OpenLLM-France/wikimedia huggingface dataset:
# Boundary Liouville theory

Liouville theory is an interesting 2d conformal field theory, which has been solved in the sense that 3pt structure constants are exactly known.
The theory is also quite well understood in the presence of a boundary, with boundary conditions described by ZZ branes or FZZT branes. These boundary conditions correspond to representations of the w:Virasoro algebra: degenerate representations and Verma modules respectively. However, this picture is only valid for central charges in $\mathbb {C} -(-\infty ,1]$. For $c\leq 1$, boundary Liouville theory is less well understood.

## Motivations

Non-perturbative effects in the Virasoro minimal string.

## Existing works

An article by Bautista and Bawane tries to construct the $c\leq 1$ counterparts of FZZT branes in the bootstrap approach.
The article on the Virasoro minimal string conjectures two families of boundary conditions with one discrete parameter, which they call half-ZZ branes.

## Challenges

Divergence of annulus partition functions.

## Possible approaches
### Limit from A-series minimal models

Liouville theory with $c\leq 1$ can be obtained as a limit of diagonal minimal models, whose central charges are dense in the half-line $c\leq 1$. In practice, it could be good to do it in two steps:
* Obtain boundary conditions in generalized minimal models.
* Obtain Liouville theory from generalized minimal models as a limit at fixed $c$.
In generalized minimal models, we expect to find boundary conditions labelled by two integers $m,n\in \mathbb {N} ^{*}$. The only subtlety is that annulus partition functions probably diverge, just like the torus partition function. Then the Liouville limit could be taken at fixed $m,n$, leading to ZZ-like boundary conditions. The further limit $m\beta -n\beta ^{-1}\to p$ with $m,n\to \infty$ and $c=13-6\beta ^{2}-6\beta ^{-2}$ would yield FZZT-like boundary conditions. It might be possible to cook up other limits, and obtain boundary conditions that are neither FZZT-like nor ZZ-like.
An advantage of this approach is that the conformal bootstrap constraints are automatically satisfied. Another advantage is that non-perturbative effects in the Virasoro minimal string might follow from the same effects in Minimal Liouville gravity, where they are understood to some extent. A weakness of the approach is that there is some freedom of having prefactors when taking the limit, making it unclear which quantities are supposed to be finite or to diverge.

### Construction from chiral CFT

In a simple diagonal CFT like Liouville theory, the structure constants are closely related to the fusion and modular kernels for Virasoro conformal blocks. This might still hold for $c\leq 1$. However, the kernels in question become less regular than for generic central charge, and less well understood.
# Fingertips/Bijections/Exercise

We want to realize as many bijective mappings as possible from the fingertips of the left hand to the fingertips of the right hand, by letting the corresponding (according to the assignment) fingertips touch each other.
* Realize the "natural“ bijection.
* Realize those bijections, where two adjacent fingertips swap their natural counterparts, and where the other three fingertips touch their natural counterparts (adjacent transposition).
* Realize those bijections, where two fingertips swap their natural counterparts, and where the other three fingertips touch their natural counterparts (transposition).
* Realize those bijections, where exactly two fingertips touch their natural counterparts.
* Realize those bijections, where exactly one fingertip touches its natural counterpart.
* Realize those bijections, where no fingertip touches its natural counterpart.
# Finger-permutation/Sign/Exercise

Determine the sign of the permutation given by the image (the left hand represents the domain, the right hand represents the codomain).
# Dark Energy

Viewing space as a system of stimulated absorption or stimulated/spontaneous emission governed by Einstein Probabilistic's Units.
Using Einstein Probabilistic Units dark energy density is defined as;
$u_{D}={\frac {A^{2}}{B}}$
$u_{D}=\left({\frac {8\pi h}{c^{3}}}\left(\ v_{A}^{2}v_{B}^{2}\right)\right)=\left({\frac {8\pi h}{c^{3}}}\left(\ v_{B}^{4}\right)\right)$
Assuming the energy density of dark energy is;
$u_{D}=6.20934*10^{-10}{\frac {J}{m^{3}}}$
Let us calculate the frequency of dark energy;
$v_{D}=1.001*10^{12}s^{-1}$
Given Rydberg constant is defined as;
$R_{\infty }={\frac {\alpha ^{2}m_{e}c}{2h}}$
Using Rydberg constant we can define dark energy frequency as;
$v_{D}={\frac {R_{\infty }\;c\;8\;\pi }{\alpha }}=\alpha \;4\pi R_{\infty }{\frac {m_{e}c^{2}}{h}}=1.0008*10^{12}s^{-1}$
From an order of magnitude perspective dark energy is related to Einstein Probabilistic Units by Rydberg constant.
# Electrical network/System of linear equations/Introduction/Example

