+a-2b,\phantom{\rule{0em}{0ex}}a,\phantom{\rule{0em}{0ex}}b\right)\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}a\in {\u2102}^{},\phantom{\rule{0em}{0ex}}b\in {\u2102}^{}\right\}$$ |

It would be instructive to finish off your study of this example by taking the general form of the solutions given in this set and substituting them into each of the three equations and verify that they are true in each case (Exercise SSLE.M40). $\u22a0$

In the next section we will describe how to use equation operations to systematically solve any system of linear equations. But first, read one of our more important pieces of advice about speaking and writing mathematics. See Technique L.

Before attacking the exercises in this section, it will be helpful to read some advice on getting started on the construction of a proof. See Technique GS.

- How many solutions does the system of equations $3x+2y=4$, $6x+4y=8$ have? Explain your answer.
- How many solutions does the system of equations $3x+2y=4$, $6x+4y=-2$ have? Explain your answer.
- What do we mean when we say mathematics is a language?

C10 Find a solution to the system in Example IS where
${x}_{3}=6$ and
${x}_{4}=2$.
Find two other solutions to the system. Find a solution where
${x}_{1}=-17$ and
${x}_{2}=14$. How
many possible answers are there to each of these questions?

Contributed by Robert Beezer

C20 Each archetype (Appendix A) that is a system of equations begins by
listing some specific solutions. Verify the specific solutions listed in the
following archetypes by evaluating the system of equations with the solutions
listed.

Archetype A

Archetype B

Archetype C

Archetype D

Archetype E

Archetype F

Archetype G

Archetype H

Archetype I

Archetype J

Contributed by Robert Beezer

C50 A three-digit number has two properties. The tens-digit and the
ones-digit add up to 5. If the number is written with the digits in the
reverse order, and then subtracted from the original number, the result is
$792$. Use
a system of equations to find all of the three-digit numbers with these properties.

Contributed by Robert Beezer Solution [55]

C51 Find all of the six-digit numbers in which the first digit is one less than the
second, the third digit is half the second, the fourth digit is three times the third
and the last two digits are the sum of the fourth and fifth. The sum of all the
digits is 24. (From The MENSA Puzzle Calendar for January 9, 2006.)

Contributed by Robert Beezer Solution [56]

C52 Driving along, Terry notices that the last four digits on his car’s odometer
are palindromic. A mile later, the last five digits are palindromic. After driving
another mile, the middle four digits are palindromic. One more mile,
and all six are palindromic. What was the odometer reading when Terry
first looked at it? Form a linear system of equations that expresses the
requirements of this puzzle. (Car Talk Puzzler, National Public Radio,
Week of January 21, 2008) (A car odometer displays six digits and a
sequence is a palindome if it reads the same left-to-right as right-to-left.)

Contributed by Robert Beezer Solution [59]

M10 Each sentence below has at least two meanings. Identify the source of the double meaning, and rewrite the sentence (at least twice) to clearly convey each meaning.

- They are baking potatoes.
- He bought many ripe pears and apricots.
- She likes his sculpture.
- I decided on the bus.

Contributed by Robert Beezer Solution [59]

M11 Discuss the diffence in meaning of each of the following three almost identical sentences, which all have the same grammatical structure. (These are due to Keith Devlin.)

- She saw him in the park with a dog.
- She saw him in the park with a fountain.
- She saw him in the park with a telescope.

Contributed by Robert Beezer Solution [60]

M12 The following sentence, due to Noam Chomsky, has a correct grammatical
structure, but is meaningless. Critique its faults. “Colorless green ideas sleep
furiously.” (Chomsky, Noam. Syntactic Structures, The Hague/Paris: Mouton,
1957. p. 15.)

Contributed by Robert Beezer Solution [60]

M13 Read the following sentence and form a mental picture of the situation.

The baby cried and the mother picked it up.

What assumptions did you make about the situation?

Contributed by Robert Beezer Solution [60]

M30 This problem appears in a middle-school mathematics textbook:
Together Dan and Diane have $20. Together Diane and Donna have $15. How
much do the three of them have in total? (Transition Mathematics,
Second Edition, Scott Foresman Addison Wesley, 1998. Problem 5–1.19.)

Contributed by David Beezer Solution [60]

M40 Solutions to the system in Example IS are given as

$$\left({x}_{1},\phantom{\rule{0em}{0ex}}{x}_{2},\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}{x}_{4}\right)=\left(-1-2a+3b,\phantom{\rule{0em}{0ex}}4+a-2b,\phantom{\rule{0em}{0ex}}a,\phantom{\rule{0em}{0ex}}b\right)$$ |

Evaluate the three equations of the original system with these expressions in
$a$ and
$b$ and
verify that each equation is true, no matter what values are chosen for
$a$ and
$b$.

