OpenLLM-France/wikimedia huggingface dataset:
# Mapping/Graph/Image/Preimage/Introduction/Section
## Definition

Let $L$ and $M$ be sets, and let
    $F\colon L\longrightarrow M$
be a mapping. Then the set
    $\Gamma =\Gamma _{F}={\left\{(x,F(x))\mid x\in L\right\}}\subseteq L\times M\,$
$F$
The graph is a concept of set theory. Whether it is possible to "visualize“ it in a picture depends on whether we can visualize the product set $L\times M$.

## Definition

Let $L$ and $M$ be sets, and let
    $F\colon L\longrightarrow M$
be a mapping. For a subset $S\subseteq L$, we call
    $F(S)={\left\{y\in M\mid {\text{there exists an }}x\in S{\text{ such that }}F(x)=y\right\}}\,$
the image of $S$ under $F$. For $S=L$,
    $F(L)=\operatorname {Im} F\,$

## Definition

Let $L$ and $M$ be sets, and let
    $F\colon L\longrightarrow M$
be a mapping. For a subset $T\subseteq M$, we call
    $F^{-1}(T)={\left\{x\in L\mid F(x)\in T\right\}}\,$
the preimage of $T$ under $F$. For a subset $T=\{y\}$ with one element, we call
    $F^{-1}(\{y\})$
$y$
# Mapping/Squaring/Image and preimage/Example

For the mapping
    $\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{2},$
the image of $[1,2]$ is the set of all squares of real numbers between $1$ and $2$, this is thus $[1,4]$. The preimage of $[1,4]$ consists of all real numbers whose square is between $1$ and $4$. This is $[-2,-1]\cup [1,2]$.
# Binary operation/Commutative/Definition

A binary operation
    $\circ \colon M\times M\longrightarrow M,(x,y)\longmapsto x\circ y,$
on a set $M$ is called commutative if for all $x,y\in M$ the equality
    $x\circ y=y\circ x\,$
holds.
# Binary operation/Associative/Definition

A binary operation
    $\circ \colon M\times M\longrightarrow M,(x,y)\longmapsto x\circ y,$
on a set $M$ is called associative if for all $x,y,z\in M$ the equality
    $(x\circ y)\circ z=y\circ (x\circ z)\,$
holds.
# Binary operation/Neutral element/Definition

Let a set $M$ and a binary operation
    $\circ \colon M\times M\longrightarrow M,(x,y)\longmapsto x\circ y,$
be given. An element $e\in M$ is called neutral element of the operation if, for all $x\in M$, the equalities $x\circ e=x=e\circ x$ hold.
# Binary operation/Inverse element/Definition

Let a set $M$ with a binary operation
    $\circ \colon M\times M\longrightarrow M,(x,y)\longmapsto x\circ y,$
and a neutral element $e\in M$ be given. For an element $x\in M$, an element $y\in M$ is called inverse element (for $x$) if the equalities
    $x\circ y=e=y\circ x\,$
hold.
# Set of mappings/Properties of composition/Example

Let $L$ be a set, and let
    $M=\operatorname {Map} \,{\left(L,L\right)}\,$
be the set of all mappings from $L$ to itself. The composition of mappings gives a binary operation on $M$, which is associative, due to fact. In general, it is not commutative. The identity on $L$ is the neutral element. A mapping $f\colon L\rightarrow L$ has an inverse element if and only if it is bijective; the inverse element is just the inverse mapping.
# Thinking Tools/Reversals