An electrical network consists of several connected wires, which we call the edges of the network in this context. In every edge $K_{j}$, there is a certain (depending on the material and the length of the edge) resistance $R_{j}$. The points $P_{n}$, where the edges meet, are called the vertices of the network. If we put to some edges of the network a certain electric tension (voltage), then we will have in every edge a certain current $I_{j}$. The goal is to determine the currents from the data of the network and the voltages.
It is helpful to assign to each edge a fixed direction in order to distinguish the direction of the current in this edge (if the current is in the opposite direction, it gets a minus sign). We call these directed edges. In every vertex of the network, the currents of the adjacent edges come together; therefore, their sum must be $0$. In an edge $K_{j}$, there is a voltage drop $U_{j}$, determined by Ohm's law to be
    $U_{j}=R_{j}\cdot I_{j}\,.$
We call a closed, directed alignment of edges in a network a mesh. For such a mesh, the sum of voltages is $0$, unless a certain voltage is enforced from "outside“.
We list these Kirchhoff's laws again.
* In every vertex, the sum of the currents equals $0$.
* In every mesh, the sum of the voltages equals $0$.
* If in a mesh, a voltage $V$ is enforced, then the sum of the voltages equals $V$.
Due to "physical reasons“, we expect that, given voltages in every edge, there should be a well-defined current in every edge. In fact, these currents can be computed if we translate the stated laws into a system of linear equations and solve this system.
In the example given by the picture, suppose that the edges $K_{1},\ldots ,K_{5}$ (with the resistances $R_{1},\ldots ,R_{5}$) are directed from left to right and that the connecting edge $K_{0}$ from $A$ to $C$ (where the voltage $V$ is applied) is directed upwards. The four vertices and the three meshes $(A,D,B),\,(D,B,C)$ and $(A,D,C)$ yield the system of linear equations
    ${\begin{matrix}I_{0}&+I_{1}&&-I_{3}&&&=&0\\&&&I_{3}&+I_{4}&+I_{5}&=&0\\-I_{0}&&+I_{2}&&-I_{4}&&=&0\\&-I_{1}&-I_{2}&&&-I_{5}&=&0\\&R_{1}I_{1}&&+R_{3}I_{3}&&-R_{5}I_{5}&=&0\\&&-R_{2}I_{2}&&-R_{4}I_{4}&+R_{5}I_{5}&=&0\\&-R_{1}I_{1}&+R_{2}I_{2}&&&&=&-V\,.\end{matrix}}$
Here the $R_{j}$ and $V$ are given numbers, and the $I_{j}$ are the unknowns we are looking for.
# K^n/Vector as linear combination of vectors/System of linear equations/Example

Let $K$ denote a field, and $m\in \mathbb {N}$. Suppose that in $K^{m}$, there are $n$ vectors (or $m$-tuples)
    $v_{1}={\begin{pmatrix}a_{11}\\a_{21}\\\vdots \\a_{m1}\end{pmatrix}},\,v_{2}={\begin{pmatrix}a_{12}\\a_{22}\\\vdots \\a_{m2}\end{pmatrix}},\ldots ,v_{n}={\begin{pmatrix}a_{1n}\\a_{2n}\\\vdots \\a_{mn}\end{pmatrix}}$
given. Let
    $w={\begin{pmatrix}c_{1}\\c_{2}\\\vdots \\c_{m}\end{pmatrix}}\,$
be another vector. We want to know whether $w$ can be written as a linear combination of the $v_{j}$. Thus, we are dealing with the question whether there are $n$ elements $s_{1},\ldots ,s_{n}\in K$ such that
    $s_{1}{\begin{pmatrix}a_{11}\\a_{21}\\\vdots \\a_{m1}\end{pmatrix}}+s_{2}{\begin{pmatrix}a_{12}\\a_{22}\\\vdots \\a_{m2}\end{pmatrix}}+\cdots +s_{n}{\begin{pmatrix}a_{1n}\\a_{2n}\\\vdots \\a_{mn}\end{pmatrix}}={\begin{pmatrix}c_{1}\\c_{2}\\\vdots \\c_{m}\end{pmatrix}}\,$
holds. This equality of vectors means identity in every component, so that this condition yields a system of linear equations
    ${\begin{matrix}a_{11}s_{1}+a_{12}s_{2}+\cdots +a_{1n}s_{n}&=&c_{1}\\a_{21}s_{1}+a_{22}s_{2}+\cdots +a_{2n}s_{n}&=&c_{2}\\\vdots &\vdots &\vdots \\a_{m1}s_{1}+a_{m2}s_{2}+\cdots +a_{mn}s_{n}&=&c_{m}.\end{matrix}}$
# Linear system of inequalities/Remark