Contributed by Robert Beezer

M70 We have seen in this section that systems of linear equations have limited possibilities for solution sets, and we will shortly prove Theorem PSSLS that describes these possibilities exactly. This exercise will show that if we relax the requirement that our equations be linear, then the possibilities expand greatly. Consider a system of two equations in the two variables $x$ and $y$, where the departure from linearity involves simply squaring the variables.

$$\begin{array}{llll}\hfill {x}^{2}-{y}^{2}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}+{y}^{2}& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$After solving this system of non-linear equations, replace the second equation in turn
by ${x}^{2}+2x+{y}^{2}=3$,
${x}^{2}+{y}^{2}=1$,
${x}^{2}-x+{y}^{2}=0$,
$4{x}^{2}+4{y}^{2}=1$ and
solve each resulting system of two equations in two variables.

Contributed by Robert Beezer Solution [60]

T10 Technique D asks you to formulate a definition of what it means for a whole
number to be odd. What is your definition? (Don’t say “the opposite of even.”) Is
$6$ odd?
Is $11$
odd? Justify your answers by using your definition.

Contributed by Robert Beezer Solution [61]

T20 Explain why the second equation operation in Definition EO requires that
the scalar be nonzero, while in the third equation operation this restriction on the
scalar is not present.

Contributed by Robert Beezer Solution [61]

C50 Contributed by Robert Beezer Statement [50]

Let $a$ be the
hundreds digit, $b$
the tens digit, and $c$
the ones digit. Then the first condition says that
$b+c=5$. The original
number is $100a+10b+c$, while the
reversed number is $100c+10b+a$.
So the second condition is

$$792=\left(100a+10b+c\right)-\left(100c+10b+a\right)=99a-99c$$ |

So we arrive at the system of equations

$$\begin{array}{llll}\hfill b+c& =5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 99a-99c& =792\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Using equation operations, we arrive at the equivalent system

$$\begin{array}{llll}\hfill a-c& =8\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill b+c& =5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$We can vary $c$ and obtain infinitely many solutions. However, $c$ must be a digit, restricting us to ten values (0 – 9). Furthermore, if $c>1$, then the first equation forces $a>9$, an impossibility. Setting $c=0$, yields $850$ as a solution, and setting $c=1$ yields $941$ as another solution.

C51 Contributed by Robert Beezer Statement [50]

Let $abcdef$
denote any such six-digit number and convert each requirement in the problem
statement into an equation.

In a more standard form this becomes

$$\begin{array}{llll}\hfill a-b& =-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -b+2c& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -3c+d& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -d+f& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill a+b+c+d+e+f& =24\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Using equation operations (or the techniques of the upcoming Section RREF), this system can be converted to the equivalent system

$$\begin{array}{llll}\hfill a-\frac{2}{3}f& =-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill b-\frac{2}{3}f& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill c-\frac{1}{3}f& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill d-f& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill e+\frac{11}{3}f& =25\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Clearly, we must choose $f$ to be a multiple of 3, and of the choices $f=0,\phantom{\rule{0em}{0ex}}3,\phantom{\rule{0em}{0ex}}6,\phantom{\rule{0em}{0ex}}9$ only $f=6$ results in a sensible (positive, single-digit) value for $e$. So with $f=6$ we have

$$\begin{array}{llllllllllllllllllll}\hfill e& =3\phantom{\rule{2em}{0ex}}& \hfill d& =6\phantom{\rule{2em}{0ex}}& \hfill c& =2\phantom{\rule{2em}{0ex}}& \hfill b& =4\phantom{\rule{2em}{0ex}}& \hfill a& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$So the only such number is $342636$. Notice that the question casts the numbers as digits, but their role as place values is not relevant.

C52 Contributed by Robert Beezer Statement [51]

1988888.

M10 Contributed by Robert Beezer Statement [51]

- Is “baking” a verb or an adjective?

Potatoes are being baked.

Those are baking potatoes. - Are the apricots ripe, or just the pears? Parentheses could indicate just
what the adjective “ripe” is meant to modify. Were there many apricots
as well, or just many pears?

He bought many pears and many ripe apricots.

He bought apricots and many ripe pears. - Is “sculpture” a single physical object, or the sculptor’s style expressed
over many pieces and many years?

She likes his sculpture of the girl.