Many innovations can be inspired and imagined by reversing the viewpoint, components, users, and other elements of an existing product or solution. Consider how the various reversals listed here might transform an old idea into a new one, identify an opportunity, or offer a new viewpoint that can help reframe a problem.
Cause ↔ Effect
Customer ↔ Supplier
Bottom up ↔ Top down
Emergent ↔ Designed
Social constructs ↔ Brute facts
Distributed ↔ Centralized
Giver ↔ Recipient
Give ↔ Take
Scarce ↔ Abundant
Awful ↔ Awesome
Payer ↔ Collector
Share ↔ Hoard
Talk ↔ Listen
Input ↔ Output
In ↔ Out
Top ↔ Bottom
Upright ↔ Sideways
Left ↔ Right
Big ↔ Small
Close ↔ Distant
Near ↔ Far
Fast ↔ Slow
Strong ↔ Weak
Forward ↔ Backward
Movable ↔ Fixed
Dynamic ↔ Static
Linear ↔ Circular
Linear ↔ Radial
Circular ↔ Spiral
Spiral ↔ Helix
Helix ↔ Möbius strip
Microscope ↔ Telescope
Focused ↔ Circumspect
Complex ↔ Simple
Black ↔ White
Objective ↔ Subjective
Light ↔ Dark
Day ↔ Night
Light ↔ Heavy
Foreground ↔ Background
Forest ↔ Trees
Headline ↔ Footnote
Push ↔ Pull
Assignment ↔ Invitation
Fall ↔ Rise
Obvious ↔ Cryptic
Obvious ↔ Incorrect
Known ↔ Unknown
Master ↔ Slave
Powerful ↔ Powerless
Own ↔ Rent
Democracy ↔ Tyranny
Comedy ↔ Tragedy
Play ↔ Work
Teacher ↔ Student
Teach ↔ Learn
Doctor ↔ Patient
Expert ↔ Novice
Predator ↔ Prey
Victim ↔ Perpetrator
Guilty ↔ Innocent
Male ↔ Female
Rich ↔ Poor
Growing ↔ Shrinking
Fearful ↔ Confident
Timid ↔ Bold
Costly ↔ Free
Constrained ↔ Free
Reader ↔ Writer
Problem ↔ Solution
Good ↔ Bad
Good ↔ Evil
Certain ↔ Unlikely
Rich ↔ Poor
Rough ↔ Smooth
Soft ↔ Hard
Sharp ↔ Round
Edgy ↔ Cozy
Frightening ↔ Comforting
Maximize ↔ Minimize
Negative ↔ Positive
Cost ↔ Benefits
Difficult ↔ Easy
Active ↔ Passive
Accuser ↔ Accused
Competition ↔ Cooperation
Win Win ↔ Zero Sum
Not ↔ Now
Specialize ↔ Generalize
Temporary ↔ Permanent
Aggregate ↔ Separate
Isolate ↔ Congregate
Divide ↔ Unite
Component ↔ System
Minority ↔ Majority
Old ↔ Young
Generous ↔ Greedy
Secret ↔ Transparent
Copyright ↔ Copy left
Kind ↔ Cruel
Helpful ↔ Hurtful
Humble ↔ Arrogant
Vindictive ↔ Compassionate
Authority ↔ Expertise
Loyalty ↔ Fairness
Me ↔ You
Ideological ↔ Empirical
Theory ↔ Practice
Rigid ↔ Flexible
Near ↔ Far
Fast ↔ Slow
Fear ↔ Curiosity
Local ↔ Global
Tribal ↔ Global
War ↔ Peace
Love ↔ Hate
Reification ↔ Instantiation
Symptom ↔ Cause
Power ↔ Reason
Passion ↔ Reason
Anecdote ↔ Representative evidence
Despair ↔ Hope
Advocate ↔ Criticize
Personal ↔ Communal
Strange ↔ Charm
Tuba ↔ Sousaphone
Left-handed ↔ Right-handed
Noun ↔ Verb
Verb ↔ Preposition
# Motivation and emotion/Assessment/Chapter/Figures
## Finding images

There are three ways of locating images for use in book chapters:
* Search Wikimedia Commons. These images can be easily embedded on Wikiversity,
* Upload free-to-use images to Wikimedia Commons and then embed on Wikiversity pages. See suggestions about how to find free-to-use images
* Create and upload your own images to Wikimedia Commons
For some starting ideas, see the motivation and emotion gallery.