A system of linear inequalities over the rational numbers or over the real numbers is a system of the form
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&\star &c_{1}\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&\star &c_{2}\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&\star &c_{m}\,,\end{matrix}}$
where $\star$ might be $\leq$ or $\geq$. It is considerably more difficult to find the solution set of such a system than in the case of equations. In general, it is not possible to eliminate the variables.
# System of linear equations/Set of variables/Homogeneous and inhomogeneous/Definition

Let $K$ denote a field, and let $a_{ij}\in K$ for $1\leq i\leq m$ and $1\leq j\leq n$. We call
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$
a (homogeneous) system of linear equations in the variables $x_{1},\ldots ,x_{n}$. A tuple $(\xi _{1},\ldots ,\xi _{n})\in K^{n}$ is called a solution of the linear system if $\sum _{j=1}^{n}a_{ij}\xi _{j}=0$ holds for all $i=1,\ldots ,m$.
If $(c_{1},\ldots ,c_{m})\in K^{m}$ is given, then
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&c_{1}\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&c_{2}\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&c_{m}\end{matrix}}$
is called an inhomogeneous system of linear equations. A tuple $(\zeta _{1},\ldots ,\zeta _{n})\in K^{n}$ is called a solution to the inhomogeneous linear system if $\sum _{j=1}^{n}a_{ij}\zeta _{j}=c_{i}$ holds for all $i$.
# Flatbread/Thaler/Exercise

Two persons, $A$ and $B$, are lying beneath a palm tree, $A$ has $2$ flatbreads, and $B$ has $3$ flatbreads. A third person $C$ joins them, $C$ has no flatbread, but $5$ thalers. They agree to distribute, against the $5$ thalers, the $5$ flatbreads uniformly among them. How many thalers gives $C$ to $A$, how many to $B$?
# Dating service/11 Minutes/Exercise

The dating service "e-Tarzan meets e-Jane“ is successful, it claims that in each eleven minutes, one of the customers falls in love. How long does it take (in rounded years) until all adult people in Germany (about $65000000$) fall in love, if this service is the only possible way.
# Family/Age/Linear system/Exercise

$M,P,S$ and $T$ are the members of one family. In this case, $M$ is three times as old as $S$ and $T$ together, $M$ is older than $P$, and $S$ is older than $T$, moreover, the age difference between $S$ and $T$ is twice as large as the difference between $M$ and $P$. Furthermore, $P$ is seven times as old as $T$, and the sum of the ages of all family members is equal to the paternal grandmother's age, which is $83$.
a) Set up a linear system of equations that expresses the conditions described.
b) Solve this system of equations.
# Matrix/Standard columns and rows/Exercise

Let $M$ be an $m\times n$- matrix. Show that the matrix product $Me_{j}$ of $M$ with the $j$-th standard vector (regarded as a column vector) is the $j$-th column of $M$. What is $e_{i}M$, where $e_{i}$ is the $i$-th standard vector (regarded as a row vector)?
# Strassen-algorithm/2x2/Exercise