She likes his sculptural style. - Was a decision made while in the bus, or was the outcome of a decision
to choose the bus. Would the sentence “I decided on the car,” have a
similar double meaning?

I made my decision while on the bus.

I decided to ride the bus.

M11 Contributed by Robert Beezer Statement [51]

We know the dog belongs to the man, and the fountain belongs to the park.
It is not clear if the telescope belongs to the man, the woman, or the
park.

M12 Contributed by Robert Beezer Statement [52]

In adjacent pairs the words are contradictory or inappropriate. Something cannot
be both green and colorless, ideas do not have color, ideas do not sleep, and it is
hard to sleep furiously.

M13 Contributed by Robert Beezer Statement [52]

Did you assume that the baby and mother are human?

Did you assume that the baby is the child of the mother?

Did you assume that the mother picked up the baby as an attempt to stop the
crying?

M30 Contributed by Robert Beezer Statement [52]

If $x$,
$y$ and
$z$
represent the money held by Dan, Diane and Donna, then
$y=15-z$ and
$x=20-y=20-\left(15-z\right)=5+z$. We can let
$z$ take on any
value from $0$
to $15$
without any of the three amounts being negative, since presumably middle-schoolers
are too young to assume debt.

Then the total capital held by the three is $x+y+z=\left(5+z\right)+\left(15-z\right)+z=20+z$. So their combined holdings can range anywhere from $20 (Donna is broke) to $35 (Donna is flush).

We will have more to say about this situation in Section TSS, and specifically Theorem CMVEI.

M70 Contributed by Robert Beezer Statement [53]

The equation ${x}^{2}-{y}^{2}=1$
has a solution set by itself that has the shape of a hyperbola when plotted. The
five different second equations have solution sets that are circles when plotted
individually. Where the hyperbola and circle intersect are the solutions to the
system of two equations. As the size and location of the circle varies, the
number of intersections varies from four to none (in the order given).
Sketching the relevant equations would be instructive, as was discussed in
Example STNE.

The exact solution sets are (according to the choice of the second equation),

$$\begin{array}{llllllll}\hfill {x}^{2}+{y}^{2}& =4:\phantom{\rule{2em}{0ex}}& \hfill & \left\{\left(\sqrt{\frac{5}{2}},\sqrt{\frac{3}{2}}\right),\phantom{\rule{0em}{0ex}}\left(-\sqrt{\frac{5}{2}},\sqrt{\frac{3}{2}}\right),\phantom{\rule{0em}{0ex}}\left(\sqrt{\frac{5}{2}},-\sqrt{\frac{3}{2}}\right),\phantom{\rule{0em}{0ex}}\left(-\sqrt{\frac{5}{2}},-\sqrt{\frac{3}{2}}\right)\right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}+2x+{y}^{2}& =3:\phantom{\rule{2em}{0ex}}& \hfill & \left\{\left(1,0\right),\phantom{\rule{0em}{0ex}}\left(-2,\sqrt{3}\right),\phantom{\rule{0em}{0ex}}\left(-2,-\sqrt{3}\right)\right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}+{y}^{2}& =1:\phantom{\rule{2em}{0ex}}& \hfill & \left\{\left(1,0\right),\phantom{\rule{0em}{0ex}}\left(-1,0\right)\right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}-x+{y}^{2}& =0:\phantom{\rule{2em}{0ex}}& \hfill & \left\{\left(1,0\right)\right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 4{x}^{2}+4{y}^{2}& =1:\phantom{\rule{2em}{0ex}}& \hfill & \left\{\right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$T10 Contributed by Robert Beezer Statement [54]

We can say that an integer is odd if when it is divided by
$2$ there is a
remainder of 1. So $6$
is not odd since $6=3\times 2+0$,
while $11$ is
odd since $11=5\times 2+1$.

T20 Contributed by Robert Beezer Statement [54]

Definition EO is engineered to make Theorem EOPSS true. If we were to allow a
zero scalar to multiply an equation then that equation would be transformed to the
equation $0=0$,
which is true for any possible values of the variables. Any restrictions on the
solution set imposed by the original equation would be lost.

However, in the third operation, it is allowed to choose a zero scalar, multiply an equation by this scalar and add the transformed equation to a second equation (leaving the first unchanged). The result? Nothing. The second equation is the same as it was before. So the theorem is true in this case, the two systems are equivalent. But in practice, this would be a silly thing to actually ever do! We still allow it though, in order to keep our theorem as general as possible.

Notice the location in the proof of Theorem EOPSS where the expression $\frac{1}{\alpha}$ appears — this explains the prohibition on $\alpha =0$ in the second equation operation.