## Examples

* Examples of captioned figures are provided in Figures 1, 2, and 3
* The captions are descriptive
* Figures 2 and 3 have been made larger than the default size, to make them easier to read
* Figures 1 and 2 were created by Motivation and emotion students in 2020 and uploaded to Wikimedia Commons. Click on the images to find out more.
* Another example of a correctly presented image is provided in the book chapter template.
* Each figure should be referred to at least once in the main body text. For example, see Figure 1.
# Motivation and emotion/Assessment/Chapter/Readability

One way to improve a chapter draft is to conduct a readability analysis.
Most readability analyses use formulas based on a combination of:
* length of sentences (shorter is better)
* length of words (shorter is better)
Try pasting the URL into the webfx readability tool. Aim for a readability score of at least 50. For example, funerals and grief work (2020) gets a score of 57.
If you would like more diagnostic detail, try pasting text into the Hemingway app or asking a genAI tool for a readability analysis.
Can you suggest other readability tools?
# Group/Inverse element/Unique/Fact

Let $(G,e,\circ )$ be a group.
$x\in G$, the element $y\in G$ fulfilling
    $x\circ y=y\circ x=e\,$
# Field/Domain/Fact/Proof

We prove this by contradiction, so we assume that $a$ and $b$ are both not $0$. Then there exist inverse elements $a^{-1}$ and $b^{-1}$, and hence $(ab){\left(b^{-1}a^{-1}\right)}=1$. On the other hand, we have $ab=0$ by the premise, and so the annihilation rule gives
    $(ab){\left(b^{-1}a^{-1}\right)}=0{\left(b^{-1}a^{-1}\right)}=0\,,$
hence $0=1$, which contradicts the field properties.
# Ring/Basic properties/Fact

Let $R$ be a ring, and let $a,b,c,a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{s}$ denote elements from $R$. Then the following statements hold.
* $a0=0$ (annihilation rule).
* (rules for sign).
    $(-a)b=-ab=a(-b)\,$
    $(-a)(-b)=ab\,.$
    $a(b-c)=ab-ac\,.$
* ${\left(\sum _{i=1}^{r}a_{i}\right)}{\left(\sum _{k=1}^{s}b_{k}\right)}=\sum _{1\leq i\leq r,\,1\leq k\leq s}a_{i}b_{k}$ (general law of distributivity).
# Algebra/Ring/Direct/Definition

A set $R$ is called a ring if there are two binary operations (called addition and multiplication)
    $+:R\times R\longrightarrow R{\text{ and }}\cdot :R\times R\longrightarrow R$
and two elements $0,1\in R$ that fulfill the following properties.
* Axioms for the addition:
** Associative law: $(a+b)+c=a+(b+c)$ holds for all $a,b,c\in R$.
** Commutative law: $a+b=b+a$ holds for all $a,b\in R$.
** $0$ is the neutral element of the addition, i.e., $a+0=a$ holds for all $a\in R$.
** Existence of the negative: For every $a\in R$, there exists an element $b\in R$ with $a+b=0$.
* Axioms of the multiplication:
** Associative law: $(a\cdot b)\cdot c=a\cdot (b\cdot c)$ holds for all $a,b,c\in R$.
** $1$ is the neutral element for the multiplication, i.e., $a\cdot 1=a$ holds for all $a\in R$.
* Distributive law: $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ holds for all $a,b,c\in R$.
# Ring/Linear algebra/Introduction/Section
## Definition

A set $R$ is called a ring if there are two binary operations (called addition and multiplication)
    $+:R\times R\longrightarrow R{\text{ and }}\cdot :R\times R\longrightarrow R$
and two elements $0,1\in R$ that fulfill the following properties.
* Axioms for the addition:
** Associative law: $(a+b)+c=a+(b+c)$ holds for all $a,b,c\in R$.
** Commutative law: $a+b=b+a$ holds for all $a,b\in R$.
** $0$ is the neutral element of the addition, i.e., $a+0=a$ holds for all $a\in R$.
** Existence of the negative: For every $a\in R$, there exists an element $b\in R$ with $a+b=0$.
* Axioms of the multiplication:
** Associative law: $(a\cdot b)\cdot c=a\cdot (b\cdot c)$ holds for all $a,b,c\in R$.
** $1$ is the neutral element for the multiplication, i.e., $a\cdot 1=a$ holds for all $a\in R$.
* Distributive law: $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ holds for all $a,b,c\in R$.