Let $2\times 2$-matrices $A={\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}}$ and $B={\begin{pmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{pmatrix}}$ be given. The product $A\cdot B$ is usually computed by the multiplication rule "row x column“; for this, we have to perform altogether $8$ multiplications in the field $K$. We describe a procedure for the matrix multiplication, in which only $7$ multiplications (but more additions) are necessary. We set
    $m_{1}={\left(a_{11}+a_{22}\right)}\cdot {\left(b_{11}+b_{22}\right)}\,,$
    $m_{2}={\left(a_{21}+a_{22}\right)}\cdot b_{11}\,,$
    $m_{3}=a_{11}\cdot {\left(b_{12}-b_{22}\right)}\,,$
    $m_{4}=a_{22}\cdot {\left(b_{21}-b_{11}\right)}\,,$
    $m_{5}={\left(a_{11}+a_{12}\right)}\cdot b_{22}\,,$
    $m_{6}={\left(a_{21}-a_{11}\right)}\cdot {\left(b_{11}+b_{12}\right)}\,,$
    $m_{7}={\left(a_{12}-a_{22}\right)}\cdot {\left(b_{21}+b_{22}\right)}\,.$
Show that the coefficients of the product matrix
    $AB=C={\begin{pmatrix}c_{11}&c_{12}\\c_{21}&c_{22}\end{pmatrix}}\,$
satisfy the equations
    $c_{11}=m_{1}+m_{4}-m_{5}+m_{7}\,,$
    $c_{12}=m_{3}+m_{5}\,,$
    $c_{21}=m_{2}+m_{4}\,,$
    $c_{22}=m_{1}-m_{2}+m_{3}+m_{6}\,.$
# Square matrices/Matrix ring/Exercise

Let $K$ be a field, and $n\in \mathbb {N} _{+}$. Show that the set of all square $n\times n$-matrices over $K$, with the addition of matrices and with the product of matrices as multiplication, forms a ring.
# Transposed matrix/Properties/Exercise

Let $K$ be a field, and $m,n,p\in \mathbb {N}$. Prove that the transpose of a matrix satisfies the following properties (where $A,B\in \operatorname {Mat} _{m\times n}(K)$, $C\in \operatorname {Mat} _{n\times p}(K)$, and $s\in K$).
    ${({A^{\text{tr}}})^{\text{tr}}}=A\,.$
    ${(A+B)^{\text{tr}}}={A^{\text{tr}}}+{B^{\text{tr}}}\,.$
    ${(sA)^{\text{tr}}}=s\cdot {A^{\text{tr}}}\,.$
    ${(A\circ C)^{\text{tr}}}={C^{\text{tr}}}\circ {A^{\text{tr}}}\,.$
# Matrix multiplication/Upper triangular matrix/4/Zero diagonal/Nilpotent/Exercise

We consider the matrix
    $M={\begin{pmatrix}0&a&b&c\\0&0&d&e\\0&0&0&f\\0&0&0&0\end{pmatrix}}\,$
over a field $K$. Show that the fourth power of $M$ is $0$, that is,
    $M^{4}=MMMM=0\,.$
# Matrix multiplication/Associativity/Exercise

Show that the multiplication of matrices is associative. More precisely: Let $K$ be a field, and let $A$ be an $m\times n$-matrix, $B$ an $n\times p$-matrix, and $C$ a $p\times r$-matrix over $K$. Show that $(AB)C=A(BC)$.
# Linear system/Simultaneously/Remark

Sometimes, we want to solve a simultaneous system of linear equations of the form
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&c_{1}&(=&d_{1},&=&e_{1},\ldots )\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&c_{2}&(=&d_{2},&=&e_{2},\ldots )\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&c_{m}&(=&d_{m},&=&e_{m},\ldots )\,.\end{matrix}}$
The goal is to find the solutions of the corresponding inhomogeneous linear systems for different vectors. In principle, we could consider independent linear systems, and solve them. However, it is smarter to perform those manipulations that we do on the left-hand side to achieve upper triangular form, simultaneously with all the vectors on the right-hand side. An important special case, for $n=m$, is when the vectors are the standard vectors, see fact.
# Linear system/Substitution method/Remark

Another method to solve a linear system is the substitution method. Here, the variables are also successively eliminated, but in another way. If we want to eliminate the variable $x_{1}$, then we look at an equation, say $G_{1}$, where $x_{1}$ occurs with a coefficient different from $0$. In this equation, we isolate $x_{1}$ and get a new equation of the form
    $G_{1}':\,\,\,x_{1}=F_{1},$
where $x_{1}$ does not occur in $F_{1}$. Then in all other equations $G_{2},\ldots ,G_{m}$, we replace the variable $x_{1}$ by $F_{1}$, and obtain (after some simplifications) a linear system $G_{2}',\ldots ,G_{m}'$ without the variable $x_{1}$, which is, together with $G_{1}'$, equivalent with the original system.
# Linear system/Superposition principle/Fact