## Definition

A ring is called commutative
For us, the most important commutative rings are the set of integer numbers $\mathbb {Z}$, the rational numbers $\mathbb {Q}$, and the real numbers $\mathbb {R}$. The real numbers (and the rational numbers) with their natural operations fulfill all these axioms, as should be known from school. An axiomatic reasoning is possible, but we will not do this here. With its addition, a ring $(R,0,+)$ (forgetting the multiplicative structure) is in particular a commutative group.
In a ring, we use the convention that multiplication ties stronger than addition. Therefore, we write $a\cdot b+c\cdot d$ instead of $(a\cdot b)+(c\cdot d)$. To simplify further the notation, we omit the product symbol. The special elements $0$ and $1$ in a ring are called the null element and the unit. For $a\in R$, we call the
(according to fact) uniquely determined element $y$ fulfilling $a+y=0$ the negative of $a$ and denote it by $-a$. We have $-(-a)=a$, since $a+(-a)=0$ shows that the element $a$ equals the uniquely determined negative of $-a$. We write $b-a$ instead of $b+(-a)$ and call this a difference. Hence, the difference is not a basic operation but is defined as the addition with the negative element.
The following properties are familiar for the real numbers; we prove them using only the axioms of a ring. So they hold for any ring.

## Lemma

Let $R$ be a ring,
$a,b,c,a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{s}$
$R$
* $a0=0$ (annihilation rule).
* (rules for sign).
    $(-a)b=-ab=a(-b)\,$
    $(-a)(-b)=ab\,.$
    $a(b-c)=ab-ac\,.$
* ${\left(\sum _{i=1}^{r}a_{i}\right)}{\left(\sum _{k=1}^{s}b_{k}\right)}=\sum _{1\leq i\leq r,\,1\leq k\leq s}a_{i}b_{k}$ (general law of distributivity).

### Proof

In the noncommutative case, we only proof one half of the statements.
* We have $a0=a(0+0)=a0+a0$. Subtracting $a0$ (meaning addition with the negative of $a0$) on both sides gives the claim.
* due to part (1). Therefore, $(-a)b$ is the (uniquely determined) negative of $ab$.
    $(-a)b+ab=(-a+a)b=0b=0\,$
* Due to (2), we have $(-a)(-b)=(-(-a))b$, and because of $-(-a)=a$ (which holds in every group) we get the claim.
* This follows from the parts proved so far.
* This follows with a double induction.
$\Box$
# Ring/Basic properties/Fact/Proof

In the noncommutative case, we only proof one half of the statements.
* We have $a0=a(0+0)=a0+a0$. Subtracting $a0$ (meaning addition with the negative of $a0$) on both sides gives the claim.
* due to part (1). Therefore, $(-a)b$ is the (uniquely determined) negative of $ab$.
    $(-a)b+ab=(-a+a)b=0b=0\,$
* Due to (2), we have $(-a)(-b)=(-(-a))b$, and because of $-(-a)=a$ (which holds in every group), we get the claim.
* This follows from the parts proved so far.
* This follows with a double induction.
# Group/Direct/Definition