Let $M={\left(a_{ij}\right)}_{1\leq i\leq m,1\leq j\leq n}$ denote a matrix over a field $K$. Let $c={\left(c_{1},\ldots ,c_{m}\right)}$ and $d={\left(d_{1},\ldots ,d_{m}\right)}$ denote two $m$-tuples, and let $y={\left(y_{1},\ldots ,y_{n}\right)}\in K^{n}$ be a solution of the linear system
    $Mx=c\,,$
and $z={\left(z_{1},\ldots ,z_{n}\right)}\in K^{n}$ a solution of the system
    $Mx=d\,.$
$y+z={\left(y_{1}+z_{1},\ldots ,y_{n}+z_{n}\right)}$ is a solution of the system
    $Mx=c+d\,.$
# Linear system/Superposition principle/Homogeneous and inhomogeneous/Fact

Let $K$ be a field, and let
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&c_{1}\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&c_{2}\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&c_{m}\end{matrix}}$
be an inhomogeneous linear system over $K$, and let
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$
be the corresponding homogeneous linear system. If ${\left(y_{1},\ldots ,y_{n}\right)}$ is a solution of the inhomogeneous system and if ${\left(z_{1},\ldots ,z_{n}\right)}$ is a solution of the homogeneous system,
${\left(y_{1}+z_{1},\ldots ,y_{n}+z_{n}\right)}$
# Linear system/Superposition principle/Section
## Theorem

Let $M={\left(a_{ij}\right)}_{1\leq i\leq m,1\leq j\leq n}$ denote a matrix over a field $K$. Let $c={\left(c_{1},\ldots ,c_{m}\right)}$ and $d={\left(d_{1},\ldots ,d_{m}\right)}$ denote two $m$-tuples, and let $y={\left(y_{1},\ldots ,y_{n}\right)}\in K^{n}$ be a solution of the linear system
    $Mx=c\,,$
and $z={\left(z_{1},\ldots ,z_{n}\right)}\in K^{n}$ a solution of the system
    $Mx=d\,.$
$y+z={\left(y_{1}+z_{1},\ldots ,y_{n}+z_{n}\right)}$ is a solution of the system
    $Mx=c+d\,.$

### Proof

$\Box$

## Corollary

Let $K$ be a field, and let
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&c_{1}\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&c_{2}\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&c_{m}\end{matrix}}$
be an inhomogeneous linear system over $K$, and let
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$
${\left(y_{1},\ldots ,y_{n}\right)}$
${\left(z_{1},\ldots ,z_{n}\right)}$
${\left(y_{1}+z_{1},\ldots ,y_{n}+z_{n}\right)}$

### Proof

This follows immediately from fact.
$\Box$
In particular, this means that when $L$ is the solution space of a homogeneous linear system, and when $y$ is one (particular) solution of an inhomogeneous linear system, then the mapping
    $L\longrightarrow L',z\longmapsto y+z,$
gives a bijection between $L$ and the solution set $L'$ of the inhomogeneous linear system.
# Linear system/Cramer's rule/Fact

Linear system inhomogeneous/Situation Suppose that the describing matrix $M={\left(a_{ij}\right)}_{ij}$ is invertible.
$x_{j}$
$x_{j}={\frac {\det[[:Template:MatrixCramer]]}{\det M}}$
# Linear system/Cramer's rule/Fact/Proof

For an invertible matrix $M$, the solution of the linear system $Mx=c$ can be found by applying $M^{-1}$, that is, $x=M^{-1}c$. Using fact, this means $x={\frac {1}{\det M}}(M^{\operatorname {adj} })c$. For the $j$-th component, this means
    $x_{j}={\frac {1}{\det M}}{\left(\sum _{k=1}^{n}(-1)^{k+j}(\det M_{kj})\cdot c_{k}\right)}\,.$
The right-hand factor is the expansion of the determinant of the matrix shown in the numerator with respect to the $j$-th column.
# Linear algebra (Osnabrück 2024-2025)/Matrix/Adjugate matrix/Formula with determinant/Fact/Factreferencenumber

Theorem 17.9
# Cramer's rule/2x2/1/Example

We solve the linear system
    ${\begin{pmatrix}4&3\\-1&5\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}={\begin{pmatrix}2\\7\end{pmatrix}}\,$
using Cramer's rule. This yields
    $x_{1}={\frac {\det {\begin{pmatrix}2&3\\7&5\end{pmatrix}}}{\det {\begin{pmatrix}4&3\\-1&5\end{pmatrix}}}}=-{\frac {11}{23}}\,$
and
    $x_{2}={\frac {\det {\begin{pmatrix}4&2\\-1&7\end{pmatrix}}}{\det {\begin{pmatrix}4&3\\-1&5\end{pmatrix}}}}={\frac {30}{23}}\,.$
# Cramer's rule/3x3/Exercise