A set $G$ together with a special element $e\in G$ and with a binary operation
    $G\times G\longrightarrow G,(g,h)\longmapsto g\circ h$
is called a group
when the following properties are fulfilled.
* The binary operation is associative, i.e., for all $f,g,h\in G$, we have
    $(f\circ g)\circ h=f\circ (g\circ h)\,.$
* The element $e$ is a neutral element, i.e., for all $g\in G$, we have
    $g\circ e=g=e\circ g\,.$
* For every $g\in G$, there exists an inverse element, i.e., there exists some $h\in G$ such that
    $h\circ g=g\circ h=e\,.$
# Axiomatic structure/Abstract and examples/Remark

Abstract structures like a set, a mapping, a binary operation, or a group have a double life. On one hand, they are really just the given formal structure; the elements are just some elements in a somehow given set, a binary operation is just any binary operation, and one should not imagine anything concrete. The symbols chosen are arbitrary and without any meaning. On the other hand, these abstract structures gain a second life in that many concrete mathematical structures obey the abstract properties. These concrete structures are examples or models for the abstract structure (and they are also a motivation to introduce the abstract structure). Both viewpoints are important, and one should always try not to confuse them.
# Group/Linear algebra/Introduction/Section

In linear algebra, we work in general over a fixed base field $K$. The most important field for us is the field of real numbers $\mathbb {R}$, which we have already used and which is introduced in analysis in an axiomatic way. A field is characterized by the existence of two binary operations fulfilling certain properties, namely addition and multiplication. Both these operations (for multiplication, one has to remove $0$) are instances of an important algebraic structure: a group.

## Definition

A set $G$ together with a special element $e\in G$ and with a binary operation
    $G\times G\longrightarrow G,(g,h)\longmapsto g\circ h$
is called a group
when the following properties are fulfilled.
* The binary operation is associative, i.e., for all $f,g,h\in G$, we have
    $(f\circ g)\circ h=f\circ (g\circ h)\,.$
* The element $e$ is a neutral element, i.e., for all $g\in G$, we have
    $g\circ e=g=e\circ g\,.$
* For every $g\in G$, there exists an inverse element, i.e., there exists some $h\in G$ such that
    $h\circ g=g\circ h=e\,.$
A group is called commutative if the operation is commutative. Important examples for commutative groups are $(\mathbb {Z} ,0,+)$, $(\mathbb {R} ,0,+)$, $(\mathbb {R} \setminus \{0\},1,\cdot )$ or $(\mathbb {R} ^{n},0,+)$ with the componentwise zero
    $0=(0,0,\ldots ,0)\,$
and componentwise addition.
In a group $(G,e,\circ )$, the neutral element is uniquely determined. For if $e'$ is another element fulfilling the characteristic property of the neutral element, meaning
    $x\circ e'=e'\circ x=x\,$
for all $x\in G$, then we can directly deduce
    $e=e\circ e'=e'\,.$

## Lemma

Let $(G,e,\circ )$ be a group. Then, for every $x\in G$, the element $y\in G$ fulfilling
    $x\circ y=y\circ x=e\,$

### Proof

Let
    $x\circ y=y\circ x=e\,$
and
    $x\circ z=z\circ x=e\,.$
Then we have
    $y=y\circ e=y\circ (x\circ z)=(y\circ x)\circ z=e\circ z=z\,.$
$\Box$
Abstract structures like a set, a mapping, a binary operation, or a group have a double life. On one hand, they are really just the given formal structure; the elements are just some elements in a somehow given set, a binary operation is just any binary operation, and one should not imagine anything concrete. The symbols chosen are arbitrary and without any meaning. On the other hand, these abstract structures gain a second life in that many concrete mathematical structures obey the abstract properties. These concrete structures are examples or models for the abstract structure (and they are also a motivation to introduce the abstract structure). Both viewpoints are important, and one should always try not to confuse them.
Group theory is a branch of mathematics on its own, which we will not develop here systematically. Instead, we work with rings, and in particular with fields.
# Field/Via ring/Definition

A commutative ring $K$ is called a field if $R\neq 0$ and if every element different from $0$ has a multiplicative inverse.
# Field/Two elements/Example

We are trying to find the structure of a field on the set $\{0,1\}$. If $0$ is supposed to be the neutral element of the addition and $1$ the neutral element of the multiplication, then everything is already determined: The equation $1+1=0$ must hold since $1$ has an inverse element with respect to the addition, and since $0\cdot 0=0$ holds, due to fact. Hence, the operation tables look like
and
With some tedious computations, one can check that this is indeed a field.
# Field/Seven elements/Introduction/Example

On the set $\{0,1,2,3,4,5,6\}$ (with seven elements), one can define a field structure using
Without any further theory, it is very tedious to so show that this is indeed a field.
# Linear equation/One variable/Solution behavior/Remark

The "mother of all systems of linear equations“ is just one linear equation in one variable of the form
    $ax=b\,,$
with given elements $a,b$ from a field $K$ and wanted $x$. We have three possibilities how the solution behavior might look like. For $a\neq 0$, we can multiply the equation with the inverse $a^{-1}$ of $a$, yielding the unique solution
    $x=ba^{-1}={\frac {b}{a}}\,.$
Computationally, one can find the solution, as long as one can find the inverse element and can perform the multiplication in the field. For $a=0$, the solution behavior depends on $b$. If $b=0$, then every $x\in K$ is a solution; if $b\neq 0$, then there is no solution.
# Planes in space/Equations 4x-2y-3z is 5, 3x-5y+2z is 1/Intersection line/Example

Suppose that two planes are given in $\mathbb {R} ^{3}$,
    $E={\left\{(x,y,z)\in \mathbb {R} ^{3}\mid 4x-2y-3z=5\right\}}\,$
and
    $F={\left\{(x,y,z)\in \mathbb {R} ^{3}\mid 3x-5y+2z=1\right\}}\,.$
How can we describe the intersecting line $G=E\cap F$? A point $P=(x,y,z)$ belongs to the intersection line if and only if it satisfies both plane equations. Therefore, both equations,
    $4x-2y-3z=5\,\,{\text{ and }}\,\,3x-5y+2z=1,$
must hold. We multiply the first equation by $3$, and subtract from that four times the second equation, and get
    $14y-17z=11\,.$
If we set $y=0$, then $z=-{\frac {11}{17}}$ and $x={\frac {13}{17}}$ must hold. This means that the point $P=\left({\frac {13}{17}},\,0,\,-{\frac {11}{17}}\right)$ belongs to $G$. In the same way, setting $z=0$, we find the point $Q=\left({\frac {23}{14}},\,{\frac {11}{14}},\,0\right)$. Therefore, the intersecting line is the line connecting these points, so
    $G={\left\{\left({\frac {13}{17}},\,0,\,-{\frac {11}{17}}\right)+t\left({\frac {209}{238}},\,{\frac {11}{14}},\,{\frac {11}{17}}\right)\mid t\in \mathbb {R} \right\}}\,.$
# Vector space/Introductory example/Mulled wine/Example

At a booth on the Christmas market, there are three different pots of mulled wine. All three contain the ingredients cinnamon, cloves, red wine, and sugar, but the compositions differ. The mixtures of the mulled wines are
    $G_{1}={\begin{pmatrix}1\\2\\11\\2\end{pmatrix}},\,G_{2}={\begin{pmatrix}2\\2\\12\\3\end{pmatrix}},\,G_{3}={\begin{pmatrix}3\\1\\20\\7\end{pmatrix}}.$
Every mulled wine is represented by a four-tuple, where the entries represent the respective shares of the ingredients. The set of all (possible) mulled wines forms a vector space, and the three concrete mulled wines are vectors in this space.
Now suppose that none of the three mulled wines meets exactly our taste; in fact, the wanted mulled wine has the mixture
    $W={\begin{pmatrix}1\\2\\20\\5\end{pmatrix}}\,.$
Is there a possibility to get the wanted mulled wine by pouring together the given mulled wines in some way? Are there numbers $a,b,c\in \mathbb {Q}$ such that
    $a{\begin{pmatrix}1\\2\\11\\2\end{pmatrix}}+b{\begin{pmatrix}2\\2\\12\\3\end{pmatrix}}+c{\begin{pmatrix}3\\1\\20\\7\end{pmatrix}}={\begin{pmatrix}1\\2\\20\\5\end{pmatrix}}\,$
holds? This vector equation can be expressed by four equations in the "variables“ $a,b,c$, where the equations come from the rows. When does there exist a solution, when none, when many? These are typical questions of linear algebra.
# Matrices/Several concepts/Section
## Definition