We consider the matrix
    $A={\begin{pmatrix}1&0&2\\2&1&3\\1&0&1\end{pmatrix}}\in \operatorname {Mat} _{3}(\mathbb {R} )\,.$
Solve the linear system $Ax=(1,0,2)^{t}$ using Cramer's rule (check first that we may apply this rule).
# Endomorphism/Homothety/Definition

Let $V$ be a vector space over a field $K$. For $a\in K$, the linear mapping
    $\varphi \colon V\longrightarrow V,v\longmapsto av,$
is called homothety (or dilation) with scaling factor $a$.
# Games of gain-ground/Ballon au poing

Ballon au poing
Chapter 5
Lesson : Games of gain-ground
Previous chapter:Real tennis
Next chapter :Balle pelote
Games of gain-ground ${\mathcal {Ballon~au~poing}}$ :
ballon au poing - championnat de France 2012 à Amiens
The ballon au poing is a popular team sport in Picardy (France). This game is played by teams of six. It is a game of gain-ground.
This sport must not be confused with the fistball.
To be able to hit the ball, players surround their hand and their wrist with a strip of canvas or leather.
Each 15 August at the Parc de la Hotoie ( Amiens ), the city receives the final stages of Ballon au poing.
Playing area ( Ballodrome )
ballodrome
Rules
The teams are separated on the ground by a line ( la corde - the rope) which is mobile during the party. This new limit is shown by a chasse.
The purpose of the game is to gain ground by moving the rope. To do it, we try to make the ball "die" in the opposite camp.
Points mark by 15, 30, 40 and game.
ballon au poing - championnat de France 2009 à Amiens
The two chasses (red and blue) :
* Notes and External links
* Lazure Marcel , Les Jeux de Balle et Ballon Picards : Ballon au Poing, Balle à la Main, Balle au Tamis, Longue Paume , Éditeu : CRDP éd Picardie, (1996).
* Fédération Française de Ballon au Poing
* {French} {PDF} archive.wikiwix.com - languepicarde.fr
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# Two points in plane/(-4,3), (5,-6)/Line equation/Exercise

Determine an equation for the line in $\mathbb {R} ^{2}$ that runs through the points $(-4,3)$ and $(5,-6)$.
# Three points in space/(1,0,0), (0,1,2), (2,3,4)/Equation for plane/Exercise

Determine a linear equation for the plane in $\mathbb {R} ^{3}$, where the three points
    $(1,0,0),\,(0,1,2),{\text{ and }}(2,3,4)$
lie.
# Linear system/Not in standard form/1/Exercise

Bring the system of linear equations
    $3x-4+5y=8z+7x\,,$
    $2-4x+z=2y+3x+6\,,$
    $4z-3x+2x+3=5x-11y+2z-8\,,$
into a standard form, and solve it.
# Linear system/3 equations/Differences/Not equivalent/Exercise

Show by an example that the linear system given by three equations I, II, III is not equivalent to the linear system given by the three equations I-II, I-III, II-III.
# Matrix of production/2/Exercise

Out of the resources $R_{1},R_{2}$, and $R_{3}$, several commodities $P_{1},P_{2},P_{3},P_{4}$ are produced. The following table shows how much of the resources are needed to produce the commodities (always in suitable units).
a) Establish a matrix that computes, applied to a four-tuple of commodities, the required resources.
b) The following table shows how much of each commodity shall be produced in a month.
What resources are necessary?
c) The following table shows how much of each resource is delivered on a certain day.
What tuples of commodities can be produced from this without waste?
# Diagonal matrix/Linear system/Solution method/Exercise

Let
    $D={\begin{pmatrix}d_{11}&0&\cdots &\cdots &0\\0&d_{22}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1\,n-1}&0\\0&\cdots &\cdots &0&d_{nn}\end{pmatrix}}\,$
be a diagonal matrix, and let $c={\begin{pmatrix}c_{1}\\\vdots \\c_{n}\end{pmatrix}}$ be an $n$-tuple over a field $K$, and let $x={\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}$ be a tuple of variables. What is specific about the system of linear equations
    $Dx=c\,,$
and how can you solve it?