The $n\times n$-matrix
    $E_{n}:={\begin{pmatrix}1&0&\cdots &\cdots &0\\0&1&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&1&0\\0&\cdots &\cdots &0&1\end{pmatrix}}\,$
The identity matrix $E_{n}$ has the property $E_{n}M=M=ME_{n}$, for an arbitrary $n\times n$-matrix $M$. Hence, the identity matrix is the neutral element with respect to matrix multiplication.

## Remark

If we multiply an $m\times n$-matrix $A=(a_{ij})_{ij}$ with a column vector $x={\begin{pmatrix}x_{1}\\x_{2}\\\vdots \\x_{n}\end{pmatrix}}$, then we get
    $Ax={\begin{pmatrix}a_{11}&a_{12}&\ldots &a_{1n}\\a_{21}&a_{22}&\ldots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&\ldots &a_{mn}\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\\vdots \\x_{n}\end{pmatrix}}={\begin{pmatrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}\\\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}\end{pmatrix}}\,.$
Hence, an inhomogeneous system of linear equations with disturbance vector ${\begin{pmatrix}c_{1}\\c_{2}\\\vdots \\c_{m}\end{pmatrix}}$ can be written briefly as
    $Ax=c\,.$
Then, the manipulations on the equations, which do not change the solution set, can be replaced by corresponding manipulations on the rows of the matrix. It is not necessary to write down the variables.

## Definition

An $n\times n$-matrix of the form
    ${\begin{pmatrix}d_{11}&0&\cdots &\cdots &0\\0&d_{22}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1\,n-1}&0\\0&\cdots &\cdots &0&d_{nn}\end{pmatrix}}$

## Definition

Let $K$ be a field, and let $M=(a_{ij})_{ij}$ be an $m\times n$-matrix over $K$. Then the $n\times m$-matrix
    ${M^{\text{tr}}}={\left(b_{ij}\right)}_{ij}{\text{ with }}b_{ij}:=a_{ji}$
$M$
The transposed matrix arises by interchanging the roles of the rows and the columns. For example, we have
    ${{\begin{pmatrix}t&n&o&d\\r&s&s&x\\a&p&e&y\end{pmatrix}}^{\text{tr}}}={\begin{pmatrix}t&r&a\\n&s&p\\o&s&e\\d&x&y\end{pmatrix}}\,.$
# Linear system/Matrix/Remark

If we multiply an $m\times n$-matrix $A=(a_{ij})_{ij}$ with a column vector $x={\begin{pmatrix}x_{1}\\x_{2}\\\vdots \\x_{n}\end{pmatrix}}$, then we get
    $Ax={\begin{pmatrix}a_{11}&a_{12}&\ldots &a_{1n}\\a_{21}&a_{22}&\ldots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&\ldots &a_{mn}\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\\vdots \\x_{n}\end{pmatrix}}={\begin{pmatrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}\\\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}\end{pmatrix}}\,.$
Hence, an inhomogeneous system of linear equations with disturbance vector ${\begin{pmatrix}c_{1}\\c_{2}\\\vdots \\c_{m}\end{pmatrix}}$ can be written briefly as
    $Ax=c\,.$
Then, the manipulations on the equations that do not change the solution set, can be replaced by corresponding manipulations on the rows of the matrix. It is not necessary to write down the variables.